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> meets 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 (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, so <math>p</math> is either <math>3</math>, <math>11</math>, or <math>33</math>. We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, <math>p</math> is <math>33</math> so 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> meets 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 (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, so <math>p</math> is either <math>3</math>, <math>11</math>, or <math>33</math>. We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, <math>p</math> is <math>33</math> so we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>. | ||
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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>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>. | ||
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Therefore, the answer is <math> \boxed{\textbf{(D) }61}</math>. | Therefore, the answer is <math> \boxed{\textbf{(D) }61}</math>. | ||
− | ==Solution | + | ==Solution 3== |
<asy> | <asy> | ||
unitsize(2); | unitsize(2); | ||
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<math>\phantom{solution and diagram by bobjoe123}</math> | <math>\phantom{solution and diagram by bobjoe123}</math> | ||
− | ==Solution | + | ==Solution 4== |
Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>BX.</math> Since <math>\triangle ABX</math> is isosceles <math>AX=AB=86,EB=EX,</math> and the answer is <math>EC+EB=EC+EX.</math> <math>(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013</math> by the Pythagorean Theorem. Only <math>EC+EX=\boxed{(D)~61}</math> is a factor of <math>2013</math> such that <math>97>EC+EX>EC-EX=\frac{2013}{EC+EX}.</math> | Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>BX.</math> Since <math>\triangle ABX</math> is isosceles <math>AX=AB=86,EB=EX,</math> and the answer is <math>EC+EB=EC+EX.</math> <math>(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013</math> by the Pythagorean Theorem. Only <math>EC+EX=\boxed{(D)~61}</math> is a factor of <math>2013</math> such that <math>97>EC+EX>EC-EX=\frac{2013}{EC+EX}.</math> |
Latest revision as of 22:20, 22 December 2024
- The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution 1 (Number Theoretic Power of a Point)
Let , , and meets the circle at and , with on . Then . Using the Power of a Point (Secant-Secant Power Theorem), we get that . We know that , so is either , , or . We also know that by the triangle inequality on . Thus, is so we get that .
Solution 2
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 .
Solution 3
We first draw the height of isosceles triangle and get two equations by the Pythagorean Theorem. First, . Second, . Subtracting these two equations, we get . We then add to both sides to get . We then complete the square to get . Because and are both integers, we get that is a square number. Simple guess and check reveals that . Because equals , therefore . We want , so we get that .
Solution 4
Let be the foot of the altitude from to Since is isosceles and the answer is by the Pythagorean Theorem. Only is a factor of such that
~dolphin7
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=f1nxu8MWWKc
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=2692
~ pi_is_3.14
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.