# 2001 AIME I Problems/Problem 5

## 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 $m+n$.

## Solution $[asy] pointpen = black; pathpen = black + linewidth(0.7); path e = xscale(2)*unitcircle; real x = -8/13*3^.5; D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */ D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle); [/asy]$

### 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 $\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*}$ 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 $y_{0}.$

It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$. Our answer is $$BC = 2*\frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.$$

### 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 $\boxed{937}$.

### Solution 3

Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$, and WLOG, we can say that $x$ is positive.

Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base): $\sqrt{x^2 + (y-1)^2} = 2x$

Square both sides, $x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$

Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$ $x^2 = 4-4y^2$ $3x^2 = 12-12y^2$

Substituting, $12-12y^2 = y^2 - 2y +1$

Moving stuff around and solving: $y = \frac{-11}{3}, 1$

The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be: $\sqrt{\frac{768}{169}}$

and so we get the desired answer of $\boxed{937}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 