Library / Mongolian Mathematical Olympiad /97 of 97

Mongolian Mathematical Olympiad

Geometry Difficulty 7.5 National Olympiad, round 2 Prove it Mongolia

A circle $\gamma$ with center $I$ and radius $R$ is inscribed in quadrilateral $ABCD$ . Another circle $\omega$ with center $O$ , ( $I \neq O$ ) and radius $r$ is situated inside the quadrilateral $ABCD$ . Circles $\gamma_B, \gamma_C, \gamma_D, \gamma_A$ inscribed in the angles $\angle ABC, \angle BCD, \angle CDA, \angle DAB$ are tangent to the circle $\omega$ at points $B_1, C_1, D_1, A_1$ respectively. If circles $\gamma_B, \gamma_C$ tangent to $\omega$ externally and circles $\gamma_D, \gamma_A$ tangent to $\omega$ internally and $AA_1 \cap DD_1 = P, BB_1 \cap CC_1 = Q$ then find the value of $\frac{IP+IQ}{PQ}$ .

Solution

By the given condition $R > r$ . Denote $H_A^k$ ...homothety with center $A$ and coefficient $k$ . Let denote $r_a$ , $r_b$ , $r_c$ , $r_d$ radius of circles $\gamma_A$ , $\gamma_B$ , $\gamma_C$ , $\gamma_D$ respectively. Then we have
$H_{B_1}^{-\frac{r_b}{r}}(\omega) = \gamma_b \text{ and } H_B^{\frac{R}{r_b}}(\gamma_b) = \gamma \Rightarrow H_B^{\frac{R}{r_b}} \circ H_{B_1}^{-\frac{r_b}{r}}(\omega) = \gamma.$
If we set $H_Q^{-\frac{R}{r_b}} = H_B^{\frac{R}{r_b}} \circ H_{B_1}^{-\frac{r_b}{r}}$ then $Q' \in BB_1$ .
Similarly, we get $Q' \in CC_1$ and $Q' \equiv Q$ . $H_Q^{-\frac{R}{r}}(\omega) = \ell \Rightarrow QI = \frac{R}{r}QO$ . If the condition $AA_1 \cap DD_1 = P$ implies $H_P^{\frac{R}{r}}(\omega) = \gamma$ .

Consequently $H_P^R(O) = I$ and $PI = \frac{R}{r}PO$ .
Thus we conclude that points $I$ , $Q$ , $O$ , $P$ are lie on a line in this order. Hence we have $IP + IQ = \frac{R}{r}(PO + QO) = \frac{R}{r}QP \Rightarrow \frac{IP + IQ}{QP} = \frac{R}{r}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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