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

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The answer is <math>(\text{B})</math>
 
The answer is <math>(\text{B})</math>
  
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==Solution 4==
  
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[asy]
 +
unitsize(2cm);
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defaultpen(linewidth(.8pt)+fontsize(10pt));
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dotfactor=4;
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pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
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pair D=(1,-2), E=(0,-2);
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draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D);
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draw(A--E); draw(B--D); draw((2,0)--(2,-1));
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draw(A--(2,-1)); draw(B--E);
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pair[] ps={A,B,C,D,E};
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dot(ps);
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label("<math>A</math>",A,N);
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label("<math>B</math>",B,N);
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label("<math>C</math>",C,W);
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label("<math>D</math>",D,S);
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label("<math>E</math>",E,S);
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label("<math>1</math>",(D--E),S);
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label("<math>1</math>",(A--B),N);
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label("<math>2</math>",(A--E),W);
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label("<math>\sqrt{5}</math>",(B--E),NW);[/asy]
  
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 +
 +
 +
 +
Let <math>L</math> be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let <math>K</math> be the top right corner of the top right unit square, where segment <math>ABL</math> is 2 units in length.
 +
 +
Because of the Pythagorean Theorem, since <math>AC = 2</math> and <math>LK</math> = 1, the diagonal of triangle <math>ALK</math> is <math>\sqrt{5}</math>.
 +
 +
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Triangle <math>ALK</math> is clearly a similar triangle to triangle <math>ABC</math>. Segment <math>AB</math> is the hypotenuse of triangle <math>ABC</math>. So, we can write down:
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<cmath>AK/AB = LK/BC</cmath>, which is equal to:
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<cmath>\frac{\sqrt{5}}/{1} = 1/BC</cmath>.  Solving this equation yields:
 +
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<cmath>BC = \frac{1}/{\sqrt{5}}</cmath>
 +
 +
By Pythagorean theorem, we can now find segment <math>AC</math>.
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<cmath>\frac{1}/{\sqrt{5}}^2 + AC^2 = 1</cmath> Solving this yields:
 +
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<cmath>AC^2 = \frac{4}/{5}</cmath>, so <math>AC = \frac{2}/{\sqrt{5}}</math>.
 +
 +
So then we can use <cmath>A = \frac{1}/{2} * b * a.</cmath>
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So  <cmath>A = \frac{1}/{2} * \frac{1}/{\sqrt{5}} *  \frac{2}/{\sqrt{5}}</cmath>
 +
 +
<cmath>= \frac{1}/{5} B)</cmath>
  
 
== See Also ==
 
== See Also ==

Revision as of 00:27, 17 October 2014

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 $(\text{B})$.

Solution 3

[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 $(\text{B})$

Solution 4

[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]



Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.

Because of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\sqrt{5}$.


Triangle $ALK$ is clearly a similar triangle to triangle $ABC$. Segment $AB$ is the hypotenuse of triangle $ABC$. So, we can write down:

\[AK/AB = LK/BC\], which is equal to: \[\frac{\sqrt{5}}/{1} = 1/BC\]. Solving this equation yields:

\[BC = \frac{1}/{\sqrt{5}}\]

By Pythagorean theorem, we can now find segment $AC$. \[\frac{1}/{\sqrt{5}}^2 + AC^2 = 1\] Solving this yields:

\[AC^2 = \frac{4}/{5}\], so $AC = \frac{2}/{\sqrt{5}}$.

So then we can use \[A = \frac{1}/{2} * b * a.\] So \[A = \frac{1}/{2} * \frac{1}/{\sqrt{5}} * \frac{2}/{\sqrt{5}}\]

\[= \frac{1}/{5} B)\]

See Also

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

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