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# 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?

- Can anyone please tell and explain to me what power of a point is?(34 votes)
- Is there a way to work out this problem that a 4th grader understands?(4 votes)
- Well, I'm sure that there are 4th graders that can understand this: however, if you are looking for a solution that the average 4th grader can understand, then no.(3 votes)

- hold up were them points(3 votes)
- At1:25, he mentions something called "power of a point", where CX times CB is CY times CZ. I still don't get it, despite watching the video many times. Can someone explain what it is?(2 votes)
- It's a special formula that can be proved: https://en.wikipedia.org/wiki/Power_of_a_point(1 vote)

- how did he make it to 61 and why not finish the answer(2 votes)
- Is there any alternative way to solve this problem?(1 vote)
- There are several other ways to solve it, but they are less straight forward: https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23(1 vote)

- umm is there a AMC 8 questions solving here in Khan Academy❔(1 vote)
- at3:10, why do you need that 11*3*61? the y was 86 and the man said that the x should be less than 183 so why 11*3*61?(1 vote)
- What is the A B circle,and the CZ and the 86 thing, because I don't get it?(1 vote)
- the distance from A to B is 86, and CZ is the distance from C to Z.(1 vote)

- What is the power of the point?(1 vote)

## 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.