# Puzzles

## 2 December

Carol draws a square with area 62. She then draws the smallest possible circle that this square is contained inside.
Next, she draws the smallest possible square that her circle is contained inside. What is the area of her second square?

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By drawing an appropriate diagram, it can be seen that the small square has half the area of the large square.

Therefore the area of the large square is **124**.

## 19 December

The diagram below shows three squares and five circles.
The four smaller circles are all the same size, and the red square's vertices are the centres of these circles.

The area of the blue square is 14 units. What is the area of the red square?

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The diagonal of the blue square is \(\sqrt{28}\). The radius \(R\) of the large circle satisfies \(R^2+R^2=(R+\sqrt{28})^2\).
Solving this, we find that \(R=\frac{\sqrt{28}}{\sqrt2-1}=\sqrt{28}(\sqrt2+1)\).

The radius \(r\) of the small circles satisfies \(r+r\sqrt2=R\), and so \(r=\frac{\sqrt{28}(\sqrt2+1)}{\sqrt2+1}=\sqrt{28}\).

The area of the square is \(4r^2=4\left(\sqrt{28}\right)^2=4\times28\). This is **112**.

## Is it equilateral?

In the diagram below, \(ABDC\) is a square. Angles \(ACE\) and \(BDE\) are both 75°.

Is triangle \(ABE\) equilateral? Why/why not?

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The triangle is equilateral.

To see why, add a copy of point \(E\) rotated by 90°. This is labelled \(F\) on the diagram below.

Angles \(BDE\) and \(CDF\) are both 75°. Therefore angles \(CDE\) and \(BDF\) are both 15°. This means that angle \(FDE\) is 60°.

Line \(AD\) is a line of symmetry of the diagram, so angles \(DFE\) and \(DEF\) are equal and both 60°. Therefore, triangle DEF is equilateral. This triangle is show in green in the diagram below.

Lines \(EF\), \(DF\) and \(BF\) are all equal length, so triangles \(BFE\) and \(BFD\) are isosceles.
Angles \(BDF\) and \(FBD\) are both 15°. Angles \(FBE\) and \(FEB\) are equal, and the angles in triangle \(BED\) add to 180°: this means that angle \(FBE\) is 15°.

Angles \(FBE\) and \(FBD\) are both 15°, and so angle \(EBD\) is 30°. Angles \(EBD\) and \(ABE\) add to 90°, and so angle \(ABE\) is 60°.

By symmetry, angle \(BAE\) is also 60°. Angle \(BEA\) must therefore also be 60°, so triangle \(ABE\) is equilateral.

## Squared circle

Each side of a square has a circle drawn on it as diameter. The square is also inscribed in a fifth circle as shown.

Find the ratio of the total area of the shaded crescents to the area
of the square.

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Let the radius of the small circles be \(r\). The are of half of one of these circles is \(\frac{1}{2}\pi r^2\).

The side of the square is \(2r\) and so the area of the square is \(4r^2\). Therefore the area of the whole shape is \((4+2\pi)r^2\).

By Pythagoras' Theorem, the radius of the large circle is \(r\sqrt{2}\). Therefore the area of the circle is \(2\pi r^2\). This means that the shaded area is \((4+2\pi)r^2 - 2\pi r^2\) or \(4r^2\).

This is the same as the area of the square, so the ratio is **1:1**.

## Square deal

This unit square is divided into four regions by a diagonal and a line that connects a vertex to the midpoint of an opposite side. What are the areas of the four regions?

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The square is unit, so \(a+b+c+d=1\). By the definitions of the lines, \(a+d=\frac{1}{2}\) and \(a+b=\frac{1}{4}\).

\(a\) and \(c\) are similar triangles. The vertical side of \(c\) is twice that of \(a\) so \(c=4a\).

Therefore we have the system of simultaneous equations:

$$a+b+c+d=1\\a+d=\frac{1}{2}\\a+b=\frac{1}{4}\\c=4a$$

These can be solved to find:

$$a=\frac{1}{12}\\b=\frac{1}{6}\\c=\frac{1}{3}\\d=\frac{5}{12}\\$$

#### Extension

What would be the areas if the lines were a diagonal and another line which divides the sides in the ratio \(x:y\)?

## Light work

"*I don't know if you are fond of puzzles, or not. If you are, try this. ... A gentleman (a nobleman let us say, to make it more interesting) had a sitting-room with only one window in it—a square window, 3 feet high and 3 feet wide. Now he had weak eyes, and the window gave too much light, so (don't you like 'so' in a story?) he sent for the builder, and told him to alter it, so as only to give half the light. Only, he was to keep it square—he was to keep it 3 feet high—and he was to keep it 3 feet wide. How did he do it? Remember, he wasn't allowed to use curtains, or shutters, or coloured glass, or anything of that sort.*"