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Difference between revisions of "2011 AIME II Problems/Problem 2"
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Problem:
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== Problem 2 ==
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On [[square]] <math>ABCD</math>, point <math>E</math> lies on side <math>AD</math> and point <math>F</math> lies on side <math>BC</math>, so that <math>BE=EF=FD=30</math>. Find the area of the square <math>ABCD</math>.
  
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On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD.
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== Solution ==
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Drawing the square and examining the given lengths,
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<asy>
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size(2inch, 2inch);
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currentpen = fontsize(8pt);
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pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
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pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
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pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
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pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
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pair E = (0, 1); dot(E); label("$E$", E, plain.W);
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pair F = (3, 2); dot(F); label("$F$", F, plain.E);
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label("$\frac x3$", E--A);
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label("$\frac x3$", F--C);
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label("$x$", A--B);
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label("$x$", C--D);
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label("$\frac {2x}3$", B--F);
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label("$\frac {2x}3$", D--E);
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label("$30$", B--E);
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label("$30$", F--E);
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label("$30$", F--D);
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draw(B--C--D--F--E--B--A--D);
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</asy>
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you find that the three segments cut the square into three equal horizontal sections. Therefore, (<math>x</math> being the side length), <math>\sqrt{x^2+(x/3)^2}=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for <math>x</math>, we get <math>x=9\sqrt{10}</math>, and <math>x^2=810.</math>
  
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Area of the square is <math>\fbox{810}</math>.
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Solution: (Needs better solution, I cannot remember exactly how I got the side length)
 
  
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Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (I dont know how to make a diagram so somebody please insert one)
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==See also==
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Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for x, we get that x=9sqrt(10), and <math>x^2</math>=810
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{{AIME box|year=2011|n=II|num-b=1|num-a=3}}
  
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Area of the square is 810.
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 17:02, 9 August 2018

Problem 2

On square $ABCD$, point $E$ lies on side $AD$ and point $F$ lies on side $BC$, so that $BE=EF=FD=30$. Find the area of the square $ABCD$.

Solution

Drawing the square and examining the given lengths, [asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("$A$", A, plain.SW); pair B = (3, 0); dot(B); label("$B$", B, plain.SE); pair C = (3, 3); dot(C); label("$C$", C, plain.NE); pair D = (0, 3); dot(D); label("$D$", D, plain.NW); pair E = (0, 1); dot(E); label("$E$", E, plain.W); pair F = (3, 2); dot(F); label("$F$", F, plain.E); label("$\frac x3$", E--A); label("$\frac x3$", F--C); label("$x$", A--B); label("$x$", C--D); label("$\frac {2x}3$", B--F); label("$\frac {2x}3$", D--E); label("$30$", B--E); label("$30$", F--E); label("$30$", F--D); draw(B--C--D--F--E--B--A--D); [/asy] you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Solving for $x$, we get $x=9\sqrt{10}$, and $x^2=810.$

Area of the square is $\fbox{810}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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