Consider a regular triangular-based right pyramid. The length of the edges meeting at the apex is $a$, and the angle between any two of these edges is $\alpha$. Calculate the volume of the pyramid. If $a$ remains constant, what should $\alpha$ be for the volume to be maximized? ${ }^{1} \sqrt{1}$[^0]


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The base of the triangle

$$
b=2 a \sin \frac{\alpha}{2}
$$

and thus the area of the base

$$
T=a^{2} \sin ^{2} \frac{\alpha}{2} \sqrt{3}
$$

while the volume

$$
V=T \frac{h}{3}
$$

where $h$ is the height of the pyramid.

But

$$
h=\sqrt{a^{2}-R^{2}}
$$

where $R$ is the radius of the circumcircle of the base triangle.

$$
\begin{gathered}
R=\frac{b}{2 \sin 60^{\circ}}=\frac{b}{\sqrt{3}} \\
R=\frac{2 a \sin \frac{\alpha}{2}}{\sqrt{3}}
\end{gathered}
$$

and thus

$$
h=a \sqrt{\frac{3-4 \sin ^{2} \frac{\alpha}{2}}{3}}
$$

Therefore

$$
3 V=a^{3} \sin ^{2} \frac{\alpha}{2} \sqrt{3-4 \sin ^{2} \frac{\alpha}{2}}
$$

Since $V$ and $V^{2}$ are maximized at the same value, let's find the value of the latter.

$$
9 V^{2}=a^{6} \sin ^{4} \frac{\alpha}{2}\left(3-4 \sin ^{2} \frac{\alpha}{2}\right)
$$

or at the value of $\alpha^{\prime}$

$$
9 V^{\prime 2}=a^{6} \sin ^{4} \frac{\alpha^{\prime}}{2}\left(3-4 \sin ^{2} \frac{\alpha^{\prime}}{2}\right)
$$

thus

$$
\begin{gathered}
\frac{1}{\alpha^{6}}\left(9 V^{2}-9 V^{\prime 2}\right)=3\left(\sin ^{4} \frac{\alpha}{2}-\sin ^{4} \frac{\alpha^{\prime}}{2}\right)-4\left(\sin ^{6} \frac{\alpha}{2}-\sin ^{6} \frac{\alpha^{\prime}}{2}\right)= \\
=3\left(\sin ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha^{\prime}}{2}\right)\left(\sin ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha^{\prime}}{2}\right)-4\left(\sin ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha^{\prime}}{2}\right)\left(\sin ^{4} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\alpha^{\prime}}{2}+\sin ^{4} \frac{\alpha^{\prime}}{2}\right)
\end{gathered}
$$

Dividing this by $\left(\sin ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha^{\prime}}{2}\right)$ and setting $\alpha=\alpha^{\prime}$, we get

$$
6 \sin ^{2} \frac{\alpha}{2}-12 \sin ^{4} \frac{\alpha}{2}=d^{1}
$$

from which

$$
\sin ^{2} \frac{\alpha}{2}=0 \text { or } 1-2 \sin ^{2} \frac{\alpha}{2}=0
$$

From the first equation, $\alpha=0$, which clearly corresponds to a minimum, while from the second equation,

$$
\sin \frac{\alpha}{2}=\sqrt{\frac{1}{2}}
$$

or

$$
\frac{\alpha}{2}=45^{\circ} \alpha=90^{\circ}
$$

In this case,

$$
V_{\max }=\frac{a^{3}}{6}
$$

that is, when $\alpha=90^{\circ}$, or a right angle.

Meitner Elemér, V. ker. föreálisk. VIII. B. o. t., Budapest.