# Puzzles

## 23 December

Today's number is the area of the largest area rectangle with perimeter 46 and whose sides are all integer length.

## 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.

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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.

## Equal lengths

The picture below shows two copies of the same rectangle with red and blue lines. The blue line visits the midpoint of the opposite side. The lengths shown in red and blue are of equal length.

What is the ratio of the sides of the rectangle?

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Let \(a\) be the height of the rectangle and \(b\) be the width.

The length of the red line is \(a+b\). The length of the blue line is \(2\sqrt{a^2+\frac{b^2}4}\). These are equal so:

\begin{align}
a+b&=2\sqrt{a^2+\frac{b^2}4}\\
(a+b)^2&=4\left(a^2+\frac{b^2}{4}\right)\\
a^2+2ab+b^2&=4a^2+b^2\\
0&=3a^2-2ab\\
0&=3a-2b\\
2b&=3a
\end{align}

Therefore the ratio of the sides is 2:3.

## 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.

## Bending a straw

Two points along a drinking straw are picked at random. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?

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A triangle will be made if none of the segments of straw is longer than the other two added together. This is the same as requiring that each segment must be less than half the straw.

Let the length of the straw be 1 unit. Call the points \(x\) and \(y\). A triangle is made if either:

- \(x\lt y\), \(x\lt\tfrac12\), \(y-x\lt\tfrac12\), \(1-y\lt\tfrac12\); or
- \(y\lt x\), \(y\lt\tfrac12\), \(x-y\lt\tfrac12\), \(1-x\lt\tfrac12\).

For the second condition, the allowable region is shown below.

This region covers \(\tfrac18\) of the whole square. By switching \(x\) and \(y\) it can be seen that the first condition's region is the same size as the second's, plus they don't overlap. Therefore the probability of making a triangle is \(\tfrac18+\tfrac18=\tfrac14\).

#### Extension

One point along a drinking straw is picked, then a coin is flipped. If the coin shows heads, a second point above the first is chosen; If tails, a second point below the first is chosen. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?

## Placing plates

Two players take turns placing identical plates on a square table. The player who is first to be unable to place a plate loses. Which player wins?

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The first player can always win by first placing a plate on the exact centre of the table. Then the first player can copy what the second player does, but rotated 180°, and hence can always place a plate if the second player could.

#### Extension

What if the two players play on a regular hexagonal table? Or a regular octagonal table? Or a regular pentagonal table? Or a regular \(n\)-gonal table?

## 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?

## 17 December

The number of degrees in one internal angle of a regular polygon with 360 sides.