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2001 AIME I Problems/Problem 5

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Problem

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.

Contents

Solution

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Solution 1

Denote the vertices of the triangle A,B, and C, where B is in quadrant 4 and C is in quadrant 3.

Note that the slope of \overline{AC} is \tan 60^\circ = \sqrt {3}. Hence, the equation of the line containing \overline{AC} is y = x\sqrt {3} + 1. This will intersect the ellipse when \begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\& = & x^{2} + 4(3x^{2} + 2x\sqrt {3... Since the triangle is symmetric with respect to the y-axis, the coordinates of B and C are now \left(\frac {8\sqrt {3}}{13},y_{0}\right) and \left(\frac { - 8\sqrt {3}}{13},y_{0}\right), respectively, for some value of y_{0}.

Since we're going to use the distance formula, the value of y_{0} is irrelevant. Our answer is BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.

Solution 2

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 and simplifying through algebra yields x^2=192/169, so 2x=\sqrt{768/169} and the answer is \boxed{937}.

See also

2001 AIME I (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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