Problem 2. Consider a regular tetrahedron as shown in the figure. If point E moves along the edge $A B$, when is the angle $\widehat{C E D}$ maximized?

![](https://cdn.mathpix.com/cropped/2024_04_26_bc48ef325a8049c5ff0ag-1.jpg?height=625&width=691&top_left_y=1969&top_left_x=725)
Solution. Suppose the tetrahedron has an edge length of 1, let $\alpha$ be the angle $C E D$ and $x$ the length of the segment $A E$.

If we apply the Law of Cosines to the triangle $A E C$, we have:

$$
E C^{2}=x^{2}+1-2 x \cos 60^{\circ}=x^{2}+1-x
$$

By symmetry, the length $E C$ is equal to the length $E D$. Again by the Law of Cosines, now for the triangle $E C D$, we have

$$
\begin{gathered}
1=\left(x^{2}-x+1\right)+\left(x^{2}-x+1\right)-2\left(\sqrt{x^{2}-x+1}\right)^{2} \cos \alpha \\
1=2\left(x^{2}-x+1\right)(1-\cos \alpha)
\end{gathered}
$$

Solving for $\cos \alpha$, we get:

$$
\cos \alpha=1-\frac{1}{2\left(x^{2}-x+1\right)}
$$

Since the cosine function is decreasing in the first quadrant, to find the maximum value of $\alpha$, we need to find the value of $x \in[0,1]$ that minimizes the function given in (1), or equivalently, that minimizes the denominator $x^{2}-x+1$. It is evident that this minimum is achieved for $x=\frac{1}{2}$.

The answer is: when the point $E$ is the midpoint of the side $A B$. We can calculate the value of $\alpha$ in this case; we have: $\alpha=\arccos \frac{1}{3}$.