Difference between revisions of "2013 AMC 12B Problems/Problem 19"
Crazyboulder (talk | contribs) (→Solution 1: Added clarification) |
|||
Line 9: | Line 9: | ||
==Solution 1== | ==Solution 1== | ||
− | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, we obtain <math>DF=\frac{16}{5}</math>, so our answer is <math>16+5=\boxed{21\,\textbf{(B)}}</math>. | + | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, triangles <math>ABF</math> and <math>ADE</math> are similar, and triangles <math>ADE</math> and <math>ADC</math> are similar. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is <math>\frac{15}{12} = \frac{4}{5}</math>, and the ratio of the hypotenuse to the shorter leg is <math>\frac{15}{9} = \frac{3}{5}</math>. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, we obtain <math>DF=\frac{16}{5}</math>, so our answer is <math>16+5=\boxed{21\,\textbf{(B)}}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 14:31, 2 January 2021
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
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, triangles and are similar, and triangles and are similar. We can easily find , , and using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is , and the ratio of the hypotenuse to the shorter leg is . It follows that . By Ptolemy's Theorem, we have . Cancelling , we obtain , so our answer is .
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, and our answer is .
Solution 3
If we draw a diagram as given, but then add point on such that in order to use the Pythagorean theorem, we end up with similar triangles and . Thus, and , where is the length of . Using the Pythagorean theorem, we now get and can be found out noting that is just through base times height (since , similar triangles gives ), and that is just . From there, Now, , and squaring and adding both sides and subtracting a 169 from both sides gives , so . Thus, the answer is .
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.