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Difference between revisions of "2017 AMC 12B Problems/Problem 18"
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==Solution==
 
==Solution==
  
−
[asy]
+
<asy>
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 
import graph; size(8.865514650638614cm);  
 
import graph; size(8.865514650638614cm);  
Line 28: Line 28:
 
  /* dots and labels */
 
  /* dots and labels */
 
dot((0.,0.),dotstyle);  
 
dot((0.,0.),dotstyle);  
−
label("<math>O</math>", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);  
+
label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);  
 
dot((-2.,0.),dotstyle);  
 
dot((-2.,0.),dotstyle);  
−
label("<math>A</math>", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);  
+
label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);  
 
dot((2.,0.),dotstyle);  
 
dot((2.,0.),dotstyle);  
−
label("<math>B</math>", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);  
+
label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);  
 
dot((5.,0.),dotstyle);  
 
dot((5.,0.),dotstyle);  
−
label("<math>D</math>", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);  
+
label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);  
 
dot((5.,5.),dotstyle);  
 
dot((5.,5.),dotstyle);  
−
label("<math>E</math>", (5.06574004507889,5.15104432757325), NE * labelscalefactor);  
+
label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor);  
 
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);  
 
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);  
−
label("<math>C</math>", (0.48271975957926694,2.100706235912847), NE * labelscalefactor);  
+
label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
  /* end of picture */
 
  /* end of picture */
−
[/asy]
+
</asy>
  
 
Let <math>O</math> be the center of the circle. Note that <math>EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}</math>. However, by Power of a Point, <math>(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}</math>, so <math>AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}</math>. Now <math>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}}</math>. Since <math>\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}}</math>.
 
Let <math>O</math> be the center of the circle. Note that <math>EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}</math>. However, by Power of a Point, <math>(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}</math>, so <math>AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}</math>. Now <math>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}}</math>. Since <math>\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}}</math>.

Revision as of 23:12, 16 February 2017

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

[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 $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.

See Also

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