All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 A Trisected Angle and the Opposing Side (Posted on 2012-04-22)
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 No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 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

 Search: Search body:
Forums (0)