Difference between revisions of "2009 AIME II Problems/Problem 3"
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<cmath>AD=100\sqrt2</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 == | == 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:17, 18 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:
=Solution 4
Draw and to form a parallelogram . Since , by the problem statement, so is right. Letting , we have and $AC=EX=\sqrt{100^2+(2y)^2$ (Error compiling LaTeX. ! Missing } inserted.). 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.