Square inscribed in a triangle

A special feature of 13-14-15 triangle is that it is made of two Pythagorean triple triangles 5-12-13 and 9-12-15. It has a height $CN$ of 12. Since $\triangle CEF$ and $\triangle ABC$ are similar, we have:

$\begin{aligned} \frac {EF}{AB} & = \frac {CM}{CN} \\ \frac s{14} & = \frac {12-s}{12} \\ 6s & = 84 - 7s \\ \implies s & = \frac {84}{13} \end{aligned}$

Therefore $p+q = 84 + 13 = \boxed{97}$ .

By Heron's Formula, the area of the triangle is $A = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = 84$ , so its height is $h = \frac{2A}{b} = \frac{2 \cdot 84}{14} = 12$ .

By Cavalieri's principle the square will also be inscribed by a right triangle of the same height.

If the right angle is at the origin, then the hypotenuse of the right triangle will be on $y = -\frac{6}{7}x + 12$ , and the diagonal of the inscribed square will be on $y = x$ .

These two equations intersect at $x = \frac{84}{13}$ , the side length of the square. Therefore, $p = 84$ , $q = 13$ , and $p + q = \boxed{97}$ .