Let circle A be in the interior of circle B and tangent to it at point M. Let chord QR of circle B be tangent to circle A at point P. Prove that angles PMQ and PMR are equal.
1993 British Mathematical Olympiad,Round 1,Problem 4.
To simplify this writing, let angle PMR=a & angle PMQ=b
We need to prove that a=b
Construct MN perpendicular to the tangent at M  this passes through the centre of both circles. Call the centre of A, O
Draw in OQ, MR, PN
Note angle MPN = 90 (in a semicircle)
And, angle NMR = angle NPR (subtended by NR) call them c
Finally, let angle MQP = d
The detail of the proof depends on whether PR crosses MN
If not:
In the isosceles triangle OMQ, angle OQM = a+c and
a+c+d = 90 (tangent/radius)
In triangle MQP, b+90+c+d = 180; i.e. a+c+d = 90
Put these together: a=b
If PR crosses MN
In the isosceles triangle MOQ
angle MQO=ac and
ac+d = 90
In triangle MPQ
b+90c+d =180; i.e. bc+d = 90
Again, putting those two together: a=b

Posted by DrBob
on 20051214 12:15:09 