In the tetrahedron $ABCD$, the medians from vertices $A$, $B$, and $C$ are pairwise perpendicular, and the edge $DA$ is perpendicular to the face $ABC$. Calculate the ratio of the lengths of the edges and medians of the tetrahedron.
I. Solution. We will show that the orthogonal projection $S^{\prime}$ of the centroid $S$ of the tetrahedron onto the plane $ABC$ is the orthocenter of the triangle $ABC$ and that the triangle is isosceles: $AB = AC$.

![](https://cdn.mathpix.com/cropped/2024_05_02_39e060f565b3f4e6eb9cg-1.jpg?height=724&width=420&top_left_y=267&top_left_x=836)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_39e060f565b3f4e6eb9cg-1.jpg?height=710&width=583&top_left_y=1095&top_left_x=760)

Figure 2

Indeed, $AS$ is perpendicular to $SB$ and $SC$, and thus to any line in the plane they define, including $BC$. Therefore, the projection $AS^{\prime}$ is also perpendicular to $BC$, because $BC$ is perpendicular to the line $SS^{\prime}$ in the plane $ASS^{\prime}$, and thus to any line in this plane.

Similarly, it can be shown that $BS$ (and $CS^{\prime}$) is an altitude of the triangle $ABC$, so $S^{\prime}$ is the orthocenter.

According to the problem, $DA$ is perpendicular to the plane $ABC$, so it is parallel to $SS^{\prime}$, and thus they lie in the same plane. Then the centroid $S_{D}$ of the triangle $ABC$ as the intersection point of the line $DS$ with the plane $ABC$, which is the intersection line of the plane $ADS^{\prime}$ and the plane $ABC$, lies on the line $AS^{\prime}$. This line is therefore both a median and an altitude, and thus a line of symmetry. Hence, $AB = AC$.

Since the centroid $S$ satisfies $S_{D}S = \frac{1}{4}S_{D}D$, we have $S_{D}S^{\prime} = \frac{1}{4}S_{D}A = \frac{1}{6}AF_{a}$, where $F_{a}$ is the midpoint of side $BC$. If we choose $S_{D}S_{1}$ as the unit of measurement, then

$$
AF_{a} = 6, \quad AS^{\prime} = AS_{D} - S^{\prime}S_{D} = 3 = \frac{1}{2}AF_{a} = S^{\prime}F_{a}
$$

The triangles $AF_{a}S$ and $BCS$ are isosceles right triangles. The latter is because the tetrahedron is symmetric with respect to the plane $ADF_{a}$, which is perpendicular to the plane $ABC$ and contains $S$; the former is right-angled at $S$ and $SS^{\prime}$ is both an altitude and a median. Therefore,

$$
\begin{gathered}
SS^{\prime} = S^{\prime}A = 3, \quad AS = 3\sqrt{2} = SF_{a} = F_{a}B, \quad BC = 2F_{a}B = 6\sqrt{2} \\
SB = SC = \frac{BC}{\sqrt{2}} = 6; \quad AB = AC = \sqrt{AS^2 + SB^2} = \sqrt{54} = 3\sqrt{6}, \quad AD = 4SS^{\prime} = 12 \\
BD = CD = \sqrt{AB^2 + AD^2} = \sqrt{198} = 3\sqrt{22}
\end{gathered}
$$

Finally, the lengths of the medians $s_{A}, s_{B}, s_{C}, s_{D}$ from the vertices $A, B, C, D$ are

$$
s_{A} = \frac{4}{3}AS = 4\sqrt{2}, \quad s_{B} = s_{C} = \frac{4}{3}SB = 8, \quad s_{D} = 4S_{D}S = 4\sqrt{S_{D}S^{\prime 2} + SS^{\prime 2}} = 4\sqrt{10}
$$

Thus, the ratios of the sought distances (taking the $\frac{1}{\sqrt{2}}$ multiple of each obtained length)

$$
\begin{aligned}
& AB : AC : AD : BC : BD : CD : s_{A} : s_{B} : s_{C} : s_{D} = \\
& = 3\sqrt{3} : 3\sqrt{3} : 6\sqrt{2} : 6 : 3\sqrt{11} : 3\sqrt{11} : 4 : 4\sqrt{2} : 4\sqrt{2} : 4\sqrt{5}
\end{aligned}
$$

(The ratio of the edges in a simpler form is $\sqrt{3} : \sqrt{3} : \sqrt{8} : 2 : \sqrt{11} : \sqrt{11}$, and the ratio of the medians is $1 : \sqrt{2} : \sqrt{2} : \sqrt{5}$.)

II. Solution. Vectors can be well used for solving computational problems. 1 Let the unit vectors in the directions of $SA, SB, SC$ be denoted by $\mathbf{i}, \mathbf{j}, \mathbf{k}$. Then the position vectors of the vertices $A, B, C$ are

$$
\mathbf{a} = \overrightarrow{SA} = \frac{3}{4}s_{A}\mathbf{i}, \quad \mathbf{b} = \overrightarrow{SB} = \frac{3}{4}s_{B}\mathbf{j}, \quad \mathbf{c} = \overrightarrow{SC} = \frac{3}{4}s_{C}\mathbf{k}
$$

It is known that the position vector of the centroid $\mathbf{s}$ is the arithmetic mean of the position vectors of the vertices:

$$
\mathbf{s} = \frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d})
$$

In our case, this is the zero vector $\mathbf{0}$. Therefore,

$$
\mathbf{d} = -\mathbf{a} - \mathbf{b} - \mathbf{c} = -\frac{3}{4}(s_{A}\mathbf{i} + s_{B}\mathbf{j} + s_{C}\mathbf{k})
$$

and its length is $\frac{3}{4}s_{D}$. The vectors of the edges starting from $A$ are:

$$
\begin{gathered}
\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \frac{3}{4}(-s_{A}\mathbf{i} + s_{B}\mathbf{j}), \quad \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = \frac{3}{4}(-s_{A}\mathbf{i} + s_{C}\mathbf{k}) \\
\overrightarrow{AD} = \mathbf{d} - \mathbf{a} = \frac{3}{4}(2s_{A}\mathbf{i} + s_{B}\mathbf{j} + s_{C}\mathbf{k})
\end{gathered}
$$

Since $AD \perp AB$ and $AD \perp AC$, their dot products are zero:

$$
\begin{array}{ll}
\overrightarrow{AB} \cdot \overrightarrow{AD} = \frac{9}{16}(-2s_{A}^2 + s_{B}^2) = 0, & s_{B} = \sqrt{2}s_{A} \\
\overrightarrow{AC} \cdot \overrightarrow{AD} = \frac{9}{16}(-2s_{A}^2 + s_{C}^2) = 0, & s_{C} = \sqrt{2}s_{A}
\end{array}
$$

Therefore,

$$
\begin{aligned}
\mathbf{d}^2 & = \frac{9}{16}(s_{A}^2 + s_{B}^2 + s_{C}^2) = \frac{45}{16}s_{A}^2 = \frac{9}{16}s_{D}^2, \quad s_{D} = \sqrt{5}s_{A} \\
AB^2 & = \mathbf{a}^2 + \mathbf{b}^2 = \frac{27}{16}s_{A}^2 = AC^2, \quad BC^2 = \mathbf{b}^2 + \mathbf{c}^2 = \frac{9}{4}s_{A}^2 \\
AD^2 & = \frac{9}{16}(4s_{A}^2 + s_{B}^2 + s_{C}^2) = \frac{9}{2}s_{A}^2
\end{aligned}
$$

From these, the final result of the first solution can be easily derived.

Béla Pataki (Budapest, István I. Gymnasium, 2nd year student) Jenő Reiczigel (Budapest, M. Fazekas Secondary School, 2nd year student)

Note. The second figure shows the inner side of the net of the pyramid. [^0]

[^0]: ${ }^{1}$ The theorems used can be found, for example, in: György Hajós, Gyula Neukomm, János Surányi: Mathematical Competition Problems, Part II, 2nd, expanded edition, pp. 28-33, Tankönyvkiadó, Budapest, 1965.

    ${ }^{2}$ See, for example, Katalin Horvay, Lóránt Pálmay: Mathematics for the 