Difference between revisions of "2012 AMC 10A Problems/Problem 15"

Revision as of 12:15, 15 October 2013

Problem

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$ ?

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2)); draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--(0,-2)); pair[] ps={A,B,C}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); [/asy]$

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Solution 1

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$1$",(D--E),S); label("$1$",(A--B),N); label("$2$",(A--E),W); label("$\sqrt{5}$",(B--E),NW); [/asy]$

$AC$ intersects $BC$ at a right angle, so $\triangle ABC \sim \triangle BED$ . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$ .

$\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC$

$\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}$

Since AC=2BC, $BC=\frac{1}{\sqrt{5}}$ . $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}$

Solution 2

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); [/asy]$

Let $\text{E}$ be the origin. Then, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$

$\widebar{EB}$ (Error compiling LaTeX. Unknown error_msg) can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$

Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$ . Thus, $x=\frac{4}{5}$ . Plugging this into the first equation gives us $y=\frac{8}{5}$ .

Since $\text{C} (0.8, 1.6)$ , $G$ is $(0.8, 2)$ ,

${AB}=1$ and ${CG}=0.4$ .

Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}$ . The answer is $(\text{B})$ .

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 10 Problems and Solutions

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

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); label("$H$",H,W); label("$I$",I,E); [/asy]$

Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$

Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$ .

Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$ .

Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}$ . The answer is $(\text{B})$

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Difference between revisions of "2012 AMC 10A Problems/Problem 15"

Revision as of 12:15, 15 October 2013

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 113: / Line 113: @@
 pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
 pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
+pair H=(0,-1), I=(0.5,-1);
 draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
 draw(A--E); draw(B--D); draw((2,0)--F);
 draw(A--F); draw(B--E); draw(C--G);
-draw(H--(0,-1); draw(I--(0.5,-1);
+draw(H--I);
-pair[] ps={A,B,C,D,E,F,G};
+pair[] ps={A,B,C,D,E,F,G, H, I};
 dot(ps);
@@ Line 128: / Line 129: @@
 label("$F$",F,E);
 label("$G$",G,N);
+label("$H$",H,W);
+label("$I$",I,E);
 </asy></center>
+Triangle <math>EAB</math> is similar to triangle <math>EHI</math>; line <math>HI = 1/2</math>
+Triangle <math>ACB</math> is similar to triangle <math>FCI</math> and the ratio of line <math>AB</math> to line <math>IF = 1 : \frac{3}{2} = 2: 3</math>.
+Based on similarity the length of the height of <math>GC</math> is thus <math>\frac{2}{5}\cdot1 = \frac{2}{5}</math>.
+Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}</math>.
+The answer is <math>(\text{B})</math>