Puzzles

Let \(f(a,b)\) be the number of \(a\)-dimensional sides on a \(b\)-dimensional hypercube. A \((b+1)\)-dimensional hypercube is formed by making two copies of a \(b\)-dimensional hypercube and adding edges between the corresponding vertices on the two copies. Therefore the \((b+1)\)-dimensional hypercube will contain two copies of each \(a\)-dimensional side in the \(b\)-dimensional hypercube; it will also contain a new \(a\)-dimensional side formed by each \((a-1)\)-dimensional side of the \(b\)-dimensional hypercube. Therefore \(f(a,b+1)=2f(a,b)+f(a-1,b)\).

(This can be seen more clearly by looking at the 1-dimensional sides (edges) and 2-dimensional sides (faces) on a 3-dimensional hypercube (cube) and a 2-dimensional hypercube (a square).)

Using the above relation, we find that there are 112 6-dimensional sides on a 8-dimensional hypercube.

Consider the top, front, and right sides of the cube.

If the top number is \(a\) more than a multiple of six, then the front and right numbers must both be \(a\) less than a multiple of 6 (so that when added to the top number they make a multiple of 6). But when the front and right numbers are added, they make \(2a\) less than a multiple of 6; but this must also be a multiple of 6.

This is only possible if \(a=0\) or \(a=3\). So the numbers must be either all multiples of 6, or all 3 more than multiples of 6.

The smallest set of numbers that are all 3 more than multiples of 6 is 3,9,15,21,27,33. The sum of these is 108. The smallest set of numbers that are all multiples of 6 is the set with each number three more than these, so 108 is the smallest possible total.

Six numbers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of \(n\).

What is the smallest possible sum of the six numbers?

798

One module is needed for each edge of the dodecahedron. Therefore 30 modules are needed.

48 tube maps are needed for the stellated rhombicuboctahedron.

Therefore today's number is 720.

\(2009=7\times 7\times 41\), so there are four possible sets of dimensions of the cuboid:

In the first three cuboids, there is a face with an area of 2009 units (\(2009\times 1\), \(7\times 287\) and \(41\times 49\) respectively) and so 2009 stickers will not be enough. Therefore the cuboid has dimensions \(7\times 7\times 41\) and a surface area of 1246, leaving 763 stickers left over

For which numbers \(n\) can a cuboid be made with \(n\) unit cube such that \(n\) unit square stickers can cover the faces of the cuboid?

Once the map is folded, it will look like this:

For the final tetrahedron to be regular, the red lengths must be equal. Let each red length be 2 (this will get rid of halves in the upcoming calculations). By drawing a vertical line in we can work out the width and height of the rectangle:

The width of the rectangle is 3 (one and a half red lengths). Using Pythagoras' Theorem in the blue triangle, we find that the height of the rectangle is \(\sqrt{3}\). Therefore, the ratio of the rectangle is \(\sqrt{3}:3\) or \(1:\sqrt{3}\).

If the ratio of the rectangle is \(1:a\), what is the ratio of the lengths of the sides of the tetrahedron?

By walking in a straight line, the people will follow a great circle. They begin by travelling parallel to each other, so the 1km they are apart at the start is the furthest they will be apart. They will be 1km apart again after walking half way around the world and will collide at one quarter and three quarters of the way around.

Hence, each person will walk one quarter of the circumference of the Earth before colliding, which will be 10,018km.

Another way to see that this is the answer is to assume that the people stand on the equator and walk North. They will meet at the North Pole, one quarter of the way round.

If two people stand 10km apart and walk in the same direction, how far will the have to walk until they collide due to the curvature of the Earth? (diameter of Earth = 12,742km)

Let the length of a side of a triangle be \(2L\). Using \(2L\) instead of \(L\) gets rid of fractions in the later calculations.

By Pythagoras' Theorem, the height of a triangle is \(L\sqrt{3}\).

Using Pythagoras' Theorem again, the height of the square-based pyramid is \(L\sqrt{2}\).

Therefore, the volume of the square-based pyramid is:

Next, we find the area of the tetrahedron.

Call the point on the base of the tetrahedron, directly below the vertex at the top, \(A\).

Using cosine in the triangle made by \(A\), the corner of the base and the midpoint of a side of the base, the distance from the corner to \(A\) is \frac{2}{\sqrt{3}}L\).

Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is \(\frac{2\sqrt{2}}{\sqrt{3}}L\).

Therefore, the volume of the tetrahedron is:

Finally, the ratio of volume of the square based pyramid to the tetrahedron is:

What would the ratio be if they were isosceles triangles?