\section*{Problem 3 - 151243}

Let \(P\) be a point inside an equilateral triangle \(A B C\), which has distances \(P A=u, P B=v, P C=w\) from the vertices of the triangle.

Calculate the side length \(s\) of the equilateral triangle \(A B C\) from \(u, v, w\).
}

First, the triangle \(A B C\) and the point \(P\) inside this triangle, with an angle of size \(\varphi=60^{\circ}\), are rotated around point \(A\). This results in the figure shown in the illustration.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1999.jpg?height=377&width=528&top_left_y=1599&top_left_x=770)

By assumption and construction, \(C=B^{\prime}\) and

\[
\begin{gathered}
\angle P A P^{\prime}=60^{\circ} \\
u=A P=A P^{\prime} \\
v=P B=P^{\prime} C
\end{gathered}
\]

Here, \(B^{\prime}\) is the image point of \(B\), and \(P^{\prime}\) is the image point of \(P\). From (2) it follows that triangle \(P A P^{\prime}\) is isosceles. From (1) and (2), by the base angle theorem:

\[
\angle A P P^{\prime}=\angle A P^{\prime} P
\]

By the interior angle theorem in the triangle:

\[
\angle P A P^{\prime}+\angle A P^{\prime} P+\angle P^{\prime} P A=180^{\circ}
\]

From (5) it follows with (1) and (4): \(\angle A P P^{\prime}=60^{\circ}\) (6). Thus, it is shown that triangle \(A P P^{\prime}\) is equilateral. Therefore, \(P P^{\prime}=u \quad(7)\).

Furthermore, set \(\angle P^{\prime} P C=\alpha\) and apply the cosine rule of plane trigonometry, taking into account (3) and (7), to triangle \(P^{\prime} P C\):

\[
\cos \alpha=\frac{u^{2}+w^{2}-v^{2}}{2 u w}
\]

Since \(P\) is an interior point of triangle \(A B C\), \(\angle A B C0\). Using the relationship \(\sin ^{2} \alpha+\cos ^{2} \alpha=1\), it follows that:

\[
\sin \alpha=\sqrt{1-\left(\frac{u^{2}+w^{2}-v^{2}}{2 u w}\right)^{2}}
\]

From this, through algebraic manipulation:

\[
\sin \alpha=\frac{1}{2 u w} \sqrt{\left(u^{2}+v^{3}+w^{2}\right)^{2}-2 u^{4}-2 v^{4}-2 w^{4}}
\]

With (8) and (10), it is now possible to apply the cosine rule to triangle \(A P C\) to express the length \(s\) of the triangle side \(A C\) in terms of \(u, v\), and \(w\). One obtains:

\[
s^{2}=u^{2}+w^{2}-2 u w \cos \left(\alpha+60^{\circ}\right)
\]

Since \(\cos \left(\alpha+60^{\circ}\right)=\cos \alpha \cos 60^{\circ}-\sin \alpha \sin 60^{\circ}, \cos 60^{\circ}=\frac{1}{2}\) and \(\sin 60^{\circ}=\frac{1}{2} \sqrt{3}\), it follows with (8), (10), and (11):

\[
s^{2}=\frac{1}{2}\left(u^{2}+v^{2}+w^{2}\right)+\sqrt{3} \cdot \sqrt{\left(u^{2}+v^{2}+w^{2}\right)^{2}-2\left(u^{4}+v^{4}+w^{4}\right)}
\]

Since the length \(s\) of the segment \(A C\) is positive, it follows from (12):

\[
s=\sqrt{\frac{1}{2}\left(u^{2}+v^{2}+w^{2}\right)+\sqrt{3} \cdot \sqrt{\left(u^{2}+v^{2}+w^{2}\right)^{2}-2\left(u^{4}+v^{4}+w^{4}\right)}}
\]

\section*{Adapted from \([3]\)}