## Task B-3.7.

The base of the pyramid is a right-angled triangle with sides of length 1, $a, a^{2}, a>1$. The apex of the pyramid is orthogonally projected onto the vertex of the right angle of the base. The acute angle opposite the side of length 1 is equal to the angle at which one lateral face is inclined to the base. Calculate the volume of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-13.jpg?height=474&width=377&top_left_y=2142&top_left_x=768)

By applying the Pythagorean theorem, we get

$$
\begin{aligned}
& a^{4}=a^{2}+1 \\
& a^{4}-a^{2}-1=0
\end{aligned}
$$

The positive solution to this biquadratic equation is

$$
a^{2}=\frac{1+\sqrt{5}}{2} \text { or } a=\sqrt{\frac{1+\sqrt{5}}{2}}
$$

If $\alpha$ is the angle opposite the leg of length 1, then

$$
\operatorname{tg} \alpha=\frac{1}{a} \text{ and } \operatorname{tg} \alpha=\frac{h}{v} \text { so } h=\frac{v}{a}
$$

In the right triangle, the length of the altitude to the hypotenuse is

$$
v=\frac{1 \cdot a}{a^{2}}=\frac{1}{a}
$$

Then

$$
\begin{aligned}
& h=\frac{v}{a}=\frac{1}{a^{2}} \\
& V=\frac{1}{3} B \cdot h=\frac{1}{3} \cdot \frac{a \cdot 1}{2} \cdot \frac{1}{a^{2}}=\frac{1}{6} \cdot \frac{1}{a} \\
& V=\frac{1}{6} \sqrt{\frac{2}{1+\sqrt{5}} \cdot \frac{1-\sqrt{5}}{1-\sqrt{5}}} \\
& V=\frac{1}{6} \sqrt{\frac{\sqrt{5}-1}{2}}
\end{aligned}
$$

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
January 5, 2016.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.