(V.Yasinsky, Ukraine) Suppose $ X$ and $ Y$ are the common points of two circles $ \omega_1$ and $ \omega_2$. The third circle $ \omega$ is internally tangent to $ \omega_1$ and $ \omega_2$ in $ P$ and $ Q$ respectively. Segment $ XY$ intersects $ \omega$ in points $ M$ and $ N$. Rays $ PM$ and $ PN$ intersect $ \omega_1$ in points $ A$ and $ D$; rays $ QM$ and $ QN$ intersect $ \omega_2$ in points $ B$ and $ C$ respectively. Prove that $ AB = CD$.
Problem
Source: Sharygin contest. The final raund. 2008. Grade 10. First day. Problem 3
Tags: geometry, trapezoid, geometric transformation, homothety, power of a point, radical axis, geometry unsolved
plane geometry
01.09.2008 05:36
A,B,P,Q are concyclic L tangent to ⊙O at M , L1 tangent to ⊙O1 at A , L2 tangent to ⊙O2 at B Then L1∥L2∥L But ∠BAP=∠BQP=∠PML => AB∥L => AB=L1=L2 => AB is the common tangent of ⊙O1 and ⊙O2 Analogously,CD is the common tangent of ⊙O1 and ⊙O2 => AB=CD
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sunken rock
20.03.2009 14:53
M being on radical axis of w1 and w2, we get AM.PM=BM.MQ, therefore ABPQ is cyclic, so ÐBAM=ÐBQP ( 1 ). From MPNQ cyclic we get ÐPMN=ÐPQN ( 2 ). The circles w and w1 are homothetic at P, hence AD||AD, so ÐMAD=ÐPMN ( 3 ). From (1), (2) and (3) we get ÐBAD = ÐMQN. ( 4 ) Analogously we have QPCD cyclic and get ÐADC = ÐMQN ( 5 ), also w and w2 homothetic and BC||MN, hence AD||BC and, with (4) and (5) ABCD is an isosceles trapezoid, done. (Here the sign Ð means "angle"). Best regards, sunken rock
djmathman
10.11.2013 02:05
I guess this is similar to sunken rock's solution, except worded differently. Consider the homothety sending $\omega$ to $\omega_1$. This sends $M$ to $A$ and $N$ to $D$, and as such $AD\parallel MN$. Similarly, $BC\parallel MN$ and therefore $AD\parallel BC$. Now, since $XY$ is the radical axis of $\omega_1$ and $\omega_2$, we have $AM\times MP=BM\times MQ$, so $AQPB$ is cyclic. Similarly, $DQPC$ is cyclic. Hence, by angle chasing with what we know, we find \begin{align*}\angle DAP=\angle NMP=\angle NQP=\angle NDC,\qquad \angle BAM=\angle MQP=\angle MNP=\angle ADP.\end{align*} Therefore $\angle BAD=\angle ADC$, so $\triangle BAD=\triangle CDA\implies AB=CD$ as desired. $\blacksquare$
stephcurry
19.04.2015 20:30
Solution that avoids messy angle chasing (too lazy to do it ): Let O_1 and O_2 be the centers of w_1 and w_2, respectively. Consider the homothety that sends w to w_1, which sends MN to AD. Thus, AD || MN. Similarly, BC || MN so AD || BC || XY. Now note that O_1O_2 is perpendicular to XY, and since AD and BC are parallel to XY, O_1O_2 is perpendicular to both AD and BC. Now, we consider a reflection of the diagram across O_1O_2. Let A be mapped to A'. Clearly, A' is a point on w_1. Since AD is perpendicular to O_1O_2, A' must be a point on AD. However, the only point that satisfies these conditions is D, so A is mapped to D. Similarly, B is mapped to C. Finally, since reflection does not change lengths, AB = CD. QED