Four swallows are catching a fly. At first, the swallows are at the four vertices of a tetrahedron, and the fly is in its interior. Their maximal speeds are equal. Prove that the swallows can catch the fly.

Two segments slide along two skew lines. On each straight line there is a segment. Consider the tetrahedron with vertices at the endpoints of the segments. Prove that the volume of the tetrahedron does not depend on the position of the segments

Find out whether it is possible to determine the volume of a tetrahedron knowing the areas of its faces and its circumradius.

(a) Prove the parallelogram law that says that in a parallelogram the sum of the squares of the lengths of the four sides equals the sum of the squares of the lengths of the two diagonals. (b) The edges of a tetrahedron have lengths $a, b, c, d, e$ and $f$. The three line segments connecting the centers of intersecting edges have lengths $x, y$ and $z$. Prove that $$4 (x^2 + y^2 + z^2) = a^2 + b^2 + c^2 + d^2 + e^2 + f^2$$

On the tetrahedron $ ABCD $ make the notation $ M,N,P,Q, $ for the midpoints of $ AB,CD,AC, $ respectively, $ BD. $ Additionally, we know that $ MN $ is the common perpendicular of $ AB,CD, $ and $ PQ $ is the common perpendicular of $ AC,BD. $ Show that $ AB=CD, BC=DA, AC=BD. $

Tetrahedron $ABCD$ has $AB=5$, $AC=3$, $BC=4$, $BD=4$, $AD=3$, and $CD=\tfrac{12}5\sqrt2$. What is the volume of the tetrahedron? $\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2$

Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\epsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\epsilon$ from $AB$ and $CD$ is equal to $k$. Compute the ratio of the volumes of the two solids obtained.

The tetrahedron $OABC$ has all angles at vertex $O$ equal to $60^o$. Prove that $$AB \cdot BC + BC \cdot CA + CA \cdot AB \ge OA^2 + OB^2 + OC^2$$

Let $a,b$ be the lengths of two nonadjacent edges of a tetrahedron with inradius $r$. Prove that \[r<\frac{ab}{2(a+b)}.\]

Given four points $ A_1, A_2, A_3, A_4$ in the plane, no three collinear, such that \[ A_1A_2 \cdot A_3 A_4 \equal{} A_1 A_3 \cdot A_2 A_4 \equal{} A_1 A_4 \cdot A_2 A_3, \] denote by $ O_i$ the circumcenter of $ \triangle A_j A_k A_l$ with $ \{i,j,k,l\} \equal{} \{1,2,3,4\}.$ Assuming $ \forall i A_i \neq O_i ,$ prove that the four lines $ A_iO_i$ are concurrent or parallel. [i]Nikolai Ivanov Beluhov, Bulgaria[/i]

Points $A, B, C, D$ lie on the sphere of radius $1$. It is known that $AB\cdot AC\cdot AD\cdot BC\cdot BD\cdot CD=\frac{512}{27}$. Prove that $ABCD$ is a regular tetrahedron.

In the tetrahedron $SABC$, points $E, F, K, L$ are the midpoints of the sides $SA , BC, AC, SB$ respectively, . The lengths of the segments $EF$ and $KL$ are equal to $11 cm$ and $13 cm$ respectively, and the length of the segment $AB$ equals to $18 cm$. Find the length of the side $SC$ of the tetrahedron.

Cube $ABCDEFGH$, labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. [asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((0,10)--(4,13)--(14,13)--(10,10)); draw((10,0)--(14,3)--(14,13)); draw((0,0)--(4,3)--(4,13), dashed); draw((4,3)--(14,3), dashed); dot((0,0)); dot((0,10)); dot((10,10)); dot((10,0)); dot((4,3)); dot((14,3)); dot((14,13)); dot((4,13)); dot((14,8)); dot((5,0)); label("A", (0,0), SW); label("B", (10,0), S); label("C", (14,3), E); label("D", (4,3), NW); label("E", (0,10), W); label("F", (10,10), SE); label("G", (14,13), E); label("H", (4,13), NW); label("M", (5,0), S); label("N", (14,8), E); [/asy]

Let $ P$ be a point inside a regular tetrahedron $ T$ of unit volume. The four planes passing through $ P$ and parallel to the faces of $ T$ partition $ T$ into 14 pieces. Let $ f(P)$ be the joint volume of those pieces that are neither a tetrahedron nor a parallelepiped (i.e., pieces adjacent to an edge but not to a vertex). Find the exact bounds for $ f(P)$ as $ P$ varies over $ T.$

On the sides $(AB)$, $(BC)$, $(CD)$ and $(DA)$ of the regular tetrahedron $ABCD$, one considers the points $M$, $N$, $P$, $Q$, respectively Prove that $$MN \cdot NP \cdot PQ \cdot QM \ge AM \cdot BN \cdot CP \cdot DQ.$$

Tetrahedron $ABCD$ has sides of lengths, in increasing order, $7, 13, 18, 27, 36, 41$. If $AB=41$, then what is the length of $CD$?

Vectors coincide with the edges of an arbitrary tetrahedron (possibly non-regular). Is it possible for the sum of these six vectors to equal the zero vector? (Problem from Leningrad)

Prove that the product of the sines of two opposite dihedrals in a tetrahedron is proportional to the product of the lengths of the edges of these dihedrals.

In a tetrahedral $ABCD$, given that $\angle ADB=\angle BDC =\angle CDA=\frac{\pi}{3}$, $AD=BD=3$, and $CD=2$. Find the radius of the circumsphere of $ABCD$.

A tetrahedron is called a [i]MEMO-tetrahedron[/i] if all six sidelengths are different positive integers where one of them is $ 2$ and one of them is $ 3$. Let $ l(T)$ be the sum of the sidelengths of the tetrahedron $ T$. (a) Find all positive integers $ n$ so that there exists a MEMO-Tetrahedron $ T$ with $ l(T)\equal{}n$. (b) How many pairwise non-congruent MEMO-tetrahedrons $ T$ satisfying $ l(T)\equal{}2007$ exist? Two tetrahedrons are said to be non-congruent if one cannot be obtained from the other by a composition of reflections in planes, translations and rotations. (It is not neccessary to prove that the tetrahedrons are not degenerate, i.e. that they have a positive volume).

Rob is helping to build the set for a school play. For one scene, he needs to build a multi-colored tetrahedron out of cloth and bamboo. He begins by fitting three lengths of bamboo together, such that they meet at the same point, and each pair of bamboo rods meet at a right angle. Three more lengths of bamboo are then cut to connect the other ends of the first three rods. Rob then cuts out four triangular pieces of fabric: a blue piece, a red piece, a green piece, and a yellow piece. These triangular pieces of fabric just fill in the triangular spaces between the bamboo, making up the four faces of the tetrahedron. The areas in square feet of the red, yellow, and green pieces are $60$, $20$, and $15$ respectively. If the blue piece is the largest of the four sides, find the number of square feet in its area.

Six altitudes are constructed from the three vertices of the base of a tetrahedron to the opposite sides of the three lateral faces. Prove that all three straight lines joining two base points of the altitudes in each lateral face are parallel to a certain plane. (IF Sharygin, Moscow)

In the tetrahedron $ABCD$, the point $E$ is the projection of the point $D$ on the plane $(ABC)$. Prove that the following statements are equivalent: a) $C = E$ or $CE \parallel AB$ b) For each point M belonging to the segment $CD$, the following equation is satisfied $$S^2_{\vartriangle ABM}= \frac{CM^2}{CD^2}\cdot S^2_{\vartriangle ABD}+\left(1- \frac{CM^2}{CD^2} \right)S^2_{\vartriangle ABC}$$ where $S_{\vartriangle XYZ}$ means the area of ​​triangle $XYZ$.

(a) Given a point $O$ inside an equilateral triangle $\vartriangle ABC$. Line $OG$ connects $O$ with the center of mass $G$ of the triangle and intersects the sides of the triangle, or their extensions, at points $A', B', C'$ . Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G} = 3.$$ (b) Point $G$ is the center of the sphere inscribed in a regular tetrahedron $ABCD$. Straight line $OG$ connecting $G$ with a point $O$ inside the tetrahedron intersects the faces at points $A', B', C', D'$. Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G}+ \frac{D'O}{D'G} = 4.$$

In the tetrahedron $SABC $ at the height $SH$ the following point $O$ is chosen, such that: $$\angle AOS + \alpha = \angle BOS + \beta = \angle COS + \gamma = 180^o, $$ where $\alpha, \beta, \gamma$ are dihedral angles at the edges $BC, AC, AB $, respectively, at this point $H$ lies inside the base $ABC$. Let ${{A} _ {1}}, \, {{B} _ {1}}, \, {{C} _ {1}} $be the points of intersection of lines and planes: ${{A} _ {1}} = AO \cap SBC $, ${{B} _ {1}} = BO \cap SAC $, ${{C} _ {1}} = CO \cap SBA$ . Prove that if the planes $ABC $ and ${{A} _ {1}} {{B} _ {1}} {{C} _ {1}} $ are parallel, then $SA = SB = SC $. (Alexey Klurman)