Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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− | + | == Problem== | |
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− | + | In <math> \bigtriangleup ABC </math>, <math> AB = 86 </math>, and <math> AC = 97 </math>. A circle with center <math> A </math> and radius <math> AB </math> intersects <math> \overline{BC} </math> at points <math> B </math> and <math> X </math>. Moreover <math> \overline{BX} </math> and <math> \overline{CX} </math> have integer lengths. What is <math> BC </math>? | |
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− | + | <math> \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 </math> | |
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1 (Diophantine PoP)=== | ||
+ | |||
+ | <asy> | ||
+ | //Made by samrocksnature | ||
+ | size(8cm); | ||
+ | pair A,B,C,D,E,X; | ||
+ | A=(0,0); | ||
+ | B=(-53.4,-67.4); | ||
+ | C=(0,-97); | ||
+ | D=(0,-86); | ||
+ | E=(0,86); | ||
+ | X=(-29,-81); | ||
+ | draw(circle(A,86)); | ||
+ | draw(E--C--B--A--X); | ||
+ | label("$A$",A,NE); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,NE); | ||
+ | label("$X$",X,dir(250)); | ||
+ | dot(A^^B^^C^^D^^E^^X); | ||
+ | </asy> | ||
+ | |||
+ | 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(CX+XB) = (97-86)(97+86)</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> | ||
+ | |||
+ | However, by the Triangle Inequality on <math>\triangle ACX,</math> we see that <math>CX>13.</math> This implies that we must have <math>CX+XB= \boxed{\textbf{(D) }61}.</math> | ||
+ | |||
+ | (Solution by unknown, latex/asy modified majorly by samrocksnature) | ||
===Solution 2=== | ===Solution 2=== | ||
+ | |||
+ | 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>AC</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 3=== | ||
+ | Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB = AX = 86</math>. | ||
+ | Then by Stewart's Theorem, | ||
+ | |||
+ | <math>xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.</math> | ||
+ | |||
+ | <math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math> | ||
+ | |||
+ | <math>x^2 + xy + 86^2 = 97^2</math> | ||
+ | |||
+ | (Since <math>y</math> cannot be equal to <math>0</math>, dividing both sides of the equation by <math>y</math> is allowed.) | ||
+ | |||
+ | <math>x(x+y) = (97+86)(97-86)</math> | ||
+ | |||
+ | <math>x(x+y) = 2013</math> | ||
+ | |||
+ | The prime factors of <math>2013</math> are <math>3</math>, <math>11</math>, and <math>61</math>. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal <math>33</math>, and <math>x+y</math> must equal <math> \boxed{\textbf{(D) }61}</math> | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc12a/357 | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/zxW3uvCQFls | ||
+ | |||
+ | ~sugar_rush | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Number theory]] | ||
+ | {{MAA Notice}} |
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 |
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