# Puzzles

## Balanced sets

A set of points in the plane is called 'balanced' if for any two points \(A\) and \(B\) in the set, there is another point \(C\) in the set such that \(AC=BC\) (here \(AC\) is the distance between \(A\) and \(C\)).

For all \(n\geq3\), find a balanced set of \(n\) points.

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If \(n\) is odd, the vertices of a regular \(n\)-gon are balanced.

If \(n\) is even, a balanced set can be constructed as follows:

If \(n=4\), this set is balanced (both the triangles are equilateral):

Draw a triangle with its centre at one of the points, going through the other three points:

For \(n>4\), repeatedly add a two new points on the circle which form an equilateral triangle with the centre. For example, for \(n=6\), this set is balanced:

And for \(n=8\), this set is balanced:

## 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**.

## Two triangles

The three sides of this triangle have been split into three equal parts and three lines have been added.

What is the area of the smaller blue triangle as a fraction of the area of the original large triangle?

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Draw on the following lines parallel to those which were added in the question.

Then a grid of copies of the smaller blue triangle has been created. Now consider the three triangles which are coloured green, purple and orange in the following diagram:

Each of these traingles covers half a parallelogram made from four blue triangles. Therefore the area of each of these triangles is twice the area of the small blue triangle.

And so the blue triangle covers one seventh of the large triangle.

#### Extension

If the sides of the triangle were split into \(n\) pieces the the lines added, what would the area of the smaller blue triangle be as a fraction of the area of the original large triangle?

## Dodexagon

In the diagram, B, A, C, D, E, F, G, H, I, J, K and L are the vertices of a regular dodecagon and B, A, M, N, O and P are the vertices of a regular hexagon.

Show that A, M and E lie on a straight line.

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The interior angle of a regular hexagon is 120°. The interior angle of a regular dodecagon is 150°. Therefore angle CAM is 30°.

Now, consider the quadrilateral ACDE. This quadrilateral is symmetric (as the dodecagon is regular) so the angles CAE and DEA are equal. Hence:

$$360 = CAE+DEA+ACD+CDE\\
= 2CAE + 2\times 150\\
2CAE = 60\\
CAE=30
$$

The angles CAM and CAE are equal, so A, M and E lie on a straight line.

#### Extension

The vertices \(P_1\), \(P_2\), ..., \(P_n\) make up a regular \(n\)-gon and \(Q_1\), \(Q_2\), ..., \(Q_m\) make up a regular \(m\)-gon, with \(P_1=Q_1\) and \(P_2=Q_2\).

The vertices \(P_2\), \(Q_3\) and \(P_5\) lie on a straight line. What is the relationship between \(m\) and \(n\)?

## Equal side and angle

In the diagram shown, the lengths \(AD = CD\) and the angles \(ABD=CBD\).

Prove that the lengths \(AB=BC\).

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By the sine rule in BCD:

$$\frac{CD}{\sin(CBD)}=\frac{BD}{\sin(BCD)}$$

By the sine rule in BAD:

$$\frac{AD}{\sin(ABD)}=\frac{BD}{\sin(BAD)}$$

\(ABD=CBD\) and \(AD=CD\), so:

$$\sin(BAD)=\frac{BD\sin(ABD)}{AD}\\
=\frac{BD\sin(CBD)}{CD}\\
=\sin(BCD)$$

Using the sine rule in ABC:

$$\frac{AB}{\sin(BCD)}=\frac{BC}{\sin(BAD)}$$

The two sines are equal and so:

$$AB=BC$$

## Arctan

Prove that \(\arctan(1)+\arctan(2)+\arctan(3)=\pi\).

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Let \(\alpha=\arctan(1)\), \(\beta=\arctan(2)\) and \(\gamma=\arctan(3)\), then draw the angles as follows:

#### Extension

Can you find any other integers \(a\), \(b\) and \(c\) such that:

$$\arctan(a)+\arctan(b)+\arctan(c)=\pi$$

## 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 \(x:y\)?

## Half an equilateral triangle

What is the shortest straight line which bisects an equilateral triangle?

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Consider a line which bisects an equilateral triangle with sides of length 1. Let the sides of the smaller triangle created be \(a\) and \(b\) as shown below:

The area of the equilateral triangle is:

$$\frac{1}{2}\times 1\times 1\times \sin(60)\\=\frac{\sqrt{3}}{4}$$

The area of the smaller triangle is:

$$\frac{1}{2}ab \sin(60)\\=\frac{ab\sqrt{3}}{4}$$

This is half the area of the equilateral triangle, so:

$$ab=\frac{1}{2}$$

The length (\(L\)) of the line bisecting the equilateral triangle is:

$$L=a^2+b^2-2ab\cos(60)\\
=a^2+\left(\frac{1}{2a}\right)^2-\frac{2a}{2a}\times \frac{1}{2}\\
=a^2+\frac{1}{4a^2}-\frac{1}{2}$$

To find the minimum, we set the derivative to 0:

$$0=\frac{dL}{da}=2a-\frac{1}{2a^3}\\
4a^4-1=0\\
a^4=\frac{1}{4}\\
a=\frac{1}{\sqrt{2}}$$

For this value of \(a\), \(b\) and \(L\) also equal to \(\frac{1}{\sqrt{2}}\).

#### Extension

What if the line does not need to be straight?