## Task 4 - 100924

A regular right triangular pyramid is a pyramid whose base is an equilateral triangle and whose foot of the height coincides with the centroid of the base. In the regular pyramid with vertices $A, B, C, D$ and apex $D$, the inclination angle between each of the three side faces and the base is $60^{\circ}$. The base has a side length of $a$.

Calculate the volume $V$ of this pyramid!

Note: If two planes have a common edge and $P$ is a point on this edge different from the endpoints, then the angle formed by two perpendicular segments erected at $P$ on the edge and lying in the two planes is equal to the inclination angle of the two planes to each other.
For the volume $V$ of the pyramid with vertices $A, B, C, D$, we have $V=\frac{1}{3} G h$, where $G$ is the area of the base and $h$ is the height of the pyramid. According to the problem, the base is the area of the equilateral triangle $\triangle A B C$. For the area $G$ of this triangle, we have $G=\frac{a^{2}}{4} \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0940.jpg?height=443&width=332&top_left_y=247&top_left_x=862)

Let $F$ be the foot of the pyramid's height. Since $F$ coincides with the centroid of the triangle $\triangle A B C$ by assumption, the ray from $C$ through $F$ intersects the side $A B$ at its midpoint, denoted by $E$. Therefore, $C E$ is a median and, due to the equilateral nature of $\triangle A B C$, also the height of this triangle. Consequently, we have

$$
|A E|=|E B| \quad \text { (1) and } \quad|F E|=\frac{1}{3}|C E|=\frac{a}{6} \sqrt{3}
$$

Since $\triangle D F A \cong \triangle D F B$ (SAS), we have $|A D|=|B D|$, so $\triangle A B D$ is isosceles. Due to (1), $D E$ is the height of this triangle.

The angle $\angle F E D$ is therefore the inclination angle between the base and a side face of the pyramid, which is given as $60^{\circ}$ in the problem. Since $\angle E F D$ is a right angle, the area of the triangle $\triangle E F D$ can be considered as half the area of an equilateral triangle. Thus, we have

$$
|D E|=2|E F|=\frac{a}{3} \sqrt{3}
$$

By the Pythagorean theorem, we get

$$
h=|D F|=\sqrt{|D E|^{2}-|E F|^{2}}=\sqrt{\frac{a^{2}}{3}-\frac{a^{2}}{12}}=\frac{a}{2}
$$

Thus, the volume $V$ of the pyramid is

$$
V=\frac{1}{3} G h=\frac{a^{3}}{24} \sqrt{3}
$$

Adapted from $[5]$

## 2. Solution:

Let $M$ be the midpoint of the segment $A B$. Then the triangle $\triangle C A M$ is a right triangle with hypotenuse $C A$ and we have

$$
|C M|=\sqrt{|C A|^{2}-|A M|^{2}}=\sqrt{a^{2}-\frac{a^{2}}{4}}=\frac{\sqrt{3}}{2} a
$$

Thus, $A_{A B C}=\frac{1}{2} \cdot|C M| \cdot|A B|=\frac{\sqrt{3}}{4} a^{2}$.

Let $S$ be the centroid of the base. Since the centroid divides the medians in the ratio 2:1 and $C M$ is a median in the base, we have $|S M|=\frac{1}{3}|C M|=\frac{1}{2 \cdot \sqrt{3}} a$.

The three points $D, S$, and $M$ form a right triangle with a right angle at $S$ and $\angle S M D=60^{\circ}$, because both $S M$ and $D M$ are perpendicular to $A B$. (The triangle $\triangle A B D$ is isosceles with $|A D|=|B D|$, so the median $D M$ on $A B$ is also the height from $D$ to $A B$, making $D M$ perpendicular to $A B$.)

By the definition of the tangent in the right triangle $\triangle D M S$, we have $\tan \angle S M D=\frac{|S D|}{|S M|}$, so $|S D|=\tan 60^{\circ} \cdot|S M|=\sqrt{3} \cdot \frac{1}{2 \cdot \sqrt{3}} a=\frac{a}{2}$.

Thus, the volume $V$ of the pyramid is

$$
V=\frac{1}{3} \cdot A_{A B C} \cdot|S D|=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} \cdot \frac{a}{2}=\frac{\sqrt{3}}{24} a^{3}
$$