Bariz - Problem

Source: Sharygin contest 2008. The correspondence round. Problem 19

Tags: ratio, geometry, parallelogram, geometry proposed

Doctor A

03.09.2008 17:04

(V.Protasov, 10-11) Given parallelogram $ ABCD$ such that $ AB = a$, $ AD = b$. The first circle has its center at vertex $ A$ and passes through $ D$, and the second circle has its center at $ C$ and passes through $ D$. A circle with center $ B$ meets the first circle at points $ M_1$, $ N_1$, and the second circle at points $ M_2$, $ N_2$. Determine the ratio $ M_1N_1/M_2N_2$.

We will prove that $ M_{1}N_{1} / M_{2}N_{2} = b/a$. Indeed, since $ M_{1}N_{1} \cdot a = 4\left[AM_{1}B\right]$, we have that $ M_{1}N_{1}=\frac{4\left[AM_{1}B\right]}{a}.$ Similarly, $ M_{2}N_{2}=\frac{4\left[BM_{2}C\right]}{b},$ and thus \[ \frac{M_{1}N_{1}}{M_{2}N_{2}}=\frac{b}{a} \cdot \frac{\left[AM_{1}B\right]}{\left[BM_{2}C\right]}.\] On the other hand, since $ AM_{1}=AD=BC$, $ M_{1}B=BM_{2}$, and $ AB=CD=CM_{2}$, the triangles $ AM_{1}B$ and $ BCM_{2}$ are congruent, and therefore $ \left[AM_{1}B\right]=\left[BM_{2}C\right]$, from which we conclude that $ M_{1}N_{1} / M_{2}N_{2} = b/a$.