Half-ellipses Intersecting

I choose a coordinate system such that $A:\ (-2, 0)$ ; $B:\ (0, 0)$ ; and so on.

Let $E_1$ be the half-ellipse from A to C; $E_2$ be the half-ellipse from B to D; $E_3$ the half-ellipse from B to E. Their equations are $\begin{cases}E_1: & x^2 + by^2 = 2^2 & \dfrac{dy}{dx} = -\dfrac{x}{by}. \\ E_2: & (x - 2)^2 + by^2 = 2^2 & \dfrac{dy}{dx} = -\dfrac{x-2}{by}. \\ E_3: & (x - 3)^2 + by^2 = 3^2 & \dfrac{dy}{dx} = -\dfrac{x-3}{by}.\end{cases}$ The value of $b$ is the same in each case because the half-ellipses are similar.

By symmetry we know that the intersection of $E_1$ and $E_2$ has $x = 1$ ; thus $1^2 + by^2 = 2^2\ \ \ \therefore\ \ \ by^2 = 3.$ Moreover, since $E_1$ and $E_2$ are symmetric and intersect at right angles, they must have $dy/dx = \pm1$ in the intersection point. Substituting this in the equations above we find $by = x = 1$ , so that $y = 3$ and $b = \tfrac13$ .

Now for the intersection of $E_1$ and $E_3$ . Subtracting their equations, we get $(x-3)^2 - x^2 = 3^2 - 2^2\ \ \ \therefore\ \ \ x = \tfrac23$ . Substituting this in the equations for $E_1$ and $E_3$ we get $y = 4\sqrt{2/3}$ . Thus $E_1$ and $E_3$ intersect in $(x,y) = (\tfrac23, 4\sqrt{2/3}).$

We will now find tangent vectors: $\vec v_1 = \left(1, \left.\frac{dy}{dx}\right|_{E_1}\right) = \left(1, -\frac{\tfrac23}{\tfrac13\cdot 4\sqrt{2/3}}\right) \sim (4\sqrt 2, -2 \sqrt 3); \\ \vec v_2 = \left(1, \left.\frac{dy}{dx}\right|_{E_2}\right) = \left(1, -\frac{\tfrac23-3}{\tfrac13\cdot 4\sqrt{2/3}}\right) \sim (4\sqrt 2, 7 \sqrt 3).$ The cosine of the intersection follows from their dot product: $\cos\theta = \frac{\vec v_1\cdot \vec v_2}{|\vec v_1|\ |\vec v_2|} = \frac{(4\sqrt 2)^2 - 2\sqrt 3\cdot 7\sqrt 3}{\sqrt{((4\sqrt 2)^2 + (2\sqrt 3)^2)((4\sqrt 2)^2 + (7 \sqrt 3)^2)}} \\ = \frac{32 - 42}{\sqrt{44\cdot 179}} = \frac{-10}{\sqrt{44\cdot 179}} = \frac{-5}{\sqrt{11\cdot 179}} = -\frac{5}{\sqrt{1969}}.$ This is the obtuse angle; for the acute angle we drop the negative sign.

Thus the answer is $5 + 1969 = \boxed{1974}$ .