## Task 6 - 311236

Consider all pyramids $A B C S$ that satisfy the following conditions (1), (2), (3):

(1) The base $A B C$ of the pyramid has an area of 1.

(2) It holds that $A B=A C=S B=S C$.

(3) It holds that $B C=S A$.

Investigate whether among all pyramids that satisfy these conditions, there is one with the largest volume. If this is the case, determine the size of the angle $\angle B A C$ for such a pyramid.
Since the area $F$ of triangle $ABC$ is to be constantly equal to one, we have

$$
V=\frac{1}{3} F h=\frac{1}{3} h
$$

Thus, the volume is maximal when the height $h$ is maximal. Let $D$ be the midpoint of the segment $BC$. Then we have:

$$
\begin{gathered}
F=\frac{1}{2} b c=1 \\
c=\frac{2}{b}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2307.jpg?height=551&width=736&top_left_y=501&top_left_x=660)

Now consider the isosceles triangle $ADS$. Its area is

$$
\frac{1}{2} b h'=\frac{1}{2} c h
$$

where

$$
h'=\sqrt{c^{2}-\frac{1}{4} b^{2}}
$$

Thus, we obtain:

$$
h=\frac{b}{c} \sqrt{c^{2}-\frac{1}{4} b^{2}}
$$

Substituting $c$ yields:

$$
\begin{array}{r}
h=\frac{b^{2}}{2} \sqrt{\frac{4}{b^{2}}-\frac{1}{4} b^{2}} \\
h=\sqrt{b^{2}-\frac{1}{16} b^{6}}
\end{array}
$$

For $h$ and thus $V$ to be maximal, the term under the square root must be maximal. We set the first derivative to zero:

$$
2 b-\frac{3}{8} b^{5}=0
$$

$b=0$ clearly does not represent a maximum, so the solution is:

$$
\begin{aligned}
& b^{4}=\frac{16}{3} \\
& b=\frac{2}{\sqrt[4]{3}}
\end{aligned}
$$

For the angle $\beta$, we have:

$$
\begin{aligned}
\beta=2 \arctan \frac{b}{2 c} & =2 \arctan \frac{b^{2}}{4}=2 \arctan \frac{1}{\sqrt{3}} \\
\beta & =2 \cdot 30^{\circ}=60^{\circ}
\end{aligned}
$$

Then $b=a$. The volume is thus maximal when the pyramid is a regular tetrahedron.