Congruent Incircles

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In this solution we use the fact that, in the case where the two incircles are congruent, the following relation holds: $AD=\sqrt{s\cdot \left( s-a \right)} \ \ \ \ \ (1)$ where $a=BC$ and $s$ is the semiperimeter of $\triangle ABC$ .

This is problem 2.2.5 in Fukagawa, Hidetoshi, and Dan Pedhoe. 1989, Japanese temple geometry problems, Winnipeg: Charles Babbage Research Centre. In this book no solution is given, but you can find one here (problem 8).

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In our triangle we have $s=\dfrac{13+14+15}{2}=21$ .

Using formula $(1)$ , we get $AD=\sqrt{21\cdot 7}=7\sqrt{3}$ .

Let $CD=x$ . Then, $DB=14-x$ and by Stewart's Theorem on $\triangle ABC$ for the cevian $AD$ we have

$CD\cdot A{{B}^{2}}+DB\cdot A{{C}^{2}}=BC\cdot \left( A{{D}^{2}}+CD\cdot DB \right)\Rightarrow x\cdot {{15}^{2}}+\left( 14-x \right)\cdot {{13}^{2}}=14\cdot \left[ {{\left( 7\sqrt{3} \right)}^{2}}+x\cdot \left( 14-x \right) \right]$ which solves to $x=5+\sqrt{3} \text{ or } x=5-\sqrt{3}$

Case 1 : $x=5+\sqrt{3}$

Denoting by ${{s}_{1}}$ and ${{s}_{2}}$ the semiperimeters of $\triangle ACD$ and $\triangle ABD$ respectively, we have ${{s}_{1}}=\dfrac{13+\left( 5+\sqrt{3} \right)+7\sqrt{3}}{2}=4\sqrt{3}+9$ and ${{s}_{2}}=\dfrac{15+\left[ 14-\left( 5+\sqrt{3} \right) \right]+7\sqrt{3}}{2}=3\sqrt{3}+12$ By Heron’s folmula $\left[ ACD \right]=\sqrt{\left( 4\sqrt{3}+9 \right)\left( 4\sqrt{3}-4 \right)\left( 3\sqrt{3}+4 \right)\left( 9-3\sqrt{3} \right)}=30+6\sqrt{3}$ and $\left[ ABD \right]=\sqrt{\left( 3\sqrt{3}+12 \right)\left( 12-4\sqrt{3} \right)\left( 4\sqrt{3}+3 \right)\left( 3\sqrt{3}-3 \right)}=54-6\sqrt{3}$ For the radii ${{r}_{1}}$ and ${{r}_{2}}$ of $\triangle ACD$ and $\triangle ABD$ respectively, we have

${{r}_{1}}=\dfrac{\left[ ACD \right]}{{{s}_{1}}}=\dfrac{30+6\sqrt{3}}{9+4\sqrt{3}}=6-2\sqrt{3}$ and ${{r}_{2}}=\dfrac{\left[ ABD \right]}{{{s}_{2}}}=\dfrac{54-6\sqrt{3}}{12+3\sqrt{3}}=6-2\sqrt{3}$ We notice that ${{r}_{1}}={{r}_{2}}$ .

Case 2 : $x=5-\sqrt{3}$

Working the same way we find that ${{r}_{1}}\ne {{r}_{2}}$ , hence this value is rejected.

To conclude, the common radius of the two circles is $r=6-2\sqrt{3}$ .
For the answer, $a+b+c=6-2+3=\boxed{7}$ .

I hadn't seen the formula in @Thanos Petropoulos ' solution before, but it simplifies things a lot.

If the common inradius is $r$ , then the area of $\Delta ABC$ is $\frac12 r(13+14+15+2AD)=r(21+AD)$

(using the fact that a triangle's area is its inradius times its semiperimeter in both $ABD$ and $ACD$ ). It's easy to verify (Heron's formula) that the area of $ABC$ is $84$ ; so $r(21+AD)=84$

From the formula Thanos quotes, $AD=\sqrt{21\cdot 7}=\sqrt{147}$

So $r=\frac{84}{21+\sqrt{147}}=6-2\sqrt3$

giving the answer $\boxed7$ .

The triangle $ABC$ is a $5,12,13$ triangle stuck to a $9,12,15$ triangle, and hence has height $12$ . Suppose that $BD = x$ and $AD = d$ . Then $ACD$ has area $6(14-x)$ and semiperimeter $\tfrac12(27-x+d)$ , while $ABD$ has area $6x$ and semiperimeter $\tfrac12(15+x+d)$ . Thus we deduce that $\begin{aligned} \frac{12(14-x)}{27 - x + d} & = \; \frac{12x}{15+x+d} \\ (14-x)(15 + x + d) & = \; x(27-x+d) \\ 210 - x + 14d - x^2 - xd & = \; 27x - x^2 + xd \\ 210 - 28x & = \; 2(x-7)d \\ 7(15 - 2x) & = \; (x-7)d \end{aligned}$ Thus we deduce that $7 < x < 7\tfrac12$ and moreover, since $\cos B = \tfrac35$ , we deduce that $d^2 = 225 + x^2 - 18x$ , so that $\begin{aligned} 49(15 - 2x)^2 & = \; (x - 7)^2(x^2 - 18x + 225) \\ 196x^2 - 2940x + 11025 & = \; x^4 - 32x^3 + 526x^2 - 4032x + 11025 \\ x^4 - 32x^3 + 330x^2 - 1092x & = \; 0 \\ x(x - 14)(x^2 - 18x + 78) & = \; 0 \\ x(x -14)\big((x-9)^2 - 3\big) & = \; 0 \end{aligned}$ Since $7 < x < 7\tfrac12$ , we deduce that $x = 9 - \sqrt{3}$ , so that $d = 7\sqrt{3}$ and hence the common inradius is $\frac{12x}{15+x+d} \; = \; 6 - 2\sqrt{3}$ making the answer $6 + (-2) + 3 = \boxed{7}$ .

By cosine rule , we have $15^2 = 13^2+14^2 - 2(13)(14)\cos C$ . $\implies \cos C = \dfrac 5{13}$ . Then altitude of $\triangle ABC$ is $13 \sin C = 12$ . Let the inradius be $r$ , $CD=x$ , and $AD = \ell$ . Then the areas of $\triangle ACD$ and $\triangle ADB$ are given by:

$\begin{cases} [ACD] = \dfrac {13+x+\ell}2r = \dfrac {12x}2 & \implies (13+x+\ell) r = 12 x &...(1) \\ [ADB] = \dfrac {29-x+\ell}2r = \dfrac {12(14-x)}2 & \implies (29-x+\ell) r = 12(14-x) &...(2) \end{cases}$

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

$\begin{aligned} \frac {13+x+\ell}{29-x+\ell} & = \frac x{14-x} \\ 182 + x + 14 \ell - x^2 - x\ell & = 29x - x^2 + x\ell \\ 91 - 14x & = (x-7) \blue \ell & \small \blue{\text{Note that }\ell = \sqrt{(x-5)^2+12^2}} \\ 91 - 14x & = (x-7) \blue{\sqrt{(x-5)^2+12^2}} \\ \implies x & = 5+\sqrt 3 \end{aligned}$

From $(2)-(1)$ :

$\begin{aligned} (16-2x)r & = 12(14-2x) \\ (8-x)r & = 12(7-x) \\ \implies r & = \frac {12(7-x)}{8-x} \\ & = \frac {12(7-5-\sqrt 3)}{8-5-\sqrt 3} \\ & = \frac {12(2-\sqrt 3)}{3-\sqrt 3} \\ & = \frac {12(2-\sqrt 3)(3+\sqrt 3)}6 \\ & = 2(3-\sqrt 3) = 6 - 2\sqrt 3 \end{aligned}$

Therefore $a+b+c = 6-2+3 = \boxed 7$ .