# Puzzles

## 2 December

Today's number is the area of the largest dodecagon that it's possible to fit inside a circle with area \(\displaystyle\frac{172\pi}3\).

## Two semicircles

The diagram shows two semicircles.

\(CD\) is a chord of the larger circle and is parallel to \(AB\). The length of \(CD\) is 8m. What is the area of the shaded region (in terms of \(\pi\))?

#### Show answer & extension

#### Hide answer & extension

The question does not fix the length of \(AB\), yet implies that there is a unique answer. Therefore we can take \(AB\) to be any length we like and expect the right answer. If \(AB\) is 8m long, then the unshaded semicircle has no area. Therefore the shaded area is \(\tfrac12\pi\times4^2=8\pi\)m.

#### Extension

How would you calculate the area if you don't assume that the length of \(AB\) doesn't affect the area?

## 1 December

What is area of the largest area rectangle which will fit in a circle of radius 10?

#### Show answer

#### Hide answer

The largest rectangle will be a square. 20 (double the radius) will be the length of its diagonal.
By Pythagoras' Theorem, the sides of the square are \(10\sqrt{2}\). Therefore the area of the square is 200.

## 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.

#### Show answer

#### Hide answer

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**.

## Dartboard

Concentric circles with radii 1, \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\), ... are drawn. Alternate donut-shaped regions are shaded.

What is the total shaded area?

#### Show answer & extension

#### Hide answer & extension

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

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

## Circles

Which is largest, the red or the blue area?

#### Show answer & extension

Let \(4x\) be the side length of the square. This means that the radius of the red circle is \(2x\) and the radius of a blue circle is \(x\). Therefore the area of the red circle is \(4\pi x^2\).

The area of one of the blue squares is \(\pi x^2\) so the blue area is \(4\pi x^2\). Therefore **the two areas are the same**.

#### Extension

Is the red or blue area larger?

## Semi circle in a triangle

This right-angled triangle above has sides of lengths 12cm, 5cm and 13cm. The diameter of the semicircle lies on the 12cm side and the 13cm side is a tangent to the circle. What is the radius of the semi circle?

#### Show answer & extension

#### Hide answer & extension

Label the triangle as follows:

EC and BC are both tangents to the circle so FEA is a right angle and the lengths EC and BC are equal, so the length of EC is 5. Let *r* be the radius of the circle.

Using Pythagoras' Theorem in triangle FEA:

$$8^2+r^2=(12-r)^2$$
$$64+r^2=144-24r+r^2$$
$$24r=80$$
$$r=\frac{80}{24}=\frac{10}{3}$$

#### Extension

What is the radius of the circle whose diameter lies on the 5cm side and to which the 12cm and 13cm sides are tangents?

What is the radius of the circle whose diameter lies on the 13cm side and to which the 12cm and 5cm sides are tangents?

**Additional observation**

For each pair of semi circles, draw a straight line between the two points where the semi circles intersect. These lines all meet at a point.