Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2019 AMC 10B Problems/Problem 16

Revision as of 03:33, 17 February 2019 by LauraZed (talk | contribs) (→‎See Also)

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

Solution 2

$AC=CD=4x$, and $DE=EB=3x$. (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as "t". Then $<ACD$ is $180-2t$ degrees, which implies that $<DCB$ is $2t - 90$ degrees. Similarly, the angle of point B is $90 - t$ degrees, which implies that $<BED$ is $2t$ degrees. This further implies that $<DEC$ is $180 - 2t$ degrees.

This may seem strange, but if you draw the diagram, the solution will work itself out like this.

Now we see that $<CDE = 180 - <ECD - <CED \Rightarrow 180 - 2x + 90 - 180 + 2x \Rightarrow 90$. Thus triangle CDE is a right triangle, with side lengths of 3x, 4x, and by the pythaogrean theorem, 5x. Now we see that AC is 4x (by definition), BC is 5x+3x = 8x, and AB is $4\sqrt{5}$x. Now, we find the cosine of 2y - this is $2cos^2x - 1$. which is $2*(\frac{1}{\sqrt{5}})^2 - 1 \Rightarrow \frac{-3}{5}$ Using law of cosines on triangle BED, and denoting the length of BD as "d", we get \[d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x)\] \[d^2 = 18x^2 + \frac{54x^2}{5} \Rightarrow {144x^2}{5}\] \[d = \frac{12x}{\sqrt{5}}\] Since this is DB, and we know AB, to find the ratio we find AD, which is $\frac{4x}{\sqrt{5}} - \frac{12x}{\sqrt{5}}$, which is $\frac{8x}{\sqrt{5}}$. Thus the answer is $\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)2:3}$

Solution 3

Draw a nice big diagram and measure. Only use as 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&oldid=103149"