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Difference between revisions of "2017 AMC 12B Problems/Problem 18"

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Thus the area of <math>\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>.
 
Thus the area of <math>\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>.
  
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== Solution 2b: Area shortcut ==
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Because <math>AE</math> is <math>\sqrt{74}</math> and <math>AB</math> is <math>4</math>, the ratio of the sides is <math>\frac{\sqrt{74}}{4}</math>, meaning the ratio of the areas is thus <math>{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}</math>. We then have the proportion <math>\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}</math>
  
 
==Solution 3: Similar triangles without Pythagorean==
 
==Solution 3: Similar triangles without Pythagorean==

Revision as of 19:06, 28 June 2019

Problem

The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$


Solution 1

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.865514650638614cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */ draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); /* draw figures */ draw(circle((0.,0.), 2.)); draw((-2.,0.)--(5.,5.)); draw((5.,5.)--(5.,0.)); draw((5.,0.)--(-2.,0.)); draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); draw((2.,0.)--(-2.,0.)); draw((2.,0.)--(5.,5.)); draw((0.,0.)--(5.,5.)); /* dots and labels */ dot((0.,0.),dotstyle); label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); dot((-2.,0.),dotstyle); label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); dot((2.,0.),dotstyle); label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); dot((5.,0.),dotstyle); label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); dot((5.,5.),dotstyle); label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor); dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}$. Now $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}$. Since $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.


Solution 2: Similar triangles with Pythagorean

$AB$ is the diameter of the circle, so $\angle ACB$ is a right angle, and therefore by AA similarity, $\triangle ACB \sim \triangle ADE$.

Because of this, $\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}$, so $AC = \frac{28}{\sqrt{74}}$.

Likewise, $\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}$, so $BC = \frac{20}{\sqrt{74}}$.

Thus the area of $\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.

Solution 2b: Area shortcut

Because $AE$ is $\sqrt{74}$ and $AB$ is $4$, the ratio of the sides is $\frac{\sqrt{74}}{4}$, meaning the ratio of the areas is thus ${(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}$. We then have the proportion $\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}$

Solution 3: Similar triangles without Pythagorean

Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:

Draw $BF \parallel ED$ with $F$ on $AE$. $BF=5\times\frac{4}{7}=\frac{20}{7}$.

$[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}$.

$AC:CB:CF=49:35:25$. ($7:5$ ratio applied twice)

$[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}$.


See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AMC 12 Problems and Solutions

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