Puzzles

There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square on a chessboard.

This can be shown by counting how many positions the top left corner of the square can sit on. For example, the top left corner of a 5×5 square can be in the first four rows and columns of the board (otherwise the square will go off the board) and 4×4=16.

64+49+36+25+16+9+4+1=204.

How many rectangles are there on a chessboard?

Let \(A\) be the area of the square (and the triangle).

The length of a side of the square is \(\sqrt{A}\), so the perimeter of the square is \(4\sqrt{A}\).

Let \(l\) be the length of a side the triangle. Then \(\frac{1}{2}l^2\sin{60}=A\), so \(l^2=\frac{4A}{\sqrt{3}}\). Therefore \(l=\frac{2\sqrt{A}}{3^\frac{1}{4}}\) and the perimeter of the triangle is \(\frac{6\sqrt{A}}{3^\frac{1}{4}}\).

Hence the ratio of the perimeters is \(\frac{6\sqrt{A}}{3^\frac{1}{4}} : 4\sqrt{A}\) which simplifies to 3^3/4:2

If an \(n\) sided regular polygon has the area \(A\), what is the length of one of its sides?

If mirrors were placed along the walls of the rectangle, the ball would appear to travel in a straight line across a grid of rectangles:

Viewed this way, the ball will still stop once it reaches a corner:

For an \(n\) by \(m\) (in above example: \(n=4\), \(m=3\)) rectangle, this will occur once the ball has travelled through \(\mathrm{lcm}(n,m)\) squares.

On its way to the corner, the ball will bounce \(b-1\) times off the top and bottom and \(a-1\) times off the sides. It can be seen that if \(a-1\) is even, then the ball will end in one of the corners on the right hand side. The complete results can be seen in the following Carroll diagram:

It can be shown that the ball will never finish in the top left (where it started) as this would require it to travel through the bottom right first. Therefore the following holds:

For which sizes of rectangle will the path of the ball make the same pattern?

Any complex number can be written in the form \(z=re^{i\theta}\).

This gives that \(z^2=r^2e^{2i\theta}\), which will have positive real and complex parts when \(0+2\pi n < 2\theta < \frac{\pi}{2}+2\pi n\).

This will occur when \(0 < \theta < \frac{\pi}{4}\) and \(\pi < \theta < \frac{5\pi}{4}\).

This can be represented on an Argand diagram:

A complex number \(z\) falls in these regions when \(|\mathrm{Re}(z)|>|\mathrm{Im}(z)|\) and \(\mathrm{sign}(\mathrm{Re}(z))=\mathrm{sign}(\mathrm{Im}(z))\).

For which complex numbers, \(z\), are \(\mathrm{Re}(z^3)\) and \(\mathrm{Re}(z^3)\) both positive?

\(A=8\), \(B=9\) and \(C=7\).

Find bases \(A\), \(B\) and \(C\) such that \(123_A+456_B=789_C\)

445 in base 10 is equal to 544 in base 7.

Find another pair of bases \(A\) and \(B\) so that there exist digits \(d\), \(e\) and \(f\) such that \(def\) in base \(A\) is equal to \(fed\) in base \(B\)?

a+(a+A)^-1 = a + A = 2.

So, a+2^-1 = 2

So, a = ³/₂

By the same method, b = 2-√2, c = -2 and d = 2-2^k

For which values of k does this converge?

If \(ab\) in base 10 is equal to \(ba\) in base 4, then \(10a+b=4b+a\).

So, \(9a=3b\).

\(a\) and \(b\) must both be less than 4, as they are digits used in base 4, so \(a=1\) and \(b=3\).

So 13 in base 10 is equal to 31 in base 4.

By the same method, we find that:

For which pairs of bases \(A\) and \(B\) can you find two digits \(g\) and \(h\) such that \(gh\) in base \(A\) is equal to \(hg\) in base \(B\)?