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IMO 1999 Bucharest geometry problem

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 $

[geo]
(point O 186 193)
(point R 85 308)
(circle t O R)
(pointoncircle M t 134 49)
(pointoncircle N t 305 290)
(segment MO M O)
(segment NO N O)
(pointonline O1 MO 175 163)
(circle t1 O1 M)

Taylor formula for p-differentiable function

$$f(x)=f(x_0)+\frac{1}{1!}df(x_0)(x-x_0)+\frac{1}{2!}d^2f(x_0)(x-x_0,x-x_0)+...\frac{1}{(p-1)!}d^{p-1}f(x_0)(x-x_0,.....,x-x_0)+$$
$$+\frac{1}{p!}d^pf(\xi)(x-x_0,x-x_0,...x-x_0)$$

=

$$f(x_0)+\sum_{i_1=1}^{i_1=n}\frac{\partial f}{\partial x_{i_1}}(x_0)(x_{i_1}-x_{0,i_1})+\frac{1}{2!}\sum_{i_1=1}^{i_1=n}\sum_{i_2=1}^{i_2=n}\frac{\partial^2 f}{\partial x_{i_1}\partial x_{i_2}}(x_0)(x_{i_1}-x_{0,i_1})(x_{i_2}-x_{0,i_2})+$$

Power of a point with respect to a circle

Two lines by a point P outside a circle cut in A,B respectively in C and D. Then

$$PA*PB=PC*PD=TP^2$$
$$\triangle PAC\sim\triangle PDB$$

as $ \angle PBD=\angle PCA\:\:;\angle PDB=\angle PAC $

$$\frac{PA}{PD}=\frac{PC}{PB}$$

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