Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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Use power of a point on point C to the circle centered at A. | Use power of a point on point C to the circle centered at A. | ||
− | So <math>CX*CB=CD*CE</math> | + | So <math>CX*CB=CD*CE=></math> |
− | <math>x(x+y)=(97-86)(97+86)</math> | + | <math>x(x+y)=(97-86)(97+86)=></math> |
<math>x(x+y)=3*11*61</math>. | <math>x(x+y)=3*11*61</math>. | ||
Revision as of 18:53, 6 September 2014
Problem
In ,
, and
. A circle with center
and radius
intersects
at points
and
. Moreover
and
have integer lengths. What is
?
Solution
Solution 1
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 D) 61.
Solution 2
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 3
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 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.