Let ABC be an equilateral triangle with K its circumcircle.

Let P be a point on K (different from A, B, and C) and m 
the tangent line to K at point P.

Let D, E, and F be points on m such that AD, BE, and CF 
are perpendicular to m.

Prove that |AD| + |BE| + |CF| equals twice the length of 
an altitude of triangle ABC.

Center the circle at the origin and let its radius be r.
The altitude is 3r/2 so twice this is 3r.
Each of A,B,C forms an angle with the positive x axis.  Call the smallest angle x, then the others are x+120 and x-120.
WLOG let point P be (0,-r) and the tangent line be y=-r

The three lengths are r sin(x) + r, r sin(x+120) + r, and r sin(x-120) + r
The sum is 3r + r [sin(x+120) + sin(x) + sin(x-120)]

So now to show sin(x+120) + sin(x) + sin(x-120) = 0

sin (x+120) = sin(x)cos(120) + cos(x)sin(120) = -.5sin(x)
sin (x-120) = sin(x)cos(120) - cos(x)sin(120) = -.5sin(x)
sin (x) = sin(x)
So the sum of these terms is zero and the sum of the three lengths is 3r.

Edited on June 11, 2010, 4:01 pm

Posted by Jer on 2010-06-11 16:01:23