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

(Created page with "== Problem == Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of <math>\triangle ABC</math>? <center><asy> unitsize(2cm); ...")
 
(Solution)
Line 23: Line 23:
 
<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>
  
== Solution ==
+
== Solution 1 ==
  
 
<center><asy>
 
<center><asy>
Line 57: Line 57:
  
 
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>
 
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>
 +
 +
== Solution 2 ==
 +
 +
<center><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></center>
 +
 +
Let <math>\text{E}</math> be the origin. Then,
 +
<math>\text{D}=(1, 0)</math>
 +
<math>\text{A}=(0, 2)</math>
 +
<math>\text{B}=(1, 2)</math>
 +
<math>\text{F}=(2, 1)</math>
 +
 +
<math>\bar{AB}</math> can be represented by the line <math>y=2x</math>
 +
Also, <math>\bar{AF}</math> can be represented by the line <math>y=-\frac{1}{2}x+2</math>
 +
 +
Subtracting the second equation from the first gives us <math>\frac{5}{2}x-2=0</math>.
 +
Thus, <math>x=\frac{4}{5}</math>.
 +
Plugging this into the first equation gives us <math>y=\frac{8}{5}</math>.
 +
 +
Since <math>\text{C} (0.8, 1.6)</math>, <math>G</math> is <math>(0.8, 2)</math>.
 +
 +
<math>\bar{AB}=1</math> and <math>\bar{CG}=0.4</math>.
 +
 +
Thus, <math>[ABC]=\frac{1}{2} \cdot \bar{AB} \cdot \bar{CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}. The answer is then </math>(\text{B})$.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2012|ab=A|num-b=14|num-a=16}}

Revision as of 22:17, 17 February 2012

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

$\bar{AB}$ can be represented by the line $y=2x$ Also, $\bar{AF}$ 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 $(0.8, 2)$.

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

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

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS