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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\(n\)) 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:

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