perplexus.info :: Geometry : Locus of Triangle

Locus of Triangle (Posted on 2008-05-15)

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

Let P be a point in the plane of triangle ABC.

Let PQ = PA + PB + PC.

What is the locus of points Q as P traces the triangle ABC ?

Submitted by Bractals

Rating: 3.5000 (2 votes)

Solution: (Hide)

Let G be the centroid of triangle ABC and A' the midpoint of BC.
PQ = PA + PB + PC = (PG + GA) + (PG + GB) + (PG + GC) = 3PG + GA + GB + GC = 3PG + GA + (GA' + A'B) + (GA' + A'C) = 3PG + (GA + 2GA') + (A'B + A'C) = 3PG + (0) + (0) = 3PG GQ = PQ - PG = 3PG - PG = 2PG = -2GP
We therefore have a similarity with an expansion of 2 and a rotation about G of 180°.

Therefore, the locus of points Q as P traces triangle ABC is
the anticomplementary triangle of ABC.

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

Subject	Author	Date
Re: Solution?	Adrian	2008-05-21 05:34:47
Might be the Solution	Praneeth	2008-05-16 03:40:34
re: Solution?	Charlie	2008-05-15 17:16:20
computer for insight; then the reason	Charlie	2008-05-15 16:44:16
Solution?	Adrian	2008-05-15 14:23:40

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