Three congruent circles G
_{1}, G
_{2}, G
_{3} have a common point P.
Further, define
G
_{2} intersect G
_{3}={A, P},
G
_{3} intersect G
_{1}={B, P},
G
_{1} intersect G
_{2}={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 G_{1}, G_{2}, G_{3}.
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) * (y3y2)/(x3x2)
= (y3^2  y2^2)/(x3^2x2^2)
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)/(x3^2x2^2)
= (x2^2  x3^2)/(x3^2  x2^2) = 1
The others two pairs can be shown similarly.

Posted by Jer
on 20130318 12:32:26 