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Three Perps (Posted on 2010-06-11) Difficulty: 2 of 5

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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Solution: (Hide)

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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.

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      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.

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      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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsMechanics ApproachHarry2010-06-11 20:07:47
Analytic solutionJer2010-06-11 16:01:23
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