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Absolutely Geometric (Posted on 2007-12-21) Difficulty: 3 of 5
In triangle PQR, the respective length of the sides PQ and PR are denoted by u and v while the length of the median PS is denoted by w. It is known that w is the geometric mean of u and v, and Angle QPR = 60o.

Determine |cos(Angle PQR) - cos(Angle QRP)|, where |x| denotes the absolute value of x.




See The Solution Submitted by K Sengupta    
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Solution Solution | Comment 2 of 6 |

Let t = |QR|, u = |PQ|, v = |PR|, and w = |PS|.
Let <AB> denote the vector from point A to point B.

  uv = w^2 = <PS> dot <PS>
        <PQ> + <PR>       <PQ> + <PR> 
     = ------------- dot -------------  
             2                 2
        u^2 + v^2 + 2uv cos(QPR)
     = --------------------------
                   4
        u^2 + v^2 + 2uv cos(60)
     = -------------------------
                   4
     or
  u^2 -3uv + v^2 = 0

       3 +- sqrt(5)
  u = -------------- v
            2

  t^2 = |QR|^2 = |PQ|^2 + |PR|^2 - 2|PQ||PR|cos(QPR)
      = u^2 + v^2 - 2uv cos(60) = u^2 + v^2 - uv
      = 2uv

  u^2 = |PQ|^2 = |PR|^2 + |QR|^2 - 2|PR||QR| cos(QRP)
      = v^2 + t^2 - 2vt cos(QRP)
      or
              v^2 + t^2 - u^2 
  cos(QRP) = -----------------
                    2vt

  v^2 = |PR|^2 = |PQ|^2 + |QR|^2 - 2|PQ||QR| cos(PRQ)
      = u^2 + t^2 - 2ut cos(PQR)
      or
              u^2 + t^2 - v^2
  cos(PQR) = -----------------
                    2ut
 

  |cos(PQR) - cos(QRP)|
          u^2 + t^2 - v^2     v^2 + t^2 - u^2
     = | ----------------- - ----------------- |
                2ut                 2vt
        3 |u - v|     3 sqrt(2)
     = ----------- = ----------- 
        sqrt(8uv)     4

     =~ 1.06066
 

Edited on December 21, 2007, 8:12 pm
  Posted by Bractals on 2007-12-21 15:09:56

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