Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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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>. | 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 2== | + | ==Solution 2 (Stewart's Theorem)== |
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>. | 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(x+y) = 2013</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> | + | 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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==See Also== | ==See Also== |
Revision as of 13:49, 27 January 2015
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 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 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 .
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.