Quadrilateral ABCD is inscribed in a circle. Diagonals AC and BD meet at point Q.Rays CB and DA meet at point P. If |CD| = |CP| = |DQ|, what is the measure of angle CAD?

2001 British Mathematical Olympiad, Round 1, Problem 2.

Well, I'll try.  Draw a picture if you want to follow along.

How about:
a) Let angle DAC = x
b) Then angle DBC = x, because the quadrilateral is inscribed in a circle and they subtend the same arc.  angle PBD = 180 - x
c) Let angle DPC = y
d) Then angle PDC = y, because triangle PDC is isosceles
e) Then angle ACD = 180 - x - y, because the three angles total to 180 degrees.
f) Then angle DQC = 180 - x - y, because triangle DCQ is isosceles.
So AQD = x + y
q) But PDQ = 180 - DPC - PBD = x - y
r) Then 180 degrees = ADQ + AQD + DAQ
    = (x-y) + (x+y) + x = 3x
s) So x = 60 degrees

Maybe I've made a mistake.  Maybe there's a simpler solution.
Anybody?

Edited on December 17, 2005, 11:38 pm

Posted by Steve Herman on 2005-12-17 17:00:05