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=a-c and

a-c+d = 90

In triangle MPQ

b+90-c+d =180; i.e. b-c+d = 90

Again, putting those two together: a=b

 

Posted by DrBob on 2005-12-14 12:15:09