Proposed Problem

See complete Problem 259 at:

gogeometry.com/problem/p259_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 259: Equilateral Triangle, Incircle, Tangency Points, Side, Distances, Squares

Labels:
distance,
equilateral,
incircle,
square,
tangency point,
triangle

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If the centre of the in-circle is taken as the origin then in-radius = a/2V3 where V=square root. Therefore, the in-circle is: x^2 + y^2 = a^2/12 ----(1).

ReplyDeleteWe can easily ascertain that E:(-a/4,a/(4V3)),F:(a/4,a/(4V3)) and G::(0,-a/(2V3)), Thus, d^2+e^2+f^2 =(x+a/4)^2+(y-a/(4V3))^2+(x-a/4)^2+(y-a/(4V3))^2+x^2+(y+a/(2V3))^2 from where we obtain: d^2+e^2+f^2= a^2/4 + 3(x^2+y^2) = a^2/4 + 3(a^2/12) by equation (1) Or d^2+e^2+f^2=a^2/4 +a^2/4 = a^2/2 QED

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