Triangular Kenmotu

Figure 1 The idea of the solution comes from @Chew-Seong Cheong’s solution on this , since it fits almost identically in this new problem. So, I copy his notation and part of his text.

Let $\angle LXM = \alpha$ , $\angle KXJ = \beta$ , and $\angle OXP = \gamma$ (figure 1). We note that $\angle XPO = 90^\circ - \dfrac \gamma 2$ , hence, $\angle MPB=180{}^\circ -\left( 90{}^\circ -\dfrac{\gamma }{2} \right)-60{}^\circ =\dfrac{\gamma }{2}+30{}^\circ$ , thus, $\angle LMX=\angle LMP-60{}^\circ =\left( \dfrac{\gamma }{2}+30{}^\circ +B \right)-60{}^\circ =B+\dfrac{\gamma }{2}-30{}^\circ$ .

Consequently, $\alpha =180{}^\circ -2\cdot \angle LMX=180{}^\circ -2 \left( B+\dfrac{\gamma }{2}-30{}^\circ \right)=240{}^\circ -2 B-\gamma$ .
Similarly, $\beta =240{}^\circ -2 A-\gamma$ .

Moreover, $\begin{aligned} \alpha + \beta + \gamma + 3 \times 60^\circ & = 360^\circ \\ \alpha + \beta + \gamma & = 180^\circ \\ 240^\circ - 2B- \gamma + 240^\circ - 2A- \gamma + \gamma & = 180^\circ \end{aligned}$

$\Rightarrow \gamma =300{}^\circ -2\left( A+B \right)=300{}^\circ -2\left( 180{}^\circ -C \right)=2C-60{}^\circ \Rightarrow \dfrac{\gamma }{2}=C-30{}^\circ$ Similarly, $\dfrac{\alpha}{2}=A-30{}^\circ$ and $\dfrac{\beta}{2}=B-30{}^\circ$ .

Using Heron’s formula we find the area of $\triangle ABC$ : $\left[ ABC \right]=84$ .

Let the side length of the equilateral triangles be $s$ 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: $\begin{aligned} \dfrac {h_a\cdot BC + h_b \cdot CA + h_c \cdot AB}2 & = [ABC] \\ \dfrac 12 \left(15 s \cos \dfrac \alpha 2 + 13 s \cos \dfrac \beta 2 + 14 s \cos \dfrac \gamma 2 \right) & = 84 \ \ \ \ \ (1)\end{aligned}$

In order to calculate $\cos \dfrac \alpha 2$ , $\cos \dfrac \beta 2$ and $\cos \dfrac \gamma 2$ , we first find the sines and cosines of the angles of $\triangle ABC$ .
We notice that $\triangle ABC$ is a compound of a $5$ - $12$ - $13$ and a $9$ - $12$ - $15$ right-angled triangle (figure 2). Figure 2

Thus, it is easy to find that $\cos A=\dfrac{5}{13}$ , $\sin A=\dfrac{12}{13}$ , $\cos B=\dfrac{9}{15}$ , $\sin B=\dfrac{12}{15}$ .

By sine law, $\dfrac{\sin C}{14}=\dfrac{\sin A}{15}\Rightarrow \sin C=\dfrac{14}{15}\cdot \dfrac{12}{13}\Rightarrow \sin C=\dfrac{56}{65}$ .

$\cos C=\sqrt{1-{{\sin }^{2}}C}=\sqrt{1-{{\left( \dfrac{56}{65} \right)}^{2}}}\Rightarrow \cos C=\dfrac{33}{65}$

Combining the results obtained, we get

$\cos \dfrac{\alpha }{2}=\cos \left( A-30{}^\circ \right)=\cos A\cdot \cos 30{}^\circ +\sin A\cdot \sin 30{}^\circ =\dfrac{5}{13}\cdot \dfrac{\sqrt{3}}{2}+\dfrac{12}{13}\cdot \dfrac{1}{2}=\dfrac{12+5\sqrt{3}}{26}$ Similarly, $\cos \dfrac{\beta }{2}=\cos \left( B-30{}^\circ \right)=\dfrac{4+3\sqrt{3}}{10}$ $\cos \dfrac{\gamma }{2}=\cos \left( C-30{}^\circ \right)=\dfrac{56+33\sqrt{3}}{130}$ Finaly, $\left( 1 \right)\Rightarrow s\left[ 15\cdot \dfrac{12+5\sqrt{3}}{26}+13\cdot \dfrac{4+3\sqrt{3}}{10}+14\cdot \dfrac{56+33\sqrt{3}}{130} \right]=84\cdot 2$ which solves to $s=\dfrac{-35280\sqrt{3}+61950}{181}$ for the answer, $a+b+c+d=-35280+3+61950+181=\boxed{26854}$ .

Nice problem!