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

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<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath>
 
<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath>
 
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math>
 
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math>
 +
 +
 +
[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).
 +
 +
Meanwhile, FE:EG=[CBE]:[ADE]=1:17.
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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 12:48, 17 February 2017

Problem

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC$ is similar to $\triangle BCD$, and $AB > BC$. There exists a point $E$ in the interior of $ABCD$ such that $\triangle ABC$ is similar to $\triangle CEB$ and the area of Triangle $AED$ is $17$ times the area of Triangle $CEB$. Find $AB/BC$. $\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}$

Solution

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//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2).

Meanwhile, FE:EG=[CBE]:[ADE]=1:17. FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), Namely x^4-18x^2+1=0.

The rest is the same with 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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