An Equilateral Dilemma: Solution

The original problem

I'd like to thank Hong Li for solving the more interesting part of the problem.

First, lets establish the existence of the outer triangle. We'll demonstrate that at least an equilateral triangle exists.

Let s be the angle depicted above, and s = s' = s'' . Then the lines that pass through the vertices of the inner triangle at the angle s will intersect each other at points P1, P2, P3. Trivially, this forms a triangle, and we are assured they meet since none of the lines generated this way are parallel. By similar triangles, the points form an equilateral triangle.

To demonstrate the triangle formed can only be equilateral, we eliminate the other possibilities.

## Isosceles triangles:

Without loss of generality: a = a' != a'' => s = s' != s''

But triangle abc = triangle a'bc, so t = t' => s'' = s', since t + s' = 120, and t' + s'' = 120. A contradiction.

## Scalene triangles:

I would like to thank Bo Hu for discovering a flaw in my argument. I have now updated it with corrections.

Without loss of generality: a < a' < a'' => s < s' < s''

Similarly, r'' < r < r', when looking at the sidelengths of the large triangle and their opposing angles.

Let's start with the Cosine Law. With the above variables we have: | |

Rearrange to isolate cos r: | |

Identify cos r - cos r': | |

Express a' relative to a: | |

Substitute: | |

Common denominator: | |

Expand: | |

Cancel common terms: | |

If b <= c (and given a, b, c, d > 0) then cos r - cos r' < 0. So cos r < cos r' so r > r' (cosine being monotonic decreasing from 0 to pi). This is a contradiction to r'' < r < r'.

However we still have to solve c < b.

Instead we turn to t. We have t + s' = 120, t' + s'' = 120, and t'' + s = 120.

So t' < t < t''

Applying the same cosine law above between t' and t" we get (after skipping identical steps): | |

With c < b this leads to cos t' - cos t'' < 0 => cos t' < cos t'' => t' > t'' from 0 to pi. This contradicts t' < t < t''.

The triangle cannot be scalene.

Therefore the outer triangle generated as described must be equilateral.

If you have comments, suggestions or questions, mail me at:

*bscriver@oneoddsock.com*