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 Inradius And Consecutive Sides (Posted on 2007-12-06)
The incircle of the triangle PQR touches its sides QR, RP and PQ respectively at the points S, T and U. It is known that the respective lengths of QS, RT and PU (in that order) are consecutive integers and the length of the inradius of of the triangle PQR is 4 units.

Determine the respective lengths of each of the sides PQ, QR and RP of the triangle.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 Solution | Comment 2 of 4 |
`Let the side lengths be p, q, and r and thesemiperimeter s = (p+q+r)/2.`
`  |QS| = k-1 = s-q  |RT| =   k = s-r  |PU| = k+1 = s-p`
`Adding these gives`
`  3k = 3s-(p+q+r) = 3s-2s = s`
`  Area(PQR) = inradius*semiperimeter`
`            = sqrt(s(s-p)(s-q)(s-r))`
`            or`
`  16s^2 = s(s-p)(s-q)(s-r)`
`            or`
`  16(3k) = k(k^2 - 1)`
`            or`
`  k = 7`
`Therefore,`
`  |QS| = 6  |RT| = 7  |PU| = 8`
`and`
`  |PQ| = |PU|+|UQ| = |PU|+|QS| = 8+6 = 14       |QR| = |QS|+|SR| = |QS|+|RT| = 6+7 = 13`
`  |RP| = |RT|+|TP| = |RT|+|PU| = 7+8 = 15`

 Posted by Bractals on 2007-12-06 11:43:01

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