2019 AMC 10B Problems/Problem 16

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
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.

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 16

Problem

Solution

See Also