3 in 3 -- try 3

In my solution of An Old Sangaku Problem by Michael Mendrin I prove that $\triangle JPI$ and $\triangle JPL$ are $3$ - $4$ - $5$ right triangles with $JP$ being the longer leg. Hence, $\dfrac{JP}{R}=\dfrac{4}{1}$ . The answer is $\boxed{4}$ .

Proof of $\triangle PIJ$ and $\triangle PLJ$ are $3$ - $4$ - $5$ right triangles given here.

Note that $\triangle PIJ$ and $\triangle PLJ$ are congruent right triangles. Let $\angle PJI = \angle PJL = \theta$ , then $\angle PIJ = 90^\circ - \theta$ , $\angle KIL = 3\theta - 90^\circ$ , and $\angle KLI = 90^\circ + \theta$ . Let $IJ = 1$ . If $R$ is the radius of the three circles, then:

$\begin{aligned} R \cot \frac {\angle PJI}2 + R \cot \frac {\angle PIJ}2 & = IJ \\ R \cot \frac \theta 2 + R \cot \left(45^\circ - \frac \theta 2\right) & = 1 & ...(1) \end{aligned}$

And

$\begin{aligned} R \cot \frac {\angle KLI}2 + R \cot \frac {\angle KIL}2 & = IL \\ R \cot \left(45^\circ + \frac \theta 2\right) + R \cot \left(\frac {3\theta}2 - 45^\circ \right) & = 2\sin \theta & ...(2) \end{aligned}$

From $\dfrac {(2)}{(1)}$ :

$\begin{aligned} \frac {\cot \left(45^\circ + \frac \theta 2\right) + \cot \left(\frac {3\theta}2 - 45^\circ \right)}{\cot \frac \theta 2 + \cot \left(45^\circ - \frac \theta 2 \right)} & = 2\sin \theta & \small \blue{\text{Let }t=\tan \frac \theta 2} \\ \frac {\frac {1-t}{1+t} + \frac {\frac {3t-t^3}{1-3t^2}+1}{\frac {3t-t^3}{1-3t^2}-1}}{\frac 1t + \frac {1+t}{1-t}} & = \frac {4t}{1+t^2} \\ \frac {\frac {1-t}{1+t} - \frac {1+3t-3t^2-t^3}{1-3t-3t^2+t^3}}{\frac {1+t^2}{t-t^2}} & = \frac {4t}{1+t^2} \\ \frac {1-t}{1+t} - \frac {(1-t)(1+4t+t^2)}{(1+t)(1-4t+t^2)} & = \frac {4t}{1+t^2} \cdot \frac {1+t^2}{t-t^2} \\ \frac {8t(t-1)}{(1+t)(1-4t+t^2)} & = \frac 4{1-t} \\ - 2t+4t^2 - 2t^3 & = 1-3t-3t^2+t^3 \\ 3t^3 - 7t^2 - t + 1 & = 0 \\ (3t-1)(t^2-2t-1) & = 0 \\ t & = \frac 13, 1 + \sqrt 2, 1 - \sqrt 2 & \small \blue{\text{For acute }\theta} \\ \implies t = \tan \frac \theta 2 & = \frac 13 \end{aligned}$

As $\tan \dfrac \theta 2 = \dfrac 13$ , $\implies \tan \theta = \dfrac {2t}{1-t^2} = \dfrac {\frac 23}{1-\frac 19} = \dfrac 34$ . This implies that $\triangle PIJ$ and $\triangle PLJ$ are $3$ - $4$ - $5$ right triangles with side lengths $0.6$ - $0.8$ - $1$ , where $PJ = 0.8$ and the inradius $R$ is given by:

$\begin{aligned} \frac {0.6+0.8+1}2 R & = \frac {0.6 \cdot 0.8}2 \\ 2.4 R & = 0.48 \\ \implies R & = 0.2 \\ \implies \frac {PJ}R & = \frac {0.8}{0.2} = \boxed 4 \end{aligned}$

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@Fletcher Mattox , The answer to the previous problem should be $\boxed {39}$ . Because $KI = KJ = \dfrac {IJ}2 sec (2\theta) \implies \dfrac {KJ}{IJ} = \dfrac {1+\tan^2 \theta}{2(1-\tan^2 \theta)} = \dfrac {1+\frac 9{16}}{2\left(1-\frac 9{16}\right)} = \dfrac {25}{14}$ . Therefore $p+q = \boxed{39}$

Solve y+z-x=2, x^2=y^2+z^2, x/(2w+2x)=cos(2u), z/y=tan(u), s=(2w+2z+x)/2, (s-w-x)(s-w)(s-2z)=s, x>0,y>0,z>0,w>0 using WolframAlpha

Where, sides, IJ=x, PJ=y, PI=z, LJ=x, KL=w. I have assumed circles with radius 1, used inscribed circle radius formula for (1) right triangle with sides x, y and z (2) triangle with sides w, w+x and 2z (3) used angle relations with angle LJI=2u and angle LJP=u.

Answer

w=55/14, x=5, y=4, z=3