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2019 AMC 10B Problems/Problem 16

Revision as of 01:08, 18 February 2019 by A87321dc0 (talk | contribs) (Undo revision 103319 by Sevenoptimus (talk))

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 1

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.

In 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 $AD:DB = \boxed{\textbf{(A) } 2:3}$.

Solution 2

Let $AC=CD=4x$, and $DE=EB=3x$. (For this solution, $A$ is above $C$, and $B$ is to the right of $C$). Also let $\angle A = t^{\circ}$, so $\angle ACD = \left(180-2t\right)^{\circ}$, which implies $\angle DCB = \left(2t - 90\right)^{\circ}$. Similarly, $\angle B = \left(90-t\right)^{\circ}$, which implies $\angle BED = 2t^{\circ}$. This further implies that $\angle DEC = \left(180 - 2t\right)^{\circ}$.

Now we see that $\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2x^{\circ} + 90^{\circ} - 180^{\circ} + 2x^{\circ} = 90^{\circ}$. Thus $\triangle CDE$ is a right triangle, with side lengths of $3x$, $4x$, and $5x$ (by the Pythagorean Theorem, or simply the Pythagorean triple $3-4-5$). Therefore $AC=4x$ (by definition), $BC=5x+3x = 8x$, and $AB=4\sqrt{5}x$. Hence $\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1$ (by the double angle formula), giving $2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}$.

By the Law of Cosines in $\triangle BED$, if $BD = d$, we have \[\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}\] Now $AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}$. Thus the answer is $\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}$.

~IronicNinja

Solution 3

Draw a nice big diagram and measure. (Note: this strategy should only be used as a last resort!)

See Also

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

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