HMMT-Feb 2017 geo, Problem 10

Let $A B C D$ be a quadrilateral with an inscribed circle $\omega$. Let $I$ be the center of $\omega$ let $I A=12$, $I B=16, I C=14$, and $I D=11$. Let $M$ be the midpoint of segment $A C$. Compute $\frac{I M}{I N}$, where $N$ is the midpoint of segment $B D$.

Solution: Let points $W, X, Y, Z$ be the tangency points between $\omega$ and lines $A B, B C, C D, D A$ respectively. Now invert about $\omega$. Then $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ are the midpoints of segments $Z W, W X, X Y, Y Z$ respectively. Thus by Varignon's Theorem $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a parallelogram. Then the midpoints of segments $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ coincide at a point $P$. Note that figure $I A^{\prime} P C^{\prime}$ is similar to figure $I C M A$ with similitude ratio $\frac{r^{2}}{I A \cdot I C}$ where $r$ is the radius of $\omega$. Similarly figure $I B^{\prime} P D^{\prime}$ is similar to figure $I D M B$ with similitude ratio $\frac{r^{2}}{I B \cdot I D}$. Therefore \[ I P=\frac{r^{2}}{I A \cdot I C} \cdot I M=\frac{r^{2}}{I B \cdot I D} \cdot I N \] which yields \[ \frac{I M}{I N}=\frac{I A \cdot I C}{I B \cdot I D}=\frac{12 \cdot 14}{16 \cdot 11}=\frac{21}{22} \] $\fbox{\frac{21}{22}}$.

Answer: \frac{21}{22}