Pentagonal Kenmotu

Generalization: After trying three Kenmoto problems involving equilateral triangle , square , and regular pentagon, I notice that we can reduce square and pentagon Kenmoto problems to an isosceles triangle Kenmoto problem, where the isosceles triangle is formed by joining the shared center vertex of the regular polygon to the two vertices touching the side of the inscribing triangle. Then the length of the leg of the isosceles triangle is given by:

$r = \frac {2\Delta}{a \cos \frac \alpha 2 + b \cos \frac \beta 2 + c \cos \frac \gamma 2} \text{ where } \begin{cases} \alpha = 2A - \theta \\ \beta = 2B - \theta \\ \gamma = 2C - \theta \end{cases}$

where $\Delta$ is the area of the inscribing triangle, $a$ , $b$ , and $c$ and $\alpha$ , $\beta$ , and $\gamma$ are the side lengths of the inscribing triangle and the angles between the isosceles triangles opposite vertices $A$ , $B$ , and $C$ respectively, and $\theta$ is the apex angle of the isosceles triangles. For equilateral triangle $\theta_3 = 60^\circ$ , for square $\theta_4 = 90^\circ$ , and regular pentagon $\theta_5 = 36^\circ$ . Putting in the values, we get

$\begin{aligned} r_3 & = \frac {61950-35280\sqrt 3}{181} \\ r_4 & = \frac {1365\sqrt 2}{463} \\ r_5 & = \frac {5460}{295(\sqrt 5 -1)+168\sqrt{2(5+\sqrt 5)}} \end{aligned}$

And the side length of the pentagon $s= \dfrac {r_5}{2 \cos 36^\circ} = \dfrac {2730}{168\sqrt{5+2\sqrt 5}=295}$ . Therefore the required answer $2730+168+295=\boxed{3195}$ .

Figure 1 Let’s give a solution similar to the one used in Triangular Kenmotu and in A Squared Triangle . The original idea of this approach, where we use only tools of elementary geometry, belongs to @Chew-Seong Cheong.

Let $\angle LXM = \alpha$ , $\angle KXJ = \beta$ , and $\angle OXP = \gamma$ (figure 1). We note that $\triangle XPO$ is isosceles, thus, $\angle XPO = 90^\circ - \dfrac \gamma 2$ , hence, $\angle MPB=180{}^\circ -\left( 90{}^\circ -\dfrac{\gamma }{2} \right)-72{}^\circ =\dfrac{\gamma }{2}+18{}^\circ$ , thus, $\angle LMX=\angle LMP-72{}^\circ =\left( \dfrac{\gamma }{2}+18{}^\circ +B \right)-72{}^\circ =B+\dfrac{\gamma }{2}-54{}^\circ$ .

In isosceles $\triangle XML$ , $\angle LXM+2\cdot \angle LMX=180{}^\circ \Rightarrow a+2\left( B+\frac{\gamma }{2}-54{}^\circ \right)=180{}^\circ \Rightarrow a=288{}^\circ -2B-\gamma.$ Similarly, $\beta =288{}^\circ -2A-\gamma$ .

Moreover,

$\begin{aligned} \alpha +\beta +\gamma +3\cdot 36{}^\circ & =360{}^\circ \\ \alpha +\beta +\gamma & =252{}^\circ \\ \left( 288{}^\circ -2B-\gamma \right)+\left( 288{}^\circ -2\Alpha -\gamma \right)+\gamma & =252{}^\circ \\ \end{aligned}$ $\Rightarrow \gamma =324{}^\circ -2\left( A+B \right)=324{}^\circ -2\left( 180{}^\circ -C \right)=2C-36{}^\circ \Rightarrow \dfrac{\gamma }{2}=C-18{}^\circ$ Similarly, $\dfrac{\alpha}{2}=A-18{}^\circ$ and $\dfrac{\beta}{2}=B-18{}^\circ$ .

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

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

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 triangles (figure 2). Figure 2

Thus, it is easy to find that $\cos A=\dfrac{5}{13}$ , $\sin A=\dfrac{12}{13}$ , $\cos B=\dfrac{9}{15}$ , $\sin A=\dfrac{12}{15}$ .
By sine law, $\dfrac{\sin C}{14}=\dfrac{\sin A}{15}\Rightarrow \sin C=\dfrac{14}{14}\cdot \dfrac{12}{13}\Rightarrow \sin C=\dfrac{56}{65}$ . $\cos C=\sqrt{1-{{\sin }^{2}}C}=\sqrt{1-{{\left( \dfrac{56}{65} \right)}^{2}}}\Rightarrow \cos C=\dfrac{33}{65}$

Combining the results obtained, we get

$\cos \frac{\alpha }{2}=\cos \left( A-18{}^\circ \right)=\cos A\cdot \cos 18{}^\circ +\sin A\cdot \sin 18{}^\circ =\frac{5}{13}\cdot \frac{\sqrt{10+2\sqrt{5}}}{4}+\frac{12}{13}\cdot \frac{\sqrt{5}-1}{4}$ Similarly, $\cos \dfrac{\beta }{2}=\cos \left( B-18{}^\circ \right)= \dfrac{9}{15}\cdot \dfrac{\sqrt{10+2\sqrt{5}}}{4}+\dfrac{12}{15}\cdot \dfrac{\sqrt{5}-1}{4}$ $\cos \dfrac{\gamma }{2}=\cos \left( C-18{}^\circ \right)= \dfrac{33}{65}\cdot \dfrac{\sqrt{10+2\sqrt{5}}}{4}+\dfrac{56}{65}\cdot \dfrac{\sqrt{5}-1}{4}$

Finally, after substitutions and simplification,

$\left( 1 \right)\Rightarrow d\left[ \dfrac{168}{65}\cdot \sqrt{10+2\sqrt{5}}+\dfrac{59}{13}\cdot \left( \sqrt{5}-1 \right) \right]=84$ which solves to
$d=\dfrac{5460}{295\left( \sqrt{5}-1 \right)+168\sqrt{2\left( 5+\sqrt{5} \right)}}$ In a regular pentagon of side $s$ and diagonal $d$ it holds $\dfrac{d}{s}=\varphi$ , where $\varphi =\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. Hence, $s=\dfrac{d}{\varphi }\Rightarrow s=\dfrac{5460}{295\left( \sqrt{5}-1 \right)+168\sqrt{2\left( 5+\sqrt{5} \right)}}\cdot \dfrac{2}{1+\sqrt{5}}$ which simplifies to
$s=\dfrac{2730}{168\sqrt{5+2\sqrt{5}}+295}$ For the answer, $a+b+c=2730+168+295=\boxed{3193}$ .

Note: In the figure we see that two of the pentagons do overlap, unlike the case of equilateral triangles and squares inscribed in the same triangle $\triangle ABC$ . Indeed,
$\beta =2co{{s}^{-1}}\left( \frac{9}{15}\cdot \frac{\sqrt{10+2\sqrt{5}}}{4}+\frac{12}{15}\cdot \frac{\sqrt{5}-1}{4} \right)\approx 70.26{}^\circ <2\cdot 36{}^\circ$ .

The common vertex is the triangle center X(3396) which is isogonal conjugate of X(1139) - OUTER PENTAGON POINT. In plain terms, the point can be constructed by creating regular pentagons outward on triangle sides and connecting outermost points to the opposite vertices of the triangle. This will produce point X(1139). Then the connecting lines need to be reflected on angle bisectors of corresponding angles. This will produce point X(3396). The Trilinear Coordinates are equivalent to $\cos(A-\pi/10) : \cos(B-\pi/10) : \cos(C-\pi/10)$

The numbers can be obtained by using formulae for Exact Trilinear Coordinates . a=2730 , b= 168,c= 295.