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Difference between revisions of "2019 AMC 10B Problems/Problem 16"
(→‎Solution 2)
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This may seem strange, but if you draw the diagram, the solution will work itself out like this.
 
This may seem strange, but if you draw the diagram, the solution will work itself out like this.
  
−
Now we see that <math><CDE = 180 - <ECD - <CED \Rightarrow 180 - 2x + 90 - 180 + 2x \Rightarrow 90</math>. 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 <math>4\sqrt{5}</math>x. Now, we find the cosine of 2y - this is <math>2cos^2x - 1</math>. which is <math>2*(\frac{1}{\sqrt{5}})^2 - 1 \Rightarrow \frac{-3}{5}</math> Using law of cosines on triangle BED, and denoting the length of BD as "d", we get <cmath>d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x)</cmath> <cmath>d^2 = 18x^2 + \frac{54x^2}{5} \Rightarrow {144x^2}{5}</cmath> <cmath>d = \frac{12x}{\sqrt{5}}</cmath> Since this is DB, and we know AB, to find the ratio we find AD, which is <math>\frac{4x}{\sqrt{5}} - \frac{12x}{\sqrt{5}}</math>, which is <math>\frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)\frac{2}{3}}</math>
+
Now we see that <math><CDE = 180 - <ECD - <CED \Rightarrow 180 - 2x + 90 - 180 + 2x \Rightarrow 90</math>. 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 <math>4\sqrt{5}</math>x. Now, we find the cosine of 2y - this is <math>2cos^2x - 1</math>. which is <math>2*(\frac{1}{\sqrt{5}})^2 - 1 \Rightarrow \frac{-3}{5}</math> Using law of cosines on triangle BED, and denoting the length of BD as "d", we get <cmath>d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x)</cmath> <cmath>d^2 = 18x^2 + \frac{54x^2}{5} \Rightarrow {144x^2}{5}</cmath> <cmath>d = \frac{12x}{\sqrt{5}}</cmath> Since this is DB, and we know AB, to find the ratio we find AD, which is <math>\frac{4x}{\sqrt{5}} - \frac{12x}{\sqrt{5}}</math>, which is <math>\frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)2:3}</math>
  
 
iron (note from me, if anyone wants to edit it to make it more clear/look better, that's fine with me - you can give yourself credit if you wish)
 
iron (note from me, if anyone wants to edit it to make it more clear/look better, that's fine with me - you can give yourself credit if you wish)

Revision as of 23:14, 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

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 $\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)2:3}$

iron (note from me, if anyone wants to edit it to make it more clear/look better, that's fine with me - you can give yourself credit if you wish)

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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