An acuteangled triangle PQR is inscribed in a circle with centre O.
S is the intersection of the bisector of P with QR
and
the perpendicular to PO through S meets the line PR in a point W interior to the segment PR.
Find PQ/PW
***Source: A problem appearing in the Italian Mathematical Olympiad, 1995
Call the intersection of the perpendicular to PO through S the the bisector of angle RQP by T
Call angle RPQ by p
Call angle PQR by q
Construct segments OQ and OR.
by the given bisector angle RPS = angle QPS = p/2
by triangle PSQ angle PSQ = 180  p/2  q
by inscribed and central angle POR = 2q
by isosceles triangle POQ angle RPO = 90q
by angle subtraction angle SOP = p/2+q90
by right triangle STP angle TSP = 180p/2q
so angles PSQ and PSW are equal
by reflexive property SP=SP
because PS bisects angle RPQ angles RPQ and QPS are equal
so by ASA ΔPSW is congruent to ΔPSQ
so PW = PQ
so
PQ/PW=1

Posted by Jer
on 20150418 22:12:46 