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

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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>.
 
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>.
 
== Solution ==
 
== Solution ==
 +
 +
=== Solution 1===
 
<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);
Line 18: Line 20:
 
</asy></center>
 
</asy></center>
 
From the problem, <math>AB=100</math> and triangle <math>FBA</math> is a right triangle. As <math>ABCD</math> is a rectangle, triangles <math>BCA</math>, and <math>ABE</math> are also right triangles. By <math>AA</math>, <math>\triangle FBA \sim \triangle BCA</math>, and <math>\triangle FBA \sim \triangle ABE</math>, so <math>\triangle ABE \sim \triangle BCA</math>. This gives <math>\frac {AE}{AB}= \frac {AB}{BC}</math>. <math>AE=\frac{AD}{2}</math> and <math>BC=AD</math>, so <math>\frac {AD}{2AB}= \frac {AB}{AD}</math>, or <math>(AD)^2=2(AB)^2</math>, so <math>AD=AB \sqrt{2}</math>, or <math>100 \sqrt{2}</math>, so the answer is <math>\boxed{141}</math>.
 
From the problem, <math>AB=100</math> and triangle <math>FBA</math> is a right triangle. As <math>ABCD</math> is a rectangle, triangles <math>BCA</math>, and <math>ABE</math> are also right triangles. By <math>AA</math>, <math>\triangle FBA \sim \triangle BCA</math>, and <math>\triangle FBA \sim \triangle ABE</math>, so <math>\triangle ABE \sim \triangle BCA</math>. This gives <math>\frac {AE}{AB}= \frac {AB}{BC}</math>. <math>AE=\frac{AD}{2}</math> and <math>BC=AD</math>, so <math>\frac {AD}{2AB}= \frac {AB}{AD}</math>, or <math>(AD)^2=2(AB)^2</math>, so <math>AD=AB \sqrt{2}</math>, or <math>100 \sqrt{2}</math>, so the answer is <math>\boxed{141}</math>.
 +
 +
=== Solution 2===
 +
Let <math>A=(0,0)</math>, <math>B=(100,0)</math>, <math>C=(100,100x)</math>, and <math>D=(0,100x)</math>. Then <math>E=(0,50x)</math>. Then the slopes of <math>\overline{AC}</math> and <math>\overline{BE}</math> must multiply to <math>-1</math>, that is,
 +
<cmath>x\cdot-\frac{x}{2}=-1,</cmath>
 +
which implies that <math>-x^2=-2</math> or <math>x=\sqrt 2</math>. Therefore <math>AD=100\sqrt 2\approx 141.42</math> so <math>\lfloor AD\rfloor=\boxed{141}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2009|n=II|num-b=2|num-a=4}}

Revision as of 20:50, 19 March 2012

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

Solution 1

[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 $BC=AD$, 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 $\boxed{141}$.

Solution 2

Let $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Then the slopes of $\overline{AC}$ and $\overline{BE}$ must multiply to $-1$, that is, \[x\cdot-\frac{x}{2}=-1,\] which implies that $-x^2=-2$ or $x=\sqrt 2$. Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$.

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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