Circles/tangents/angles proof

This question comes from the BMO round 1 paper.

By David Loeffler (P865) on Saturday, January 20, 2001 - 10:48 pm :

S is a circle lying inside circle T and touching it at A. P is a point on T; PQ and PR are the chords of T through P tangent to S, at X and Y respectively. Show that angle QAR = 2 x angle XAY.
(NB. P is not A.)

Diagram

David

By Kerwin Hui (Kwkh2) on Tuesday, January 23, 2001 - 02:11 pm :

This is trivial.

[All geometry is trivial to Kerwin. - The Editor]

Using opposite angles in cyclic quadilateral are supplementary, angle QAR = pi - angle P.
By the alternate segment theorem , angle PXY = angle XAY = angle PYX. The result comes out immediately.

Kerwin

By James Lingard (Jchl2) on Tuesday, January 23, 2001 - 02:27 pm :

That may be 'trivial' to you Kerwin but it certainly isn't to me! (Although forgetting the Alternate Segment Theorem I suppose would create a problem for getting that solution.)

James.

By Brad Rodgers (P1930) on Tuesday, January 23, 2001 - 08:14 pm :

How do you prove that opposite angles in a cyclic quad are supplementary?

Brad

By Michael Doré (Md285) on Wednesday, January 24, 2001 - 01:00 am :

Good question. I wish I knew...

By Michael Doré (Md285) on Wednesday, January 24, 2001 - 01:14 am :

Aha... Actually it's easy to prove. Let the four corners of the cyclic quad by $A$ , $B$ , $C$ , $D$ (in clockwise order) and the centre of the circle be $O$ . Now $O A = O B = O C = O D$ so the trianges $O A B$ , $O B C$ , $O C D$ , $O D A$ are isosceles. Therefore:

angle $O A B$ = angle $O B A$ [1]

angle $O B C$ = angle $O C B$ [2]

angle $O C D$ = angle $O D C$ [3]

angle $O D A$ = angle $O A D$ [4]

Define the angles on each side of equations [1]-[4] to be $p$ , $q$ , $r$ , $s$ respectively. Now using the fact that angles in a triangle add to $π$ , we get:

angle $A O B = π - 2 p$

angle $B O C = π - 2 q$

angle $C O D = π - 2 r$

angle $D O A = π - 2 s$

These four angles must add to $2 π$ . So:

$2 π = 4 π - 2 (p + q + r + s)$

$p + q + r + s = π$

So angle $A D C$ + angle $A B C = π$ .

So the two opposite angles in a cyclic quad add to $π$ .

I assume that's what supplementary means... Diagram