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

Revision as of 20:17, 11 April 2009

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

$[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("$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 $BD=AD$ , 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 $\boxed{141}$ .

@@ Line 1: / Line 1: @@
-==Solution==
+== Problem ==
+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 ==
 <center><asy>
 pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5);

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

Revision as of 20:17, 11 April 2009

Problem

Solution