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Difference between revisions of "2017 AMC 12B Problems/Problem 24"

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==Problem==
 
==Problem==
  
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Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC</math> is similar to <math>\triangle BCD</math>, and <math>AB > BC</math>. There exists a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC</math> is similar to <math>\triangle CEB</math> and the area of Triangle <math>AED</math> is <math>17</math> times the area of Triangle <math>CEB</math>. Find <math>AB/BC</math>.
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Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>?
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<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math>
 
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math>
  
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==Solution==
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==Solution 1==
  
 
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:
 
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:
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[SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2).  
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==Solution 2==
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Draw line <math>FG</math> through <math>E</math>, with <math>F</math> on <math>BC</math> and <math>G</math> on <math>AD</math>, <math>FG \parallel AB</math>. WLOG let <math>CD=1</math>, <math>CB=x</math>, <math>AB=x^2</math>. By weighted average <math>FG=\frac{1+x^4}{1+x^2}</math>.
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Meanwhile, <math>FE:EG=[\triangle CBE]:[\triangle ADE]=1:17</math>.
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<math>FE=\frac{x^2}{1+x^2}</math>. We obtain <math>\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}</math>,
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namely <math>x^4-18x^2+1=0</math>.
  
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Meanwhile, FE:EG=[CBE]:[ADE]=1:17.
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The rest is the same as Solution 1.
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FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2),
 
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Namely x^4-18x^2+1=0.
 
  
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The rest is the same with solution 1.
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:01, 19 February 2017

Problem

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\tfrac{AB}{BC}$?

$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}$


Solution 1

Let $CD=1$, $BC=x$, and $AB=x^2$. Note that $AB/BC=x$. By the Pythagorean Theorem, $BD=\sqrt{x^2+1}$. Since $\triangle BCD \sim \triangle ABC \sim \triangle CEB$, the ratios of side lengths must be equal. Since $BC=x$, $CE=\frac{x^2}{\sqrt{x^2+1}}$ and $BE=\frac{x}{\sqrt{x^2+1}}$. Let F be a point on $\overline{BC}$ such that $\overline{EF}$ is an altitude of triangle $CEB$. Note that $\triangle CEB \sim \triangle CFE \sim \triangle EFB$. Therefore, $BF=\frac{x}{x^2+1}$ and $CF=\frac{x^3}{x^2+1}$. Since $\overline{CF}$ and $\overline{BF}$ form altitudes of triangles $CED$ and $BEA$, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle $BEC$ can be calculated, as it is a right triangle. Solving for each of these yields: \[[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))\] \[[ABCD]=[AED]+[DEC]+[CEB]+[BEA]\] \[(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]\] \[(x^3+x)/2=(20x^3)/(2(x^2+1))\] \[(x)(x^2+1)=20x^3/(x^2+1)\] \[(x^2+1)^2=20x^2\] \[x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5\] Therefore, the answer is $\boxed{\textbf{(D) } 2+\sqrt{5}}$


Solution 2

Draw line $FG$ through $E$, with $F$ on $BC$ and $G$ on $AD$, $FG \parallel AB$. WLOG let $CD=1$, $CB=x$, $AB=x^2$. By weighted average $FG=\frac{1+x^4}{1+x^2}$.

Meanwhile, $FE:EG=[\triangle CBE]:[\triangle ADE]=1:17$.

$FE=\frac{x^2}{1+x^2}$. We obtain $\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}$, namely $x^4-18x^2+1=0$.

The rest is the same as Solution 1.


See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 12 Problems and Solutions

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