2009 AIME II Problems/Problem 3

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 $\lfloor AD\rfloor=\boxed{141}$ .

2009 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2009 AIME II Problems/Problem 3

Contents

Problem

Solution

Solution 1

Solution 2

See Also