Friday, February 27, 2009

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

Proposed Problem

See complete Problem 260 at:
gogeometry.com/problem/p260_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

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

2. If the sides of the equilateral triangle ABC is

AB=BC=AC= a

then from problem 258 we get :

n^2+m^2+q^2=(5/4)a^2

(1.)
4(n^2+m^2+q^2)=5a^2

and from problem 259 we get :

(d^2+e^2+f^2)=(1/2)a^2

(2.)
10(d^2+e^2+f^2)=5a^2

From equation 1 and 2 we get

4(n^2+m^2+q^2)=10(d^2+e^2+f^2)

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

3. Just combine the result of problem 258 & 259