Difference between revisions of "2011 AIME II Problems/Problem 2"
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Revision as of 22:48, 4 July 2013
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,
![[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]](http://latex.artofproblemsolving.com/9/7/e/97e1902d72e5ce4625cf85ddc9e4d4562650ec51.png)
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. 
