Question

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 $\sqrt{\frac{m}{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Since one altitude lies along the $y$ -axis, the other two vertices are symmetric: $(x, y)$ and $(-x, y)$ with $x \gt 0.$ The side from $(0,1)$ to $(x,y)$ makes a $120^\circ$ angle with the positive $x$ -axis, so it lies on the line $y = -\sqrt{3}\,x + 1.$

Substituting into $x^2 + 4y^2 = 4$ gives $x^2 + 4(1 - \sqrt{3}x)^2 = 4,$ which simplifies to $13x^2 - 8\sqrt{3}\,x = 0,$ so $x = \frac{8\sqrt{3}}{13}.$

The side length is $2x = \frac{16\sqrt{3}}{13},$ whose square is $\frac{768}{169}.$ Since $\gcd(768, 169) = 1,$ the answer is $768 + 169 = 937.$

2001 AIME I Problem 5

Solution:

Problem 5 in Other Years