Difference between revisions of "2009 AIME II Problems/Problem 3"
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<cmath>x\cdot-\frac{x}{2}=-1,</cmath> | <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>. | 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> | ||
== See Also == | == See Also == |
Revision as of 20:14, 16 August 2017
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:
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.