Proposed Problem

See complete Problem 260 at:

gogeometry.com/problem/p260_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Friday, February 27, 2009

### Problem 260: Equilateral Triangle, Incircle, Tangency Points, Vertices, Distances, Square

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

Subscribe to:
Post Comments (Atom)

WLOG,let G be (0,0) and A:(-1,0), C;(1,0). Therefore, B is (0,V3) where V = square root.

ReplyDeleteIn-radius of Tr. ABC = 1/V3. Equation of the in-circle, x^2 + (y-1/V3)^2 = 1/3 or x^2+y^2-2y/V3+1/3=1/3 or x^2+y^2-2y/V3 =0

or 3x^2+3y^2=2V3y ---(1)

Now m^2+n^2+q^2=(x+1)^2+y^2+(x-1)^2+y^2+x^2+y-V3)^2 = 3x^2++3y^2-2V3y+5 = 5 using (1)

Now E is (-1/2,V3/2) while F is (1/2,V3/2)

Thus, d^2+e^2+f^2 = (x+1/2)^2+(y-V3/2)^2+x^+y^2+(x-1/2)^2+(y-V3/2)^2 = 3x^2+3y^2-2V3y+8/4=2 using(1)as before.

Thus, (m^2+n^2+q^2)/(d^2+e^2+f^2) = 5/2

QED

Ajit: ajitathle@gmail.com