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

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 $\boxed{141}$ .

@@ Line 25: / Line 25: @@
 <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>.
-Nice solution
 === Solution 3 ===

2009 AIME II (Problems • Answer Key • Resources)
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