Find the Heronian triangle one of whose altitudes and its three sides are consecutive numbers.

Bonus: Prove that it is unique.

The answer is a 13-14-15 triangle, one of whose altitudes is 12.

I determined this by looking up "

Heronian Triangles" in Wikipedia. It has a list of "almost equilateral" Heronian triangles. The first of these, 3-4-5, has an area of 6, so the altitudes are 4, 3 and 12/5, which do not work. The next is 13-14-15, with an area of 84. Altitudes are 168/13, 168/14 (ie, 12), and 168/15.

**Almost-bonus:**

Triangles of the form (n-1) - n - (n+1) get closer and closer to equilateral as n increases. That means that the altitude gets closer and closer to the altitude of an equilateral triangle, namely n*sqrt(3)/2 = n*.866. If this is one of 4 consecutive numbers, then it is greater than or equal to n-2.

Solving (n-2) <= n*.866 gives n*.134 <= 2 and then n <= 14.9.

So, larger n will not work. They result in altitudes that are too small to be consecutive. So, there are a very small number of triangles that are candidates for a solution to this problem. And the only two of those candidates are Heronians, and those are the 3-4-5 and the 13-14-15. There are no Heronians of the form (n-2) - n - (n+1), but if there were we would only need to test up to n = 14 or so.

*Edited on ***July 30, 2017, 8:01 am**