This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 619

1970 IMO Shortlist, 5

Let $M$ be an interior point of the tetrahedron $ABCD$. Prove that \[ \begin{array}{c}\ \stackrel{\longrightarrow }{MA} \text{vol}(MBCD) +\stackrel{\longrightarrow }{MB} \text{vol}(MACD) +\stackrel{\longrightarrow }{MC} \text{vol}(MABD) + \stackrel{\longrightarrow }{MD} \text{vol}(MABC) = 0 \end{array}\] ($\text{vol}(PQRS)$ denotes the volume of the tetrahedron $PQRS$).

1970 IMO Longlists, 45

Let $M$ be an interior point of tetrahedron $V ABC$. Denote by $A_1,B_1, C_1$ the points of intersection of lines $MA,MB,MC$ with the planes $VBC,V CA,V AB$, and by $A_2,B_2, C_2$ the points of intersection of lines $V A_1, VB_1, V C_1$ with the sides $BC,CA,AB$. [b](a)[/b] Prove that the volume of the tetrahedron $V A_2B_2C_2$ does not exceed one-fourth of the volume of $V ABC$. [b](b)[/b] Calculate the volume of the tetrahedron $V_1A_1B_1C_1$ as a function of the volume of $V ABC$, where $V_1$ is the point of intersection of the line $VM$ with the plane $ABC$, and $M$ is the barycenter of $V ABC$.

1992 AIME Problems, 7

Faces $ABC$ and $BCD$ of tetrahedron $ABCD$ meet at an angle of $30^\circ$. The area of face $ABC$ is $120$, the area of face $BCD$ is $80$, and $BC=10$. Find the volume of the tetrahedron.

1967 IMO Shortlist, 1

Prove that a tetrahedron with just one edge length greater than $1$ has volume at most $ \frac{1}{8}.$

1967 Bulgaria National Olympiad, Problem 4

Outside of the plane of the triangle $ABC$ is given point $D$. (a) prove that if the segment $DA$ is perpendicular to the plane $ABC$ then orthogonal projection of the orthocenter of the triangle $ABC$ on the plane $BCD$ coincides with the orthocenter of the triangle $BCD$. (b) for all tetrahedrons $ABCD$ with base, the triangle $ABC$ with smallest of the four heights that from the vertex $D$, find the locus of the foot of that height.

1970 IMO Longlists, 17

In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \[ (AB+BC+CA)^2\le6(AD^2+BD^2+CD^2). \] When do we have equality?

1986 IMO Longlists, 17

We call a tetrahedron right-faced if each of its faces is a right-angled triangle. [i](a)[/i] Prove that every orthogonal parallelepiped can be partitioned into six right-faced tetrahedra. [i](b)[/i] Prove that a tetrahedron with vertices $A_1,A_2,A_3,A_4$ is right-faced if and only if there exist four distinct real numbers $c_1, c_2, c_3$, and $c_4$ such that the edges $A_jA_k$ have lengths $A_jA_k=\sqrt{|c_j-c_k|}$ for $1\leq j < k \leq 4.$

1988 Bulgaria National Olympiad, Problem 5

The points of space are painted in two colors. Prove that there is a tetrahedron such that all its vertices and its centroid are of the same color.

1992 Bulgaria National Olympiad, Problem 1

Through a random point $C_1$ from the edge $DC$ of the regular tetrahedron $ABCD$ is drawn a plane, parallel to the plane $ABC$. The plane constructed intersects the edges $DA$ and $DB$ at the points $A_1,B_1$ respectively. Let the point $H$ is the midpoint of the altitude through the vertex $D$ of the tetrahedron $DA_1B_1C_1$ and $M$ is the center of gravity (barycenter) of the triangle $ABC_1$. Prove that the measure of the angle $HMC$ doesn’t depend on the position of the point $C_1$. [i](Ivan Tonov)[/i]

1967 IMO Longlists, 40

Prove that a tetrahedron with just one edge length greater than $1$ has volume at most $ \frac{1}{8}.$

1971 IMO Longlists, 28

All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$. [b]a.)[/b] If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length; [b]b.)[/b] If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.

2013 Online Math Open Problems, 44

Suppose tetrahedron $PABC$ has volume $420$ and satisfies $AB = 13$, $BC = 14$, and $CA = 15$. The minimum possible surface area of $PABC$ can be written as $m+n\sqrt{k}$, where $m,n,k$ are positive integers and $k$ is not divisible by the square of any prime. Compute $m+n+k$. [i]Ray Li[/i]

1983 IMO Longlists, 67

The altitude from a vertex of a given tetrahedron intersects the opposite face in its orthocenter. Prove that all four altitudes of the tetrahedron are concurrent.

2021 Math Prize for Girls Problems, 19

Let $T$ be a regular tetrahedron. Let $t$ be the regular tetrahedron whose vertices are the centers of the faces of $T$. Let $O$ be the circumcenter of either tetrahedron. Given a point $P$ different from $O$, let $m(P)$ be the midpoint of the points of intersection of the ray $\overrightarrow{OP}$ with $t$ and $T$. Let $S$ be the set of eight points $m(P)$ where $P$ is a vertex of either $t$ or $T$. What is the volume of the convex hull of $S$ divided by the volume of $t$?

1989 Poland - Second Round, 3

Given is a trihedral angle $ OABC $ with a vertex $ O $ and a point $ P $ in its interior. Let $ V $ be the volume of a parallelepiped with two vertices at points $ O $ and $ P $, whose three edges are contained in the rays $ \overrightarrow{OA} $, $ \overrightarrow{OB} $, $ \overrightarrow{OC} $. Calculate the minimum volume of a tetrahedron whose three faces are contained in the faces of the trihedral angle $OABC$ and the fourth face contains the point $P$.

1966 IMO Longlists, 56

In a tetrahedron, all three pairs of opposite (skew) edges are mutually perpendicular. Prove that the midpoints of the six edges of the tetrahedron lie on one sphere.

1966 IMO, 3

Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

1986 China Team Selection Test, 2

Given a tetrahedron $ABCD$, $E$, $F$, $G$, are on the respectively on the segments $AB$, $AC$ and $AD$. Prove that: i) area $EFG \leq$ max{area $ABC$,area $ABD$,area $ACD$,area $BCD$}. ii) The same as above replacing "area" for "perimeter".

1973 IMO Longlists, 2

Let $OX, OY$ and $OZ$ be three rays in the space, and $G$ a point "[i]between these rays[/i]" (i. e. in the interior of the part of the space bordered by the angles $Y OZ, ZOX$ and $XOY$). Consider a plane passing through $G$ and meeting the rays $OX, OY$ and $OZ$ in the points $A, B, C$, respectively. There are infinitely many such planes; construct the one which minimizes the volume of the tetrahedron $OABC$.

2003 Croatia National Olympiad, Problem 3

In a tetrahedron $ABCD$, all angles at vertex $D$ are equal to $\alpha$ and all dihedral angles between faces having $D$ as a vertex are equal to $\phi$. Prove that there exists a unique $\alpha$ for which $\phi=2\alpha$.

1988 IMO Shortlist, 6

In a given tedrahedron $ ABCD$ let $ K$ and $ L$ be the centres of edges $ AB$ and $ CD$ respectively. Prove that every plane that contains the line $ KL$ divides the tedrahedron into two parts of equal volume.

1994 Tournament Of Towns, (432) 2

Prove that one can construct two triangles from six edges of an arbitrary tetrahedron. (VV Proizvolov)

1975 Vietnam National Olympiad, 3

Let $ABCD$ be a tetrahedron with $BA \perp AC,DB \perp (BAC)$. Denote by $O$ the midpoint of $AB$, and $K$ the foot of the perpendicular from $O$ to $DC$. Suppose that $AC = BD$. Prove that $\frac{V_{KOAC}}{V_{KOBD}}=\frac{AC}{BD}$ if and only if $2AC \cdot BD = AB^2$.

1974 Poland - Second Round, 3

Prove that the orthogonal projections of the vertex $ D $ of the tetrahedron $ ABCD $ onto the bisectors of the internal and external dihedral angles at the edges $ \overline{AB} $, $ \overline{BC} $ and $ \overline{CA} $ belong to one plane .

1973 IMO Shortlist, 1

Let a tetrahedron $ABCD$ be inscribed in a sphere $S$. Find the locus of points $P$ inside the sphere $S$ for which the equality \[\frac{AP}{PA_1}+\frac{BP}{PB_1}+\frac{CP}{PC_1}+\frac{DP}{PD_1}=4\] holds, where $A_1,B_1, C_1$, and $D_1$ are the intersection points of $S$ with the lines $AP,BP,CP$, and $DP$, respectively.