BMO1 2006/07 Question 4 Geometry Problem $4.$ Two touching circles $S$ and $T$ share a common tangent which meets
$S$ at $A$ and $T$ at $B$. Let $AP$ be a diameter of $S$ and let the tangent
from $P$ to $T$ touch it at $Q$. Show that $AP = PQ$.
I'm unsure of the configuration. If $P$ is on the opposite side of the circle to $A$, does that not suggest the tangent to that point produces a line parallel to the other tangent. Then the other circle would have to be of equal size and the statement of $AP=PQ$ becomes trivial and I just don't see it being so easy. So to anyone who can come up with a hint or full solution if you like, thank you in advance.
 A: Let the center of the circles be S and T. Let r be the radius of S and R the radius of T. 
$PQ^2=PT^2-R^2$ (pythagoraean theorem)
$=r^2+(R+r)^2-2r(R+r)cos(<TSP) -R^2$ (law of cosines)
Drop a perpendicular from S to BT to see that $cos( <TSP) = \frac{R-r}{R+r}$, so 
$PQ^2=r^2+(R+r)^2-2r(R-r) -R^2$ 
$=4r^2$ so PQ=2r=AP. 
Sorry I don't have a diagram here. 
A: If we call the point where the two circles meet $X$, then clearly $AXP$ and $ABP$ are similar. From this we deduce $\frac{AP}{PB}=\frac{PX}{AP}$, therefore $AP^2=PX.PB$. Similarly, $QXP$ and $QBP$ are similar and this time we obtain that $\frac{PQ}{PB}=\frac{PX}{PQ}$, and so we have that $PQ^2=PX.PB$ and so $AP=PQ$. 
I've left out a few details to be concise, but most of this is just simple angle chasing and use of the alternate angle theorem.
A: *

*if S,T are same size, AP=PQ

*if S larger than T, it is impossible

*if T larger than S, AP is not equal to PQ
(if we assume the point where the two circles meet is X, QX's extension cord intersect AB at Y, triangle PQX congruent to triangle XYA, so triangle PAX must be a isosceles right triangle. AP is not equal to PQ)

