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
PQS = PQR - SQR = (OQP + OQR) - SQR = 2x - 60
PQS is an isosceles triangle. Therefore,
QPS = QSP = 90 - PQS/2 = 120 - x
RST = 180 - PSQ - QSR = x
PQRT is a cyclic quadrilateral. Therefore,
QRT = 180 - QPT = 180 - QPS = 60 + x
SRT = QRT - QRS = x
SRT = RST implies ST = RT
ORT = QRT - QRO = 60
The base angle of the isosceles triangle ROT is 60.
Therefore, ROT is an equilateral triangle with
RT = OT
ST = RT = OT
ST equals the radius of circle L.
Posted by Bractals
on 2008-01-29 17:30:11