We attached three rods (whose thickness can be neglected) pairwise perpendicularly to each other at a common point at one of their ends. The lengths of the rods are 1, 2, and 3. We placed the resulting structure on the table so that the free ends of the rods touch the table surface. Determine the exact height at which the attachment point is above the table.

We attached three rods (whose thickness can be neglected) pairwise perpendicularly to each other at a common point at one of their ends. The lengths of the rods are 1, 2, and 3. We placed the resulting structure on the table so that the free ends of the rods touch the table surface. Determine the exact height at which the attachment point is above the table.
Solution. Together with the tabletop, the rods form a pyramid with a triangular base. The sides of the triangle defined by the points of contact with the tabletop can be calculated using the Pythagorean theorem due to the pairwise perpendicular rods:

$$
\sqrt{1^{2}+2^{2}}=\sqrt{5}, \quad \sqrt{1^{2}+3^{2}}=\sqrt{10}, \quad \sqrt{2^{2}+3^{2}}=\sqrt{13}
$$

The volume of a pyramid formed by pairwise perpendicular edges of lengths $a, b, c$ can be calculated as follows. The area of one of the faces is: $T=\frac{a b}{2}$, and the volume with the third edge as the height is: $V=\frac{T c}{3}=\frac{a b c}{6}$. In our case: $V=\frac{1 \cdot 2 \cdot 3}{6}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_016c9425cc5694069230g-1.jpg?height=204&width=442&top_left_y=489&top_left_x=825)

On the other hand, we calculate the area of the triangle with side lengths $\sqrt{5}, \sqrt{10}, \sqrt{13}$. We determine the cosine of the angle opposite the side of length $\sqrt{13}$ using the cosine rule:

$$
13=5+10-2 \sqrt{50} \cos \varphi, \quad \text { or } \quad \cos \varphi=\frac{1}{\sqrt{50}}
$$

Since $\sin ^{2} x+\cos ^{2} x=1$, and in a triangle $\sin x>0$, we have

$$
\sin \varphi=\sqrt{1-\frac{1}{50}}=\sqrt{\frac{49}{50}}=\frac{7}{\sqrt{50}}
$$

The area of the triangle is:

$$
t=\frac{\sqrt{5} \cdot \sqrt{10} \cdot \frac{7}{\sqrt{50}}}{2}=\frac{7}{2}
$$

The volume of the pyramid is: $V=\frac{t \cdot m}{3}$, from which the height is:

$$
m=\frac{3 V}{t}=\frac{3 \cdot 1}{\frac{7}{2}}=\frac{6}{7}
$$

Therefore, the fixation point is $\frac{6}{7}$ units above the table.