# 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.

• Note that $PQ$ is not a tangent to both circles, just one of them. – Arthur May 21 '15 at 13:19
• That seems to imply two things then. That $T$ must be larger than $S$, otherwise they will not intersect and that the tangent to $P$ must intersect $T$ at two points, making it unclear which point $Q$ is. Am I assume it to be the closest point? – MadChickenMan May 21 '15 at 13:25
• You've got it backwards: $Q$ must be a tangent point, but $P$ doesn't have to be. The text says "let the tangent from $P$ to $T$ touch it at $Q$". If the circles aren't of the same size, then the line will intersect $S$ at two points: between $P$ and $Q$ if $S$ is the larger circle, and on the far side of $P$ if $T$ is the larger circle. – Arthur May 21 '15 at 13:55
• Ah I see now. Thanks, now to actually try and prove it. – MadChickenMan May 21 '15 at 14:07
• Any ideas how to prove it? I can't really see anything? – MadChickenMan May 21 '15 at 15:22

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.

• Thanks for your answer. I did manage to come up with a solution of my own in the end, but it was quite dissimilar to yours, just goes to show how problems can be approached in any number of ways. – MadChickenMan May 22 '15 at 17:03
• I would love to see it! You can always click the 'answer own question' box – Faraz Masroor May 22 '15 at 18:05

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.

• I like it. I think it would be helpful to note that <AXB=90 (if you dont see this, prove it! very quick) and a little bit more explanation on QXP and QBP's similarity. But great solution that's not considered bashy! Power of a point is a very commonly used tactic on Olympiads – Faraz Masroor May 23 '15 at 0:06
1. if S,T are same size, AP=PQ
2. if S larger than T, it is impossible
3. 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)