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IMO 1999 Bucharest geometry problem

Posted January 8th, 2009 by Structure

Two circles $ G_1 $ and $ G_2 $ are contained inside the circle G, and are tangent to G at the
distinct points M and N, respectively. $ G_1 $ passes through the center of $ G_2 $. The line
passing through the two points of intersection of $ G_1 $ and $ G_2 $ meets G at A and B.
The lines MA and MB meet $ G_1 $ at C and D, respectively.
Prove that CD is tangent to $ G_2 $

Here it is a proof without (many) words.
Homotopy of center M which maps C in A transform $ G_1 $ in $ G $ .so $ D $ passes in $ B $ so line $ CD $ passes in line $ AB $ and $ AB $ is parallel to $ CD $. As $ AB $ is orthogonal to $ O_1O_2 $ we have the same relation $ CD $ orthogonal to $ O_1O_2 $.
Inversion of pole B which maps P in Q , transforms $ G_1 $ in $ G_1 $, transforms $ G_2 $ in $ G_2 $ and transforms $ G $ to common tangent line $ ED $ so $ $O_1O_2ED $ is a trapezoid with $ O_1O_2=O_1D $ so $ O_2F=HD=O_2E=r_1 $ so $ CD $ is tangent to $ G_2 $

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Comments

At the Olympics held in

On July 29th, 2010 madalinar says:

At the Olympics held in Bucharest , took place a program of sightseeing in Romania, he even went to Bran Castle

Geoometry problem

On January 26th, 2009 mthom078 says:

$ \angle P $ and $ \angle Q $ are supplementary
$ m (\angle P) p=x^2 $
$ m (\angle Q) =8x $

find each angle

please show me how you worked this problem

The solution is on the

On January 27th, 2009 Structure says:

The solution is on the forum:

Find angles

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