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 Absolutely Geometric (Posted on 2007-12-21)
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 No Rating

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 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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