2012 AMC 10A Problems/Problem 15
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)
Coordinate geo
Making the bottom right corner the origin, the will be the intersection of
and
solving for
we get
. This means that the height of the triangle is
. Because the base is 1 the area is
.-harsha1234
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/3/f/2/3f2962535eca035d3778df987a78828ddda1ef94.png)
intersects
at a right angle, so
. The hypotenuse of right triangle BED is
.
Since ,
.
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,
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
Solution 4
Let be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let
be the top right corner of the top right unit square, where segment
is 2 units in length.
Because of the Pythagorean Theorem, since and
= 1, the diagonal of triangle
is
.
Triangle is clearly a similar triangle to triangle
. Segment
is the hypotenuse of triangle
. So, we can write down:
, which is equal to:
Solving this equation yields:
By Pythagorean theorem, we can now find segment
Solving this yields:
, so
So then we can use
So
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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