201. Segments defining an equilateral triangle. Three segments of lengths 3, 4, and 5 cm connect an internal point \( P \) of an equilateral triangle to its vertices. What is the length of the side of this triangle?

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-046.jpg?height=391&width=460&top_left_y=564&top_left_x=798)
201. In triangle $A B C$, the ratio of the given three segments is $P C: P A: P B=3: 4: 5$. Construct an equilateral triangle $P C F$ such that points $P$ and $F$ are on opposite sides of $A C$. Draw segment $A F$. Drop a perpendicular $A E$ from point $A$ to the extension of segment $C P$.

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-141.jpg?height=388&width=326&top_left_y=888&top_left_x=868)

Notice that $\angle P C B=60^{\circ}-\angle P C A=\angle A C F$; therefore, triangle $P C B$ is congruent to triangle $F C A$, and $A F=B P=5$. Hence, triangle $A P F$ is a right triangle, from which $\angle A P E=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}$. Consequently, $A E=2$, and $E P=2 \sqrt{3}$. Therefore,

$$
A C=\sqrt{(2)^{2}+(3+2 \sqrt{3})^{2}}=\sqrt{25+12 \sqrt{3}} \approx 6.7664 \text{ cm}
$$

$[S . S . M ., \mathbf{3 3}, 450$ (April 1933).]