Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>BC</math>. Then <math>AZ = AY = 86</math>. Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math> | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>BC</math>. Then <math>AZ = AY = 86</math>. Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math> | ||
− | + | ==Solution 2== | |
Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86. | Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86. | ||
Then by Stewart's Theorem, | Then by Stewart's Theorem, |
Revision as of 11:12, 9 February 2013
Contents
[hide]Problem
In ,
, and
. A circle with center
and radius
intersects
at points
and
. Moreover
and
have integer lengths. What is
?
Solution 1
Let ,
, and
meet the circle at
and
, with
on
. Then
. Using the Power of a Point, we get that
. We know that
, and that
by the triangle inequality on
. Thus, we get that
Solution 2
Let represent
, and let
represent
. Since the circle goes through
and
,
=
= 86.
Then by Stewart's Theorem,
(Since cannot be equal to 0, dividing both sides of the equation by
is allowed.)
The prime factors of 2013 are 3, 11, and 61. Obviously, . In addition, by the Triangle Inequality,
, so
. Therefore,
must equal 33, and
must equal 61.
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 |