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Locus of Triangle (Posted on 2008-05-15) Difficulty: 3 of 5
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 ?

See The Solution Submitted by Bractals    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Solution? | Comment 3 of 5 |
(In reply to Solution? by Adrian)

When P moves from (0,0) to (xb,yb), PQ does indeed go from (xb+xc,yb+yc) to (xc-2*xb,yc-2*yb).

However, PQ is a vector; it isn't the point Q. The vector takes the point P, now at (xb, yb), to the point Q at (xc-xb,yc-yb), and so the traced line of Q for this segment is from (xb+xc,yb+yc) to (xc-xb,yc-yb), and thus is twice (not three times) the length of the segment traced by P.

Symmetry shows that if it were three times, the opposite vertex would fall 1/3 of the way along the generated segment; but 1/3 from which direction?  ... there's no asymmetry to favor one or the other.

  Posted by Charlie on 2008-05-15 17:16:20
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