Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=14</math>, and <math>CA=15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD}\perp\overline{BC}</math>, <math>\overline{DE}\perp\overline{AC}</math>, and <math>\overline{AF}\perp\overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=14</math>, and <math>CA=15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD}\perp\overline{BC}</math>, <math>\overline{DE}\perp\overline{AC}</math>, and <math>\overline{AF}\perp\overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | ||
− | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math> |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
Revision as of 22:57, 10 March 2015
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
Problem
In triangle , , , and . Distinct points , , and lie on segments , , and , respectively, such that , , and . The length of segment can be written as , where and are relatively prime positive integers. What is ?
Solution 1
Since , quadrilateral is cyclic. It follows that . In addition, since , triangles and are similar. It follows that . By Ptolemy, we have . Cancelling , the rest is easy. We obtain
Solution 2
Using the similar triangles in triangle gives and . Quadrilateral is cyclic, implying that = 180°. Therefore, , and triangles and are similar. Solving the resulting proportion gives . Therefore,
See also
2013 AMC 12B (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 |
2013 AMC 10B (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.