Dynamic Geometry: P53

Label the Heronian triangle $ABC$ with side lengths $a$ , $b$ , and $c$ ; and the moving point $P$ . From the given altitudes we have $20 a = \dfrac {144}5 c = \dfrac {720}{29}b = \Delta$ , where $\Delta$ is the area of $\triangle ABC$ . Since a Heronian triangle has integer side lengths and area, we can assume $b=29$ , then $\Delta = 720$ , $a = 36$ , and $c = 25$ .

Using sine rule and cosine rule we can find the measures of angles $A$ , $B$ , and $C$ and using the formula $\tan \dfrac \theta 2 = \dfrac {1-\cos \theta}{\sin \theta}$ , we can find all the tangents of half-angles of $A$ , $B$ , and $C$ .

$\begin{cases} \tan A = \frac {144}{17} & \implies \tan \frac A2 = \frac 89 \\ \tan B = \frac {116}{87} & \implies \tan \frac B2 = \frac 12 \\ \tan C = \frac {20}{21} & \implies \tan \frac C2 = \frac 25 \end{cases}$

Let the radius of the incircle of $\triangle ABC$ be $r_0$ . Then we have:

$\begin{aligned} r_0 \cot \frac A2 + r_0 \cot \frac B2 & = 25 \\ \left(\frac 98 + 2 \right) r_0 = 25 \\ \implies r_0 & = 8 \end{aligned}$

Similarly, let the radii of the cyan and pink circles be $r_1$ and $r_2$ respectively, $\angle PAB = \theta$ , and $t = \tan \dfrac \theta 2$ . Using half-angle tangent substitution , we have:

$\begin{aligned} r_1 \cot \frac \theta 2 + r_1 \cot \frac B2 & = 25 \\ \frac {r_1}t + 2 r_1 & = 25 \\ \implies r_1 & = \frac {25}{\frac 1t + 2} = \frac {25t}{2t+1} \end{aligned}$

$\begin{aligned} r_2 \cot \left(\frac A2 - \frac \theta 2\right) + r_2 \cot \frac C2 & = 29 \\ \frac {1+\frac 89t}{\frac 89 - t} r_2 + \frac 52 r_2 & = 29 \\ \frac {58-29t}{2(8-9t)} & = 29 \\ \implies r_2 & = \frac {2(8-9t)}{2-t} \end{aligned}$

Now $r_1$ , $r_2$ , and $r_3$ are in a arithmetic progression, we have:

$\begin{aligned} r_1 + r_0 & = 2r_2 \\ \frac {25t}{2t+1} + 8 & = \frac {4(8-9t)}{2-t} \\ \frac {41t+8}{2t+1} & = \frac {4(9t-8)}{t-2} \\ 41t^2 - 74t-16 & = 72t^2 - 28t-32 \\ 31t^2 + 46t - 16 & = 0 & \small \blue{\text{Since }\frac \theta 2 < 90^\circ} \\ \implies t & = \frac {5\sqrt{41}-23}{31} \end{aligned}$

Then the ratio of radii:

$\begin{aligned} \frac \blue{r_2}{r_1} & = \frac \blue{r_1 + 8}{\blue 2 r_1} & \small \blue{\text{Note that }r_1 + 8 = 2r_2} \\ & = \frac 12 + \frac 4{r_1} \\ & = \frac 12 + \frac {4(2t+1)}{25t} \\ & = \frac 12 + \frac 8{25} + \frac 4{25t} \\ & = \frac {41}{50} + \frac {124}{25(5\sqrt{41}-23)} \\ & = \frac {41}{50} + \frac {124(5\sqrt{41}+23)}{25\cdot 496} \\ & = \frac {41}{50} + \frac {5\sqrt{41}+23}{100} \\ & = \frac {5\sqrt{41}+105}{100} \\ & = \frac {\sqrt{41}+21}{20} \end{aligned}$

Therefore $l+m+n = 41+20+21 = \boxed{82}$ .