Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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− | Let circle <math>A</math> intersect <math>AC</math> at <math>D</math> and <math>E</math> as shown. We apply Power of a Point on point <math>C</math> with respect to circle <math>A.</math> | + | Let circle <math>A</math> intersect <math>AC</math> at <math>D</math> and <math>E</math> as shown. We apply Power of a Point on point <math>C</math> with respect to circle <math>A.</math> This yields the diophantine equation |
− | + | <cmath>CX \cdot CB = CD \cdot CE</cmath> | |
− | |||
− | <cmath>CX \cdot CB = CD \cdot | ||
<cmath>CX(CX+XB) = (97-86)(97+86)</cmath> | <cmath>CX(CX+XB) = (97-86)(97+86)</cmath> | ||
− | <cmath>CX(CX+XB) = 3 \cdot 11 \cdot 61</cmath> | + | <cmath>CX(CX+XB) = 3 \cdot 11 \cdot 61.</cmath> |
Since lengths cannot be negative, we must have <math>CX+XB \ge CX.</math> This generates the four solution pairs for <math>(CX,CX+XB)</math>: <cmath>(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).</cmath> | Since lengths cannot be negative, we must have <math>CX+XB \ge CX.</math> This generates the four solution pairs for <math>(CX,CX+XB)</math>: <cmath>(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).</cmath> |
Latest revision as of 15:24, 19 September 2021
Contents
Problem
In ,
, and
. A circle with center
and radius
intersects
at points
and
. Moreover
and
have integer lengths. What is
?
Solution
Solution 1 (Diophantine PoP)
Let circle intersect
at
and
as shown. We apply Power of a Point on point
with respect to circle
This yields the diophantine equation
Since lengths cannot be negative, we must have This generates the four solution pairs for
:
However, by the Triangle Inequality on we see that
This implies that we must have
(Solution by unknown, latex/asy modified majorly by samrocksnature)
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
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/357
~dolphin7
Video Solution
~sugar_rush
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.