2013 AMC 10 A #23 / AMC 12 A #19

We've got a geometry problem here, so you know where we're going to start. We're going to draw the diagram. Got a triangle, couple side lengths. Have a circle centered at one of the vertices of the triangle. And the radius is one of the side lengths of the triangle. So it's going to go through one of the vertices. Well, I'm going to start with the circle, because circles are hard for me to draw. And once I have the circle, then maybe I can fit everything else in. So there's AB right there. Center A. Radius AB. There's my circle. And let's see. AC is 97. It's longer than AB. AB is 86. So C is going to be outside the circle. And then I see that BC is going to intersect the circle at B, of course, and then also at another point. So I know that C-- I've got to put C somewhere so that BC will hit the circle. I'm going to put it right there. Call that C. Call this X. And I'll go ahead and finish our triangle right there. We know that this is 86, because it's a radius. And this piece out here is 97 minus 86. That's 11. Now, let's see. We're looking for BC. We're looking for the length of this secant out here. We've got a bunch of lengths. We're looking for the length of a secant out here. This problem just screams power of a point to me. Now what's power of a point? Tells us that CX times CB equals this length right here. I'm going to call that point Y. CY times what you get if you continue CY all the way through the circle and hit the other side, CZ. And if you don't buy that, try to prove it. Break out some similar triangles. Try to find some similar triangles in the diagram. You'll have to draw a few more lines and see if you can prove this relationship. Very powerful relationship that I always think of when I have lengths of a secant, or chords, or something like that-- what we've got in this problem right here. Well, now we know CX times CB equals CY is 11. And CZ coming all the way across, 11. Plus 86-- this is also a radius, is 86. 86 plus 97 gives us 183. Now we might jump in and say, OK, we can just let CX be 11 and CB be 183. But this wouldn't be much of a triangle if this is 183, because 86 plus 97, that's 183. That would mean our triangle was actually a line segment, because of the triangle inequality. So BC can't be 183. It can't be greater than 183. We know this has to be less than 183. We also, of course, know that CX has to be less than CB. So what we're looking for is a way to break this up into the product of two integers-- because we know BX and CX are integers, so CB has to be an integer as well-- so actually, the larger of the two integers is less than 183. We've created a number theory problem. We'll go ahead and write the prime factorization of this. It's clearly divisible by 3. It's 3 times 61. And now, we see that we can write this as just 33 times 61. It's the only way we can write this as the product of two integers where both integers are less than 183. We know that BC has to be the larger of the two. It's right there. It's 61. And we're done.