Relationship Between Orthic Triangle And Original Triangle?

The lengths of the legs of the orthic triangle of a triangle $\Delta ABC$ with legs $a,b,c$ are given by $a' = a|\cos A|, \quad b' = b|\cos B|, \quad c' = c|\cos C|$ so the perimeter of this orthic triangle is applying cosine rule : $13 \cdot (\frac{14^2 + 15^2 - 13^2}{2\cdot 14 \cdot 15}) + 14 \cdot (\frac{13^2 + 15^2 - 14^2}{2\cdot 13 \cdot 15}) + 15 \cdot (\frac{13^2 + 14^2 - 15^2}{2\cdot 13 \cdot 14}) = \frac{1344}{65}...$

We use three formulae to solve this problem. Try to prove them all, as they are very handy in the geometer's arsenal.

By Heron's formula, the area of a triangle is

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$s=\dfrac{13+14+15}{2} = 21$ , so we have

$A = \sqrt{21(21-13)(21-14)(21-15)} = 84$

We then use the following formula to determine the circumradius

$\begin{aligned} R &= \dfrac {abc}{4A}\\ &= \dfrac {13 \times 14 \times 15}{4 \times 84}\\ &= \dfrac {65}{8} \end{aligned}$

We finally use the formula to determine the perimeter of the orthic triangle

$A = \dfrac {PR}{2}$

where $P$ is the perimeter of the orthic triangle. Thus, $P=\frac {2A}{R} = \frac {168 \times 8}{65} = \frac {1344}{65}$ . Thus, the perimeter of the orthic triangle is $\frac {1344}{65}$ , so the answer is $\boxed{1409}$ .