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

Video transcript

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.