Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
− | From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>15 - \frac{16}{5} = \frac{48}{5}</math> .Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = \angle{EFA}</math>, and triangles <math>AEF</math> and <math>ADB </math> are similar. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math> and our answer is <math>\boxed{\textbf{(B)}}</math>. | + | From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{16}{5} = \frac{48}{5}</math> .Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = \angle{EFA}</math>, and triangles <math>AEF</math> and <math>ADB </math> are similar. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math> and our answer is <math>\boxed{\textbf{(B)}}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 22:15, 27 March 2023
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
[hide]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 ?
Diagram
Solution 1
Since , quadrilateral is cyclic. It follows that , so are similar. In addition, . We can easily find , , and using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is , and the ratio of the shorter leg to the hypotenuse is . It follows that .
Let . By Ptolemy's Theorem, we have Dividing by we get so our answer is .
~Edits by BakedPotato66
Solution 2
From solution 1, we know that and . Since , we can figure out that . We also know what is so we can figure what is: .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 .
Solution 4 (Power of a Point)
First, we find , , and via the Pythagorean Theorem or by using similar triangles. Next, because is an altitude of triangle , . Using that, we can use the Pythagorean Theorem and similar triangles to find and .
Points , , , and all lie on a circle whose diameter is . Let the point where the circle intersects be . Using power of a point, we can write the following equation to solve for : Using that, we can find , and using , we can find that .
We can use power of a point again to solve for : Thus, .
Video Solution
https://youtu.be/XZBKnobK-JU?t=3064
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.