Three congruent circles G₁, G₂, G₃ have a common point P.
Further, define
G₂ intersect G₃={A, P},
G₃ intersect G₁={B, P},
G₁ intersect G₂={C, P}.

(1) Prove that the point P is the orthocenter of triangle ABC.
(2) Prove that the circumcircle of triangle ABC is congruent to the given circles G₁, G₂, G₃.

Using synthetic geometry:
P = (0,0)
Center of G1= (x1,y1)
Center of G2= (x2,y2)
Center of G3= (x3,y3)
Trying to find point A led to some expressions that really don't want to reduce. 

Using good old graph paper I am able to make the conjecture that if P=(0,0) then ABC is a rotation image of G1G2G3 by 180 degrees about a center point ((x1+x2+x3)/2 , (y1+y2+y3)/2)

If this conjecture is true then the rest is relatively easy.
As broll pointed out part 2 is clearly true since the triangles are congruent.  As a bonus the circumcenter of ABC = (x1+x2+x3,y1+y2+y3)

Based on the conjecture it is simple to show the points
A = (x2+x3,y2+y3)
B = (x1+x3,y1+y3)
C = (x1+x2,y1+y2)

For part 1 we must show segments
AP and BC are perpendicular,
BP and AC are perpendicular, and
CP and AB are perpendicular.

Slope AP * slope BC = (y2+y3)/(x2+x3) * (y3-y2)/(x3-x2)
= (y3^2 - y2^2)/(x3^2-x2^2)
But since P is equidistant from G3 and G2, x2^2+y2^2=x3^2+y3^2
or y3^2 = x2^2 + y2^2 - x3^2
Substituting this gives
(x2^2 + y2^2 - x3^2 - y2^2)/(x3^2-x2^2)
= (x2^2 - x3^2)/(x3^2 - x2^2) = -1
The others two pairs can be shown similarly.

Posted by Jer on 2013-03-18 12:32:26