# Difference between revisions of "2019 AMC 10B Problems/Problem 16"

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− | In <math>\triangle ABC</math> with a right angle at <math>C | + | ==Problem== |

+ | In <math>\triangle ABC</math> with a right angle at <math>C</math>, point <math>D</math> lies in the interior of <math>\overline{AB}</math> and point <math>E</math> lies in the interior of <math>\overline{BC}</math> so that <math>AC=CD,</math> <math>DE=EB,</math> and the ratio <math>AC:DE=4:3</math>. What is the ratio <math>AD:DB?</math> | ||

<math>\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2</math> | <math>\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2</math> | ||

+ | |||

+ | ==Solution== | ||

+ | Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a 3-4-5 triangle with <math>CE = 5</math>. | ||

+ | |||

+ | Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a 1-2-<math>\sqrt{5}</math> triangle. | ||

+ | |||

+ | On isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>. | ||

+ | |||

+ | -scrabbler94 |

## Revision as of 14:25, 14 February 2019

## Problem

In with a right angle at , point lies in the interior of and point lies in the interior of so that and the ratio . What is the ratio

## Solution

Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a 3-4-5 triangle with .

Then , and is a 1-2- triangle.

On isosceles triangles and , drop altitudes from and onto ; denote the feet of these altitudes by and respectively. Then by AAA similarity, so we get that , and . Similarly we get , and .

-scrabbler94