Proposed Problem

See complete Problem 258 at:

gogeometry.com/problem/p258_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Thursday, February 26, 2009

### Problem 258: Equilateral Triangle, Incircle, Point, Vertices, Side, Distances, Squares

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The sum of the distances between D and the sides of the triangle ABC are constant and could be expressed depending on a.

ReplyDeleteLet the midpoint of AC be(0,0). So we've:

ReplyDeleteA:(-a/2,0), C: (a/2,0) and B:((0,aV3/2). The inradius =a/2V3 while the incentre:(0,a/2V3)

Hence, incircle is: x^2+(y-a/2V3)^2 =a^2/12

or x^2+y^2-ay/V3 = 0 or 3x^2+3y^2-ayV3 = 0----(1)

d^2+e^2+f^2=(x-a/2)^2+(x+a/2)^2+2y^2+x^2+(y-aV3/2)^2=3x^2+3y^2-ayV3+5a^2/4 = 0 + 5a^2/4 using (1). Hence, d^2+e^2+f^2=5a^2/4 QED

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