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A Trisected Angle and the Opposing Side (Posted on 2012-04-22) Difficulty: 3 of 5
Triangle ABC has points P and Q on BC in the order B,P,Q,C. Lines AP and AQ trisect angle BAC.

Part 1: Given BP=3, PQ=4, and QC=6, determine the lengths of AB and AC.

Part 2: More generally, express AB and AC in terms of BP, PQ, and QC.

No Solution Yet Submitted by Brian Smith    
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Solution Part 2 then Part 1 Comment 1 of 1

Part 2:

Let b = |AB|, c = |AC|, p = |AP|, q = |AQ|,
    x = |BP|, y = |PQ|, z = |QC|

AP bisects angle BAQ and AQ bisects angle PAC.
Therefore,

   p = c*y/z                                  (1)   

     and   

   q = b*y/x                                  (2)

Applying the cosine law to triangles BAP, PAQ,
and QAC gives

   x^2 = b^2 + p^2 - 2*b*p*cos(angle BAP)  

       and

   y^2 = p^2 + q^2 - 2*p*q*cos(angle PAQ)

       and

   z^2 = q^2 + c^2 - 2*q*c*cos(angle QAC)

Angle BAP = angle PAQ. Therefore,

    b^2 + p^2 - x^2     p^2 + q^2 - y^2
   ----------------- = -----------------      (3)
           b                   q 

Angle QAC = angle PAQ. Therefore,

    c^2 + q^2 - z^2     p^2 + q^2 - y^2
   ----------------- = -----------------      (4)
           c                   p 

Substituting p and q from (1) and (2) into (3)
gives

   b^2*z^2 - c^2*x*y = x^2*z^2                (5)

Substituting p and q from (1) and (2) into (4)
gives

   -b^2*y*z + c^2*x^2 = x^2*z^2               (6)

Solving (5) and (6) for b^2 and c^2

   b^2 = x^2*z*(x + y)/(x*z - y^2)

   c^2 = x*z^2*(y + z)/(x*z - y^2)

             or

   |AB| = |BP|*sqrt(|BQ||QC|/|UV|^2)

   |AC| = |QC|*sqrt(|BP||PC|/|UV|^2)

   where |UV|^2 = |BP||QC| - |PQ|^2

QED for Part 2

Part 1:

   |BP| = 3, |PQ| = 4, and |QC| = 6

   |BQ| = |BP| + |PQ| = 3 + 4 = 7

   |PC| = |PQ| + |QC| = 4 + 6 = 10

   |UV|^2 = 3*6 - 4^2 = 2

   |AB| = 3*sqrt(7*6/2) = 3*sqrt(21)

        ~= 13.7477

   |AC| = 6*sqrt(3*10/2) = 6*sqrt(15)

        ~= 23.2379

QED for Part 1



  Posted by Bractals on 2012-04-22 17:57:41
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