Difference between revisions of "2009 AIME II Problems/Problem 3"
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− | Let <math>x</math> be the ratio of <math>BC</math> to <math>AB</math>. <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>, and the slopes of <math>\overline{AC}</math> and <math>\overline{BE}</math> must multiply to <math>-1</math>. In other words, | + | 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>, and the slopes of <math>\overline{AC}</math> and <math>\overline{BE}</math> must multiply to <math>-1</math>. In other words, |
<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>. |
Revision as of 19:51, 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
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 , and the slopes of and must multiply to . In other words, 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 |