Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution 3== | ==Solution 3== | ||
− | If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math> | + | If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math>3/5 * x</math> and <math>DG</math> is <math>4/5 *x</math>. Using Pythagorean theorem, we now get |
<math>BF = \sqrt{(4/5 *x + 5)^2 + (3/5 * x)^2}</math> | <math>BF = \sqrt{(4/5 *x + 5)^2 + (3/5 * x)^2}</math> | ||
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From there, <math>AF</math> is <math>\sqrt{((36/5 - x)^2 + (48/5)^2)}</math> | From there, <math>AF</math> is <math>\sqrt{((36/5 - x)^2 + (48/5)^2)}</math> | ||
− | + | Now, <math>BF^2 + AF^2 = 169</math>, and adding both sides and subtracting a 169 from both sides gives: | |
+ | |||
+ | <math>2x^2 - 32/5x = 0</math>, so | ||
x = <math>(16/5)</math>. | x = <math>(16/5)</math>. |
Revision as of 19:02, 3 February 2017
- 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, 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,
Solution 3
If we draw a diagram as given, but then add as an altitude to use the Pythagorean theorem, we end up with similar triangles and . Thus, is and is . Using Pythagorean theorem, we now get
and can be found out noting that is just 48/5 through area times height (12(9) = 15 * , similar triangles gives AE = ), and that is just .
From there, is
Now, , and adding both sides and subtracting a 169 from both sides gives:
, so
x = .
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.