# Difference between revisions of "2013 AMC 12B Problems/Problem 19"

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If we draw a diagram as given, but then add point <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{FG}\perp\overline{BC}</math> in order to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG=\tfrac35x</math> and <math>DG=\tfrac45x</math>, where <math>x</math> is the length of <math>\overline{DF}</math>. Using the Pythagorean theorem, we now get <cmath>BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}</cmath> and <math>AF</math> can be found out noting that <math>AE</math> is just <math>\tfrac{48}5</math> through base times height (since <math>12\cdot 9 = 15 \cdot \tfrac{36}5</math>, similar triangles gives <math>AE = \tfrac{48}5</math>), and that <math>EF</math> is just <math>\tfrac{36}5 - x</math>. From there, <cmath>AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.</cmath> Now, <math>BF^2 + AF^2 = 169</math>, and squaring and adding both sides and subtracting a 169 from both sides gives <math>2x^2 - \tfrac{32}5x = 0</math>, so <math>x = \tfrac{16}5</math>. Thus, the answer is <math>\boxed{\textbf{(B)}}</math>. | If we draw a diagram as given, but then add point <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{FG}\perp\overline{BC}</math> in order to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG=\tfrac35x</math> and <math>DG=\tfrac45x</math>, where <math>x</math> is the length of <math>\overline{DF}</math>. Using the Pythagorean theorem, we now get <cmath>BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}</cmath> and <math>AF</math> can be found out noting that <math>AE</math> is just <math>\tfrac{48}5</math> through base times height (since <math>12\cdot 9 = 15 \cdot \tfrac{36}5</math>, similar triangles gives <math>AE = \tfrac{48}5</math>), and that <math>EF</math> is just <math>\tfrac{36}5 - x</math>. From there, <cmath>AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.</cmath> Now, <math>BF^2 + AF^2 = 169</math>, and squaring and adding both sides and subtracting a 169 from both sides gives <math>2x^2 - \tfrac{32}5x = 0</math>, so <math>x = \tfrac{16}5</math>. Thus, the answer is <math>\boxed{\textbf{(B)}}</math>. | ||

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## Revision as of 20:10, 28 September 2020

*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, triangles and are similar. 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 .