A Squared Triangle

Let $\angle LXM = \alpha$ , $\angle KXJ = \beta$ , and $\angle OXP = \gamma$ . We note that $\angle XPO = 90^\circ - \dfrac \gamma 2$ , $\implies \angle XPM = 135^\circ - \dfrac \gamma 2$ , $\angle MPB = 45^\circ + \dfrac \gamma 2$ , $\angle PML = 45^\circ + \dfrac \gamma 2 + B$ , $\angle XML = \dfrac \gamma 2 + B$ . Therefore $\angle LXM = \alpha = 180^\circ - \gamma - 2B$ . Similarly, $\beta = 180^\circ - \gamma - 2A$ . Also note that:

$\begin{aligned} \alpha + \beta + \gamma + 3 \times 90^\circ & = 360^\circ \\ \alpha + \beta + \gamma & = 90^\circ \\ 180^\circ - \gamma - 2B + 180^\circ - \gamma - 2A + \gamma & = 90^\circ \end{aligned}$

$\begin{aligned} \implies \gamma & = 270^\circ - 2(A+B) = 270^\circ - 2(180^\circ - C) = 2C - 90^\circ \end{aligned}$

Therefore we can find the values of $\alpha$ , $\beta$ , and $\gamma$ . A special feature of $13$ - $14$ - $15$ triangle is that the altitude from $C$ to $AB$ is $12$ . Then the area of $\triangle ABC$ is $84$ . Let the side length of the squares be $r$ and the altitudes from $X$ to $BC$ , $CA$ , and $AB$ be $h_a$ , $h_b$ , and $h_c$ respectively. Then the area of $\triangle ABC$ is also given by:

$\begin{aligned} \frac {h_a\cdot BC + h_b \cdot CA + h_c \cdot AB}2 & = [ABC] \\ \frac 12 \left(15 r \cos \frac \alpha 2 + 13 r \cos \frac \beta 2 + 14 r \cos \frac \gamma 2 \right) & = 84 \end{aligned}$

We find that $\tan A = \dfrac {12}5$ , $\tan B = \dfrac 43$ , and $\tan C = \dfrac {\frac 5{12}+\frac 34}{1-\frac 5{12} \cdot \frac 34} = \dfrac {56}{33}$ and

$\begin{aligned} \tan \frac \gamma 2 & = \tan (C-45^\circ) = \frac {\frac {56}{33}-1}{\frac {56}{33}+1} = \frac {23}{89} & \implies \cos \frac \gamma 2 = \frac {89}{65\sqrt 2} \\ \tan \frac \alpha 2 & = \tan \left(90^\circ - \frac \gamma 2 - B\right) = \cot \left(\frac \gamma 2 + B\right) = \frac {1-\frac {23}{89}\cdot \frac 43}{\frac {23}{89}+\frac 43} = \frac 7{17} & \implies \cos \frac \alpha 2 = \frac {17}{13\sqrt 2} \\ \tan \frac \beta 2 & = \tan \left(90^\circ - \frac \gamma 2 - A\right) = \cot \left(\frac \gamma 2 + A\right) = \frac {1-\frac {23}{89}\cdot \frac {12}5}{\frac {23}{89}+\frac {12}5} = \frac 17 & \implies \cos \frac \beta 2 = \frac 7{5\sqrt 2} \end{aligned}$

Therefore,

$\begin{aligned} \left(\frac {17}{13\sqrt 2} \cdot 15 + \frac 7{5\sqrt 2} \cdot 13 + \frac {89}{65\sqrt 2} \cdot 14 \right) r & = 84 \cdot 2 \\ \frac {3704}{65 \sqrt 2} r & = 84 \cdot 2 \\ \implies r & = \frac {1365\sqrt 2}{463} \end{aligned}$

$\implies a+b+c = 1365 + 2 + 463 = \boxed{1830}$

Formula for square edge for the first Kenmotu point is given here . After plugging the values in we get a=1365, b=2, c=463. a+b+c=1830