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

Revision as of 01:18, 26 September 2020

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 $\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}$ . 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 $\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}$ .

Solution 3

WLOG, let $AC=CD=4$ , and $DE=EB=3$ . $\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}$ . Because of this, $\triangle DEC$ is a 3-4-5 right triangle. Draw the altitude $DF$ of $\triangle DEC$ . $DF$ is $\frac{12}{5}$ by the base-height triangle area formula. $\triangle ABC$ is similar to $\triangle DBF$ (AA). So $\frac{DF}{AC} = \frac{BD}{AB} = \frac35$ . $DB$ is $\frac35$ of $AB$ . Therefore, $AD:DB$ is $\boxed{\textbf{(A) } 2:3}$ .

~Thegreatboy90

Video Solution

https://youtu.be/_0YaCyxiMBo

~IceMatrix

@@ Line 22: / Line 22: @@
 ~Thegreatboy90
+==Video Solution==
+https://youtu.be/_0YaCyxiMBo
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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