Transform by inversion of the circumcircle

Posted January 19th, 2009 by Structure

Consider the inversion of pole I , the incenter of a triangle ABC , with leaves invariant the incircle of ABC. Let D,E,F the tangent points of the incircle to the sides of ABC.
Circles of diameter IE,ID,IF intersect in K,L,M. Then circle by K,L,M is the transform by inversion of the circumcircle of triangle ABC.

[geo]
(point A 118 38)
(point B 53 327)
(point C 332 328)
(circle3points t1 A B C)
(incenter I A B C)
(segment d1 B C)
(segment d2 A B)
(segment d3 A C)
(projection D I d3)
(projection E I d1)
(projection F I d2)
(middle G D I)
(middle H I E)
(middle J I F)

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 $

[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

Posted December 18th, 2008 by Structure
$$f(x)=f(x_0)+\frac{1}{1!}df(x_0)(x-x_0)+\frac{1}{2!}d^2f(x_0)(x-x_0,x-x_0)+...\frac{1}{(p-1)!}d^{p-1}f(x_0)(x-x_0,.....,x-x_0)+$$
$$+\frac{1}{p!}d^pf(\xi)(x-x_0,x-x_0,...x-x_0)$$

=

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