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 (Number theoretic power of a point)=== | ||
− | + | Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>. | |
− | Let <math> | + | Use power of a point on point C to the circle centered at A. |
− | <math> | + | So <math>CX*CB=CD*CE=></math> |
+ | <math>x(x+y)=(97-86)(97+86)=></math> | ||
+ | <math>x(x+y)=3*11*61</math>. | ||
− | <math> | + | Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>. |
+ | By the Triangle Inequality, only<math> x+y=61</math> yields a possible length of <math>BX+CX=BC</math>. | ||
+ | |||
+ | Therefore, the answer is '''D) 61.''' | ||
− | + | ===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== | + | ===Solution 3=== |
− | Let <math>x</math> represent <math> | + | 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, | Then by Stewart's Theorem, | ||
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<math>x^2 + xy + 86^2 = 97^2</math> | <math>x^2 + xy + 86^2 = 97^2</math> | ||
− | (Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.) | + | (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) = (97+86)(97-86)</math> | ||
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<math>x(x+y) = 2013</math> | <math>x(x+y) = 2013</math> | ||
− | The prime factors of 2013 are 3, 11, and 61. 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 33, and <math>x+y</math> must equal | + | 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> |
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+ | |||
+ | ==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}} |
Revision as of 01:36, 25 November 2020
Contents
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution
Solution 1 (Number theoretic power of a point)
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
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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