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
SPRS+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
RSSP
= 
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 20071008 14:26:24 