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

Revision as of 22:48, 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 $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$ .

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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Difference between revisions of "2017 AMC 12B Problems/Problem 18"

Revision as of 22:48, 16 February 2017

Problem

Solution

See Also

@@ Line 5: / Line 5: @@
 ==Solution==
-WORK IN PROGRESS
+[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("<math>O</math>", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);
+dot((-2.,0.),dotstyle);
+label("<math>A</math>", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);
+dot((2.,0.),dotstyle);
+label("<math>B</math>", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);
+dot((5.,0.),dotstyle);
+label("<math>D</math>", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);
+dot((5.,5.),dotstyle);
+label("<math>E</math>", (5.06574004507889,5.15104432757325), NE * labelscalefactor);
+dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);
+label("<math>C</math>", (0.48271975957926694,2.100706235912847), NE * labelscalefactor);
+clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
+ /* end of picture */
+[/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>.
 ==See Also==
 {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}
 {{MAA Notice}}