Ellipse and Circle

From $16x^2 + 25y^2 = 400$ $\implies \dfrac {x^2}{5^2} + \dfrac {y^2}{4^2} = 1$ . Therefore the major and minor semi-axes are $a=5$ and $b=4$ respectively. Let the distance between the center and a focus be $c$ . Then $c = \sqrt{a^2-b^2} = \sqrt{5^2-4^2} = 3$ . Therefore $F_1F_2 = 2c = 6$ .

Since $F_1F_2$ is a diameter of circle $B$ , then $\angle F_1BF_2 = 90^\circ$ as shown in the figure below.

Let $AB = x$ and $AF_1 = y$ , then $\cos (\angle F_1AF_2) = \dfrac xy = \dfrac mn$ . By the definition of ellipse :

$\begin{aligned} AF_1+AF_2 & = 2a = 10 \\ 5+x + y & = 10 \\ \implies x + y & = 5 & ...(1) \end{aligned}$

By Pythagorean theorem :

$\begin{aligned} AF_1^2 - AB^2 & = F_1F_2^2 - BF_2^2 \\ y^2 - x^2 & = 6^2 - 5^2 \\ (y-x)\color{#3D99F6}(y+x) & = 11 & \small \color{#3D99F6} \text{Note that }(1): \ x + y = 5 \\ \implies y - x & = \frac {11}5 \quad ... (2) \end{aligned}$

From $(1) + (2): \ 2y = 5 + \dfrac {11}5$ $\implies y = \dfrac {18}5$ and $x = \dfrac 75$ . $\implies \cos \angle F_1AF_2 = \dfrac xy = \dfrac 7{18}$ . Therefore, $m+n + mn = 7 + 18 + 7\times 18 = \boxed{151}$ .

By putting the ellipse in standard form, the semi-minor axis is $4$ and the semi-major axis is $5$ . The foci are a length of $3$ away from the center.

A Triangle can be constructed by connecting the upper endpoint of the semi-minor axis, the right focus, and the center. This results in the length from the upper endpoint of the semi-minor axis to the right focus being $5$ because of the Pythagorean theorem as the minor axis and major axis are perpendicular. By doing the same to the other focus, the length from the upper endpoint of the semi-minor axis to the left focus is also $5$ . Since the sum of lengths from a point on the ellipse to the two foci are the same, $AF1 + AF2 = 5 + 5 = 10$ .

Next, the construction in the following diagram can be drawn. Angle $F1BF2=90$ because $BF1$ and $BF2$ connect the endpoints of a diameter. Let $AF2=b$ and $AF1=a$ . Then, $AB=b-5$ since $BF2=5$ . Also, $F1F2=3+3=6$ since the length from one focus to the center is $3$ . By the pythagorean theorem, $BF1=√11$ . Because $AF1+AF2=a+b=10$ , the pythagorean theorem can be used again to determine the lengths of a and b since $11+(b-5)^2=a^2=(10-b)^2$ . By solving this, $b=6.4, b-5=6.4-5=1.4$ and $a=3.6$ . Since $ABF1$ is a right triangle, $\cos(θ)=\frac{1.4}{3.6}$ $=$ $\frac{7}{18}$ . As a result, $m=7$ and $n=18$ so $(7+18+(7*18))=151$ which is the final answer.