Difference between revisions of "2001 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
− | {{ | + | Solving for y in terms of x gives <math>y=\sqrt{4-x^2}/2</math>, so the two other points of the triangle are <math>(x,\sqrt{4-x^2}/2)</math> and <math>(-x,\sqrt{4-x^2}/2)</math>, which are a distance of 2x apart. Thus 2x equals the distance between <math>(x,\sqrt{4-x^2}/2)</math> and (0,1), so by the distance formula we have <math>2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}</math>. Squaring both sides, letting <math>x^2=n</math>, and simplifying gives <math>n=192/169</math>, so <math>2x=\sqrt{768/169}</math> and the answer is 937. |
== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=4|num-a=6}} | {{AIME box|year=2001|n=I|num-b=4|num-a=6}} |
Revision as of 19:57, 17 March 2008
Problem
An equilateral triangle is inscribed in the ellipse whose equation is . One vertex of the triangle is , one altitude is contained in the y-axis, and the length of each side is , where and are relatively prime positive integers. Find .
Solution
Solving for y in terms of x gives , so the two other points of the triangle are and , which are a distance of 2x apart. Thus 2x equals the distance between and (0,1), so by the distance formula we have . Squaring both sides, letting , and simplifying gives , so and the answer is 937.
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |