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

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<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 BED is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>.
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<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>

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
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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