Difference between revisions of "2012 AMC 10A Problems/Problem 15"
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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>(\text{B})</math>. | ||
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== Solution 3 == | == Solution 3 == | ||
Revision as of 13:24, 15 October 2013
Problem
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of ?
![[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]](http://latex.artofproblemsolving.com/3/4/4/344ad68ce1c966777c56c860c4fc9b0da8bb06cb.png)
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]](http://latex.artofproblemsolving.com/9/6/a/96a4225d39bd0f0c94ba3c7de1676da3c06c84d4.png)
intersects
at a right angle, so
. The hypotenuse of right triangle BED is
.
Since AC=2BC, .
is a right triangle so the area is just
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]](http://latex.artofproblemsolving.com/3/9/d/39df2e0f6418671630036f076cb8d2c2353be73f.png)
Let be the origin. Then,
$\widebar{EB}$ (Error compiling LaTeX. Unknown error_msg) can be represented by the line
Also,
can be represented by the line
Subtracting the second equation from the first gives us .
Thus,
.
Plugging this into the first equation gives us
.
Since ,
is
,
and
.
Thus, . The answer is
.
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]](http://latex.artofproblemsolving.com/9/d/e/9de7e0273e04ad8bd29e5a7990f53d6bafc21be6.png)
Triangle is similar to triangle
; line
Triangle is similar to triangle
and the ratio of line
to line
.
Based on similarity the length of the height of is thus
.
Thus, .
The answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.