## Problem A2

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


## Solution

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 $\cos \mathrm{BPC}=\cos \mathrm{APC} \cos \mathrm{BPC}-\sin \mathrm{APC} \sin$ BPC, so $\left(113-x^{2}\right) / 112=\left(74-x^{2}\right) / 79\left(89-x^{2}\right) / 80-\sqrt{ }\left(\left(1-\left(74-x^{2}\right)^{2} / 70^{2}\right)\left(1-\left(89-x^{2}\right)^{2} / 80^{2}\right)\right)$.

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 $\mathrm{x}=\sqrt{ } 129$.

## Alternative solution by Johannes Tang

Rotate the triangle about $\mathrm{C}$ through $60^{\circ}$. Let $\mathrm{P}$ go to $\mathrm{P}^{\prime}$. We have $\mathrm{AP}^{\prime}=7, \mathrm{CP}^{\prime}=8$ and angle $\mathrm{PCP}^{\prime}=60^{\circ}$, so $\mathrm{PP}^{\prime} \mathrm{C}$ is equilateral. Hence angle $\mathrm{CPP}^{\prime}=60^{\circ}$. Also $\mathrm{PP}^{\prime}=8$. Using the cosine formula on triangle $\mathrm{APP}^{\prime}$ we find angle $\mathrm{APP}^{\prime}=60^{\circ}$. Hence angle $\mathrm{APC}=120^{\circ}$. Now applying cosine formula to triangle APC, we get result.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-02.jpg?height=534&width=554&top_left_y=270&top_left_x=251)

Find the roots $r_{1}, r_{2}, r_{3}, r_{4}$ of the equation $4 x^{4}-a x^{3}+b x^{2}-$ $\mathrm{cx}+5=0$, given that they are positive reals satisfying $\mathrm{r}_{1} / 2+\mathrm{r}_{2} / 4+\mathrm{r}_{3} / 5+\mathrm{r}_{4} / 8=1$.

