If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

2013 AMC 10 A #23 / AMC 12 A #19

Video by Art of Problem Solving.  Problem from the MAA American Mathematics Competitions. Created by Art of Problem Solving.

Want to join the conversation?

Video transcript

- We've get a geometry problem here, so you know where we're gonna start, we're gonna draw the diagram. Got a triangle, couple of 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 gonna go through one of the vertices. Ah, I'm gonna start with the circle 'cause 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, that's longer than AB, AB is 86. So, C is gonna 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, gotta put C somewhere, so that BC'll hit the circle. I'm gonna 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, 'cause 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 Power of a Point tells us, that CX times CB equals this length right here, I'm gonna call that point Y, CY times, what you get if you continue CY all the way through the circle, and hit at 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 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, we 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, our radius is 86, 86 plus 97 gives us 183. Now, we might jump in and say okay, 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's actually a line segment, 'cause 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 then CB, so what we're looking for is a way to break this up, into the product of two integers, 'cause we know BX and CX are integers. So CB has to be an integer as well. So, it's just 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 three. It's three 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.