### Instruction: 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 square of the length of each side is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.### Response: To solve this problem, I will use some properties of ellipses, equilateral triangles, and coordinate geometry.
First, I will find the other two vertices of the triangle by using the fact that the ellipse is symmetric about both axes.
Since one vertex is $(0,1)$, the other two must be $(\pm a, -\frac{1}{2})$, where $a$ is some positive number.
I can find $a$ by plugging these points into the equation of the ellipse.
If I plug in $(a, -\frac{1}{2})$ into $x^2+4y^2=4$, I get $a^2+4(\frac{1}{4})=4$, which simplifies to $a^2=3$.
Therefore, $a=\sqrt{3}$ and the other two vertices are $(\pm \sqrt{3}, -\frac{1}{2})$.
Next, I will use the fact that the altitude contained in the y-axis is the perpendicular bisector of the opposite side.
This means that the midpoint of the side with endpoints $(\pm \sqrt{3}, -\frac{1}{2})$ must lie on the y-axis.
The midpoint is $(0, -\frac{1}{2})$.
Now, I can use the distance formula to find the length of the side of the triangle.
The distance between $(0,1)$ and $(\pm \sqrt{3}, -\frac{1}{2})$ is the same, and it is equal to $\sqrt{(\pm \sqrt{3}-0)^2+( -\frac{1}{2}-1)^2}=\sqrt{3+\frac{9}{4}}=\sqrt{\frac{21}{4}}$.
Therefore, the square of the length of each side is $\frac{21}{4}$.
This means that $m=21$ and $n=4$.
Finally, I add $m$ and $n$ to get the answer.
$m+n=21+4=25$.