Difference between revisions of "2009 AIME II Problems/Problem 3"
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− | ==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 == | ||
+ | |||
+ | === 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> | + | 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>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, | ||
+ | <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>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | 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> | ||
+ | <cmath> 2x^2-10000=0\rightarrow x^2=5000</cmath> | ||
+ | |||
+ | Substituting AD/2 for x: | ||
+ | <cmath>(AD/2)^2=5000\rightarrow AD^2=20000</cmath> | ||
+ | <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); | ||
+ | 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></center> | ||
+ | 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. | ||
+ | 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 | ||
+ | <cmath> 100^2+ 100^2 + y^2 + 4y^2 = 9y^2</cmath> | ||
+ | <cmath> 2\cdot 100^2 = 4y^2</cmath> | ||
+ | <cmath>\frac{100^2}{2}=y^2</cmath> | ||
+ | <cmath>\frac{100}{\sqrt{2}}=y</cmath> | ||
+ | <cmath>\frac{100\sqrt{2}}{2}=y</cmath> | ||
+ | <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}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:47, 2 August 2023
Contents
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 be the ratio of
to
. On the coordinate plane, plot
,
,
, and
. Then
. Furthermore, the slope of
is
and the slope of
is
. They are perpendicular, so they multiply to
, that is,
which implies that
or
. Therefore
so
.
Solution 3
Similarly to Solution 2, let the positive x-axis be in the direction of ray and let the positive y-axis be in the direction of ray
. Thus, the vector
and the vector
are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:
Substituting AD/2 for x:
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]](http://latex.artofproblemsolving.com/3/1/5/3159bf6d6d813a950fc9c58e5e2e5b31158752e8.png)
Draw and
to form a parallelogram
. Since
,
by the problem statement, so
is right.
Letting
, we have
and
. Since
,
. Solving this, we have
, so the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.