Let $ABC$ be at triangle with incircle $\Gamma$. Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ be three circles inside $\triangle ABC$ each of which is tangent to $\Gamma$ and two sides of the triangle and their radii are $1,4,9$. Find the radius of $\Gamma$.
Problem
Source: India Postals 2015
Tags: incircle, trigonometry, geometry
Luis González
15.11.2015 18:42
Label $r,r_1,r_2,r_3$ the radii of $\Gamma,\Gamma_1,\Gamma_2,\Gamma_3,$ respectively. $\Gamma \equiv (I)$ touches $AB$ at $F$ and $Y$ is the projection of $I_1$ on $IF.$ Thus $\sin \frac{A}{2}=\frac{IY}{II_1}=\frac{r-r_1}{r+r_1} \Longrightarrow \frac{r_1}{r}=\frac{1-\sin \frac{A}{2}}{1+\sin \frac{A}{2}}=\tan^2 \left (45^{\circ}-\frac{A}{4} \right) \Longrightarrow$ $\sqrt{\frac{r_1}{r}}=\tan \left (45^{\circ}-\frac{A}{4} \right)=\tan \left (\frac{90^{\circ}-\frac{A}{2}}{2} \right)=\tan \frac{\lambda_A}{2}.$ Since $\lambda_A+\lambda_B+\lambda_C=180^{\circ},$ then we have the well-known trigonometric identiy $\tan \frac{\lambda_B}{2} \cdot \tan \frac{\lambda_C}{2}+\tan \frac{\lambda_C}{2} \cdot \tan \frac{\lambda_A}{2}+\tan \frac{\lambda_A}{2} \cdot \tan \frac{\lambda_B}{2}=1$ $\Longrightarrow$ $\sqrt{\frac{r_1}{r}} \cdot \sqrt{\frac{r_2}{r}}+ \sqrt{\frac{r_2}{r}} \cdot \sqrt{\frac{r_3}{r}}+\sqrt{\frac{r_3}{r}} \cdot \sqrt{\frac{r_1}{r}}=1 \Longrightarrow r=\sqrt{r_1 \cdot r_2}+\sqrt{r_2 \cdot r_3}+\sqrt{r_3 \cdot r_1}$ $\Longrightarrow r=\sqrt{1 \cdot 4}+\sqrt{4 \cdot 9}+\sqrt{9 \cdot 1}=11.$