Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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We then add <math>k^2</math> to both sides to get <math>k^2 + 2km + m^2 = 2013 + k^2</math>. | We then add <math>k^2</math> to both sides to get <math>k^2 + 2km + m^2 = 2013 + k^2</math>. | ||
We then complete the square to get <math>(k + m)^2 = 2013 + k^2</math>. Because <math>k</math> and <math>m</math> are both integers, we get that <math>2013 + k^2</math> is a square number. Simple guess and check reveals that <math>k = 14</math>. | We then complete the square to get <math>(k + m)^2 = 2013 + k^2</math>. Because <math>k</math> and <math>m</math> are both integers, we get that <math>2013 + k^2</math> is a square number. Simple guess and check reveals that <math>k = 14</math>. | ||
− | Because <math>k</math> equals <math>14</math>, therefore <math>m = 33</math>. We want <math>\overline{BC} = 2k + m</math>, so we get that <math>\overline{BC} = \boxed{(B) 61}</math> | + | Because <math>k</math> equals <math>14</math>, therefore <math>m = 33</math>. We want <math>\overline{BC} = 2k + m</math>, so we get that <math>\overline{BC} = \boxed{(B)~61}</math> |
+ | <math>\phantom{solution by bobjoe123}</math> | ||
==See Also== | ==See Also== |
Revision as of 01:44, 30 January 2018
- The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.
Contents
[hide]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
.
Solution 3
Let . Let the circle intersect
at
and the diameter including
intersect the circle again at
.
Use power of a point on point C to the circle centered at A.
So
.
Obviously so we have three solution pairs for
.
By the Triangle Inequality, only
yields a possible length of
.
Therefore, the answer is .
Solution 4
We first draw the height of isosceles triangle ABD, and get two equations by the Pythagorean Theorem.
First, . Second,
.
Subtracting these two equations, we get
.
We then add
to both sides to get
.
We then complete the square to get
. Because
and
are both integers, we get that
is a square number. Simple guess and check reveals that
.
Because
equals
, therefore
. We want
, so we get that
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.