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
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.