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Difference between revisions of "2009 AIME II Problems/Problem 3"

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==Solution==
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== Problem ==
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In rectangle <math>ABCD</math>, <math>AB=100</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that line <math>AC</math> and line <math>BE</math> are perpendicular, find the greatest integer less than <math>AD</math>.
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== Solution ==
 
<center><asy>
 
<center><asy>
 
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5);
 
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5);

Revision as of 21:17, 11 April 2009

Problem

In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$.

Solution

[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(F\)",F,W); label("\(100\)",Q,W); [/asy]

From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}{BC}$. $AE=\frac{AD}{2}$ and $BD=AD$, so $\frac {AD}{2AB}= \frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \sqrt{2}$, or $100 \sqrt{2}$, so the answer is $\boxed{141}$.

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