P, Q and R are three points located on a circle L with diameter 4 and satisfying PQ = QR. Point S is located inside L in such a manner that QR = RS = SQ. The line passing through P and S intersects L at the point T.
Determine the length of ST.
QRS is an equilateral triangle,
QRS = RSQ = SQR = 60
Therefore,
PQS = PQR  SQR = (OQP + OQR)  SQR = 2x  60
PQS is an isosceles triangle. Therefore,
QPS = QSP = 90  PQS/2 = 120  x
Therefore,
RST = 180  PSQ  QSR = x
PQRT is a cyclic quadrilateral. Therefore,
QRT = 180  QPT = 180  QPS = 60 + x
Therefore,
SRT = QRT  QRS = x
Therefore,
SRT = RST implies ST = RT
Therefore,
ORT = QRT  QRO = 60
The base angle of the isosceles triangle ROT is 60.
Therefore, ROT is an equilateral triangle with
RT = OT
Hence,
ST = RT = OT
ST equals the radius of circle L.

Posted by Bractals
on 20080129 17:30:11 