Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
Line 3: Line 3:
  
 
<math>\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}</math>
 
<math>\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}</math>
 +
  
 
==Solution 1==
 
==Solution 1==
Line 46: Line 47:
  
  
−
==Solution 2: Similar triangles==
+
==Solution 2: Similar triangles with Pythagorean==
 
<math>AB</math> is the diameter of the circle, so <math>\angle ACB</math> is a right angle, and therefore by AA similarity, <math>\triangle ACB \sim \triangle ADE</math>.
 
<math>AB</math> is the diameter of the circle, so <math>\angle ACB</math> is a right angle, and therefore by AA similarity, <math>\triangle ACB \sim \triangle ADE</math>.
  
−
Because of this, <math>\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{4}{\sqrt{7^2 + 5^2}}</math>, so <math>AC = \frac{28}{\sqrt{74}}</math>.
+
Because of this, <math>\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}</math>, so <math>AC = \frac{28}{\sqrt{74}}</math>.
  
−
Likewise, <math>\frac{CB}{DE} = \frac{AB}{AE} \Longrightarrow \frac{CB}{5} = \frac{4}{\sqrt{74}}</math>, so <math>CB = \frac{20}{\sqrt{74}}</math>.
+
Likewise, <math>\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{CB}{5} = \frac{4}{\sqrt{74}}</math>, so <math>CB = \frac{20}{\sqrt{74}}</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>.
 
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>.
  
  
 +
==Solution 3: Similar triangles without Pythagorean==
 
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:
 
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:
  
−
Draw BF// ED with D on AE. AE=5×(4/7)=20/7.
+
Draw <math>BF \parallel ED</math> with <math>F</math> on <math>AE</math>. <math>BF=5\times\frac{4}{7}=\frac{20}{7}</math>.
−
[ABF]=2×20/7=40/7.
+
 
−
AC:CB:CF=49:35:25.  (7/5 ration applied twice)
+
<math>[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}</math>.
−
[ABC]=49/(49+25)[ABF]=140/37.
+
 
 +
<math>AC:CB:CF=49:35:25</math>.  (<math>7:5</math> ratio applied twice)
 +
 
 +
<math>[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}</math>.
 +
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:46, 19 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 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, $\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{CB}{5} = \frac{4}{\sqrt{74}}$, so $CB = \frac{20}{\sqrt{74}}$.

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 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&oldid=84112"