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.

Let I be the incenter of triangle SPR and A
the point of tangency of the incircle with
side SP. Then

          |SP|+|PR|+|RS|
  |PA| = ---------------- - |RS|
                2

          |SP|-|RS|+|PQ|
       = ----------------
                2

Let E be the excenter of triangle SQR (with
respect 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