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,</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>
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==Problem==
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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>
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==Solution==
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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>.
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Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a 1-2-<math>\sqrt{5}</math> triangle.
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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>.
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-scrabbler94

Revision as of 14:25, 14 February 2019

Problem

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

$\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2$

Solution

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

Then $CB = 5+3 = 8$, and $\triangle ABC$ is a 1-2-$\sqrt{5}$ triangle.

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

-scrabbler94

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