Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | + | 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>? | |
− | + | $ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ | |
==Solution== | ==Solution== | ||
− | + | ||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} |
Revision as of 17:13, 22 February 2013
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 ?
$ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\
Solution
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 |