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Triangle PQR is equilateral with PQ = QR = RP = 2. The line QP is extended to meet at point S such that P lies between S and Q.

Y is length of the inradius of Triangle SPR while Z is the length of the exradius of Triangle SQR with respect to the side QR.

Determine Y+Z.

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

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 Solution | Comment 2 of 4 |
`Let I be the incenter of triangle SPR and Athe point of tangency of the incircle withside SP. Then`
`          |SP|+|PR|+|RS|  |PA| = ---------------- - |RS|                2`
`          |SP|-|RS|+|PQ|       = ----------------                2`
`Let E be the excenter of triangle SQR (withrespect to side QR) abd B the point of tangency of the excircle with side SQ extended.Then`
`          |SQ|+|QR|+|RS|  |QB| = ---------------- - |SQ|                2`
`          |RS|-|SP|       = -----------              2`
`Therefore,`
`  Y+Z = |AI|+|BE| = (|PA|+|QB|)*sqrt(3)`
`         |PQ|      = ------ * sqrt(3)           2`
`Which is the altitude of triangle PQR.`
`For our problem,`
`  Y+Z = sqrt(3)`
`     `

 Posted by Bractals on 2007-10-08 14:26:24

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