perplexus.info :: Geometry : Collinear Intersections

Collinear Intersections (Posted on 2007-09-13)

Three circles A, B, and C have non-collinear centers, unequal radii, and pairwise the distance between their centers is greater than the sum of their radii.

Let P, Q, and R be the intersections of the external tangents to circles A&B, A&C, and B&C respectively.

Let L and M be the intersections of the internal tangents to circles A&B and A&C respectively.

Prove that P, Q, and R are collinear.

Prove that L, M, and R are collinear.

Submitted by Bractals

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Solution: (Hide)

Let XY denote the vector from point X to point Y.

If points X, Y, and Z are non-collinear and XW = aXY + bXZ, then
W lies on line YZ if and only if a + b = 1.

Let X and Y be the centers of circles with radii x and y ( |XY| > x + y ).
If E and I are the intersections of the external and internal tangents respectively, then
x x XE = ----- XY and XI = ----- XY x-y x+y
For our problem we have
a a AP = ----- AB and AL = ----- AB a-b a+b a a AQ = ----- AC and AM = ----- AC a-c a+c b BR = ----- BC b-c ----------------------------------------------------- b AR = AB + BR = AB + ----- BC b-c b = AB + ----- (AC - AB) b-c c b = ----- AB + ----- AC c-b b-c c(a-b) b(a-c) = -------- AP + -------- AQ a(c-b) a(b-c) c(a+b) b(a+c) = -------- AL + -------- AM a(c-b) a(b-c) -----------------------------------------------------
Since
c(a-b) b(a-c) -------- + -------- = 1 a(c-b) a(b-c)
P, Q, and R are collinear. Since
c(a+b) b(a+c) -------- + -------- = 1 a(c-b) a(b-c)
L, M, and R are collinear.

Comments: ( You must be logged in to post comments.)

	Subject	Author	Date
	Monge's Circle Theorem	Brian Smith	2009-11-07 12:11:10

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