Hopping a Triangular Path

The solution to Fagnano's problem is the orthic triangle . As the given triangle is acute the perimeter of the orthic triangle is $a\cos(A) + b\cos(B) + c\cos(C)$ where $a,b,c$ are the side lengths and $A,B,C$ the angles opposite sides $a,b,c$ respectively.

With $(a,b,c) = (4,5,6)$ , we then make use of the cosine law to find that

$\cos(A) = \dfrac{b^{2} + c^{2} - a^{2}}{2bc} = \dfrac{3}{4}, \cos(B) = \dfrac{a^{2} + c^{2} - b^{2}}{2ac} = \dfrac{9}{16}$ and

$\cos(C) = \dfrac{a^{2} + b^{2} - c^{2}}{2ab} = \dfrac{1}{8}$ , resulting in a perimeter of

$4*\dfrac{3}{4} + 5*\dfrac{9}{16} + 6*\dfrac{1}{8} = \dfrac{105}{16}$ , and so $m + n = 105 + 16 = \boxed{121}$ .