Difference between revisions of "2011 AIME II Problems/Problem 2"
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− | Problem | + | == Problem 2 == |
+ | 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>. | ||
− | + | == 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, (<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> | ||
− | + | Area of the square is <math>\fbox{810}</math>. | |
− | |||
− | + | ==See also== | |
− | + | {{AIME box|year=2011|n=II|num-b=1|num-a=3}} | |
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 16:02, 9 August 2018
Problem 2
On square , point lies on side and point lies on side , so that . Find the area of the square .
Solution
Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections. Therefore, ( being the side length), , or . Solving for , we get , and
Area of the square is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.