Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
− | Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB | + | Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</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> |
Revision as of 22:12, 9 February 2013
Contents
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution 1
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
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 |