A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.
The answer is $\sqrt{169 + 60\sqrt(3)}$.
If $s$ is the edge length of the triangle, and if $x$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 12, and if $y$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 13, according to the Law of Cosines, \begin{equation*} s^{2} = 169 - 120\cos(x) \end{equation*} \begin{equation*} s^{2} = 194 - 130\cos(y) \end{equation*} and since $\cos(360 - (x + y)) = \cos(x + y)$, \begin{equation*} s^{2} = 313 - 312\cos(x+y) \end{equation*} So, \begin{align*} \cos{x} = - \frac{s^{2} - 169}{120} , \\ \cos{y} = - \frac{s^{2} - 194}{130} , \\ \cos(x+y) = - \frac{s^{2} - 313}{312} . \end{align*} For any real numbers $\theta$ and $\phi$, \begin{equation*} \bigl[\cos{\theta}\cos{\phi} - \cos(\theta + \phi)\bigr]^{2} = \sin^{2}\theta\sin^{2}\phi = \bigl(1 - \cos^{2}\theta\bigr)\bigl(1 - \cos^{2}\phi\bigr) . \end{equation*} So, \begin{align*} &\left[\frac{s^{2} - 169}{120} \cdot \frac{s^{2} - 194}{130} + \frac{s^{2} - 313}{312}\right]^{2} \\ &\qquad \qquad = \left(1 - \left(\frac{s^{2} - 169}{120}\right)^{2}\right)\left(1 - \left(\frac{s^{2} - 194}{130}\right)^{2}\right) \end{align*} or equivalently, by multiplying by $(120^{2})(130^{2})312$, \begin{align*} &\Bigl[312(s^{2} - 169)(s^{2} - 194) + (120^{2})(130^{2})(s^{2} - 313)\Bigr]^{2} \\ &\qquad \qquad = 312 \Bigl((120^{2})(130^{2}) - 130^{2}(s^{2} - 194)\Bigr) \Bigl((120^{2})(130^{2}) - 120^{2}(s^{2} - 194)\Bigr) . \end{align*}
How is the quartic equation in the variable $s$ to be solved?