 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Four Coin Theorem (Posted on 2013-03-17) 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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Almost there Comment 2 of 2 | 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 Please log in:

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