tag:blogger.com,1999:blog-63319094866600929102024-02-20T21:47:18.681+00:00Matt Scroggs Blogs"The best blog I've seen all day."Matthttp://www.blogger.com/profile/01590672558655167096noreply@blogger.comBlogger82125tag:blogger.com,1999:blog-6331909486660092910.post-59114885647617687572014-09-22T08:16:00.000+01:002014-09-22T08:16:00.028+01:00Sunday Afternoon Maths XXIX Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/09/sunday-afternoon-maths-xxix.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Three Squares</h4>
<div style="text-align: justify; text-indent: 20px;">Draw three more squares and add these lines (I have coloured the angles to make equal angles clearer):</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhR8lfbsqFJvhUh2Upkc4Rrk_rZx8twwbTjWz3FdwGVgFlX-M6jBa277byp5w0Wmz_tYfHlbsG9wJlt0B-U__bnX61JuafsHaDEWuB-BRcpl5W8cC8vYAsJjyJBAip_kdXwLYoux6Oyd8/s1600/three-ans1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhR8lfbsqFJvhUh2Upkc4Rrk_rZx8twwbTjWz3FdwGVgFlX-M6jBa277byp5w0Wmz_tYfHlbsG9wJlt0B-U__bnX61JuafsHaDEWuB-BRcpl5W8cC8vYAsJjyJBAip_kdXwLYoux6Oyd8/s320/three-ans1.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Triangles \(ACE\), \(LDK\) and \(IKE\) are congruent, so angle \(KDL\) is equal to \(\beta\).</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEig6wPCftCqNU3T5cyDFKE3S0TGoSjqE8NhYb_7PDVXGkGxNA07GwPGvuMPBPT5tVyIkREtWEwW8mKG58wICcTSd4SEArf9OqjZo2sbV6_d50_00gQCpCYt-1tjm7arjPleLaRjmHNSVuw/s1600/three-ans2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEig6wPCftCqNU3T5cyDFKE3S0TGoSjqE8NhYb_7PDVXGkGxNA07GwPGvuMPBPT5tVyIkREtWEwW8mKG58wICcTSd4SEArf9OqjZo2sbV6_d50_00gQCpCYt-1tjm7arjPleLaRjmHNSVuw/s320/three-ans2.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The congruence of these triangles tells us that angles \(DKL\) and \(EKI\) add up to a right angle, so angle \(EKD\) is also a right angle.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVqeuRQ_hCUvb4it9iL7CUz3uRGqDj4bvfwN_D-9TZVxFTW-CC0L1zWUsgqZSAgqsp41sc8j-wbYqZEdPstB1gOhwH2exPNCW3X0A_FJQqcFTUyJKWymKf15SBeFX1BTxrjafZHb_uamA/s1600/three-ans3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVqeuRQ_hCUvb4it9iL7CUz3uRGqDj4bvfwN_D-9TZVxFTW-CC0L1zWUsgqZSAgqsp41sc8j-wbYqZEdPstB1gOhwH2exPNCW3X0A_FJQqcFTUyJKWymKf15SBeFX1BTxrjafZHb_uamA/s320/three-ans3.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The congruence of the triangles also tells us that \(KD\) and \(KE\) are the same length and so angle \(EDK\) is the angle in an isosceles right-angled triangle. \(\alpha\) is also the angle in an isosceles right-angled triangle, so these two angles are equal.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDylaawvx2mDWyQuqJmbPr1xcPWnG4NvCBnNIQQy4G4ATEa2P6XGnJgC2q1LLa6bGsVFNOjm4VgTxlfsechSQVodV9N8vNfLvDjh-mXDN_4hdxgHP-oPgnBFxk5xciNnm6TB1NFvjy_fY/s1600/three-ans4.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDylaawvx2mDWyQuqJmbPr1xcPWnG4NvCBnNIQQy4G4ATEa2P6XGnJgC2q1LLa6bGsVFNOjm4VgTxlfsechSQVodV9N8vNfLvDjh-mXDN_4hdxgHP-oPgnBFxk5xciNnm6TB1NFvjy_fY/s320/three-ans4.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Therefore \(\alpha+\beta+\gamma=90^\circ\).</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">The diagram shows three squares with diagonals drawn on and three angles labelled.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1hyahrmc28cTb45A9EYLFrFw6FTZ9mThQ8Czg5ami3lXpjNwAC2xcLCNh6ohuWS3bVTZBhRnyoXMpmJQZAuhOuKVb9DS4ktE0Qq-wvVgjWWoRFFfsDWFBcMK4382gw8gsOXRSbudABA8/s1600/three+rhombus.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1hyahrmc28cTb45A9EYLFrFw6FTZ9mThQ8Czg5ami3lXpjNwAC2xcLCNh6ohuWS3bVTZBhRnyoXMpmJQZAuhOuKVb9DS4ktE0Qq-wvVgjWWoRFFfsDWFBcMK4382gw8gsOXRSbudABA8/s320/three+rhombus.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What is the value of \(\alpha+\beta+\gamma\)?</div>
<h4 style="margin-bottom: 0px;">Equal Opportunity</h4>
<div style="text-align: left"><small><a href='http://www.futilitycloset.com/2014/08/27/equal-opportunity-3/' target='new'>Source: Futility Closet</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Let \(p_1\), \(p_2\), ..., \(p_6\) be the probabilities of getting 1 to 6 on one die and \(q_1\), ..., \(q_6\) on the other. The probability of getting a total of 2 is \(p_1q_1\) and the probabilty of getting a total of 12 is \(p_6q_6\). Therefore \(p_1q_1=p_6q_6\).</div>
<div style="text-align: justify; text-indent: 20px;">If \(p_1\geq p_6\) then \(q_1\leq q_6\) (and vice-versa) as otherwise the above equality could not hole. Therefore:</div>
$$(p_1-p_6)(q_1-q_6)\leq 0$$
$$p_1q_1-p_6q_1-p_1q_6+p_6q_6\leq 0$$
$$p_1q_1+q_6p_6\leq p_1q_6+p_6q_1$$
<div style="text-align: justify; text-indent: 20px;">The probability of rolling a total of 7 is \(p_1q_6+p_2q_5+...+p_6q_1\). This is larger than \(p_1q_6+p_6q_1\), which is larger than (or equal to) \(p_1q_1+q_6p_6\), which is larger than \(p_1q_1\).</div>
<div style="text-align: justify; text-indent: 20px;">Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Can two \(n\)-sided dice be weighted so that the probability of each of the numbers 2, 3, …, 2\(n\) is the same?</div>
<div style="text-align: justify; text-indent: 20px;">Can a \(n\)-sided die and a \(m\)-sided die be weighted so that the probability of each of the numbers 2, 3, …, \(n+m\) is the same?</div>
<h4 style="margin-bottom: 0px;">Double Derivative</h4>
<div style="text-align: left"><small><a href='http://www.peterliljedahl.com/wp-content/uploads/Mason-26-Years.pdf' target='new'>Source: <i>Twenty-Six Years of Problem Solving</i> compiled by John Mason</a></small></div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(i) \(\frac{dy}{dx}=1\), so \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=0\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(ii) Differentiating \(y=x^2\) with respect to \(x\) \(\frac{dy}{dx}=2x\). Let \(g=\frac{dy}{dx}\). By the chain rule:</div>
$$\frac{dg}{dy}=\frac{dg}{dx}\frac{dx}{dy}$$
$$=2\frac{1}{2x}$$
$$=\frac{1}{x}$$
<div style="text-align: justify; text-indent: 20px;margin-left:10px">So \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{1}{x}\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(iii) By the same method, \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{2}{x}\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(iv) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{n-1}{x}\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(v) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=1\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(vi) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=-\tan(x)\)</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: left"><small><a href='http://www.peterliljedahl.com/wp-content/uploads/Mason-26-Years.pdf' target='new'>Source: <i>Twenty-Six Years of Problem Solving</i> compiled by John Mason</a></small></div>
<div style="text-align: justify; text-indent: 20px;">What is</div>
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
<div style="text-align: justify; text-indent: 20px;">when \(y=f(x)\)?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-9278712691690829472014-09-21T12:15:00.000+01:002014-09-21T12:15:00.757+01:00Sunday Afternoon Maths XXIX<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>. This Tuesday is <a target="new" href="http://www.mathsjam.com">Maths Jam</a>, make sure you're there!</div>
<h4 style="margin-bottom: 0px;">Three Squares</h4>
<div style="text-align: left"><small><a href='https://www.youtube.com/watch?v=m5evLoL0xwg' target='new'>Source: Numberphile</a></small></div>
<div style="text-align: justify; text-indent: 20px;">The diagram shows three squares with diagonals drawn on and three angles labelled.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFhjeLxT8oack1Ueze8ZUaCYgiKWjPK7CvwCAm-EZYZP80tQ5_sRxiaOL0UC65pcmUAXrd19i268SzZzJ_oj_9AKEgx8hlYji0pOHfJoddJNyLFl6TQPokRacaexG4TJl_t7FTPHbMhoE/s1600/three+squares.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFhjeLxT8oack1Ueze8ZUaCYgiKWjPK7CvwCAm-EZYZP80tQ5_sRxiaOL0UC65pcmUAXrd19i268SzZzJ_oj_9AKEgx8hlYji0pOHfJoddJNyLFl6TQPokRacaexG4TJl_t7FTPHbMhoE/s320/three+squares.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What is the value of \(\alpha+\beta+\gamma\)?</div>
<h4 style="margin-bottom: 0px;">Equal Opportunity</h4>
<div style="text-align: left"><small><a href='http://www.futilitycloset.com/2014/08/27/equal-opportunity-3/' target='new'>Source: Futility Closet</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Can two (six-sided) dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?</div>
<h4 style="margin-bottom: 0px;">Double Derivative</h4>
<div style="text-align: left"><small><a href='http://www.peterliljedahl.com/wp-content/uploads/Mason-26-Years.pdf' target='new'>Source: <i>Twenty-Six Years of Problem Solving</i> compiled by John Mason</a></small></div>
<div style="text-align: justify; text-indent: 20px;">What is</div>
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
<div style="text-align: justify; text-indent: 20px;">when:</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(i) \(y=x\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(ii) \(y=x^2\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(iii) \(y=x^3\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(iv) \(y=x^n\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(v) \(y=e^x\)</div>
<div style="text-align: justify; text-indent: 20px;margin-left:10px">(vi) \(y=\sin(x)\)?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-19299060974023202542014-09-15T07:41:00.000+01:002014-09-15T14:38:50.345+01:00Sunday Afternoon Maths XXVIII Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/09/sunday-afternoon-maths-xxviii.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">2009</h4>
<div style="text-align: justify; text-indent: 20px;">\(2009=7\times 7\times 41\), so there are four possible sets of dimensions of the cuboid:</div>
$$1\times 1\times 2009$$
$$1\times 7\times 287$$
$$1\times 41\times 49$$
$$7\times 7\times 41$$
<div style="text-align: justify; text-indent: 20px;">In the first three cuboids, there is a face with an area of 2009 units (\(2009\times 1\), \(7\times 287\) and \(41\times 49\) respectively) and so 2009 stickers will not be enough. Therefore the cuboid has dimensions \(7\times 7\times 41\) and a surface area of 1246, leaving <b>764</b> stickers left over</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">For which numbers \(n\) can a cuboid be made with \(n\) unit cube such that \(n\) unit square stickers can cover the faces of the cuboid.</div>
<h4 style="margin-bottom: 0px;">3\(n\)+1</h4>
<div style="text-align: justify; text-indent: 20px;"><b>(i)</b> Let \(a,b\in S\). Then \(\exists \alpha,\beta\in \mathbb{N}\) such that \(a=3\alpha+1\) and \(b=3\beta+1\).</div>
<div style="text-align: justify; text-indent: 20px;"><small>(This says that if \(a\) and \(b\) are in \(S\) then they can be written as a multiple of three plus one.)</small></div>
$$a\times b=(3\alpha+1)\times (3\beta+1)$$
$$=9\alpha\beta+3\alpha+3\beta+1$$
$$=3(3\alpha\beta+\alpha+\beta)+1$$
<div style="text-align: justify; text-indent: 20px;">This is a multiple of three plus one, so \(a\times b\in S\).</div>
<div style="text-align: justify; text-indent: 20px;"><b>(ii)</b> No, as \(36\times 22=4\times 253\) and 36,22,4 and 253 are all irreducible.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: left"><small><a href='http://www.peterliljedahl.com/wp-content/uploads/Mason-26-Years.pdf' target='new'>Source: <i>Twenty-Six Years of Problem Solving</i> compiled by John Mason</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Try the task again with \(S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}\).</div>
<h4 style="margin-bottom: 0px;">The Ace of Spades</h4>
<div style="text-align: justify; text-indent: 20px;">The problem can be expressed using the following probability tree:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2HdGiYQx990kwKQj9Cku8Ktzj8ttkKwCAXaXiKwynJB8aB67VPSz8-cer7mDB0W9Sbn0nlXeSFELBGayz3j4fV3QhBwzWWhGKa5OXQZMUOnw9AkZIRL1OVv7CxX9lhpwrIPiMTnccO94/s1600/prob+tree.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2HdGiYQx990kwKQj9Cku8Ktzj8ttkKwCAXaXiKwynJB8aB67VPSz8-cer7mDB0W9Sbn0nlXeSFELBGayz3j4fV3QhBwzWWhGKa5OXQZMUOnw9AkZIRL1OVv7CxX9lhpwrIPiMTnccO94/s320/prob+tree.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The probability that the card turned over from C is an Ace of Spades is:</div>
$$\frac{1\times 2\times 2+1\times 50\times 1+50\times 1\times 2+50\times 51\times 1}{51\times 52\times 53}$$
$$=\frac{52}{51\times 53}$$Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com2tag:blogger.com,1999:blog-6331909486660092910.post-30348322246885350812014-09-14T11:40:00.000+01:002014-09-14T11:40:13.753+01:00Sunday Afternoon Maths XXVIII<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">2009</h4>
<div style="text-align: left"><small><a href='http://teachfurthermaths.weebly.com/puzzle-of-the-month/september-puzzle' target='new'>Source: Teaching Further Maths blog</a></small></div>
<div style="text-align: justify; text-indent: 20px;">2009 unit cubes are glued together to form a cuboid. A pack, containing 2009 stickers, is opened, and there are enough stickers to place 1 sticker on each exposed face of each unit cube.</div>
<div style="text-align: justify; text-indent: 20px;">How many stickers from the pack are left?</div>
<h4 style="margin-bottom: 0px;">3\(n\)+1</h4>
<div style="text-align: left"><small><a href='http://www.peterliljedahl.com/wp-content/uploads/Mason-26-Years.pdf' target='new'>Source: <i>Twenty-Six Years of Problem Solving</i> compiled by John Mason</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Let \(S=\{3n+1:n\in\mathbb{N}\}\) be the set of numbers one more than a multiple of three.</div>
<div style="text-align: justify; text-indent: 20px;"><b>(i)</b> Show that \(S\) is closed under multiplication.</div>
<div style="text-align: justify; text-indent: 20px;">ie. Show that if \(a,b\in S\) then \(a\times b\in S\).</div>
<div style="text-align: justify; text-indent: 20px;">Let \(p\in S\) be irreducible if \(p\not=1\) and the only factors of \(p\) in \(S\) are \(1\) and \(p\). <small>(This is equivalent to the most commonly given definition of prime.)</small></div>
<div style="text-align: justify; text-indent: 20px;"><b>(ii)</b> Can each number in \(S\) be uniquely factorised into irreducibles?</div>
<h4 style="margin-bottom: 0px;">The Ace of Spades</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Week-end-Problems-Book-Hubert-Phillips/dp/0715635336' target='new'>Source: <i>The Week-End Problems</i> by Hubert Phillips</a></small></div>
<div style="text-align: justify; text-indent: 20px;">I have three packs of playing cards with identical backs. Call the packs A, B and C.</div>
<div style="text-align: justify; text-indent: 20px;">I draw a random card from pack A and shuffle it into pack B.</div>
<div style="text-align: justify; text-indent: 20px;">I now turn up the top card of pack A, revealing the Queen of Hearts.</div>
<div style="text-align: justify; text-indent: 20px;">Next, I draw a card at random from pack B and shuffle it into pack C. Then, I turn up the top card of pack B, revealing another Queen of Hearts.</div>
<div style="text-align: justify; text-indent: 20px;">I now draw a random card from pack C and place it at the bottom of pack A.</div>
<div style="text-align: justify; text-indent: 20px;">What is the probability that the card at the top of pack C is the Ace of Spades?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-81928622097247720342014-09-08T08:54:00.000+01:002014-09-08T08:54:00.027+01:00Sunday Afternoon Maths XXVII Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/09/sunday-afternoon-maths-xxvii.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Triangles Between Squares</h4>
<div style="text-align: justify; text-indent: 20px;">Let \(T_a\) represent the \(a\)<sup>th</sup> triangle number. This means that \(T_a=\frac{1}{2}a(a+1)\).</div>
<div style="text-align: justify; text-indent: 20px;">Suppose that for some integer \(n\), \(n^2 \leq T_a <(n+1)^2\). This means that:</div>
$$n^2 \leq T_a$$
$$n^2 \leq \frac{1}{2}a(a+1)$$
$$2n^2 \leq a^2+a$$
<div style="text-align: justify; text-indent: 20px;">But for every positive integer \(a \leq a^2\), so:</div>
$$2n^2 \leq 2a^2$$
$$n^2 \leq a^2$$
<div style="text-align: justify; text-indent: 20px;">\(n\) and \(a\) are both positive integers, so:</div>
$$n \leq a$$
<div style="text-align: justify; text-indent: 20px;">Now consider \(T_{a+2}\):</div>
$$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$
$$=\frac{1}{2}(a^2+5a+6)$$
$$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$
$$=\frac{1}{2}a(a+1)+2a+3$$
$$=T_a+2a+3$$
<div style="text-align: justify; text-indent: 20px;">We know that \(a \geq n\) and \(T_a \geq n^2\), so:</div>
$$T_a+2a+3 \geq n^2+2n+3$$
$$>n^2+2n+1 = (n+1)^2$$
<div style="text-align: justify; text-indent: 20px;">And so \(T_{a+2}\) is not between \(n^2\) and \((n+1)^2\). So if a triangle number \(T_a\) is between \(n^2\) and \((n+1)^2\) then the next but one triangle number \(T_{a+2}\) cannot also be between \(n^2\) and \((n+1)^2\). So there cannot be more than two triangle numbers between \(n^2\) and \((n+1)^2\).</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Given an integer \(n\), how many triangle numbers are there between \(n^2\) and \((n+1)^2\)?</div>
<h4 style="margin-bottom: 0px;">Sine</h4>
<div style="text-align: justify; text-indent: 20px;">Cosine can be drawn the same way as sine but starting B at the top of the circle.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEih38Gp-zronbsgfqNGt2Pmbi-N7AsIcrQaxKfp-rCQDptqfKOhVCkd4rsycK3R0C9pBuIWBKsrgaUBT9qHKT8GND6JBkekKk2OH2xJqz6RbSnx564VBS8bB83GFQW2TSPmeEnyE8VOD58/s1600/cosine.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEih38Gp-zronbsgfqNGt2Pmbi-N7AsIcrQaxKfp-rCQDptqfKOhVCkd4rsycK3R0C9pBuIWBKsrgaUBT9qHKT8GND6JBkekKk2OH2xJqz6RbSnx564VBS8bB83GFQW2TSPmeEnyE8VOD58/s320/cosine.gif" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Tangent can be drawn by giving the following instructions:</div>
<div style="text-align: justify; text-indent: 20px;">A. Stand on the spot.</div>
<div style="text-align: justify; text-indent: 20px;">B. Walk around A in a circle, holding this string to keep you the same distance away.</div>
<div style="text-align: justify; text-indent: 20px;">C. Make a straight line with A and B, staying on the line tangent to the circle through B's starting point.</div>
<div style="text-align: justify; text-indent: 20px;">D. Walk in a straight line perpendicular to C's line.</div>
<div style="text-align: justify; text-indent: 20px;">E. Stay in line with C and D.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim2e7OCT93DIbc4J0XCgbFd4BDlDLmqDF-oz32bwVQWdDfzNV4mgYKi39VuiUz5HoSD08HYwFwiALtJyn9dQEVdexGyDCjQVh3zne8lzNkxKPmQzuVvbiBJ5-5lNgBswws_z3w9wWet4E/s1600/tangent.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim2e7OCT93DIbc4J0XCgbFd4BDlDLmqDF-oz32bwVQWdDfzNV4mgYKi39VuiUz5HoSD08HYwFwiALtJyn9dQEVdexGyDCjQVh3zne8lzNkxKPmQzuVvbiBJ5-5lNgBswws_z3w9wWet4E/s320/tangent.gif" /></a></div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Could people be used to draw graphs of secant, cosecant and cotangent?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-72736069656728760832014-09-07T12:36:00.000+01:002014-09-07T12:36:14.245+01:00Sunday Afternoon Maths XXVII<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Triangles Between Squares</h4>
<div style="text-align: justify; text-indent: 20px;">Prove that there are never more than two triangle numbers between two consecutive square numbers.</div>
<h4 style="margin-bottom: 0px;">Sine</h4>
<div style="text-align: justify; text-indent: 20px;">A sine curve can be created with five people by giving the following instructions to the five people:</div>
<div style="text-align: justify; text-indent: 20px;">A. Stand on the spot.</div>
<div style="text-align: justify; text-indent: 20px;">B. Walk around A in a circle, holding this string to keep you the same distance away.</div>
<div style="text-align: justify; text-indent: 20px;">C. Stay in line with B, staying on this line.</div>
<div style="text-align: justify; text-indent: 20px;">D. Walk in a straight line perpendicular to C's line.</div>
<div style="text-align: justify; text-indent: 20px;">E. Stay in line with C and D. E will trace the path of a sine curve as shown here:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiG71lmuCG_ZWthHIW7VFUjLLfHA78cISvrsCI3Vt1r1poZ_xmYjGtrV08QBjmcz7iieWCdTWpUtMJDDl8RV9ZLVHuAro7_HNY3VcdJHMDLXPqlW7Ra7xIT1nBs2ZMMtA48KtqJdyF1ml8/s1600/sine4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiG71lmuCG_ZWthHIW7VFUjLLfHA78cISvrsCI3Vt1r1poZ_xmYjGtrV08QBjmcz7iieWCdTWpUtMJDDl8RV9ZLVHuAro7_HNY3VcdJHMDLXPqlW7Ra7xIT1nBs2ZMMtA48KtqJdyF1ml8/s320/sine4.gif" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What instructions could you give to five people to trace a cos(ine) curve?</div>
<div style="text-align: justify; text-indent: 20px;">What instructions could you give to five people to trace a tan(gent) curve?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-59413731123075836872014-09-04T21:35:00.000+01:002014-09-06T18:30:12.854+01:00Electromagnetic Field Talk<div style="text-align: justify; text-indent: 20px;">Last weekend, I attended <a target='new' href='https://www.emfcamp.org'>Electromagnetic Field</a>, a camp for hackers, geeks, makers and the interested. On the Sunday, I gave a talk on four mathematical ideas/tasks which I have encountered over the past few years: Flexagons, Folding Tube Maps, Braiding and Sine Curves. I'd love to see photos, hear stories, etc from anyone who tries these activities: either comment on here or tweet <a href='https://www.twitter.com/mscroggs' target='new'>@mscroggs</a>.</div>
<h4 style="margin-bottom: 0px;">Flexagons</h4>
<div style="text-align: justify; text-indent: 20px;">It's probably best to start by showing you what a flexagon is...</div>
<iframe width="420" height="315" src="//www.youtube.com/embed/SnnvovjcC_o?rel=0" frameborder="0" allowfullscreen></iframe>
<div style="text-align: justify; text-indent: 20px;">What you saw there is called a <i>trihexaflexagon</i>. <i>Tri-</i> because it has three faces; <i>-hexa-</i> because it is a hexagon; and <i>-flexagon</i> because it can be flexed to reveal the other faces.</div>
<div style="text-align: justify; text-indent: 20px;">The story goes that, in 1939, Arthur H. Stone, who was an Englishman studying mathematics at Harvard, was trimming the edges off his American paper to fit in his English folder. He was fiddling with the offcuts and found that if he folded the paper under itself in a loop, he could make a hexagon; and when this hexagon was folded up as we saw, it would open out to reveal a different face.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgco8bjVdiu2S93FrSUX3-shB-iB0mVxfTWcx2aapLn-FwOwqmpgj176nQScjzL9GTSwgCuaahytynuqZtjM46z1fxLdJjNmD0Sans2j6D6KAmJLAHu9yU7NB7XJ-x-h6cGNd0wjCmqhFQ/s1600/flexagon+slide.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgco8bjVdiu2S93FrSUX3-shB-iB0mVxfTWcx2aapLn-FwOwqmpgj176nQScjzL9GTSwgCuaahytynuqZtjM46z1fxLdJjNmD0Sans2j6D6KAmJLAHu9yU7NB7XJ-x-h6cGNd0wjCmqhFQ/s320/flexagon+slide.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The way it flexes can be show on a diagram: In the circles, the colour on either side of the flexagon is shown and the lines show flexes which can be made.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfR-awECVar7W8qZcClmRwq7ZEWYZMziq5Sy8ISrsYVl3mk1NU0c-MIftvtVpRsxT5619RrGkSjUQuszW_DMk2Xp-KVoC8kzoiK-XosZFVgYDjfo3vDzDvjK8cNgcQOIbGpI4Wj3ESyhQ/s1600/trihexa.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfR-awECVar7W8qZcClmRwq7ZEWYZMziq5Sy8ISrsYVl3mk1NU0c-MIftvtVpRsxT5619RrGkSjUQuszW_DMk2Xp-KVoC8kzoiK-XosZFVgYDjfo3vDzDvjK8cNgcQOIbGpI4Wj3ESyhQ/s320/trihexa.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">When Stone showed his flexagon to other students at Harvard, they were equally amazed by it, and they formed what they called 'The Flexagon Committee'. Members of the committee included Richard Feynman, who was then still a graduate student. The committee could meet regularly and soon discovered other flexagons, the first of which was the hexahexaflexagon: Again shaped like a hexagon, but this time with six faces.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiygcxKboqnNrs2-yA9ebbS2kw1ay18z4MdiZCKVaxDsCn5Ibdok-Kmc6dCNJGFaemaIXn7XY31G1VjsC2P0rP0ZkXFT6T1BshnXDk0NlbqauSQJrAScUF-2ELHHLvpBwgV30B1sQVqk-c/s1600/hexahexa+instr.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiygcxKboqnNrs2-yA9ebbS2kw1ay18z4MdiZCKVaxDsCn5Ibdok-Kmc6dCNJGFaemaIXn7XY31G1VjsC2P0rP0ZkXFT6T1BshnXDk0NlbqauSQJrAScUF-2ELHHLvpBwgV30B1sQVqk-c/s320/hexahexa+instr.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">A hexahexaflexagon is created by taking a longer strip of paper and rolling it around itself like this. The shorter strip at the end is then folded and glued in the same way the trihexaflexagon was. Once made, the hexahexaflexagon can be flexed. From some positions, the flexagon can be flexed in different ways to reveal different faces. Due to this, finding some of the faces can be quite difficult.
The committee went on to find other flexagons which could be made, again made by first folding into a shorter strip, then folding up like the trihexaflexagon.</div>
<div style="text-align: justify; text-indent: 20px;">The committee later found that hexaflexagons with any number of faces could be made by starting with a certain shaped strip, rolling it up then folding it like a trihexaflexagon.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Resources & Further Reading</h5>
<div style="text-align: left"><small>An excellent article by Martin Gardner on flexagons can be found in <a href='http://www.amazon.co.uk/Mathematical-Puzzles-Diversions-Pelican-Gardner/dp/0140207139' target='new'>this book</a>.</small></div>
<div style="text-align: left"><small>Trihexaflexagon templates (click to enlarge then print):</small></div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYjZEonFbLVA9CYEmZGAViDZ-7bhx55PWkny-rEAWoxVXVde_zcJpuCplPE2xY8pH5OnXqF04hkfaRr_u5xdiVrhTXUHp8nxKAjlUh3pzLPf9gGnj7fSI4OWcHKYoYGlPjQATLgTa0Lso/s1600/flexagon.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYjZEonFbLVA9CYEmZGAViDZ-7bhx55PWkny-rEAWoxVXVde_zcJpuCplPE2xY8pH5OnXqF04hkfaRr_u5xdiVrhTXUHp8nxKAjlUh3pzLPf9gGnj7fSI4OWcHKYoYGlPjQATLgTa0Lso/s200/flexagon.png" /></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_L-FeO9D9bdpiMyIWUkm7qyif_jABnPZDrKfKY4HZeTjwuQ9VtowXEpCVUNw7ddIYU7ZEA_pCDUhHxd2WqMYoxzm_5SUYmP2km9mFgUIH1YuIBQCcnKa_sURCCz4Rn8EgwbhgxX4iT3k/s1600/flexagon2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_L-FeO9D9bdpiMyIWUkm7qyif_jABnPZDrKfKY4HZeTjwuQ9VtowXEpCVUNw7ddIYU7ZEA_pCDUhHxd2WqMYoxzm_5SUYmP2km9mFgUIH1YuIBQCcnKa_sURCCz4Rn8EgwbhgxX4iT3k/s200/flexagon2.png" /></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5wb-WCi0TosBeMz1g-XGA8ZJIxDclsJsNe4VUL99ZjEdcORRVnrhhYBYiPB6COsnhv62M9ZT9fh_YjZVNWkM_VfEas7DIUWNrJLvl_kv5bu0hRMYWmxqkGDvmlTViiPbhhr-xAQ_yfMg/s1600/flexagon-christmas.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5wb-WCi0TosBeMz1g-XGA8ZJIxDclsJsNe4VUL99ZjEdcORRVnrhhYBYiPB6COsnhv62M9ZT9fh_YjZVNWkM_VfEas7DIUWNrJLvl_kv5bu0hRMYWmxqkGDvmlTViiPbhhr-xAQ_yfMg/s200/flexagon-christmas.png" /></a></div>
<div style="text-align: left"><small><a href='http://thinkzone.wlonk.com/Flexagon/Hexahexaflexagon.pdf' target='new'>Hexahexaflexagon template</a></small></div>
<div style="text-align: left"><small><a href='http://www.flexagon.net/' target='new'>Other templates</a></small></div>
<h4 style="margin-bottom: 0px;">Folding Tube Maps</h4>
<div style="text-align: justify; text-indent: 20px;">Our second story starts with me sitting on the tube reading <a target='new' href='http://www.amazon.co.uk/Alexs-Adventures-Numberland-Alex-Bellos/dp/1408809591/'>Alex's Adventures in Numberland</a> by Alex Bellos on the tube. In his book, Alex describes how to fold a tetrahedron, or triangle-based pyramid, from two business cards. With no business cards to hand, I picked up two tube maps and followed the steps: first, I folded it corner to corner; then I folded the overlaps over.I made another one of these, but the second a mirror image of the first, slotted them together and I had my tetrahedron.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbwrSye-m23BzM2HxoYNR4GjVMp0ROzjMM1nJ54NbxgOr4GpjyMT8dkg-G0giZUbh6bgTpBZHW_2OZiH5SuHxnH0s3jQJmd4ae9k2qqUtQvtUqd8NipyA5Pa36Bhr2Sahg47je2YfdJ-4/s1600/fold1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbwrSye-m23BzM2HxoYNR4GjVMp0ROzjMM1nJ54NbxgOr4GpjyMT8dkg-G0giZUbh6bgTpBZHW_2OZiH5SuHxnH0s3jQJmd4ae9k2qqUtQvtUqd8NipyA5Pa36Bhr2Sahg47je2YfdJ-4/s200/fold1.jpg" /></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwTAm7xYqMe0tZhuki2CPCbwTSdO7REYX0Zc5y2IV9DnEN39hBZOVy5aYOLWy-pZ3hFGNVQ8CS1A5n7UGKTl6stu5cE2zhWZB8rZxlzVeQB5FeWh5sx1dr5aIQ9_AIYX5jlRIz-RcRrow/s1600/fold2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwTAm7xYqMe0tZhuki2CPCbwTSdO7REYX0Zc5y2IV9DnEN39hBZOVy5aYOLWy-pZ3hFGNVQ8CS1A5n7UGKTl6stu5cE2zhWZB8rZxlzVeQB5FeWh5sx1dr5aIQ9_AIYX5jlRIz-RcRrow/s200/fold2.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnmYb3O0esaT9jXP3ixjYRFa_6GXZ_dcItZacFlNdukVJ8c3X5fYdmV5d72a3NZETxtjTWaXdarpQe3KR7-Tm1irtJoLamQZ_D4NyKXAnb-4JgFRVQoAoACkdRAzD1ASUeV9dfwnR_iJI/s1600/tetrahedron.jpeg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnmYb3O0esaT9jXP3ixjYRFa_6GXZ_dcItZacFlNdukVJ8c3X5fYdmV5d72a3NZETxtjTWaXdarpQe3KR7-Tm1irtJoLamQZ_D4NyKXAnb-4JgFRVQoAoACkdRAzD1ASUeV9dfwnR_iJI/s200/tetrahedron.jpeg" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Then I made a tube map cube by making six squares like so and slotting them together.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgV0HVoR_Sh1o0cheO8PyX8KrI8tb_QJ7_KijzAC-y6erWuI3oAYMIlTyPj3wzAU_dWt-QrHYEhUzsaYfO3hdtMSQp1NwJjBTK_yPl9ep2j7m2FIooE65Kg5GkYm5lcceso8CDYIDGVrpc/s1600/20121006_123346.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgV0HVoR_Sh1o0cheO8PyX8KrI8tb_QJ7_KijzAC-y6erWuI3oAYMIlTyPj3wzAU_dWt-QrHYEhUzsaYfO3hdtMSQp1NwJjBTK_yPl9ep2j7m2FIooE65Kg5GkYm5lcceso8CDYIDGVrpc/s200/20121006_123346.jpg" /></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5E3cIZbnWAxs43L533TigrIxTC0Zov7twu6Ww6RwjTUnm6mdNc31uF6YIi4zoMhmeZW43y3r5WrniHo3ND1AZFSfUZ6oNmFkWLC9Cw_QqYxp6hnrfiKgKC1f2kSZPB5ZBuzNdEaSjzuE/s1600/20121006_123414.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5E3cIZbnWAxs43L533TigrIxTC0Zov7twu6Ww6RwjTUnm6mdNc31uF6YIi4zoMhmeZW43y3r5WrniHo3ND1AZFSfUZ6oNmFkWLC9Cw_QqYxp6hnrfiKgKC1f2kSZPB5ZBuzNdEaSjzuE/s200/20121006_123414.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgszGUQqWzRxqakt6sVOpokbmPI2NbGGUjf28C8Wz6ifXj-QkW9XzE_BTCME6PYzmInTgoGJxHtbB-h93DZit6W4NBoksmUvUG3C7O5eMA86nk4g7ZBvVY75IL8xRpz_lYfHkSQWCX5ZBw/s1600/cube.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgszGUQqWzRxqakt6sVOpokbmPI2NbGGUjf28C8Wz6ifXj-QkW9XzE_BTCME6PYzmInTgoGJxHtbB-h93DZit6W4NBoksmUvUG3C7O5eMA86nk4g7ZBvVY75IL8xRpz_lYfHkSQWCX5ZBw/s200/cube.jpg" /></a></div>
<div style="text-align: justify; text-indent: 20px;">
While making these shapes, I discovered an advantage of tube maps over business cards: Due to the pages, folded tube maps have slots to tuck the tabs into, so the solids are pretty sturdy.</div>
<div style="text-align: justify; text-indent: 20px;">Making these shapes got me wondering: what other Platonic solids could I make?</div>
<div style="text-align: justify; text-indent: 20px;">In 2D, we have regular shapes: shapes with all the sides of the same length and all angles equal. Platonic solids are sort of the 3D equivalent of this: they are 3D shapes where every face is the same regular shape and at each vertex the same number of faces meet.</div>
<div style="text-align: justify; text-indent: 20px;">
For example, our tetrahedron is a Platonic solid because every side is a regular triangle, and three triangles meet at every vertex. Our cube is a Platonic solid because every side is a square (which is a regular shape) and three squares meet at every vertex.</div>
<div style="text-align: justify; text-indent: 20px;">In order to fold all the Platonic solids, we must first find out how many there are.</div>
<div style="text-align: justify; text-indent: 20px;">To do this, we're going to start with a triangle, as it is the 2D shape with the smallest number of sides, and make Platonic solids.</div>
<div style="text-align: justify; text-indent: 20px;">If we try to put two triangles at each vertex, then they'll squash flat; so that's no good. We've seen that three triangles at each vertex makes a tetrahedron. If we put four triangles at each vertex then we get an octahedron.</div>
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<div style="text-align: justify; text-indent: 20px;">Five triangles at each vertex gives us an icosahedron.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCLWDZOhyiOID3jQMAwlO51KMxXuw4O7vvUkPM_TbGZP5GjOHY9yVF9_s07q36q1ASWECRYbeMoWK3HW_jXaj30PsOa5KHfehShRk08KazUl3c9MFaQEl1uViQTuHh00io_i12hheaD-o/s1600/icosahedron.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCLWDZOhyiOID3jQMAwlO51KMxXuw4O7vvUkPM_TbGZP5GjOHY9yVF9_s07q36q1ASWECRYbeMoWK3HW_jXaj30PsOa5KHfehShRk08KazUl3c9MFaQEl1uViQTuHh00io_i12hheaD-o/s320/icosahedron.jpg" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Each angles in an equilateral triangle is 60°. So if we put six triangles at each vertex the angles add up to 360°, a full turn. This means that the triangles will lie flat, giving us a nice pattern for a kitchen floor, but not a solid. Any more than 6 triangles will add up to more than 360 and also not give a solid. So we have found all the Platonic solids whose faces are triangles.</div>
<div style="text-align: justify; text-indent: 20px;">Next, four sided faces. Three squares at each vertex gives us a cube. Four squares at each vertex will add up to 4 times 90°... 360° again, so another kitchen floor and as before we have all the Platonic solids whose faces are squares.</div>
<div style="text-align: justify; text-indent: 20px;">
Now moving up again to five sided faces. Three pentagons at each vertex will gives us a dodecahedron, which looks like this.</div>
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<div style="text-align: justify; text-indent: 20px;">This is the best I could do.</div>
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<div style="text-align: justify; text-indent: 20px;">(After the talk, I was shown a few better ways to fold pentagons. Watch this space for my attempts...) Now if we try four pentagons around a vertex: the internal angle in a pentagon is 108°. 4 times 108° is 432°. This is more than a full turn, so we don't get a solid.</div>
<div style="text-align: justify; text-indent: 20px;">Moving up again, if we take three hexagons we get another tessellation. Shapes with more than six sides will all have larger angles than this so three make more than a full turn.
Therefore, we have found and folded all the Platonic solids.</div>
<div style="text-align: justify; text-indent: 20px;">In 2012, I posted this on my blog and got the following comment:</div>
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<div style="text-align: justify; text-indent: 20px;">
I'm pretty sure this was a joke, but one hour, 48 tube maps and a lot of glue later:</div>
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<h5 style="margin-top:8px;margin-bottom: 0px;">Resources & Further Reading</h5>
<div style="text-align: left"><small><a target='new' href='http://www.amazon.co.uk/Alexs-Adventures-Numberland-Alex-Bellos/dp/1408809591/'>Alex's Adventures in Numberland</a> by Alex Bellos introduced business card folding and takes it further, finishing with a business card Menger sponge.</small></div>
<h4 style="margin-bottom: 0px;">Braiding</h4>
<div style="text-align: justify; text-indent: 20px;">A few months ago, my mother showed my a way to make braids using a cardboard octagon with a slot cut on each side and a hole in the middle.</div>
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<div style="text-align: justify; text-indent: 20px;">To make a braid, seven strands of wool are tied together, fed through the hole, then one tucked into each slot.</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiv0sDNyDwbDZXMm9EnvrRbgOpM2COyzDi8TjAuc7nxei1Ap9xzKPNNeE-efXxrNDRdPUx7TLNi_qc69iuehMaTrLNyUSg9AJDX4HIfUXryS4SjEwTDM1Lnfn_a10kanCm23H6PRt0tZDE/s1600/brai2.png" /></div>
<div style="text-align: justify; text-indent: 20px;">Now, we jump over two strands, pick the third strand and move it to the vacant slot. So first, we jump over the orange and green and move the red strand.</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimgbTfXe47r0PrT4XkpmggGcrS3eYfF733r7YDgLl7PAhnQoHEoYuQMMPL5Tz3GPr1GgTPdhgm043tMeJJKL-mvy3R87v-aJjRGDdnA8jk4JR647AS_cWeUGvXRaaLt1kpGYqzalC4YDg/s1600/brai3.png" /></div>
<div style="text-align: justify; text-indent: 20px;">Then we jump the light blue and yellow and move the dark blue.</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgb6ZOOVhAfhCIaq_rwr0koNWHnVRUXhQfLez04NWgDgUjJa94BylpQTvRtg4McmM4LAZF8QaKKNRallZ0TNvnI04R0a2tfVeCkxMEVWYkwlgSh4wrItBFY-7qRNHgTgrce4yibLu-I0aU/s1600/brai4.png" /></div>
<div style="text-align: justify; text-indent: 20px;">And so on..</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgM02tEvkPx1t7piIApHjx5Bi4QkddQ7Y8qBO_aaO3jX41YlSUrQXvvvRbNSs5FwnjGHEXFoMeHet21vM2oFbv7ydSIYXKqLiwaTpxmg6U6_6b_hJB0a_2H49pp_W0KtJOSX_GmWvyH5U8/s1600/brai5.png" /><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQRsvShCocy9auw03rGmyrsy9D9i4eYWEf3uKhNHryMMJjOw8QRuihf7wZD8XoEvk8oXeYlfaU-afUkWY1UOmvUV2_m5SbvC3IwZq56mr21X8IEQ7_kxWR9if8Yy88a-FNA6RG8vlGghQ/s1600/brai6.png" /></div>
<div style="text-align: justify; text-indent: 20px;">After a while, the braid looks like this:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBN5jxjQaJgnsk1RwPQZDqRBnWWWAGTcBWCdTzApuIiZ_hAwcN-SyEKKpeT74ASz2Kw0jZb1exbz_5afeJdoxnRPW-7TEHUWq2aUNP_CuThvvZ4vsbHGYhzs3v9nEmnveBb6z6TyiaTs4/s1600/braid.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBN5jxjQaJgnsk1RwPQZDqRBnWWWAGTcBWCdTzApuIiZ_hAwcN-SyEKKpeT74ASz2Kw0jZb1exbz_5afeJdoxnRPW-7TEHUWq2aUNP_CuThvvZ4vsbHGYhzs3v9nEmnveBb6z6TyiaTs4/s320/braid.jpg" /></a></div>
<div style="text-align: justify; text-indent: 20px;">Once I'd made a few braids, I began to wonder which other numbers of threads could be used to make braids like this. To investigate this I found it useful to represent braids by drawing connections to show where a thread is moved. This shows the first move:</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDVAJpdWB-ytoZakW-uAk_rFxjOEsjbpyCeQqCvmc2xE2Ks7MrZALYen6wJMPkMxRJYrIVoaRlFYneTegwh1djeIwjLN51-qg3WzMZh7oU157sUCj1wGJ0prY3bSZGvfmbUI0ZLFqM17I/s1600/brai+the+second1.png" /></div>
<div style="text-align: justify; text-indent: 20px;">Then the second move:</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjURt2lfvx9cdIWwfzy9m4ytSZ_gTw8OTulD9f6am5Dz4BrXI5T2c2F-v4oLc91u02Eh4e9ca7alRATQgT3RuIWjJTB9h8enSnrHgfLtj9WldJi9BzJPYbxvEg81xhhy5-fYUbLd_FcwfA/s1600/brai+the+second2.png" /></div>
<div style="text-align: justify; text-indent: 20px;">And so on until you get:</div>
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<div style="text-align: justify; text-indent: 20px;">After the octagon, I tried braiding on a hexagon, moving the second thread each time. Here's what happened:</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixR0Ku3WWUI9ETuqXVsCjVvYx9KpqNhK-dPj-26kt8JDjoTbuiNWK1mBLPji9AJtYW_oc1ksA2iLzmd1WwrtrTulyvfKgCiAqaXdOJko8JG4OrsQNh27_eF-d1qCn8oNMKEYdFx3zMwcU/s1600/brai+the+third1.png" /><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5IU3A76k1emaRoK2NgqKBUyOHM4XYw5PrW4B0KKdQUzbCi4hAvk0ngR90xK48PEW6ViyksfreMN_08QNhDXzFobsSn3z6hhFUWNGgpsFBB691onxU9Kmve3RMnQnI1jc_-sOFBDbCUV0/s1600/brai+the+third2.png" /><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX60dHqP4u1Zw97rZfpQ3NRY-Y2EueKX5eKZBp5fzwM71Qw3rk7CKy_5fTysZQyw_hSqG_33I-1bQpRDx_O8fIIi0YA6aqCAoAkelFTwlJfu6fFbzS0VhA44YLGYSNjCp3LXcdI8Aozj0/s1600/brai+the+third3.png" /><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYxsqDGSyoFADIoponBYR63yms3OiKp51jBAHJQUfF4oZkC-c_wLLKQrYBLY6TqR3nglbe4Y7E8JolZ2tyEzNfrOnlXMlIxhNswNpD1RPlrQlpkhiQjpYmNexWDcUwnM0A9ySf1Almw_U/s1600/brai+the+third4.png" /></div>
<div style="text-align: justify; text-indent: 20px;">I only moved the yellow and green threads and nothing interesting happened. When I drew this out as before, it demonstrated what had gone wrong: three slots are missed so three threads are never moved.</div>
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<div style="text-align: justify; text-indent: 20px;">
So we need to find out when slots are missed and when all the slots are hit. To do this, let's call the number of slots \(a\), and let \(b\) be the number thread we pick each time. For example, in the first braid that worked \(a\) was 8 and \(b\) was 3. <small>This question is very similar to the <a target='new' href='http://legavrik.blogspot.co.uk/2014/07/sunday-afternoon-maths-xxi.html'>Wool Circles Problem from Sunday Afternoon Maths XXI</a>. You might like to have a go at it before you read on.</small></div>
<div style="text-align: justify; text-indent: 20px;">
First we'll label the slots. Label the slot which starts empty 0, then number anti-clockwise. This numbering puts all the multiples of \(a\) at the bottom slot.</div>
<div style="text-align: justify; text-indent: 20px;">Now let's look at which slots we visit. We start at0, then visit \(b\), then \(2b\), then \(3b\) and so on. We visit all the multiples of \(b\).</div>
<div style="text-align: justify; text-indent: 20px;">
Therefore we will reach the bottom slot again and finish our loop when we reach a common multiple of \(a\) and \(b\). The first time this happens will be at the lowest common multiple, or:</div>
$$\mbox{lcm}(a,b)$$
<div style="text-align: justify; text-indent: 20px;">
On our way to this slot, we visited one slot for every \(a\) we passed, so the number of slots we have visited is</div>
$$\frac{\mbox{lcm}(a,b)}{a}$$
<div style="text-align: justify; text-indent: 20px;">
and we will visit every slot if</div>
$$\frac{\mbox{lcm}(a,b)}{a}=b$$
<div style="text-align: justify; text-indent: 20px;">
or, equivalently if</div>
$$\mbox{lcm}(a,b)=ab.$$
<div style="text-align: justify; text-indent: 20px;">
This is true when, \(a\) and \(b\) have no common factors, or in other words are coprime; which can be written</div>
$$\mbox{hcf}(a,b)=1.$$
<div style="text-align: justify; text-indent: 20px;">
So we've found that if \(a\) is the number of slots and \(b\) is the jump then the braid will not work unless \(a\) and \(b\) are coprime.</div>
<div style="text-align: justify; text-indent: 20px;">For example, if \(a\) is 6 and \(b\) is 2 then 2 is a common factor so the braid fails. And, if \(a\) is 8 and \(b\) is 3 then there are no common factors and the braid works. And, if \(a\) is 12 and \(b\) is 5 then there are no common factors and the braid works.</div>
<div style="text-align: justify; text-indent: 20px;">But, if \(a\) is 5 and \(b\) is 2 then there are no common factors but the braid fails.</div>
<div style="text-align: justify; text-indent: 20px;">The rule I've explained is still correct, and explains why some braids fail. But if \(a\) and \(b\) are coprime, we need more rules to decide whether or not the braid works.</div>
<div style="text-align: justify; text-indent: 20px;">And that's as far as I've got, so I'm going to finish braiding with two open questions: Why does the 5 and 2 braid fail? And for which numbers \(a\) and \(b\) does the braid work?</div>
<h4 style="margin-bottom: 0px;">Sine Curves</h4>
<div style="text-align: justify; text-indent: 20px;">For the last part of the talk, I did a practical demonstration of how to draw a sine curve using five people.</div>
<div style="text-align: justify; text-indent: 20px;">I told the first person to stand on the spot and the second person to stand one step away, hold a length of string and walk.</div>
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<div style="text-align: justify; text-indent: 20px;">The third person was instructed to stay in line with the second person, while staying on a vertical line.</div>
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<div style="text-align: justify; text-indent: 20px;">The fourth person was told to walk in a straight line at a constant speed.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglIrbk6cR8nESC-Jn8Bg6IJGlxBaknJ2ScMDz08qWZtC2APKFo8LF6MupQsp5MZHOCHpQ6WeelnKM3YlmiAUXL7b5BmmnbA4m2FfgpW0Rp-nGwCbu1szUqujLTIuzrKrVzceUK5OgfHfI/s1600/sine3.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglIrbk6cR8nESC-Jn8Bg6IJGlxBaknJ2ScMDz08qWZtC2APKFo8LF6MupQsp5MZHOCHpQ6WeelnKM3YlmiAUXL7b5BmmnbA4m2FfgpW0Rp-nGwCbu1szUqujLTIuzrKrVzceUK5OgfHfI/s320/sine3.gif" /></a></div>
<div style="text-align: justify; text-indent: 20px;">And the fifth person had to stay in line with both the third and fourth people. This led them to trace a sine curve.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQL5inruNiSoCZf1t6e7hISeXsyzVgVKxhx2YjgxpJXEPA3aszlllSKkr_7i-3PGQuJkrOsU2JWcknlPAEuzqjYCdGT1Rk80q5ieJjYe7XA4DjtS8nW1UWkRVUPlqZw32aK2GtfHljMuc/s1600/sine4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQL5inruNiSoCZf1t6e7hISeXsyzVgVKxhx2YjgxpJXEPA3aszlllSKkr_7i-3PGQuJkrOsU2JWcknlPAEuzqjYCdGT1Rk80q5ieJjYe7XA4DjtS8nW1UWkRVUPlqZw32aK2GtfHljMuc/s320/sine4.gif" /></a></div>
<div style="text-align: justify; text-indent: 20px;">To explain why this is a sine curve, consider the following triangle:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghBHs7RymlhEJC9vRsG-GNmtoe-1QrudXh3ciWkZzE1lKFqwUxnZR-Mt6r4au39psDBS-6HOX7n24Ey-BP30fjTP_cdYZskzO66fyzLYUqUCdKJkOCtt-bf5QH3YRm73vH9-FB6rNFOJE/s1600/sine1+and+a+half.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghBHs7RymlhEJC9vRsG-GNmtoe-1QrudXh3ciWkZzE1lKFqwUxnZR-Mt6r4au39psDBS-6HOX7n24Ey-BP30fjTP_cdYZskzO66fyzLYUqUCdKJkOCtt-bf5QH3YRm73vH9-FB6rNFOJE/s320/sine1+and+a+half.gif" /></a></div>
<div style="text-align: justify; text-indent: 20px;">As our first two people are one step apart, the hypotenuse of this triangle is 1. And so the opposite (vertical) side is equal to the sine of the angle.</div>
<div style="text-align: justify; text-indent: 20px;">I like to finish with a challenge, and this task leads nicely into two challenge questions:</div>
<div style="text-align: justify; text-indent: 20px;">1. How could you draw a cosine curve with five people?</div>
<div style="text-align: justify; text-indent: 20px;">2. How could you draw a tan(gent) curve with five people?</div>
<div style="text-align: justify; text-indent: 20px;"><small>(These questions will be one of <a href='http://legavrik.blogspot.com/2014/09/sunday-afternoon-maths-xxvii.html'>this week's Sunday Afternoon Maths</a> puzzles, so I'll post the answer on Monday.)</small></div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Resources & Further Reading</h5>
<div style="text-align: left"><small><a href='http://www.atm.org.uk/Maths-Book-Reviews/people-maths-hidden-depths/67757' target='new'>People Maths: Hidden Depths</a> is full of this kind of dynamic task involving moving people.</small></div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com1tag:blogger.com,1999:blog-6331909486660092910.post-43188527522694143842014-08-25T07:31:00.000+01:002014-08-25T07:31:00.231+01:00Sunday Afternoon Maths XXVI Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/08/sunday-afternoon-maths-xxvi.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Twenty-One</h4>
<div style="text-align: justify; text-indent: 20px;">Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.</div>
<div style="text-align: justify; text-indent: 20px;">Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about \(n\)?</div>
<div style="text-align: justify; text-indent: 20px;">(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about \(m\) to \(l\)?</div>
<div style="text-align: justify; text-indent: 20px;">(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are \(k\) people?</div>
<h4 style="margin-bottom: 0px;">Odd and Even Outputs</h4>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
\(g(n,m)=1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
\(g(n,m)=n\times m + n + m\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
\(g(n,m)=n\times m +n+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
\(g(n,m)=m\)
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
\(g(n,m)=n\times m+m+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
\(g(n,m)=n\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
\(g(n,m)=n+m+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
\(g(n,m)=n\times m\)
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
\(g(n,m)=n\times m+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
\(g(n,m)=n+m\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
\(g(n,m)=n+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
\(g(n,m)=n\times m+n\)
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
\(g(n,m)=m+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
\(g(n,m)=n\times m+n\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
\(g(n,m)=n\times m+n+m+1\)
<br /><br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
\(g(n,m)=2\)
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Can you find functions \(h:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) (call the inputs \(n\), \(m\) and \(l\)) to give the following outputs:</div>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'><tr><td>
<center>\(l\) <b>odd</b></center>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td><td>
<center>\(l\) <b>even</b></center>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td></tr></table>
<br />
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'><tr><td>
<center>\(l\) <b>odd</b></center>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td><td>
<center>\(l\) <b>even</b></center>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
</td></tr></table>
<br />
<big>etc</big>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-44780329317776811572014-08-24T11:49:00.000+01:002014-08-24T11:49:00.228+01:00Sunday Afternoon Maths XXVI<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<div style="text-align: justify; text-indent: 20px;">It's no late to get <a href='http://www.emfcamp.org/tickets/choose' target='new'>tickets</a> for <a href='http://www.emfcamp.org' target='new'>EMF Camp</a>, where I will be giving a talk on <a href='http://legavrik.blogspot.co.uk/search/label/Flexagons' target='new'>flexagons</a>, <a href='http://legavrik.blogspot.co.uk/search/label/Folding Tube Maps' target='new'>folding tube maps</a> and braiding. </div>
<h4 style="margin-bottom: 0px;">Twenty-One</h4>
<div style="text-align: justify; text-indent: 20px;">Scott and Virgil are playing a game. In the game the first player says 1, 2 or 3, then the next player can add 1, 2 or 3 to the number and so on. The player who is forced to say 21 or above loses. The first game went like so:</div>
<div style="text-align: left; margin-left: 40px">Scott: 3<br />Virgil: 4<br />Scott: 5<br />Virgil: 6<br />Scott: 9<br />Virgil: 12<br />Scott: 15<br />Virgil 17<br />Scott: 20<br />Virgil: 21<br />Virgil loses.
</div>
<div style="text-align: justify; text-indent: 20px;">To give him a better chance of winning, Scott lets Virgil choose whether to go first or second in the next game. What should Virgil do?</div>
<h4 style="margin-bottom: 0px;">Odd and Even Outputs</h4>
<div style="text-align: justify; text-indent: 20px;">Let \(g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) be a function.</div>
<div style="text-align: justify; text-indent: 20px;">This means that \(g\) takes two natural number inputs and gives one natural number output. For example if \(g\) is defined by \(g(n,m)=n+m\) then \(g(3,4)=7\) and \(g(10,2)=12\).</div>
<div style="text-align: justify; text-indent: 20px;">The function \(g(n,m)=n+m\) will give an even output if \(n\) and \(m\) are both odd or both even and an odd output if one is odd and the other is even. This could be summarised in the following table:</div>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
<div style="text-align: justify; text-indent: 20px;">Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:</div>
<table>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>odd</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>odd</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td>
</tr>
<tr>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>odd</td><td>even</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>odd</td></tr>
</table>
</td>
<td>
<table border=1 style='border-collapse:collapse;border-top:0px;border-left:0px'>
<tr><td colspan=2 rowspan=2></td><td colspan=2 align=center>\(n\)</td></tr>
<tr><td><b>odd</b></td><td><b>even</b></td>
<tr><td rowspan=2>\(m\)</td><td><b>odd</b></td><td>even</td><td>even</td></tr>
<tr><td><b>e</b></td><td>even</td><td>even</td></tr>
</table>
</td>
</tr>
</table>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-56576490456162067532014-08-18T07:29:00.000+01:002014-08-18T07:29:00.338+01:00Sunday Afternoon Maths XXV Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/08/sunday-afternoon-maths-xxv.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Whist</h4>
<div style="text-align: justify; text-indent: 20px;">Represent each player by a compass point. Let B, D, A and S represent Messrs. Banker, Dentist, Apothecary and Scrivener respectively and b, d, a and s represent the four jobs.</div>
<div style="text-align: justify; text-indent: 20px;">Mr. Banker (B) partners the scrivener (s). Let B sit at West. This means s sits at East. As no other player can also be B or s, the table looks like this:</div>
<style type='text/css'>
table.poss, table.poss td {text-align:center;border:1px solid grey;}
table.poss td.str {background-color:grey; text-decoration:line-through}
</style>
<table align=center>
<tr><td></td><td><table class='poss'><tr><td colspan=4>North</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td>S</td></tr><tr><td>b</td><td>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
<tr><td><table class='poss'><tr><td colspan=4>West</td></tr><tr><td>B</td><td class='str'>D</td><td class='str'>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td><td><table class='poss'><tr><td colspan=4>East</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td class='str'>d</td><td class='str'>a</td><td>s</td></tr></table></td></tr>
<tr><td></td><td><table class='poss'><tr><td colspan=4>South</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td>S</td></tr><tr><td>b</td><td>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
</table>
<div style="text-align: justify; text-indent: 20px;">The dentist (d) sits on Mr. Scrivener's (S) right. East cannot be S, so North cannot be d. East cannot be d, so South cannot be S.</div>
<table align=center>
<tr><td></td><td><table class='poss'><tr><td colspan=4>North</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td>S</td></tr><tr><td>b</td><td class='str'>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
<tr><td><table class='poss'><tr><td colspan=4>West</td></tr><tr><td>B</td><td class='str'>D</td><td class='str'>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td><td><table class='poss'><tr><td colspan=4>East</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td class='str'>d</td><td class='str'>a</td><td>s</td></tr></table></td></tr>
<tr><td></td><td><table class='poss'><tr><td colspan=4>South</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td class='str'>S</td></tr><tr><td>b</td><td>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
</table>
<div style="text-align: justify; text-indent: 20px;">By elimination, only North can be S. This means that d must sit to the right of North (at West):</div>
<table align=center>
<tr><td></td><td><table class='poss'><tr><td colspan=4>North</td></tr><tr><td class='str'>B</td><td class='str'>D</td><td class='str'>A</td><td>S</td></tr><tr><td>b</td><td class='str'>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
<tr><td><table class='poss'><tr><td colspan=4>West</td></tr><tr><td>B</td><td class='str'>D</td><td class='str'>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td>d</td><td class='str'>a</td><td class='str'>s</td></tr></table></td><td></td><td><table class='poss'><tr><td colspan=4>East</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td class='str'>d</td><td class='str'>a</td><td>s</td></tr></table></td></tr>
<tr><td></td><td><table class='poss'><tr><td colspan=4>South</td></tr><tr><td class='str'>B</td><td>D</td><td>A</td><td class='str'>S</td></tr><tr><td>b</td><td class='str'>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
</table>
<div style="text-align: justify; text-indent: 20px;">A and a are partners. This is only possible if A is South and a is North:</div>
<table align=center>
<tr><td></td><td><table class='poss'><tr><td colspan=4>North</td></tr><tr><td class='str'>B</td><td class='str'>D</td><td class='str'>A</td><td>S</td></tr><tr><td class='str'>b</td><td class='str'>d</td><td>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
<tr><td><table class='poss'><tr><td colspan=4>West</td></tr><tr><td>B</td><td class='str'>D</td><td class='str'>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td>d</td><td class='str'>a</td><td class='str'>s</td></tr></table></td><td></td><td><table class='poss'><tr><td colspan=4>East</td></tr><tr><td class='str'>B</td><td>D</td><td class='str'>A</td><td class='str'>S</td></tr><tr><td class='str'>b</td><td class='str'>d</td><td class='str'>a</td><td>s</td></tr></table></td></tr>
<tr><td></td><td><table class='poss'><tr><td colspan=4>South</td></tr><tr><td class='str'>B</td><td class='str'>D</td><td>A</td><td class='str'>S</td></tr><tr><td>b</td><td class='str'>d</td><td class='str'>a</td><td class='str'>s</td></tr></table></td><td></td></tr>
</table>
<div style="text-align: justify; text-indent: 20px;">Therefore, <b>Mr. Banker the dentist</b> sits to the left of the banker.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?</div>
<h4 style="margin-bottom: 0px;">Polya Strikes Out</h4>
<div style="text-align: left;margin-left:30px">1, <s>3</s>, 7, <s>12</s>, 19, ...</div>
<div style="text-align: left;margin-left:30px">1, 7, 19, ...</div>
<div style="text-align: left;margin-left:30px">1=1; 1+7=8; 1+7+19=27; ...</div>
<div style="text-align: left;margin-left:30px">1, 8, 27, ...</div>
<div style="text-align: justify; text-indent: 20px;">The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.</div>
<div style="text-align: justify; text-indent: 20px;">Cross out every third number:</div>
<div style="text-align: left;margin-left:30px">1, 2, <s>3</s>, 4, 5, <s>6</s>, ..., <s>3n</s>, \(3n+1\), \(3n+2\), ...</div>
<div style="text-align: left;margin-left:30px">1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...</div>
<div style="text-align: justify; text-indent: 20px;">Find the cumulative sums:</div>
$$1=1$$
$$1+2=1+2=3$$
$$1+2+4=1+2+3+4-3=7$$
$$1+2+4+5=1+2+3+4+5-3=12$$
$$...$$
$$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$
$$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$
$$=3n^2+3n+1$$
$$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$
$$=3n^2+6n+3$$
$$...$$
<div style="text-align: left;margin-left:30px">1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...</div>
<div style="text-align: justify; text-indent: 20px;">Cross out every second number, starting with the second:</div>
<div style="text-align: left;margin-left:30px">1, <s>3</s>, 7, <s>12</s>, ..., \(3n^2+3n+1\), <s>3n2+6n+3</s>, ...</div>
<div style="text-align: left;margin-left:30px">1, 7, ..., \(3n^2+3n+1\), ...</div>
<div style="text-align: justify; text-indent: 20px;">Find the cumulative sums. The \(m\)<sup>th</sup> sum is:</div>
$$\sum_{n=0}^{m}3n^2+3n+1$$
$$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$
$$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$
$$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+4m+2)$$
$$=(m+1)(m^2+2m+1)$$
$$=(m+1)(m+1)^2$$
$$=(m+1)^3$$
<div style="text-align: justify; text-indent: 20px;">Hence the numbers obtained are the cube numbers.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)<sup>th</sup> number starting at the \(m\)<sup>th</sup>?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-59518715087545862452014-08-17T11:53:00.000+01:002014-08-17T11:53:00.602+01:00Sunday Afternoon Maths XXV<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>. This Tuesday is <a target="new" href="http://www.mathsjam.com">Maths Jam</a>, make sure you're there!</div>
<div style="text-align: justify; text-indent: 20px;">It's less than two weeks 'til <a href='http://www.emfcamp.org' target='new'>EMF Camp</a>, where <a href='http://www.emfcamp.org/talks' target='new'>there will be talks on maths and science</a> and a whole load of <a href='http://www.emfcamp.org/about' target='new'>other awesome stuff</a>. <a href='http://www.emfcamp.org/tickets/choose' target='new'>Tickets here</a></div>
<h4 style="margin-bottom: 0px;">Whist</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Puzzles-Logic-Reasoning-Dover-Recreational/dp/0486201198' target='new'>Source: <i>My Best Puzzles in Logic and Reasoning</i> by Hubert Phillips</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Messrs. Banker, Dentist, Apothecary and Scrivener played whist last night. <small>(whist is a four player card game where partners sit opposite each other.)</small> Each of these gentlemen is the namesake of another's vocation.</div>
<div style="text-align: justify; text-indent: 20px;">Last night, the apothecary partnered Mr. Apothecary; Mr. Banker's partner was the scrivener; on Mr. Scrivener's right sat the dentist.</div>
<div style="text-align: justify; text-indent: 20px;">Who sat on the banker's left?</div>
<h4 style="margin-bottom: 0px;">Polya Strikes Out</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Thinking-Mathematically-John-Mason/dp/0201102382' target='new'>Source: <i>Thinking Mathematically</i> by John Mason, Leone Burton & Kaye Stacey</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Write the numbers 1, 2, 3, ... in a row. Strike out every third number beginning with the third. Write down the cumulative sums of what remains:</div>
<div style="text-align: left;margin-left:30px">1, 2, 3, 4, 5, 6, 7, ...</div>
<div style="text-align: left;margin-left:30px">1, 2, <s>3</s>, 4, 5, <s>6</s>, 7, ...</div>
<div style="text-align: left;margin-left:30px">1, 2, 4, 5, 7, ...</div>
<div style="text-align: left;margin-left:30px">1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...</div>
<div style="text-align: left;margin-left:30px">1, 3, 7, 12, 19, ...</div>
<div style="text-align: justify; text-indent: 20px;">Now strike out every second number beginning with the second. Write down the cumulative sums of what remains. What is the final sequence? Why do you get this sequence?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-66848404923402668802014-08-11T07:14:00.000+01:002014-08-11T07:14:00.500+01:00Sunday Afternoon Maths XXIV Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/08/sunday-afternoon-maths-xxiv.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Square Cross</h4>
<div style="text-align: justify; text-indent: 20px;">Label and orient the shape as follows:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1MMStmbY7TuiZ_DbhN-IZf5RaTbwLkGXU18_u7AJ_t3917dCmDiQc611meo3KQcZjGW_x6hUmK_yJ3crjaILCGGyQ8RB2O4mCi4vTUm8AMIlyY2Qp3o9QkoJpt0anLey-KHle-UGUWfw/s1600/cross+square.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1MMStmbY7TuiZ_DbhN-IZf5RaTbwLkGXU18_u7AJ_t3917dCmDiQc611meo3KQcZjGW_x6hUmK_yJ3crjaILCGGyQ8RB2O4mCi4vTUm8AMIlyY2Qp3o9QkoJpt0anLey-KHle-UGUWfw/s320/cross+square.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.</div>
<div style="text-align: justify; text-indent: 20px;">The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation \(y=2x-4\) and ON has equation \(y=\frac{5}{2}-\frac{1}{2}x\). These lines intersect at \((\frac{13}{5},\frac{6}{5})\), these are the co-ordinates of N. By the same method, the co-ordinates of P are \((-\frac{8}{5},-\frac{1}{5})\).</div>
<div style="text-align: justify; text-indent: 20px;">By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is <b>9.8</b>.</div>
<h4 style="margin-bottom: 0px;">Exact Change</h4>
<div style="text-align: justify; text-indent: 20px;">In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.</div>
<div style="text-align: justify; text-indent: 20px;">In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: left"><small><a href='https://twitter.com/AlexDBolton/status/494682185467981824' target='new'>Source: @AlexDBolton on Twitter</a></small></div>
<div style="text-align: justify; text-indent: 20px;">In a far away country, the unit of currency is the #, which is split into 100@ <small>(# is like £ or $; @ is like p or ¢)</small>.</div>
<div style="text-align: justify; text-indent: 20px;">Let C be the number of coins less than #1. Let P be the number of coins needed to make any value between 1@ and 99@. Which coins should be the country mint to minimise the value of P+C?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-37509434949316234632014-08-10T12:13:00.000+01:002014-08-10T12:13:00.897+01:00Sunday Afternoon Maths XXIV<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Square Cross</h4>
<div style="text-align: left"><small><a href='http://teachfurthermaths.weebly.com/puzzle-of-the-month/august-puzzle' target='new'>Source: Teach Further Maths Blog</a></small></div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggDGISRyhG-PmkrZ60Yh9xoaHc2Np8aS8EsV-hBbmBduKzj9kPDB5ULsTzT5C8YGjdJ1d7XA5Odoy1AWHx9sskqa6dGzrLC0GX1vdu_nOO8woBJ-nxWV06CAMjQ-_m7k3ZqsXLy3pizs8/s320/959058_orig.png" /></div>
<div style="text-align: justify; text-indent: 20px;">A figure in the shape of a cross is made from five 1 x 1 squares, as shown. The cross is inscribed in a large square whose sides are parallel to the dashed square, formed by four vertices of the cross.</div>
<div style="text-align: justify; text-indent: 20px;">What is the area of the large outer square?</div>
<h4 style="margin-bottom: 0px;">Exact Change</h4>
<div style="text-align: left"><small><a href='https://twitter.com/AlexDBolton/status/494681380177989632' target='new'>Source: @AlexDBolton on Twitter</a></small></div>
<div style="text-align: justify; text-indent: 20px;">In the UK, the coins less than £1 are 1p, 2p, 5p, 10p, 20p and 50p. How many coins would I need to carry in my pocket so that I could make any value from 1p to 99p?</div>
<div style="text-align: justify; text-indent: 20px;">In the US, the coins less than $1 are 1¢, 5¢, 10¢, 25¢. How many coins would I need to carry in my pocket so that I could make any value from 1¢ to 99¢?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-42743982716702986662014-08-04T07:04:00.000+01:002014-08-04T07:04:00.824+01:00Sunday Afternoon Maths XXIII Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/08/sunday-afternoon-maths-xxiii.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Let the Passenger Train Through!</h4>
<div style="text-align: justify; text-indent: 20px;">The goods train reverses into the siding, leaving three trucks, then reverses out of the way.</div>
<div style="text-align: justify; text-indent: 20px;">The passenger train attaches the three trucks to its front then reverses out of the way.</div>
<div style="text-align: justify; text-indent: 20px;">The goods train backs into the siding.</div>
<div style="text-align: justify; text-indent: 20px;">The passenger train drives past the siding.</div>
<div style="text-align: justify; text-indent: 20px;">The goods train drives forwards out of the way.</div>
<div style="text-align: justify; text-indent: 20px;">The passenger train reverses, then drives forward into the siding, leaving the three trucks. It then drives forward out of the way.</div>
<div style="text-align: justify; text-indent: 20px;">The goods train reverses into the siding, picking up the trucks. The trains have now passed each other.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">1. If the goods train can drive forwards into the siding, how could it let the passenger train through?</div>
<div style="text-align: justify; text-indent: 20px;">2. If the goods train had 8 trucks, how could it let the passenger train through?</div>
<h4 style="margin-bottom: 0px;">Cooked Turkey</h4>
<div style="text-align: justify; text-indent: 20px;">Let the first digit be \(a\) and the final digit be \(b\). The cost of the 72 turkeys (in pence) is \(10000a+6790+b\). This must be divisible by 72.</div>
$$10000a+6790+b=(138\times 72+64)a+(94\times 72+22)+b$$
$$=(138a+94)\times 72+64a+22+b$$
<div style="text-align: justify; text-indent: 20px;">\(10000a+6790+b\) is divisible by 72, so \(64a+22+b\) is divisible by 72. This means that \(b\) must be even (as 64,22,72 are all even). Let \(b=2c\).</div>
<div style="text-align: justify; text-indent: 20px;">Dividing by two, we find that \(32a+11+c\) is divisible by 36. \(c\) must be odd, so that \(32a+11+c\) is even. Let \(c=2d+1\) and so \(b=4d+2\).</div>
<div style="text-align: justify; text-indent: 20px;">Dividing by two again, we find that \(16a+6+d\) is divisible by 18. \(d\) must be even, so that \(16a+6+d\) is even. Let \(d=2e\) and so \(b=8e+2\). But \(b\) is a single digit number, so \(e=0\), \(b=2\) and \(16a+6\) is divisible by 18.</div>
<div style="text-align: justify; text-indent: 20px;">Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).</div>
<div style="text-align: justify; text-indent: 20px;">Dividing by three, we find that \(8f+1\) is divisible by 3. \(8f+1=6f+2f+1\) so \(2f+1\) is divisible by 3. This will be true when \(f=1,4,7,10,...\). \(a\) must be a single digit, so \(f=1\) and \(a=3\). And so the price of the 72 turkeys is £367.92, and one turkey will cost £5.11.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?</div>
<h4 style="margin-bottom: 0px;">Mrs. Coldcream</h4>
<div style="text-align: justify; text-indent: 20px;">Let \(G\) be the number of girls, and £\(x\) be the amount each girl was originally to receive. This means that there are \(19-G\) boys who were to receive £\(x+30\). The total scholarship is £1000, so:</div>
$$xG+(x+30)(19-G)=1000$$
<div style="text-align: justify; text-indent: 20px;">This simplifies to:</div>
$$19x-30G=430$$
<div style="text-align: justify; text-indent: 20px;">\(G\) is less than 19, so the only integer solution of this equation is \(x=40\) and \(G=11\).</div>
<div style="text-align: justify; text-indent: 20px;">After the funds are redistributed, each girl therefore receives £48. Let \(B\) be the amount each boy will now receive. The total is still £1000, so:</div>
$$48\times 11+8B=1000$$
<div style="text-align: justify; text-indent: 20px;">Which can be solved to find \(B=59\).</div>
<div style="text-align: justify; text-indent: 20px;">Each girl receives £48 and each boy receives £59.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Mrs. Coldcream still feels that this is unfair. How much more should be added to each girl's scholarship (resizing the boys' accordingly) so that the girls get at least as much as the boys?</div>
<h4 style="margin-bottom: 0px;">Ten Digit Number</h4>
<div style="text-align: justify; text-indent: 20px;">6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Are there any more numbers like this? Prove that you have them all.</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com1tag:blogger.com,1999:blog-6331909486660092910.post-91255063714522560762014-08-03T12:02:00.000+01:002014-08-03T12:02:00.590+01:00Sunday Afternoon Maths XXIII<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Let the Passenger Train Through!</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/The-Moscow-Puzzles-Mathematical-Recreational/dp/0486270785' target='new'>Source: <i>The Moscow Puzzles</i> by Boris A. Kordemsky</a></small></div>
<div style="text-align: justify; text-indent: 20px;">A goods train, made up of a locomotive and 5 trucks, stops at a small station. The small station has a siding which the goods train can reverse into. The siding can hold an engine and two trucks or three trucks.</div>
<div style="text-align: justify; text-indent: 20px;">A passenger train arrives travelling in the opposite direction. How can they let it through? <small>(The passenger train is too long to fit in the siding.)</small></div>
<h4 style="margin-bottom: 0px;">Cooked Turkey</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Penguin-Curious-Interesting-Puzzles-science/dp/0140148752' target='new'>Source: <i>The Penguin Book of Curious and Interesting Puzzles</i> by David Wells</a></small></div>
<div style="text-align: justify; text-indent: 20px;">An old invoice showed that seventy-two turkeys had been purchased for "—67.9—". The first and last digits were illegible.</div>
<div style="text-align: justify; text-indent: 20px;">How much did one turkey cost?</div>
<h4 style="margin-bottom: 0px;">Mrs. Coldcream Objected</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Best-Puzzles-Mathematics-Hubert-Phillips/dp/0486200914' target='new'>Source: <i>My Best Puzzles in Mathematics</i> by Hubert Phillips</a></small></div>
<div style="text-align: justify; text-indent: 20px;">"I object," said Councillor Mrs. Coldcream. "I see no reason why the boys should be so favoured at the expense of the girls."</div>
<div style="text-align: justify; text-indent: 20px;">This was at a meeting of the Holmshire Education Committee. It had been proposed to award 19 scholarships totalling £1000 to boys and girls of the county. It had been proposed that each girl receive a set amount and each boy receive £30 more than each girl.</div>
<div style="text-align: justify; text-indent: 20px;">Mrs. Coldcream pressed her point with such fervour that it was decided to reallocate the money. Each girl would receive £8 more than originally proposed, with the boys' scholarships scaled down accordingly.</div>
<div style="text-align: justify; text-indent: 20px;">How much did each boy and each girl receive?</div>
<h4 style="margin-bottom: 0px;">Ten Digit Number</h4>
<div style="text-align: left"><small><a href='http://richardwiseman.wordpress.com/2014/08/01/its-the-friday-puzzle-270/' target='new'>Source: Richard Wiseman's Blog</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Can you create a 10-digit number, where the first digit is how many zeros in the number, the second digit is how many 1s in the number etc. until the tenth digit which is how many 9s in the number?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-75166399682885663482014-07-28T07:21:00.000+01:002014-07-28T07:21:00.284+01:00Sunday Afternoon Maths XXII Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/07/sunday-afternoon-maths-xxii.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Coming and Going</h4>
<div style="text-align: justify; text-indent: 20px;">Add up the number of doors leaving each room; call the sum \(S\). As the number in each room is even, \(S\) will be even. Each interior door has been counted twice (as they can be seen in two rooms) and each exterior door has been counted once. Let \(I\) be the number of interior doors and \(E\) be the number of exterior doors. We have:</div>
$$S=2I+E$$$$E=S-2I$$
<div style="text-align: justify; text-indent: 20px;">\(S\) and \(2I\) are even, so \(E\) must be even.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">If the number of doors in each room is odd, is the number of exterior doors odd or even?</div>
<h4 style="margin-bottom: 0px;">Overlapping Triangles</h4>
<div style="text-align: justify; text-indent: 20px;">Let \(S\) be the area of the large square, \(T\) be the area of one of the large triangles, \(U\) be one of the red overlaps and \(V\) be the uncovered blue square. We can write:</div>
$$S=4T-4U+V$$
<div style="text-align: justify; text-indent: 20px;">as the area of the square is the total of the four triangles, take away the overlaps as they have been double counted, add the blue square as it has been missed.</div>
<div style="text-align: justify; text-indent: 20px;">We know that \(4U=V\), so:</div>
$$S=4T-V+V$$$$S=4T$$
<div style="text-align: justify; text-indent: 20px;">Therefore one of the triangles covers one quarter of the square.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Five congruent triangles are drawn in a regular pentagon. The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large pentagon does each triangle take up?</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOWGyuGnso0BhFBFkpsdsR4AxReyD_eYaG1Jd9gZJE9CegvSsMj3HSePXZ3soJuROkyy1MM9_y7LI9TTobmWga7IHJ-aQ8kPOdJK7cI8mV8FR0IS4zIkQFlaELFUR6tAkx7wbqde1ae9M/s1600/pent.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOWGyuGnso0BhFBFkpsdsR4AxReyD_eYaG1Jd9gZJE9CegvSsMj3HSePXZ3soJuROkyy1MM9_y7LI9TTobmWga7IHJ-aQ8kPOdJK7cI8mV8FR0IS4zIkQFlaELFUR6tAkx7wbqde1ae9M/s200/pent.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">\(n\) congruent triangles are drawn in a regular \(n\) sided polygon. The total area which the triangles overlap is equal to the area they don't cover. What proportion of the area of the large \(n\) sided polygon does each triangle take up?</div>Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-83328371581762931232014-07-27T11:02:00.000+01:002014-07-28T08:42:55.918+01:00Sunday Afternoon Maths XXII<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Coming and Going</h4>
<div style="text-align: left"><small><a href='http://www.futilitycloset.com/2014/07/26/coming-and-going-7/' target='new'>Source: Futility Closet</a></small></div>
<div style="text-align: justify; text-indent: 20px;">In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?</div>
<h4 style="margin-bottom: 0px;">Overlapping Triangles</h4>
<div style="text-align: left"><small><a href='http://www.kcl.ac.uk/mathsschool' target='new'>Source: King's College London GCSE Enrichment Programme</a></small></div>
<div style="text-align: justify; text-indent: 20px;">Four congruent triangles are drawn in a square.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh52ryB-653vjhtIfHQtayqB89pN8M8prlKMb0iijBiASkgDf4qhB_mzQUMWD536t8RyXqtnCoA91cSBZWmDXsvpPQXqkqSy8QS2-ez2k4KD7trHXos5biusUOomYZtmPqVMSmH0sVFE3U/s1600/tringles+in+squares+1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh52ryB-653vjhtIfHQtayqB89pN8M8prlKMb0iijBiASkgDf4qhB_mzQUMWD536t8RyXqtnCoA91cSBZWmDXsvpPQXqkqSy8QS2-ez2k4KD7trHXos5biusUOomYZtmPqVMSmH0sVFE3U/s320/tringles+in+squares+1.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large square does each (purple) triangle take up?</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8byDdGEu-T__GMcEKkd_Y7X8IXx2tZFSgkBF-C-aYOCfkH7IajrYCDpJLlGtNkprXzMdIcanml3_1lAI9Y17JVRMkGHXGhQ-t0cOC16fC-BqKPhWsocwCYAbElfbbLrReEJviMtLxFvw/s1600/tringles+in+squares+2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8byDdGEu-T__GMcEKkd_Y7X8IXx2tZFSgkBF-C-aYOCfkH7IajrYCDpJLlGtNkprXzMdIcanml3_1lAI9Y17JVRMkGHXGhQ-t0cOC16fC-BqKPhWsocwCYAbElfbbLrReEJviMtLxFvw/s320/tringles+in+squares+2.png" /></a></div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com1tag:blogger.com,1999:blog-6331909486660092910.post-71413356978265432042014-07-21T07:17:00.000+01:002014-07-21T07:17:00.155+01:00Sunday Afternoon Maths XXI Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/07/sunday-afternoon-maths-xxi.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Sum Equals Product</h4>
<div style="text-align: justify; text-indent: 20px;">If \(a+b=a\times b\), then:</div>
$$ab-b=a$$
$$b(a-1)=a$$
$$b=\frac{a}{a-1}$$
<div style="text-align: justify; text-indent: 20px;">This will work for any \(a \not= 1\) (\(a=1\) will not work as this will mean division by zero).</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">(i) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{b}{a}\)?</div>
<div style="text-align: justify; text-indent: 20px;">(ii) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{a}{b}\)?</div>
<div style="text-align: justify; text-indent: 20px;">(iii) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{b}{a}\)?</div>
<div style="text-align: justify; text-indent: 20px;">(iv) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{a}{b}\)?</div>
<h4 style="margin-bottom: 0px;">Wool Circles</h4>
<div style="text-align: justify; text-indent: 20px;">Starting with the person who starts with the wool and going anti-clockwise, number the people \(0,1,2,3,4,...\). As the wool is passed, it will be held by people with numbers:</div>
$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
<div style="text-align: justify; text-indent: 20px;">For example, if \(n=10\) and \(a=3\):</div>
<div style="text-align: justify; text-indent: 20px;">The first person will have the wool again when:</div>
$$k(a+1)\equiv 0 \mod n$$
<div style="text-align: justify; text-indent: 20px;">or</div>
$$k(a+1)=ln$$
<div style="text-align: justify; text-indent: 20px;">This will first occur when (hcf is <a target='new' href='https://en.wikipedia.org/wiki/Highest_common_factor'>highest common factor</a>):</div>
$$l=\frac{a+1}{\mbox{hcf}(a+1,n)} \mbox{ and } k=\frac{n}{\mbox{hcf}(a+1,n)}$$
<div style="text-align: justify; text-indent: 20px;">\(k\) is also the number of people who are holding the wool. So the number of different coloured balls needed is:</div>
$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">The ball is passed around the circle of \(n\) people again. This time, the number of people missed alternates between \(a\) and \(b). How many different coloured balls of wool are now needed?</div>
$$y=(1-\sqrt{2x})^2$$Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-16483360978754690822014-07-20T12:19:00.002+01:002014-09-04T21:03:03.287+01:00Sunday Afternoon Maths XXI<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions <a href='http://legavrik.blogspot.co.uk/2014/07/sunday-afternoon-maths-xxi-answers.html' target='new'>here</a>. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>. This Tuesday is <a target='new' href='http://www.mathsjam.com'>Maths Jam</a>, make sure you're there!</div>
<h4 style="margin-bottom: 0px;">Sum Equals Product</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Penguin-Curious-Interesting-Puzzles-science/dp/0140148752' target='new'>Source: <i>The Penguin Book of Curious and Interesting Puzzles</i> by David Wells</a></small></div>
<div style="text-align: justify; text-indent: 20px;">\(3\) and \(1.5\) are a special pair of numbers, as \(3+1.5=4.5\) and \(3\times 1.5=4.5\) so \(3+1.5=3\times 1.5\).</div>
<div style="text-align: justify; text-indent: 20px;">Given a number \(a\), can you find a number \(b\) such that \(a+b=a\times b\)?</div>
<h4 style="margin-bottom: 0px;">Wool Circles</h4>
<div style="text-align: justify; text-indent: 20px;">\(n\) people stand in a circle. The first person takes a ball of wool, holds the end and passes the ball to his right, missing \(a\) people. Each person who receives the wool holds it and passes the ball on to their right, missing \(a\) people. Once the ball returns to the first person, a different coloured ball of wool is given to someone who isn't holding anything and the process is repeated. This is done until everyone is holding wool.</div>
<div style="text-align: justify; text-indent: 20px;">For example, if \(n=10\) and \(a=3\):</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1zAeGaORI6IwKt3bBN2bFGNAzShdP632kP0641__c__vrin9KbaE3JdP6go1qhvNcyqNObhO1R6cRbzW0uw9nJfur3EXCfyBvCr4HiC4Tw_OfOWkms2hFbCZO4zqiNiU53TN6orXseT4/s1600/woolcircles.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1zAeGaORI6IwKt3bBN2bFGNAzShdP632kP0641__c__vrin9KbaE3JdP6go1qhvNcyqNObhO1R6cRbzW0uw9nJfur3EXCfyBvCr4HiC4Tw_OfOWkms2hFbCZO4zqiNiU53TN6orXseT4/s320/woolcircles.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">In this example, two different coloured balls of wool are needed.</div>
<br />
<div style="text-align: justify; text-indent: 20px;">In terms of \(n\) and \(a\), how many different coloured balls of wool are needed?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-27854541470593170072014-07-14T09:52:00.000+01:002014-07-21T06:42:55.075+01:00Sunday Afternoon Maths XX Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/07/sunday-afternoon-maths-xx.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Folding A4 Paper</h4>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0goGndM4EiQHWCqwdIHGaunRbQAUUwsk_VS2CnvPBd_dX7XOueNi9tIUSV2LDq35CQZ767LmY6O2K_cMwEMR1TgC_x1e1mdvho-7ZXCMKVKgmpUaOhibIRD7KFBFk-imJQTXm6QU9YT0/s1600/kite.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0goGndM4EiQHWCqwdIHGaunRbQAUUwsk_VS2CnvPBd_dX7XOueNi9tIUSV2LDq35CQZ767LmY6O2K_cMwEMR1TgC_x1e1mdvho-7ZXCMKVKgmpUaOhibIRD7KFBFk-imJQTXm6QU9YT0/s320/kite.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">The sides of A4 paper are in the ratio \(1:\sqrt{2}\). Let the width of the paper be 1 unit. This means that the height of the paper is \(\sqrt{2}\) units.</div>
<div style="text-align: justify; text-indent: 20px;">Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).</div>
<div style="text-align: justify; text-indent: 20px;">\(BF=DF=\sqrt{2}-1\) so by Pythagoras' Theorem again, \(BD=2-\sqrt{2}\). \(CD=1-(\sqrt{2}-1)=2-\sqrt{2}\). Hence, \(CD=BD\) and so the shape is a kite.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a <a target='new' href='http://en.wikipedia.org/wiki/Cyclic_quadrilateral'>cyclic quadrilateral</a>.</div>
<h4 style="margin-bottom: 0px;">Bézier Curves</h4>
<div style="text-align: justify; text-indent: 20px;">If</div>
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
<div style="text-align: justify; text-indent: 20px;">then</div>
$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$
$$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$
<div style="text-align: justify; text-indent: 20px;">and so</div>
$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$
<div style="text-align: justify; text-indent: 20px;">which means that</div>
$$x=t^2$$$$y=(1-t)^2$$
<div style="text-align: justify; text-indent: 20px;">or</div>
$$y=(1-\sqrt{x})^2$$
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation</div>
$$y=(1-\sqrt{2x})^2$$Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-66268249751618596952014-07-13T11:49:00.000+01:002014-07-13T11:49:00.156+01:00Sunday Afternoon Maths XX<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Folding A4 Paper</h4>
<div style="text-align: left"><small><a href='http://www.atm.org.uk/write/MediaUploads/Journals/MT241/MT241-14-05.pdf' target='new'>Source: <i>ATM Mathematics Teaching 241 (July 2014)</i></a></small></div>
<div style="text-align: justify; text-indent: 20px;">A Piece of <a target='new' href='http://en.wikipedia.org/wiki/A4_paper#A_series'>A4</a> paper is folded as shown:</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgO5ZakzOh8625hO0QmBOo7jPfTpa92MGKTrtHsi0U8kFmCH0xzjk55Y1Ee7CEVqymTcic8TQRubD2mOFbdQKIHFT6LgKp2WwwgkED-nlwKTeQ0JKpZGhBjvG1YxaKaSNGNCvUWWqa9Wmc/s1600/folding+a4+paper.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgO5ZakzOh8625hO0QmBOo7jPfTpa92MGKTrtHsi0U8kFmCH0xzjk55Y1Ee7CEVqymTcic8TQRubD2mOFbdQKIHFT6LgKp2WwwgkED-nlwKTeQ0JKpZGhBjvG1YxaKaSNGNCvUWWqa9Wmc/s320/folding+a4+paper.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What shape is made?</div>
<h4 style="margin-bottom: 0px;">Bézier Curve</h4>
<div style="text-align: justify; text-indent: 20px;">A <a target='new' href='en.wikipedia.org/wiki/Bézier_curve'>Bézier curve</a> is created as follows:</div>
<div class="separator" style="clear: both; text-align: center;"><img border="0" src="http://upload.wikimedia.org/wikipedia/commons/a/a4/B%C3%A9zier_4_big.gif" /><br /><small><small>Image source: <a href='en.wikipedia.org/wiki/Portal:Mathematics/Selected_picture/9#mediaviewer/File:Bézier_4_big.gif' target='new'>WikiPedia</a></small></small></div>
<div style="text-align: justify; text-indent: 20px;">1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).</div>
<div style="text-align: justify; text-indent: 20px;">2) A set of points \(Q_0\), ..., \(Q_{n-1}\) are defined by \(Q_i=t P_{i+1}+(1-t) P_i\) (shown in green).</div>
<div style="text-align: justify; text-indent: 20px;">3) A set of points \(R_0\), ..., \(R_{n-2}\) are defined by \(R_i=t Q_{i+1}+(1-t) Q_i\) (shown in blue).</div>
<div style="text-align: justify; text-indent: 20px;">.</div>
<div style="text-align: justify; text-indent: 20px;">.</div>
<div style="text-align: justify; text-indent: 20px;">.</div>
<div style="text-align: justify; text-indent: 20px;">\(n\)) After repeating the process \(n\) times, there will be one point. The Bézier curve is the path traced by this point at \(t\) varies between 0 and 1.</div>
<br />
<div style="text-align: justify; text-indent: 20px;">What is the Cartesian equation of the curve formed when:</div>
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-27462858446587591172014-06-30T07:29:00.000+01:002014-06-30T07:29:00.177+01:00Sunday Afternoon Maths XIX Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/06/sunday-afternoon-maths-xix.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Dartboard</h4>
<div style="text-align: justify; text-indent: 20px;">The shaded area is:</div>
$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$
$$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$
$$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$
$$=\pi\left(\frac{\pi^2}{12}\right)$$
$$=\frac{\pi^3}{12}$$
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">Prove that</div>
$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$
<h4 style="margin-bottom: 0px;">Multiples of Three</h4>
<div style="text-align: justify; text-indent: 20px;">If \(10A+B=3n\), then:</div>
$$100A+B = 100A-10A+10A+B$$$$=90A+3n$$$$=3(30A+n)$$
<div style="text-align: justify; text-indent: 20px;">So \(A0B\div3=30A+n\).</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">What are \(A00B\div3\), \(A000B\div3\), \(A0000B\div3\), etc?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-78144671855079496432014-06-29T10:27:00.000+01:002014-06-29T10:29:20.705+01:00Sunday Afternoon Maths XIX<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Dartboard</h4>
<div style="text-align: justify; text-indent: 20px;">Concentric circles with radii 1, \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\), ... are drawn. Alternate donut-shaped regions are shaded.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhh5fjp3ua9TKPkh0tr6r8KbNK250Sa1k3c65o50_y3cvvxHDFUfXeJtGaAThR8qZEUpYW74BXCB7B3NdPXwydQBKNa86ytgA8UnNHjmOh6470Z0U74yWJpEB3XRAdXG1A3fONgvGZOsZM/s1600/dartboard.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhh5fjp3ua9TKPkh0tr6r8KbNK250Sa1k3c65o50_y3cvvxHDFUfXeJtGaAThR8qZEUpYW74BXCB7B3NdPXwydQBKNa86ytgA8UnNHjmOh6470Z0U74yWJpEB3XRAdXG1A3fONgvGZOsZM/s320/dartboard.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What is the total shaded area?</div>
<h4 style="margin-bottom: 0px;">Multiples of Three</h4>
<div style="text-align: justify; text-indent: 20px;">If the digits of a number add up to a multiple of three, then the number is a multiple of three. Therefore if a two digit number, \(AB\) (first digit \(A\), second digit \(B\); not \(A\times B\)), is a multiple of three, then \(A0B\) is also a multiple of three.</div>
<div style="text-align: justify; text-indent: 20px;">If \(AB\div 3=n\), then what is \(A0B\div 3\)?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-7956420513522327552014-06-23T07:03:00.000+01:002014-06-29T10:29:48.515+01:00Sunday Afternoon Maths XVIII Answers & Extensions<div style="text-align: justify; text-indent: 20px;">This post contains the answers to <a href='http://legavrik.blogspot.com/2014/06/sunday-afternoon-maths-xviii.html'>this week's Sunday Afternoon Maths</a> and some extension problems based around the originals.</div>
<h4 style="margin-bottom: 0px;">Parabola</h4>
<div style="text-align: justify; text-indent: 20px;">The co-ordinates of the points where the lines intersect the parabola are \((a,a^2)\) and \((-b,b^2)\). Hence the gradient of the line between them is:</div>
$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$
<div style="text-align: justify; text-indent: 20px;">Therefore the y-coordinate is:</div>
$$b^2 + b(a-b) = ba$$
<div style="text-align: justify; text-indent: 20px;">Ferdinand Möbius, who discovered this property called the curve a <i>Multiplicationsmaschine</i> or 'multipliction machine' as it could be used to perform multiplication.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">How could you use the graph of \(y=x^2\) to divide 100 by 7?</div>
<h4 style="margin-bottom: 0px;">Seven Digits</h4>
<div style="text-align: justify; text-indent: 20px;">Let's call Dr. Dingo's number \(n\). If the number is squared twice then multiplied by \(n\), we get \(n^5\).</div>
<div style="text-align: justify; text-indent: 20px;">
For all integers \(n\), the final digit of \(n^5\) is the same as the final digit of \(n\). In other words:</div>
$$n^5\equiv n \mod 10$$
<div style="text-align: justify; text-indent: 20px;">Therefore, the final digit of Dr. Dingo's number is 7.</div>
$$7^5=16807$$
$$17^5=1419857$$
$$27^5=14348907$$
<div style="text-align: justify; text-indent: 20px;">So, in order for the answer to have seven digits, Dr. Dingo's number was 17.</div>
<h5 style="margin-top:8px;margin-bottom: 0px;">Extension</h5>
<div style="text-align: justify; text-indent: 20px;">For which integers \(m\) does there exist an integer \(n\) such that for all integers \(x\):</div>
$$x^n\equiv x \mod m$$Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com0tag:blogger.com,1999:blog-6331909486660092910.post-27218791467042405422014-06-22T11:07:00.000+01:002014-06-22T11:07:00.193+01:00Sunday Afternoon Maths XVIII<div style="text-align: justify; text-indent: 20px;">Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using <a target='new' href='https://twitter.com/search?q=%23SundayAfternoonMaths&src=hash'>#SundayAfternoonMaths</a> or on <a target='new' href='http://www.reddit.com/r/math/search?q=Sunday+Afternoon&sort=new&restrict_sr=on'>Reddit</a>.</div>
<h4 style="margin-bottom: 0px;">Parabola</h4>
<div style="text-align: left"><small><a href='http://www.amazon.co.uk/Alex-Through-Looking-Glass-Reflects-Numbers/dp/1408817772/' target='new'>Source: <i>Alex Through the Looking-Glass: How Life Reflects Numbers and Numbers Reflect Life</i> by Alex Bellos</a></small></div>
<div style="text-align: justify; text-indent: 20px;">On a graph of \(y=x^2\), two lines are drawn at \(x=a\) and \(x=-b\) (for \(a,b>0\). The points where these lines intersect the parabola are connected.</div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYJ60E4WpMK_ergte1Rtj8wW_nsNH2hR937hS_uxDeAJOUaZvMyCH_amq_bmtI4GhFNad_Ot5t99YtwTyItm5sOvh3K4xwJD6z2SuDY0jRepwYQabypoiICdg-9_-KfhN42DZV8sa8_ig/s1600/parabola.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYJ60E4WpMK_ergte1Rtj8wW_nsNH2hR937hS_uxDeAJOUaZvMyCH_amq_bmtI4GhFNad_Ot5t99YtwTyItm5sOvh3K4xwJD6z2SuDY0jRepwYQabypoiICdg-9_-KfhN42DZV8sa8_ig/s400/parabola.png" /></a></div>
<div style="text-align: justify; text-indent: 20px;">What is the y-coordinate of the point where this line intersects the y-axis?</div>
<h4 style="margin-bottom: 0px;">Seven Digits</h4>
<div style="text-align: left"><small><a href="http://www.amazon.co.uk/Best-Puzzles-Mathematics-Hubert-Phillips/dp/0486200914" target="new">Source: <i>My Best Puzzles in Mathematics</i> by Hubert Phillips</a></small></div>
<div style="text-align: justify; text-indent: 20px;">"I'm thinking of a number. I've squared it. I've squared the square. And I've multiplied the second square by the original number. So I now have a number of seven digits whose final digit is a 7," said Dr. Dingo to his daughter.</div>
<div style="text-align: justify; text-indent: 20px;">Can you work out Dr. Dingo's number?</div>
Matthewhttp://www.blogger.com/profile/06708426822135844990noreply@blogger.com1