Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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==Solution 1 (Power of a Point)== | ==Solution 1 (Power of a Point)== | ||
− | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> | + | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meets the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>AC</math>. Then <math>AZ = AY = 86</math>. Using the Power of a Point (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, so <math>p</math> is either 3,11, or 33. We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>. <math>p</math> is 33. Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>. |
==Solution 2 (Stewart's Theorem)== | ==Solution 2 (Stewart's Theorem)== |
Revision as of 01:11, 10 October 2017
Problem
In ,
, and
. A circle with center
and radius
intersects
at points
and
. Moreover
and
have integer lengths. What is
?
Solution 1 (Power of a Point)
Let ,
, and
meets the circle at
and
, with
on
. Then
. Using the Power of a Point (Secant-Secant Power Theorem), we get that
. We know that
, so
is either 3,11, or 33. We also know that
by the triangle inequality on
.
is 33. Thus, we get that
.
Solution 2 (Stewart's Theorem)
Let represent
, and let
represent
. Since the circle goes through
and
,
.
Then by Stewart's Theorem,
(Since cannot be equal to
, dividing both sides of the equation by
is allowed.)
The prime factors of are
,
, and
. Obviously,
. In addition, by the Triangle Inequality,
, so
. Therefore,
must equal
, and
must equal
.
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.