Difference between revisions of "2009 AIME II Problems/Problem 3"
Aimesolver (talk | contribs) (→Solution) |
(→Solution) |
||
Line 2: | Line 2: | ||
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>. | 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 == | ||
+ | |||
+ | === 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 18: | Line 20: | ||
</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>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>. | 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>. | ||
+ | |||
+ | === Solution 2=== | ||
+ | Let <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>. Then the slopes of <math>\overline{AC}</math> and <math>\overline{BE}</math> must multiply to <math>-1</math>, that is, | ||
+ | <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>. | ||
== 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}} |
Revision as of 20:50, 19 March 2012
Problem
In rectangle ,
. Let
be the midpoint of
. Given that line
and line
are perpendicular, find the greatest integer less than
.
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]](http://latex.artofproblemsolving.com/b/d/a/bda5479bee464bcc5f5c02a387f2f7ed6129f333.png)
From the problem, and triangle
is a right triangle. As
is a rectangle, triangles
, and
are also right triangles. By
,
, and
, so
. This gives
.
and
, so
, or
, so
, or
, so the answer is
.
Solution 2
Let ,
,
, and
. Then
. Then the slopes of
and
must multiply to
, that is,
which implies that
or
. Therefore
so
.
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 |