Equilateral in an Ellipse

Let $x = AB = AC = BC$ , and draw $F_1F_2$ and $F_1C$ .

By symmetry, $\triangle BF_1F_2$ is also an equilateral triangle, so $\angle BF_1F_2 = \angle F_1F_2B = F_1BF_2 = 60°$ , and since the distance between focal points is $5$ , $F_1F_2 = BF_1 = BF_2 = 5$ .

Since $BC = x$ and $BF_2 = 5$ , $CF_2 = x - 5$ .

Since the sum of the distances between a point on an ellipse and the two focal points is a constant, $BF_1 + BF_2 = CF_1 + CF_2$ , or $5 + 5 = CF_1 + x - 5$ , which means $CF_1 = 15 - x$ .

By the law of cosines on $\triangle BCF_1$ , $(15 - x)^2 = x^2 + 5^2 - 2 \cdot x \cdot 5 \cdot \cos 60°$ , which solves to $x = 8$ , so that the perimeter is $P = 3x = 3 \cdot 8 = \boxed{24}$ .

Let the semi-major axis of the ellipse be $a$ , the semi-minor axis be $b$ and the eccentricity be $e$ . Then $2ae=5\implies ae=\dfrac {5}{2}$ .

$\tan 60\degree=\sqrt 3=\dfrac{b}{ae}=\dfrac{b}{\frac{5}{2}}\implies$

$b=\dfrac{5\sqrt 3}{2}\implies a=5\implies$ the equation of the ellipse is $3x^2+4y^2=75$ .

If the position coordinates of one end of the base of the triangle be $(p, q)$ , then

$3p^2+4\times 3(p+\frac{5}{2})^2=75\implies$

$p=0, q=\dfrac{5\sqrt 3}{2}$ and $p=-4,q=-\dfrac{3\sqrt 3}{2}$ .

Hence, length of each side of the triangle is

$\sqrt {16+(\frac{-3\sqrt 3}{2}-\frac{5\sqrt 3}{2})^2}=8$ ,

and the perimeter is $3\times 8=\boxed {24}$ .

Choose a coordinate system so that the ellipse has an equation of the form (x/a)^2 + (y/b)^2 = 1. Then, since the foci are 5 units apart, their coordinates are at (2.5, 0) and (-2.5, 0). Therefore, trigonometry yields b = 2.5 tan(60 degrees) = 2.5 sqrt(3). From this, the coordinates of B are (0, 2.5 sqrt(3)). The slope of the line containing BC is thus = (2.5 sqrt(3) - 0)/(0-(2.5)) = -sqrt(3). Then, the equation of the line containing segment BC is y = -sqrt(3) (x-2.5). Utilizing the equation for the foci of an ellipse allows us to discover that a = 5, so that the ellipse has equation (x/5)^2 + (y/(2.5 sqrt(3))^2 = 1. Substituting the equation of this line into the equation of the ellipse and simplifying shows that the coordinates of point C are (4, -1.5 sqrt(3)). By symmetry, the coordinates of A are (-4, -1.5 sqrt(3)). Therefore, the distance between A and C is 8 units, and the perimeter is 8*3 = 24 units.