$\mathrm{P}$ is a point inside the equilateral triangle $\mathrm{ABC}$ such that $\mathrm{PA}=5, \mathrm{~PB}=7, \mathrm{PC}=8$. Find $\mathrm{AB}$.
Answer: $\sqrt{ } 129$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-01.jpg?height=363&width=389&top_left_y=1469&top_left_x=248)

Let the side length be $\mathrm{x}$. Using the cosine formula, we have $\cos \mathrm{APB}=\left(74-\mathrm{x}^{2}\right) / 70, \cos \mathrm{APC} = \left(89-x^{2}\right) / 80, \cos \mathrm{BPC}=\left(113-\mathrm{x}^{2}\right) / 112$. But $\cos \mathrm{BPC}=\cos \mathrm{APC} \cos \mathrm{BPC}-\sin \mathrm{APC} \sin \mathrm{BPC}$, so $\left(113-x^{2}\right) / 112=\left(74-x^{2}\right) / 70 \cdot \left(89-x^{2}\right) / 80 - \sqrt{\left(1-\left(74-x^{2}\right)^{2} / 70^{2}\right)\left(1-\left(89-x^{2}\right)^{2} / 80^{2}\right)}$.

We isolate the square root term, then square. We multiply through by 25.256 .49 and, after some simplification, we get $x^{6}-138 x^{4}+1161 x^{2}=0$. Hence $x=0, \pm 3, \pm \sqrt{129}$. We discard the zero and negative solutions. $\mathrm{x}=3$ corresponds to a point $\mathrm{P}$ outside the triangle. So the unique solution for a point $\mathrm{P}$ inside the triangle is $\mathrm{x}=\sqrt{ } 129$.