Three Perps (Posted on 2010-06-11)

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.

	Submitted by Bractals
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-------------------------------------------------------------- G E O M E T R Y S O L U T I O N -------------------------------------------------------------- Let O be the center of circle K. The ray AO intersects side BC at point N and the circle again at point M. Let Q and R be points on line m such that NQ and MR are perpendicular to m. Lemma: Let WXYZ be a trapezoid with WX parallel to YZ. Let a line parallel to WX intersect sides WZ and XY in points S and T respectively. If |WS| = |SZ|, then |WX| + |YZ| = 2|ST|. Using the lemma, |AO| = |OM| ==> |AD| + |MR| = 2|OP| (1) |BN| = |NC| ==> |BE| + |CF| = 2|NQ| (2) |ON| = |NM| ==> |OP| + |MR| = 2|NQ| (3) Combining (1) and (2) we get |AD| + |BE| + |CF| = 2|OP| - |MR| + 2|NQ| (4) Combining (3) and (4) we get |AD| + |BE| + |CF| = 3|OP| = 3|OA| = 2|AN| twice the length of an altitude of ABC. -------------------------------------------------------------- V E C T O R S O L U T I O N -------------------------------------------------------------- Let the circle K have unit radius and center O. Let ST denote the vector from point S to point T, 0 the zero vector, i the unit vector in the direction PO, and j the unit vector perpendicular to i. Thus, PO + OA + AD + DP = 0 PO + OB + BE + EP = 0 PO + OC + CF + FP = 0 or i + OA - |AD|i + dj = 0 i + OB - |BE|i + ej = 0 i + OC - |EF|i + fj = 0 Adding these together, { 3 - (|AD| + |BE| + |CF|)}i + {OA + OB + OC} + {d + e + f}j = 0 Therefore, |AD| + |BE| + |CF| = 3 = 2(3/2) twice the length of an altitude of ABC. -------------------------------------------------------------- T R I G O N O M E T R Y S O L U T I O N -------------------------------------------------------------- Let the circle K have unit radius and center O. Let the measure of angle POA be t. Then, |AD| = 1 - cos(t) |BE| = 1 - cos(t+120) |CF| = 1 - cos(t+240) Adding these gives |AD| + |BE| + |CF| = 3 - cos(t) - cos(t+120) - cos(t+240) = 3 = 2(3/2) twice the length of an altitude of ABC.

	Subject	Author	Date
	Mechanics Approach	Harry	2010-06-11 20:07:47
	Analytic solution	Jer	2010-06-11 16:01:23