2021 WSMO Speed Round Problems/Problem 7

Problem

Consider triangle $ABC$ with side lengths $AB=13,AC=14,BC=15$ and incircle $\omega$ . A second circle $\omega_2$ is drawn which is tangent to $AB,AC$ and externally tangent to $\omega$ . The radius of $\omega_2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$ , where $\gcd{(a,b,d)}=1$ and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ .

Solution

Note that the length of the $A$ -angle bisector is $\frac{\sqrt{(b+c-a)(b+c+a)bc}}{b+c}.$ Now, let $I$ be the incenter of triangle $ABC.$ This means that $AI=\frac{b+c}{a+b+c}\cdot\frac{\sqrt{b+c-a}(b+c+a)bc}{b+c}=\frac{\sqrt{(b+c-a)(b+c+a)bc}}{a+b+c}=\frac{\sqrt{12\cdot42\cdot13\cdot14}}{42}=2\sqrt{13}.$ Now, from Heron's formula, we find that the area of triangle $ABC$ is $\sqrt{\left(\frac{\left(13+14+15\right)}{2}\right)\left(\frac{\left(13+14+15\right)}{2}-13\right)\left(\frac{\left(13+14+15\right)}{2}-14\right)\left(\frac{\left(13+14+15\right)}{2}-15\right)}=\sqrt{21\cdot6\cdot7\cdot8}=84.$ Since the area of a triangle is the product of the semi-perimeter and the inradius, we find that the length of the inradius of triangle $ABC$ is $\frac{84}{\frac{13+14+15}{2}}=\frac{84}{21}=4.$ Now, let the radius of $\omega_2$ be $r.$ From similar triangles, we find that $\frac{r}{4}=\frac{2\sqrt{13}-4-r}{2\sqrt{13}}\Leftrightarrow r=\frac{4(2\sqrt{13}-4)}{2\sqrt{13}+4}=\frac{4(2\sqrt{13}-4)^2}{(2\sqrt{13}-4)(2\sqrt{13}+4)}=\frac{4(68-16\sqrt13)}{(2\sqrt{13})^2-4^2}=\frac{4(68-16\sqrt{13})}{36}=\frac{68-16\sqrt{13}}{9}\Longrightarrow68+16+13+9=\boxed{106}.$

~pinkpig

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2021 WSMO Speed Round Problems/Problem 7

Problem

Solution