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Difference between revisions of "2009 AIME II Problems/Problem 3"
m (fixed typo)
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<cmath>AD=100\sqrt2</cmath>
 
<cmath>AD=100\sqrt2</cmath>
  
−
===Solution 4==
+
===Solution 4===
 
<center><asy>
 
<center><asy>
 
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0);
 
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0);
Line 53: Line 53:
 
</asy></center>
 
</asy></center>
 
Draw <math>CX</math> and <math>EX</math> to form a parallelogram <math>AEXC</math>. Since <math>EX \parallel AC</math>, <math>\angle BEX=90^\circ</math> by the problem statement, so <math>\triangle BEX</math> is right.  
 
Draw <math>CX</math> and <math>EX</math> to form a parallelogram <math>AEXC</math>. Since <math>EX \parallel AC</math>, <math>\angle BEX=90^\circ</math> by the problem statement, so <math>\triangle BEX</math> is right.  
−
Letting <math>AE=y</math>, we have <math>BE=\sqrt{100^2+y^2}</math> and <math>AC=EX=\sqrt{100^2+(2y)^2</math>. Since <math>CX=EA</math>, <math>\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2</math>. Solving this, we have  
+
Letting <math>AE=y</math>, we have <math>BE=\sqrt{100^2+y^2}</math> and <math>AC=EX=\sqrt{100^2+(2y)^2}</math>. Since <math>CX=EA</math>, <math>\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2</math>. Solving this, we have  
 
<cmath> 100^2+ 100^2 + y^2 + 4y^2 = 9y^2</cmath>
 
<cmath> 100^2+ 100^2 + y^2 + 4y^2 = 9y^2</cmath>
 
<cmath> 2\cdot 100^2 = 4y^2</cmath>
 
<cmath> 2\cdot 100^2 = 4y^2</cmath>
Line 60: Line 60:
 
<cmath>\frac{100\sqrt{2}}{2}=y</cmath>
 
<cmath>\frac{100\sqrt{2}}{2}=y</cmath>
 
<cmath>100\sqrt{2}=2y=AD</cmath>, so the answer is <math>\boxed{141}</math>.
 
<cmath>100\sqrt{2}=2y=AD</cmath>, so the answer is <math>\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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:18, 18 August 2017

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 $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$. They are perpendicular, so they 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 $\lfloor AD\rfloor=\boxed{141}$.

Solution 3

Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let the positive y-axis be in the direction of ray $BA$. Thus, the vector $BE=(x,100)$ and the vector $AC=(2x,-100)$ are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:

\[x\cdot2x+(100)\cdot(-100)=2x^2-10000=0\] \[2x^2-10000=0\rightarrow x^2=5000\]

Substituting AD/2 for x: \[(AD/2)^2=5000\rightarrow AD^2=20000\] \[AD=100\sqrt2\]

Solution 4

[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); draw(C--X--E); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(X\)",X,S); label("\(100\)",Q,W); [/asy]

Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \parallel AC$, $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right. Letting $AE=y$, we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have \[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\] \[2\cdot 100^2 = 4y^2\] \[\frac{100^2}{2}=y^2\] \[\frac{100}{\sqrt{2}}=y\] \[\frac{100\sqrt{2}}{2}=y\] \[100\sqrt{2}=2y=AD\], so the answer is $\boxed{141}$.

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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