Twenty-five spheres are dropped into a cone.
Each sphere above the first touches the sides of the cone, and the top of
the sphere below it.

The radius of the first sphere is 1, and the radius of the 25th sphere
is 25. What is the radius of the 13th sphere?

Let the radii of the spheres be r_{1}, r_{2}, ..., r_{25}.
We are given r_{1} = 1 and r_{25} = 25.

Consider the diagram, which shows three of the spheres, and let their radii
be a, b and c.

The small triangles are similar, so taking the ratios of the heights
over the hypotenuses, we have:

(b - a) / (b + a) |
= |
(c - b) / (c + b) |

bc + b^{2} - ac - ab |
= |
bc - b^{2} + ac - ab |

2b^{2} |
= |
2ac |

b/a |
= |
c/b |

Hence, the ratio of consecutive radii is a constant, say K, and we have
r_{i} = K r_{i-1}.

By iteration, 25 = r_{25} = K^{24}r_{1} = K^{24},
and so K = 25^{1/24}.

Finally, r_{13} = K^{12}r_{1} = 25^{12/24} =
25^{1/2} = 5,
and so the 13th sphere has radius 5.