[ [Volume helps solve the problem]

Does a triangular pyramid exist whose heights are 1, 2, 3, and 6?

#
Let's prove that the area of any face of a tetrahedron is less than the sum of the areas of the other three faces. Indeed, the orthogonal projections of the other three faces of the tetrahedron onto the plane of the given face completely cover this face. The area of the orthogonal projection of a triangle onto a plane that is not parallel to it is equal to the area of the projected triangle multiplied by the cosine of the angle between the plane and the plane of the projected triangle. Therefore, the area of a face of the tetrahedron is less than the sum of the areas of the other three faces. Suppose now that there exists a tetrahedron with heights

$$
h_1=1, h_2=2, h_3=3, h_4=6
$$

Let \( S_1, S_2, S_3 \) and \( S_4 \) be the areas of the faces of the tetrahedron onto which these heights are dropped, and \( V \) be the volume of the tetrahedron. Then

$$
V=\frac{1}{3} S_1 \cdot h_1=\frac{1}{3} S_1, V=\frac{1}{3} S_2 \cdot h_2=\frac{2}{3} S_2,
$$

$$
V=\frac{1}{3} S_3 \cdot h_3=S_3, V=\frac{1}{3} S_4 \cdot h_4=2 S_4
$$

From these equalities, we find that

$$
S_1=6 S_4, S_2=3 S_4, S_3=2 S_4.
$$

Therefore,

$$
S_1=6 S_4=S_4+2 S_4+3 S_4=S_4+S_3+S_2
$$

which is impossible.

## Answer

No.