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 $m+n.$
1. **Transform the ellipse to a circle:**
   The given ellipse is \( x^2 + 4y^2 = 4 \). We can rewrite it as:
   \[
   \frac{x^2}{4} + y^2 = 1
   \]
   This represents an ellipse with semi-major axis \( a = 2 \) and semi-minor axis \( b = 1 \).

   To transform this ellipse into a circle, we perform a scaling transformation. Let \( x' = \frac{x}{2} \) and \( y' = y \). Substituting these into the ellipse equation, we get:
   \[
   \left(\frac{x'}{1}\right)^2 + y'^2 = 1
   \]
   which is the equation of a circle with radius 1.

2. **Properties of the equilateral triangle:**
   The equilateral triangle is inscribed in the ellipse, and one vertex is at \( (0, 1) \). After the transformation, this vertex maps to \( (0, 1) \) in the circle.

3. **Area considerations:**
   The area of the original ellipse is:
   \[
   E = \pi \cdot a \cdot b = \pi \cdot 2 \cdot 1 = 2\pi
   \]
   The area of the transformed circle is:
   \[
   E' = \pi \cdot 1^2 = \pi
   \]
   The area of the equilateral triangle inscribed in the ellipse is:
   \[
   A = \frac{s^2 \sqrt{3}}{4}
   \]
   After transformation, the area of the isosceles triangle inscribed in the circle is:
   \[
   A' = \frac{A}{2} = \frac{s^2 \sqrt{3}}{8}
   \]

4. **Finding the side length \( s \):**
   The isosceles triangle inscribed in the circle has a base \( b \) and height \( b \sqrt{3} \). The circumradius of the circle is 1, so we can form a right triangle with the circumradius and the base. The height of that triangle is \( b \sqrt{3} - 1 \) and the hypotenuse is 1.

   Using the Pythagorean theorem:
   \[
   (b \sqrt{3} - 1)^2 + b^2 = 1
   \]
   Expanding and simplifying:
   \[
   3b^2 - 2b \sqrt{3} + 1 + b^2 = 1
   \]
   \[
   4b^2 - 2b \sqrt{3} = 0
   \]
   \[
   2b (2b - \sqrt{3}) = 0
   \]
   Solving for \( b \):
   \[
   b = \frac{\sqrt{3}}{2}
   \]

5. **Calculating the side length \( s \):**
   The side length of the equilateral triangle is \( 2b \):
   \[
   s = 2b = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
   \]

6. **Expressing \( s \) in the form \( \sqrt{\frac{m}{n}} \):**
   \[
   s = \sqrt{3} = \sqrt{\frac{3}{1}}
   \]
   Here, \( m = 3 \) and \( n = 1 \).

7. **Finding \( m + n \):**
   \[
   m + n = 3 + 1 = 4
   \]

The final answer is \( \boxed{4} \)