Sunday Afternoon Maths XXIX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Three Squares

Draw three more squares and add these lines (I have coloured the angles to make equal angles clearer):

Triangles \(ACE\), \(LDK\) and \(IKE\) are congruent, so angle \(KDL\) is equal to \(\beta\).

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.

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.

Therefore \(\alpha+\beta+\gamma=90^\circ\).

Extension

The diagram shows three squares with diagonals drawn on and three angles labelled.

What is the value of \(\alpha+\beta+\gamma\)?

Equal Opportunity

Source: Futility Closet

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 \(p_1q_1=p_6q_6\).

If \(p_1\geq p_6\) then \(q_1\leq q_6\) (and vice-versa) as otherwise the above equality could not hole. Therefore:

$$(p_1-p_6)(q_1-q_6)\leq 0$$ $$p_1q_1-p_6q_1-p_1q_6+p_6q_6\leq 0$$ $$p_1q_1+q_6p_6\leq p_1q_6+p_6q_1$$

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 \(p_1q_1\).

Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.

Extension

Can two \(n\)-sided dice be weighted so that the probability of each of the numbers 2, 3, …, 2\(n\) is the same?

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?

Double Derivative

Source: Twenty-Six Years of Problem Solving compiled by John Mason

(i) \(\frac{dy}{dx}=1\), so \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=0\)

(ii) Differentiating \(y=x^2\) with respect to \(x\) \(\frac{dy}{dx}=2x\). Let \(g=\frac{dy}{dx}\). By the chain rule:

$$\frac{dg}{dy}=\frac{dg}{dx}\frac{dx}{dy}$$ $$=2\frac{1}{2x}$$ $$=\frac{1}{x}$$

So \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{1}{x}\)

(iii) By the same method, \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{2}{x}\)

(iv) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{n-1}{x}\)

(v) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=1\)

(vi) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=-\tan(x)\)

Extension

Source: Twenty-Six Years of Problem Solving compiled by John Mason

What is

$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$

when \(y=f(x)\)?

Sunday Afternoon Maths XXIX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!

Three Squares

Source: Numberphile

The diagram shows three squares with diagonals drawn on and three angles labelled.

What is the value of \(\alpha+\beta+\gamma\)?

Equal Opportunity

Source: Futility Closet

Can two (six-sided) dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?

Double Derivative

Source: Twenty-Six Years of Problem Solving compiled by John Mason

What is

$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$

when:

(i) \(y=x\)

(ii) \(y=x^2\)

(iii) \(y=x^3\)

(iv) \(y=x^n\)

(v) \(y=e^x\)

(vi) \(y=\sin(x)\)?

Sunday Afternoon Maths XXVIII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

2009

\(2009=7\times 7\times 41\), so there are four possible sets of dimensions of the cuboid:

$$1\times 1\times 2009$$ $$1\times 7\times 287$$ $$1\times 41\times 49$$ $$7\times 7\times 41$$

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 764 stickers left over

Extension

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.

3\(n\)+1

(i) Let \(a,b\in S\). Then \(\exists \alpha,\beta\in \mathbb{N}\) such that \(a=3\alpha+1\) and \(b=3\beta+1\).

(This says that if \(a\) and \(b\) are in \(S\) then they can be written as a multiple of three plus one.)

$$a\times b=(3\alpha+1)\times (3\beta+1)$$ $$=9\alpha\beta+3\alpha+3\beta+1$$ $$=3(3\alpha\beta+\alpha+\beta)+1$$

This is a multiple of three plus one, so \(a\times b\in S\).

(ii) No, as \(36\times 22=4\times 253\) and 36,22,4 and 253 are all irreducible.

Extension

Source: Twenty-Six Years of Problem Solving compiled by John Mason

Try the task again with \(S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}\).

The Ace of Spades

The problem can be expressed using the following probability tree:

The probability that the card turned over from C is an Ace of Spades is:

$$\frac{1\times 2\times 2+1\times 50\times 1+50\times 1\times 2+50\times 51\times 1}{51\times 52\times 53}$$ $$=\frac{52}{51\times 53}$$

Sunday Afternoon Maths XXVIII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

2009

Source: Teaching Further Maths blog

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.

How many stickers from the pack are left?

3\(n\)+1

Source: Twenty-Six Years of Problem Solving compiled by John Mason

Let \(S=\{3n+1:n\in\mathbb{N}\}\) be the set of numbers one more than a multiple of three.

(i) Show that \(S\) is closed under multiplication.

ie. Show that if \(a,b\in S\) then \(a\times b\in S\).

Let \(p\in S\) be irreducible if \(p\not=1\) and the only factors of \(p\) in \(S\) are \(1\) and \(p\). (This is equivalent to the most commonly given definition of prime.)

(ii) Can each number in \(S\) be uniquely factorised into irreducibles?

The Ace of Spades

Source: The Week-End Problems by Hubert Phillips

I have three packs of playing cards with identical backs. Call the packs A, B and C.

I draw a random card from pack A and shuffle it into pack B.

I now turn up the top card of pack A, revealing the Queen of Hearts.

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.

I now draw a random card from pack C and place it at the bottom of pack A.

What is the probability that the card at the top of pack C is the Ace of Spades?

Sunday Afternoon Maths XXVII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Triangles Between Squares

Let \(T_a\) represent the \(a\)^th triangle number. This means that \(T_a=\frac{1}{2}a(a+1)\).

Suppose that for some integer \(n\), \(n^2 \leq T_a <(n+1)^2\). This means that:

$$n^2 \leq T_a$$ $$n^2 \leq \frac{1}{2}a(a+1)$$ $$2n^2 \leq a^2+a$$

But for every positive integer \(a \leq a^2\), so:

$$2n^2 \leq 2a^2$$ $$n^2 \leq a^2$$

\(n\) and \(a\) are both positive integers, so:

$$n \leq a$$

Now consider \(T_{a+2}\):

$$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$ $$=\frac{1}{2}(a^2+5a+6)$$ $$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$ $$=\frac{1}{2}a(a+1)+2a+3$$ $$=T_a+2a+3$$

We know that \(a \geq n\) and \(T_a \geq n^2\), so:

$$T_a+2a+3 \geq n^2+2n+3$$ $$>n^2+2n+1 = (n+1)^2$$

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 \((n+1)^2\).

Extension

Given an integer \(n\), how many triangle numbers are there between \(n^2\) and \((n+1)^2\)?

Sine

Cosine can be drawn the same way as sine but starting B at the top of the circle.

Tangent can be drawn by giving the following instructions:

A. Stand on the spot.

B. Walk around A in a circle, holding this string to keep you the same distance away.

C. Make a straight line with A and B, staying on the line tangent to the circle through B's starting point.

D. Walk in a straight line perpendicular to C's line.

E. Stay in line with C and D.

Extension

Could people be used to draw graphs of secant, cosecant and cotangent?

Sunday Afternoon Maths XXVII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Triangles Between Squares

Prove that there are never more than two triangle numbers between two consecutive square numbers.

Sine

A sine curve can be created with five people by giving the following instructions to the five people:

A. Stand on the spot.

B. Walk around A in a circle, holding this string to keep you the same distance away.

C. Stay in line with B, staying on this line.

D. Walk in a straight line perpendicular to C's line.

E. Stay in line with C and D. E will trace the path of a sine curve as shown here:

What instructions could you give to five people to trace a cos(ine) curve?

What instructions could you give to five people to trace a tan(gent) curve?

Sunday Afternoon Maths XXVI

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

It's no late to get tickets for EMF Camp, where I will be giving a talk on flexagons, folding tube maps and braiding.

Twenty-One

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:

Scott: 3
Virgil: 4
Scott: 5
Virgil: 6
Scott: 9
Virgil: 12
Scott: 15
Virgil 17
Scott: 20
Virgil: 21
Virgil loses.

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?

Odd and Even Outputs

Let \(g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) be a function.

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 \(g(10,2)=12\).

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:

		\(n\)
		odd	even
\(m\)	odd	even	odd
	e	odd	even

Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:

\(n\) odd even \(m\) odd odd odd e odd odd	\(n\) odd even \(m\) odd odd odd e odd even	\(n\) odd even \(m\) odd odd odd e even odd	\(n\) odd even \(m\) odd odd odd e even even
\(n\) odd even \(m\) odd odd even e odd odd	\(n\) odd even \(m\) odd odd even e odd even	\(n\) odd even \(m\) odd odd even e even odd	\(n\) odd even \(m\) odd odd even e even even
\(n\) odd even \(m\) odd even odd e odd odd	\(n\) odd even \(m\) odd even odd e odd even	\(n\) odd even \(m\) odd even odd e even odd	\(n\) odd even \(m\) odd even odd e even even
\(n\) odd even \(m\) odd even even e odd odd	\(n\) odd even \(m\) odd even even e odd even	\(n\) odd even \(m\) odd even even e even odd	\(n\) odd even \(m\) odd even even e even even

Sunday Afternoon Maths XXV Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Whist

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.

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:

	North B D A S b d a s
West B D A S b d a s		East B D A S b d a s
	South B D A S b d a s

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.

	North B D A S b d a s
West B D A S b d a s		East B D A S b d a s
	South B D A S b d a s

By elimination, only North can be S. This means that d must sit to the right of North (at West):

	North B D A S b d a s
West B D A S b d a s		East B D A S b d a s
	South B D A S b d a s

A and a are partners. This is only possible if A is South and a is North:

	North B D A S b d a s
West B D A S b d a s		East B D A S b d a s
	South B D A S b d a s

Therefore, Mr. Banker the dentist sits to the left of the banker.

Extension

If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?

Polya Strikes Out

1, 3, 7, 12, 19, ...

1, 7, 19, ...

1=1; 1+7=8; 1+7+19=27; ...

1, 8, 27, ...

The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.

Cross out every third number:

1, 2, 3, 4, 5, 6, ..., 3n, \(3n+1\), \(3n+2\), ...

1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...

Find the cumulative sums:

$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$...$$ $$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$...$$

1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...

Cross out every second number, starting with the second:

1, 3, 7, 12, ..., \(3n^2+3n+1\), 3n2+6n+3, ...

1, 7, ..., \(3n^2+3n+1\), ...

Find the cumulative sums. The \(m\)^th sum is:

$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$

Hence the numbers obtained are the cube numbers.

Extension

What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)^th number starting at the \(m\)^th?

Sunday Afternoon Maths XXV

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!

It's less than two weeks 'til EMF Camp, where there will be talks on maths and science and a whole load of other awesome stuff. Tickets here

Whist

Source: My Best Puzzles in Logic and Reasoning by Hubert Phillips

Messrs. Banker, Dentist, Apothecary and Scrivener played whist last night. (whist is a four player card game where partners sit opposite each other.) Each of these gentlemen is the namesake of another's vocation.

Last night, the apothecary partnered Mr. Apothecary; Mr. Banker's partner was the scrivener; on Mr. Scrivener's right sat the dentist.

Who sat on the banker's left?

Polya Strikes Out

Source: Thinking Mathematically by John Mason, Leone Burton & Kaye Stacey

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:

1, 2, 3, 4, 5, 6, 7, ...

1, 2, 3, 4, 5, 6, 7, ...

1, 2, 4, 5, 7, ...

1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...

1, 3, 7, 12, 19, ...

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?

Sunday Afternoon Maths XXIV Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Square Cross

Label and orient the shape as follows:

The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.

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 \((-\frac{8}{5},-\frac{1}{5})\).

By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is 9.8.

Exact Change

In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.

In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.

Extension

Source: @AlexDBolton on Twitter

In a far away country, the unit of currency is the #, which is split into 100@ (# is like £ or $; @ is like p or ¢).

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?

Sunday Afternoon Maths XXIV

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Square Cross

Source: Teach Further Maths Blog

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.

What is the area of the large outer square?

Exact Change

Source: @AlexDBolton on Twitter

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?

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¢?

Sunday Afternoon Maths XXIII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Let the Passenger Train Through!

The goods train reverses into the siding, leaving three trucks, then reverses out of the way.

The passenger train attaches the three trucks to its front then reverses out of the way.

The goods train backs into the siding.

The passenger train drives past the siding.

The goods train drives forwards out of the way.

The passenger train reverses, then drives forward into the siding, leaving the three trucks. It then drives forward out of the way.

The goods train reverses into the siding, picking up the trucks. The trains have now passed each other.

Extension

1. If the goods train can drive forwards into the siding, how could it let the passenger train through?

2. If the goods train had 8 trucks, how could it let the passenger train through?

Cooked Turkey

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.

$$10000a+6790+b=(138\times 72+64)a+(94\times 72+22)+b$$ $$=(138a+94)\times 72+64a+22+b$$

\(10000a+6790+b\) 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 \(b=2c\).

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 \(b=4d+2\).

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.

Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).

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.

Extension

Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?

Mrs. Coldcream

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:

$$xG+(x+30)(19-G)=1000$$

This simplifies to:

$$19x-30G=430$$

\(G\) is less than 19, so the only integer solution of this equation is \(x=40\) and \(G=11\).

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:

$$48\times 11+8B=1000$$

Which can be solved to find \(B=59\).

Each girl receives £48 and each boy receives £59.

Extension

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?

Ten Digit Number

6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.

Extension

Are there any more numbers like this? Prove that you have them all.

Sunday Afternoon Maths XXIII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Let the Passenger Train Through!

Source: The Moscow Puzzles by Boris A. Kordemsky

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.

A passenger train arrives travelling in the opposite direction. How can they let it through? (The passenger train is too long to fit in the siding.)

Cooked Turkey

Source: The Penguin Book of Curious and Interesting Puzzles by David Wells

An old invoice showed that seventy-two turkeys had been purchased for "—67.9—". The first and last digits were illegible.

How much did one turkey cost?

Mrs. Coldcream Objected

Source: My Best Puzzles in Mathematics by Hubert Phillips

"I object," said Councillor Mrs. Coldcream. "I see no reason why the boys should be so favoured at the expense of the girls."

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.

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.

How much did each boy and each girl receive?

Ten Digit Number

Source: Richard Wiseman's Blog

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?

Sunday Afternoon Maths XXII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Coming and Going

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:

$$S=2I+E$$$$E=S-2I$$

\(S\) and \(2I\) are even, so \(E\) must be even.

Extension

If the number of doors in each room is odd, is the number of exterior doors odd or even?

Overlapping Triangles

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:

$$S=4T-4U+V$$

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.

We know that \(4U=V\), so:

$$S=4T-V+V$$$$S=4T$$

Therefore one of the triangles covers one quarter of the square.

Extension

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?

\(n\) 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?

Sunday Afternoon Maths XXII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Coming and Going

Source: Futility Closet

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?

Overlapping Triangles

Source: King's College London GCSE Enrichment Programme

Four congruent triangles are drawn in a square.

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?

Sunday Afternoon Maths XXI Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Sum Equals Product

If \(a+b=a\times b\), then:

$$ab-b=a$$ $$b(a-1)=a$$ $$b=\frac{a}{a-1}$$

This will work for any \(a \not= 1\) (\(a=1\) will not work as this will mean division by zero).

Extension

(i) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{b}{a}\)?

(ii) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{a}{b}\)?

(iii) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{b}{a}\)?

(iv) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{a}{b}\)?

Wool Circles

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:

$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$

For example, if \(n=10\) and \(a=3\):

The first person will have the wool again when:

$$k(a+1)\equiv 0 \mod n$$

or

$$k(a+1)=ln$$

This will first occur when (hcf is highest common factor):

$$l=\frac{a+1}{\mbox{hcf}(a+1,n)} \mbox{ and } k=\frac{n}{\mbox{hcf}(a+1,n)}$$

\(k\) is also the number of people who are holding the wool. So the number of different coloured balls needed is:

$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$

Extension

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?

$$y=(1-\sqrt{2x})^2$$

Sunday Afternoon Maths XXI

Here's this week's collection. Answers & extensions here. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!

Sum Equals Product

Source: The Penguin Book of Curious and Interesting Puzzles by David Wells

\(3\) and \(1.5\) are a special pair of numbers, as \(3+1.5=4.5\) and \(3\times 1.5=4.5\) so \(3+1.5=3\times 1.5\).

Given a number \(a\), can you find a number \(b\) such that \(a+b=a\times b\)?

Wool Circles

\(n\) 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.

For example, if \(n=10\) and \(a=3\):

In this example, two different coloured balls of wool are needed.

In terms of \(n\) and \(a\), how many different coloured balls of wool are needed?

Sunday Afternoon Maths XX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Folding A4 Paper

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.

Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).

\(BF=DF=\sqrt{2}-1\) 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.

Extension

Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a cyclic quadrilateral.

Bézier Curves

If

$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$

then

$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$ $$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$

and so

$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$

which means that

$$x=t^2$$$$y=(1-t)^2$$

or

$$y=(1-\sqrt{x})^2$$

Extension

What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation

$$y=(1-\sqrt{2x})^2$$

Sunday Afternoon Maths XX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Folding A4 Paper

Source: ATM Mathematics Teaching 241 (July 2014)

A Piece of A4 paper is folded as shown:

What shape is made?

Bézier Curve

A Bézier curve is created as follows:

Image source: WikiPedia

1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).

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).

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).

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.

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\(n\)) 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.

What is the Cartesian equation of the curve formed when:

$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$

Sunday Afternoon Maths XIX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Dartboard

The shaded area is:

$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$ $$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$ $$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$ $$=\pi\left(\frac{\pi^2}{12}\right)$$ $$=\frac{\pi^3}{12}$$

Extension

Prove that

$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$

Multiples of Three

If \(10A+B=3n\), then:

$$100A+B = 100A-10A+10A+B$$$$=90A+3n$$$$=3(30A+n)$$

So \(A0B\div3=30A+n\).

Extension

What are \(A00B\div3\), \(A000B\div3\), \(A0000B\div3\), etc?