Puzzles

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)

CG is equal to AF as they are both diagonals in a rectangle.

AF is a radius of the triangle, so AF = 10.

Therefore CG = 10.

Calculate the lengths of CI and CJ.

Calculate the lengths of HG.

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?

Call the blue area \(B\) and the red area \(R\).

Prove that the red and blue areas are equal.

The two poles can be any distance apart; the distance does not affect the heights.

If the heights of the two poles were 12m and 24m tall, what height would the intersection of the lines be?

If the heights of the two poles were \(a\)m and \(b\)m tall, what height would the intersection of the lines be?