Difference between revisions of "2011 AIME II Problems/Problem 2"

Line 1: Line 1:
Problem:
+
== 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.
 
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 ==
Solution: (Needs better solution, I cannot remember exactly how I got the side length)
+
Drawing the square and examining the given lengths,
 
+
<asy>
Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (see geoboard diagram [that thing is actually pretty cool])
+
size(2inch, 2inch);
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
+
currentpen = fontsize(8pt);
 
+
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
Area of the square is 810.
+
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>
  
<geogebra>bb56f64482bd0dabba65a4dcad9d52a05d440f19</geogebra>
+
Area of the square is <math>\fbox{810.}</math>

Revision as of 00:05, 3 April 2011

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 $\fbox{810.}$

Invalid username
Login to AoPS