An equilateral triangle with unknown side length

I haven't looked at the inspiration video yet, but an obvious (to me) solution is via coordinate geometry.

Let $B = (0,0),\enspace C = (s,0),\enspace A= (s/2, s\sqrt{3}/2)$

Then $P$ must lie at the intersection of the circles centered at the vertices:

B: $\quad x^2 + y^2 = 13^2$

C: $\quad(x-s)^2 + y^2 = 14^2$

A: $\quad(x-s/2)^2 + (y - \sqrt{3}s/2)^2 = 15^2$

Three equations, three unknowns. solve simultaneously for $s$ :

$s = \sqrt{295 + 168\sqrt{3}},\quad$ so $295+168+3=\boxed{466}$

Now let's see how brutish this solution is compared to the inspiration video ...

Similar solution with @Fletcher Mattox 's but easier computation.

Let the centroid of equilateral $\triangle ABC$ be the origin $O(0,0)$ . Then $A \left(0, \frac s{\sqrt 3} \right)$ , $B\left(-\frac s2, - \frac s{2\sqrt 3} \right)$ , and $C\left(\frac s2, - \frac s{2\sqrt 3} \right)$ . Let $P(x,y)$ . Then

$\begin{cases} \begin{aligned} x^2 + \left(y-\frac s{\sqrt 3}\right)^2 & = 15^2 & ...(1) \\ \left(x + \frac s2 \right)^2 + \left(y+\frac s{2\sqrt 3}\right)^2 & = 13^2 & ...(2) \\ \left(x - \frac s2 \right)^2 + \left(y+\frac s{2\sqrt 3}\right)^2 & = 14^2 & ...(3) \end{aligned} \end{cases}$

$\begin{cases} (2)-(3): & - 2sx = 27 & \implies x = - \dfrac {27}{2s} \\ 2\times (1) - (2)-(3): & - 2\sqrt 3 sy = 85 & \implies y = - \dfrac {85}{2\sqrt 3s}\end{cases}$

From $(1)$ :

$\begin{aligned} \frac {27^2}{4s^2} + \left(-\frac {85}{2\sqrt 3s} - \frac s{\sqrt 3} \right)^2 & = 225 \\ \frac {2353}{3s^2} - \frac {85}3 + \frac {s^2}3 & = 225 \\ \frac {2353}{s^2} - 590 + s^2 & = 0 \\ s^4 - 590s^2 + 2353 & = 0 \end{aligned}$

$\begin{aligned} \implies s^2 & = 295 \pm 168\sqrt 3 & \small \blue{\text{Since }295 - 168\sqrt 3 \approx 4.015 \text{ is too small.}} \\ & = 295 + 168\sqrt 3 \end{aligned}$

Therefore $a+b+c = 295 + 168 + 3 = \boxed{466}$ .

As hinted in the video, rotate the entire diagram $60°$ clockwise:

Since $\angle BCP = \angle ACP'$ , and $\angle ACB = 60°$ from equilateral $\triangle ABC$ , that means $\angle PCP' = 60°$ . Also, since $CP = CP' = 14$ , $\triangle CPP'$ is an isosceles triangle, and since its vertex angle is $\angle PCP' = 60°$ , it is also an equilateral triangle, which means $PP' = CP = CP' = 14$ .

Let $\theta = \angle APP'$ . By the law of cosines on $\triangle APP'$ , $\cos \theta = \cfrac{PP'^2 + AP^2 - AP'^2}{2 \cdot PP' \cdot AP} = \cfrac{14^2 + 15^2 - 13^2}{2 \cdot 14 \cdot 15} = \cfrac{3}{5}$ , which means $\sin \theta = \sqrt{1 - \cos^2 \theta} = \cfrac{4}{5}$ .

That means $\cos \angle APC = \cos (\theta + 60°) = \cos \theta \cos 60° - \sin \theta \sin 60° = \frac{1}{10}(3 - 4\sqrt{3})$ .

Finally, by the law of cosines on $\triangle APC$ , $s^2 = CP^2 + AP^2 - 2 \cdot CP \cdot AP \cdot \cos \angle APC = 14^2 + 15^2 - 2 \cdot 14 \cdot 15 \cdot \frac{1}{10}(3 - 4\sqrt{3}) = 295 + 168\sqrt{3}$ .

Therefore, $a = 295$ , $b = 168$ , $c = 3$ , and $a + b + c = \boxed{466}$ .