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

Revision as of 14:50, 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

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}
+{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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

Revision as of 14:50, 14 February 2019

Problem

Solution

See Also