2001 AIME I Problems/Problem 5

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An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the length of each side is $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


Solving for y in terms of x gives $y=\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$, which are a distance of 2x apart. Thus 2x equals the distance between $(x,\sqrt{4-x^2}/2)$ and (0,1), so by the distance formula we have $2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}$. Squaring both sides, letting $x^2=n$, and simplifying gives $n=192/169$, so $2x=\sqrt{768/169}$ and the answer is 937.

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions
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