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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>.
 +
== 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 15: Line 19:
 
label("\(100\)",Q,W);
 
label("\(100\)",Q,W);
 
</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>BD=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>.
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 +
=== Solution 2===
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Let <math>x</math> be the ratio of <math>BC</math> to <math>AB</math>. On the coordinate plane, plot <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>. Furthermore, the slope of <math>\overline{AC}</math> is <math>x</math> and the slope of <math>\overline{BE}</math> is <math>-x/2</math>. They are perpendicular, so they multiply to <math>-1</math>, that is,
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<cmath>x\cdot-\frac{x}{2}=-1,</cmath>
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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>.
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 +
=== Solution 3 ===
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Similarly to Solution 2, let the positive x-axis be in the direction of ray <math>BC</math> and let the positive y-axis be in the direction of ray <math>BA</math>. Thus, the vector <math>BE=(x,100)</math> and the vector <math>AC=(2x,-100)</math> are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:
 +
 
 +
<cmath> x\cdot2x+(100)\cdot(-100)=2x^2-10000=0 </cmath>
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<cmath> 2x^2-10000=0\rightarrow x^2=5000</cmath>
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 +
Substituting AD/2 for x:
 +
<cmath>(AD/2)^2=5000\rightarrow AD^2=20000</cmath>
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<cmath>AD=100\sqrt2</cmath>
 +
 
 +
===Solution 4===
 +
<center><asy>
 +
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0);
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draw (A--B--C--D--cycle);
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pair E=(7,10);
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draw (B--E);
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draw (A--C);
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draw(C--X--E);
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label("\(E\)",E,N);
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label("\(A\)",A,NW);
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label("\(B\)",B,SW);
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label("\(C\)",C,SE);
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label("\(D\)",D,NE);
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label("\(X\)",X,S);
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label("\(100\)",Q,W);
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</asy></center>
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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.
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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
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<cmath> 100^2+ 100^2 + y^2 + 4y^2 = 9y^2</cmath>
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<cmath> 2\cdot 100^2 = 4y^2</cmath>
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<cmath>\frac{100^2}{2}=y^2</cmath>
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<cmath>\frac{100}{\sqrt{2}}=y</cmath>
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<cmath>\frac{100\sqrt{2}}{2}=y</cmath>
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<cmath>100\sqrt{2}=2y=AD</cmath>, so the answer is <math>\boxed{141}</math>.
 +
 
 +
== See Also ==
 +
 
 +
{{AIME box|year=2009|n=II|num-b=2|num-a=4}}
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[[Category: Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 15:47, 2 August 2023

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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