# Puzzles

## 5 December

28 points are spaced equally around the circumference of a circle. There are 3276 ways to pick three of these points.
The three picked points can be connected to form a triangle. Today's number is the number of these triangles that are isosceles.

#### Show answer

#### Hide answer

Pick one of the 28 points and imagine a line to the point opposite it. There will be 13 points on each side of this line. Picking one of these 13 points and then picking the
corresponding point on the other side of the line gives an isosceles triangle. Therefore there are 13 isosceles triangles with the first chosen point as the point where the two equal sides meet.

There were 28 choices for the first point, and so the total number of isosceles triangles will be 13×28=**364**.

## 2 December

You have 15 sticks of length 1cm, 2cm, ..., 15cm (one of each length). How many triangles can you make by picking three sticks and joining their ends?

Note: Three sticks (eg 1, 2 and 3) lying on top of each other does not count as a triangle.

Note: Rotations and reflections are counted as the same triangle.

#### Show answer

#### Hide answer

In order to make a valid triangle, the sum of the two shorter sides must be larger than the longest side.
There is
1 triangle with longest side 4cm (2,3,4);
2 with longest side 5cm (2,4,5 and 3,4,5);
4 with longest side 6cm (2,5,6; 3,5,6; 4,5,6 and 3,4,6);
6 with longest side 7cm (2,6,7; 3,6,7; 4,6,7; 5,6,7; 3,5,7 and 4,5,7);
9 with longest side 8cm;
12 with longest side 9cm;
16 with longest side 10cm;
20 with longest side 11cm;
25 with longest side 12cm;
30 with longest side 13cm;
36 with longest side 14cm; and
42 with longest side 15cm.

In total this makes **203** triangles.

## 12 December

There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.

These three vertices form a right angled triangle.

Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.

#### Show answer

#### Hide answer

The vertices of the 26-gon lie on a circle. The triangle is therefore right-angled if (and only if) the longest side is a diameter of the circle.
In other words, the triangle is right angled if (and only if) two of its vertices are opposite vertices of the 26-gon.

There are 13 different pairs of opposite points on the 26-gon. For each of these, there are 24 remaining vertices that could be the third vertex of the triangle.
Therefore there are 13×24=**312** different right angled triangles.

## 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?

#### Show answer

#### Hide answer

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.

## 20 December

Start with a line of length 2. Draw a line of length 17 perpendicular to it. Connect the ends to make a right-angled triangle.
The length of the hypotenuse of this triangle will be a non-integer.

Draw a line of length 17 perpendicular to the hypotenuse and make another right-angled triangle. Again the new hypotenuse will have a non-integer length.
Repeat this until you get a hypotenuse of integer length. What is the length of this hypotenuse?

## Cutting corners

The diagram below shows a triangle \(ABC\). The line \(CE\) is perpendicular to \(AB\) and the line \(AD\) is perpedicular to \(BC\).

The side \(AC\) is 6.5cm long and the lines \(CE\) and \(AD\) are 5.6cm and 6.0cm respectively.

How long are the other two sides of the triangle?

#### Show answer

#### Hide answer

Let \(A\), \(B\) and \(C\) represent the angles at points \(A\), \(B\) and \(C\). Looking at triangle \(ACE\) gives:

$$\sin A=\frac{5.6}{6.5}$$

And triangle \(ACD\) gives:

$$\sin C=\frac{6.0}{6.5}$$

Using Pythagoras' Theorem to find the missing sides in these triangles also gives:

$$\cos A=\frac{3.3}{6.5}$$
$$\cos C=\frac{2.5}{6.5}$$

We can use these to find \(\sin B\):

$$\begin{array}{rl}
\sin B&=\sin(\pi-A-C)\\
&=\sin(A+B)\\
&=\sin A\cos C + \cos A\sin C\\
&=\frac{5.6}{6.5}\frac{2.5}{6.5}+\frac{6.0}{6.5}\frac{3.3}{6.5}\\
&=\frac{33.8}{6.5^2}
\end{array}$$

Now the missing sides can be found using the sine rule:

$$\begin{array}{rl}
CB&=\frac{CA\sin A}{\sin B}\\
&=\frac{6.5\times\frac{5.6}{6.5}}{\frac{33.8}{6.5^2}}\\
&=\frac{6.5^2\times5.6}{33.8}\\
&=7\text{cm}
\end{array}$$
$$\begin{array}{rl}
AB&=\frac{CA\sin C}{\sin B}\\
&=\frac{6.5\times\frac{6}{6.5}}{\frac{33.8}{6.5^2}}\\
&=\frac{6.5^2\times6}{33.8}\\
&=7.5\text{cm}
\end{array}$$

## 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?

#### Show answer & extension

#### Hide answer & extension

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?

## Equal side and angle

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

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

#### Show answer

#### Hide answer

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:

$$AB=BC$$