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

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<math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math>
 
<math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math>
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
−
 
 
== Solution 1 ==
 
== Solution 1 ==
  
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pair[] ps={A,B,C,D,E};
 
pair[] ps={A,B,C,D,E};
 
dot(ps);
 
dot(ps);
−
 
 
label("$A$",A,N);
 
label("$A$",A,N);
 
label("$B$",B,N);
 
label("$B$",B,N);
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</asy></center>
 
</asy></center>
  
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<math>AC</math> intersects <math>BC</math> at a right angle, so <math>\triangle ABC \sim \triangle BED</math>. The hypotenuse of right triangle BED is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>.
+
<math>AC</math> intersects <math>BC</math> at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so <math>\triangle ABC \sim \triangle BED</math>. The hypotenuse of right triangle <math>BED</math> is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>.
  
 
<cmath>\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC</cmath>
 
<cmath>\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC</cmath>
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<cmath>\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}</cmath>
 
<cmath>\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}</cmath>
  
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Since AC=2BC, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math>
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Since <math>AC=2BC</math>, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math>
  
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== Solution 2 ==
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== Solution 2 (coordbash)==
  
 
<center><asy>
 
<center><asy>
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<math>\text{F}=(2, 1)</math>
 
<math>\text{F}=(2, 1)</math>
  
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<math>\widebar{EB}</math> can be represented by the line <math>y=2x</math>
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<math>{EB}</math> can be represented by the line <math>y=2x</math>
 
Also, <math>{AF}</math> can be represented by the line <math>y=-\frac{1}{2}x+2</math>
 
Also, <math>{AF}</math> can be represented by the line <math>y=-\frac{1}{2}x+2</math>
  
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<math>{AB}=1</math> and <math>{CG}=0.4</math>.
 
<math>{AB}=1</math> and <math>{CG}=0.4</math>.
  
−
Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}</math>. The answer is <math>(\text{B})</math>.
+
Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}</math>. The answer is <math>\boxed{\textbf{(B)}\ \frac15}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
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Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{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>.  
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The answer is <math>(\text{B})</math>
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The answer is <math>\boxed{\textbf{(B)}\ \frac15}</math>
  
 
==Solution 4==
 
==Solution 4==
  
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[asy]
 
−
unitsize(2cm);
 
−
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);
 
−
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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<cmath>AK/AB = LK/BC</cmath>, which is equal to:
 
<cmath>AK/AB = LK/BC</cmath>, which is equal to:
−
<cmath>\frac{\sqrt{5}}/{1} = 1/BC</cmath>.  Solving this equation yields:
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<cmath>\frac{\sqrt{5}}{1} = \frac{1}{BC}</cmath> Solving this equation yields:
 +
 
 +
<cmath>BC = \frac{1}{\sqrt{5}}</cmath>
 +
 
 +
By Pythagorean theorem, we can now find segment <math>AC</math>
 +
<cmath>(\frac{1}{\sqrt{5}})^2 + AC^2 = 1</cmath> Solving this yields:
 +
 
 +
<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>
 +
So  <cmath>A = \frac{1}{2} * \frac{1}{\sqrt{5}} *  \frac{2}{\sqrt{5}}</cmath>
  
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<cmath>BC = \frac{1}/{\sqrt{5}}</cmath>
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<cmath>= \boxed{\textbf{(B)}\ \frac15}</cmath>
  
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By Pythagorean theorem, we can now find segment <math>AC</math>.
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==Video Solution==
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<cmath>\frac{1}/{\sqrt{5}}^2 + AC^2 = 1</cmath> Solving this yields:
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https://youtu.be/HVesU8cTjRU
  
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<cmath>AC^2 = \frac{4}/{5}</cmath>, so <math>AC = \frac{2}/{\sqrt{5}}</math>.
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~savannahsolver
  
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So then we can use <cmath>A = \frac{1}/{2} * b * a.</cmath>
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== Video Solution by OmegaLearn ==
−
So  <cmath>A = \frac{1}/{2} * \frac{1}/{\sqrt{5}} *  \frac{2}/{\sqrt{5}}</cmath>
+
https://youtu.be/4_x1sgcQCp4?t=1717
  
−
<cmath>= \frac{1}/{5} B)</cmath>
+
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:16, 23 November 2024

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, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) 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 (coordbash)

[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, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$

${EB}$ 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 $\boxed{\textbf{(B)}\ \frac15}$.

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 $\boxed{\textbf{(B)}\ \frac15}$

Solution 4

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: \[\frac{\sqrt{5}}{1} = \frac{1}{BC}\] 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 \[A = \frac{1}{2} * \frac{1}{\sqrt{5}} * \frac{2}{\sqrt{5}}\]

\[= \boxed{\textbf{(B)}\ \frac15}\]

Video Solution

https://youtu.be/HVesU8cTjRU

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=1717

~ pi_is_3.14

See Also

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

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