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

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<math>\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</math>
 
<math>\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</math>
  
==Solution==
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==Solution 1==
Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a 3-4-5 triangle with <math>CE = 5</math>.
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Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>.
  
Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a 1-2-<math>\sqrt{5}</math> triangle.
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Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a <math>1-2-\sqrt{5}</math> triangle.
  
On isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>.
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In isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>.
  
 
==Solution 2==
 
==Solution 2==
<math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as "t". Then <math><ACD</math> is <math>180-2t</math> degrees, which implies that <math><DCB</math> is <math>2t - 90</math> degrees. Similarly, the angle of point B is <math>90 - t</math> degrees, which implies that <math><BED</math> is <math>2t</math> degrees. This further implies that <math><DEC</math> is <math>180 - 2t</math> degrees.  
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Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>.  
  
This may seem strange, but if you draw the diagram, the solution will work itself out like this.
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Now we see that <math>\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}</math>. Thus <math>\triangle CDE</math> is a right triangle, with side lengths of <math>3x</math>, <math>4x</math>, and <math>5x</math> (by the Pythagorean Theorem, or simply the Pythagorean triple <math>3-4-5</math>). Therefore <math>AC=4x</math> (by definition), <math>BC=5x+3x = 8x</math>, and <math>AB=4\sqrt{5}x</math>. Hence <math>\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1</math> (by the double angle formula), giving <math>2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}</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>
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By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\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}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>.
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==Solution 3==
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WLOG, let <math>AC=CD=4</math>, and <math>DE=EB=3</math>. <math>\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}</math>. Because of this, <math>\triangle DEC</math> is a 3-4-5 right triangle. Draw the altitude <math>DF</math> of <math>\triangle DEC</math>. <math>DF</math> is <math>\frac{12}{5}</math> by the base-height triangle area formula. <math>\triangle ABC</math> is similar to <math>\triangle DBF</math> (AA).  So <math>\frac{DF}{AC} = \frac{BD}{AB} = \frac35</math>. <math>DB</math> is <math>\frac35</math> of <math>AB</math>. Therefore, <math>AD:DB</math> is <math>\boxed{\textbf{(A) } 2:3}</math>.
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~Thegreatboy90
 +
 
 +
==Video Solution==
 +
https://youtu.be/_0YaCyxiMBo
 +
 
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest 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

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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