Difference between revisions of "2012 AMC 10A Problems/Problem 15"
(→Solution 3) |
() |
||
Line 137: | Line 137: | ||
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>. | ||
The answer is <math>(\text{B})</math> | The answer is <math>(\text{B})</math> | ||
+ | == Solution 3 == | ||
+ | |||
+ | <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); | ||
+ | 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></center> | ||
+ | |||
+ | Triangle <math>EAB</math> is similar to triangle <math>EHI</math>; line <math>HI = 1/2</math> | ||
+ | |||
+ | Triangle <math>ACB</math> is similar to triangle <math>FCI</math> and the ratio of line <math>AB</math> to line <math>IF = 1 : \frac{3}{2} = 2: 3</math>. | ||
+ | |||
+ | Based on similarity the length of the height of <math>GC</math> is thus <math>\frac{2}{5}\cdot1 = \frac{2}{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>. | ||
+ | The answer is <math>(\text{B})</math> | ||
+ | |||
+ | |||
+ | |||
+ | |||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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 ?
Solution 1
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
Let be the origin. Then,
$\widebar{EB}$ (Error compiling LaTeX. ! Undefined control sequence.) 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 .
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
Solution 3
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.