## Thursday, February 26, 2009

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

Proposed Problem

See complete Problem 258 at:
gogeometry.com/problem/p258_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

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

2. Let the midpoint of AC be(0,0). So we've:
A:(-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
Ajit: ajitathle@gmail.com

3. Let P be the point of tangency with the circle on the line AB.
Let Q be the point of tangency with the circle on the line BC.
Let R be the point of tangency with the circle on the line AC.

If we apply apollonius theorem on triangle ADB :

e^2+d^2=2PD^2+a^2/2
(1.)e^2+d^2-a^2/2=2PD^2

If we apply apollonius theorem on triangle BDC :

e^2+f^2=2DQ^2+a^2/2
(2)e^2+f^2-a^2/2=2BQ^2

If we apply apollonius theorem on triangle ADC :

f^2+d^2=2DR^2+a^2/2
(3.)f^2+d^2-a^2/2=2DR^2

If we add equations 1 ,2 and 3 :

(4.)2(PD^2+DQ^2+DR^2)=2e^2+2d^2+2f^2-(3/2)a^2

The triangle PQR is equilateral with side a/2 therefore from problem 257 we have :

PD^2+DQ^2+DR^2=2(a^2/4)
(5.)PD^2+DQ^2+DR^2=a^2/2

From equation 4 and 5 we get :

2e^2+2d^2+2f^2=a^2
2e^2+2d^2+2f^2=a+(3/2)a^2
2e^2+2d^2+2f^2=5a^2/2

d^2+e^2+f^2=5a^2/4