February, 2017

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$ . Let $I$ be the center of $\omega$ and let $IA = 12$ , $IB = 16$ , $IC = 14$ , and $ID = 11$ . Let $M$ be the midpoint of segment $AC$ . Compute $\frac{IM}{IN}$ , where $N$ is the midpoint of segment $BD$ .

Solution

Solution:

Let points $W, X, Y, Z$ be the tangency points between $\omega$ and lines $AB, BC, CD, DA$ respectively. Now invert about $\omega$ . Then $A'$ , $B'$ , $C'$ , $D'$ are the midpoints of segments $ZW, WX, XY, YZ$ respectively. Thus by Varignon's Theorem $A'B'C'D'$ is a parallelogram. Then the midpoints of segments $A'C'$ and $B'D'$ coincide at a point $P$ . Note that figure $IA'PC'$ is similar to figure $ICMA$ with similitude ratio $\frac{r^2}{IA \cdot IC}$ where $r$ is the radius of $\omega$ . Similarly, figure $IB'PD'$ is similar to figure $IDMB$ with similitude ratio $\frac{r^2}{IB \cdot ID}$ . Therefore
$IP = \frac{r^2}{IA \cdot IC} \cdot IM = \frac{r^2}{IB \cdot ID} \cdot IN$
which yields
$\frac{IM}{IN} = \frac{IA \cdot IC}{IB \cdot ID} = \frac{12 \cdot 14}{16 \cdot 11} = \frac{21}{22}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.