A point $P$ in the interior of a convex polyhedron in Euclidean space is called a [i]pivot point[/i] of the polyhedron if every line through $P$ contains exactly $0$ or $2$ vertices of the polyhedron. Determine, with proof, the maximum number of pivot points that a polyhedron can contain.

In space are given two tetrahedra with the same barycenter such that one of them contains the other. For each tetrahedron, we consider the octahedron whose vertices are the midpoints of the tetrahedron's edges. Prove that one of this octahedra contains the other.

Prove that a convex polyhedron all of whose faces are equilateral triangles has at most $30$ edges.

A [i]cuboctahedron[/i], shown below, is a polyhedron with 8 equilateral triangle faces and 6 square faces. Each edge has the same length and each of the 24 vertices borders 2 squares and 2 triangles. An \textit{octahedron} is a regular polyhedron with 6 vertices and 8 equilateral triangle faces. Compute the sum of the volumes of an octahedron with side length 5 and a cuboctahedron with side length 5. [img]http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi82LzBmNjM1OTM2M2ExYTQzOTFhODEwODkwM2FiYmM1MTljOGQzNmJhLmpwZw==&rn=Q3Vib2N0YWhlZHJvbi5qcGc=[/img] [i]2016 CCA Math Bonanza Team #7[/i]

Find out the maximum value of the numbers of edges of a solid regular octahedron that we can see from a point out of the regular octahedron.(We define we can see an edge $AB$ of the regular octahedron from point $P$ outside if and only if the intersection of non degenerate triangle $PAB$ and the solid regular octahedron is exactly edge $AB$.

There is three-dimensional space. For every integer $n$ we build planes $ x \pm y\pm z = n$. All space is divided on octahedrons and tetrahedrons. Point $(x_0,y_0,z_0)$ has rational coordinates but not lies on any plane. Prove, that there is such natural $k$ , that point $(kx_0,ky_0,kz_0)$ lies strictly inside the octahedron of partition.

Compute the volume of a regular octahedron circumscribed about a sphere of radius $1$.

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron? $ \textbf{(A) } \frac 18 \qquad \textbf{(B) } \frac 16 \qquad \textbf{(C) } \frac 14 \qquad \textbf{(D) } \frac 13 \qquad \textbf{(E) } \frac 12$

The dimensions of a wooden octahedron are natural numbers. We painted all its surface (the six faces), cut it by planes parallel to the cubed faces of an edge unit and observed that exactly half of the cubes did not have any painted faces. Prove that the number of octahedra with such property is finite. (It may be useful to keep in mind that $\sqrt[3]{\frac{1}{2}}=1,79 ... <1,8$). [hide=original wording] Las dimensiones de un ortoedro de madera son enteras. Pintamos toda su superficie (las seis caras), lo cortamos mediante planos paralelos a las caras en cubos de una unidad de arista y observamos que exactamente la mitad de los cubos no tienen ninguna cara pintada. Probar que el número de ortoedros con tal propiedad es finito[/hide]

Let $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ be the eight vertices of a $30 \times30\times30$ cube as shown. The two figures $ACFH$ and $BDEG$ are congruent regular tetrahedra. Find the volume of the intersection of these two tetrahedra. [asy] import graph; size(12.57cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -3.79, xmax = 8.79, ymin = 0.32, ymax = 4.18; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen ffzzzz = rgb(1,0.6,0.6); pen zzzzff = rgb(0.6,0.6,1); draw((6,3.5)--(8,1.5), zzzzff); draw((7,3)--(5,1), blue); draw((6,3.5)--(7,3), blue); draw((6,3.5)--(5,1), blue); draw((5,1)--(8,1.5), blue); draw((7,3)--(8,1.5), blue); draw((4,3.5)--(2,1.5), ffzzzz); draw((1,3)--(2,1.5), ffqqtt); draw((2,1.5)--(3,1), ffqqtt); draw((1,3)--(3,1), ffqqtt); draw((4,3.5)--(1,3), ffqqtt); draw((4,3.5)--(3,1), ffqqtt); draw((-3,3)--(-3,1), linewidth(1.6)); draw((-3,3)--(-1,3), linewidth(1.6)); draw((-1,3)--(-1,1), linewidth(1.6)); draw((-3,1)--(-1,1), linewidth(1.6)); draw((-3,3)--(-2,3.5), linewidth(1.6)); draw((-2,3.5)--(0,3.5), linewidth(1.6)); draw((0,3.5)--(-1,3), linewidth(1.6)); draw((0,3.5)--(0,1.5), linewidth(1.6)); draw((0,1.5)--(-1,1), linewidth(1.6)); draw((-3,1)--(-2,1.5)); draw((-2,1.5)--(0,1.5)); draw((-2,3.5)--(-2,1.5)); draw((1,3)--(1,1), linewidth(1.6)); draw((1,3)--(3,3), linewidth(1.6)); draw((3,3)--(3,1), linewidth(1.6)); draw((1,1)--(3,1), linewidth(1.6)); draw((1,3)--(2,3.5), linewidth(1.6)); draw((2,3.5)--(4,3.5), linewidth(1.6)); draw((4,3.5)--(3,3), linewidth(1.6)); draw((4,3.5)--(4,1.5), linewidth(1.6)); draw((4,1.5)--(3,1), linewidth(1.6)); draw((1,1)--(2,1.5)); draw((2,3.5)--(2,1.5)); draw((2,1.5)--(4,1.5)); draw((5,3)--(5,1), linewidth(1.6)); draw((5,3)--(6,3.5), linewidth(1.6)); draw((5,3)--(7,3), linewidth(1.6)); draw((7,3)--(7,1), linewidth(1.6)); draw((5,1)--(7,1), linewidth(1.6)); draw((6,3.5)--(8,3.5), linewidth(1.6)); draw((7,3)--(8,3.5), linewidth(1.6)); draw((7,1)--(8,1.5)); draw((5,1)--(6,1.5)); draw((6,3.5)--(6,1.5)); draw((6,1.5)--(8,1.5)); draw((8,3.5)--(8,1.5), linewidth(1.6)); label("$ A $",(-3.4,3.41),SE*labelscalefactor); label("$ D $",(-2.16,4.05),SE*labelscalefactor); label("$ H $",(-2.39,1.9),SE*labelscalefactor); label("$ E $",(-3.4,1.13),SE*labelscalefactor); label("$ F $",(-1.08,0.93),SE*labelscalefactor); label("$ G $",(0.12,1.76),SE*labelscalefactor); label("$ B $",(-0.88,3.05),SE*labelscalefactor); label("$ C $",(0.17,3.85),SE*labelscalefactor); label("$ A $",(0.73,3.5),SE*labelscalefactor); label("$ B $",(3.07,3.08),SE*labelscalefactor); label("$ C $",(4.12,3.93),SE*labelscalefactor); label("$ D $",(1.69,4.07),SE*labelscalefactor); label("$ E $",(0.60,1.15),SE*labelscalefactor); label("$ F $",(2.96,0.95),SE*labelscalefactor); label("$ G $",(4.12,1.67),SE*labelscalefactor); label("$ H $",(1.55,1.82),SE*labelscalefactor); label("$ A $",(4.71,3.47),SE*labelscalefactor); label("$ B $",(7.14,3.10),SE*labelscalefactor); label("$ C $",(8.14,3.82),SE*labelscalefactor); label("$ D $",(5.78,4.08),SE*labelscalefactor); label("$ E $",(4.6,1.13),SE*labelscalefactor); label("$ F $",(6.93,0.96),SE*labelscalefactor); label("$ G $",(8.07,1.64),SE*labelscalefactor); label("$ H $",(5.65,1.90),SE*labelscalefactor); dot((-3,3),dotstyle); dot((-3,1),dotstyle); dot((-1,3),dotstyle); dot((-1,1),dotstyle); dot((-2,3.5),dotstyle); dot((0,3.5),dotstyle); dot((0,1.5),dotstyle); dot((-2,1.5),dotstyle); dot((1,3),dotstyle); dot((1,1),dotstyle); dot((3,3),dotstyle); dot((3,1),dotstyle); dot((2,3.5),dotstyle); dot((4,3.5),dotstyle); dot((4,1.5),dotstyle); dot((2,1.5),dotstyle); dot((5,3),dotstyle); dot((5,1),dotstyle); dot((6,3.5),dotstyle); dot((7,3),dotstyle); dot((7,1),dotstyle); dot((8,3.5),dotstyle); dot((8,1.5),dotstyle); dot((6,1.5),dotstyle); [/asy]

Is it possible to cover the surface of a regular octahedron by several regular hexagons without gaps and overlaps? (A regular octahedron has $6$ vertices, each face is an equilateral triangle, each vertex belongs to $4$ faces.)

Let $ P$ be a polyhedron where every face is a regular polygon, and every edge has length $ 1$. Each vertex of $ P$ is incident to two regular hexagons and one square. Choose a vertex $ V$ of the polyhedron. Find the volume of the set of all points contained in $ P$ that are closer to $ V$ than to any other vertex.

From a regular octahedron with edge $1$, cut off a pyramid about each vertex. The base of each pyramid is a square with edge $\frac 13$. Can copies of the polyhedron so obtained, whose faces are either regular hexagons or squares, be used to tile space?

What is the shortest possible side length of a four-dimensional hypercube that contains a regular octahedron with side 1?

Some edges are painted in red. We say that a coloring of this kind is [i]good[/i], if for each vertex of the polyhedron, there exists an edge which concurs in that vertex and is not painted red. Moreover, we say that a coloring where some of the edges of a regular polyhedron is [i]completely good[/i], if in addition to being [i]good[/i], no face of the polyhedron has all its edges painted red. What regular polyhedrons is equal the maximum number of edges that can be painted in a [i]good[/i] color and a [i]completely good[/i]? Explain your answer.

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

A [i]regular octahedron[/i] has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$? [asy] // Note: This diagram was not made by me. import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (6.92820323028,0); DB = (3.46410161514,6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label("$7$",trf*DC); label("$Q$",trf*DC+DA-DB); label("$5$",trf*DC-DB); label("$3$",trf*DC-DA-DB); label("$6$",trf*CD); label("$4$",trf*CD-DA); label("$2$",trf*CD-DA-DB); label("$1$",trf*CD-2DA); [/asy] $\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5\qquad$

Prove that there exists an ellipsoid touching all edges of an octahedron if and only if the octahedron's diagonals intersect. (Here an octahedron is a polyhedron consisting of eight triangular faces, twelve edges, and six vertices such that four faces meat at each vertex. The diagonals of an octahedron are the lines connecting pairs of vertices not connected by an edge).

Find out the maximum value of the numbers of edges of a solid regular octahedron that we can see from a point out of the regular octahedron.(We define we can see an edge $AB$ of the regular octahedron from point $P$ outside if and only if the intersection of non degenerate triangle $PAB$ and the solid regular octahedron is exactly edge $AB$.

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) [asy]import three; import math; size(180); defaultpen(linewidth(.8pt)); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7));[/asy]$ \textbf{(A)}\ 210 \qquad \textbf{(B)}\ 560 \qquad \textbf{(C)}\ 840 \qquad \textbf{(D)}\ 1260 \qquad \textbf{(E)}\ 1680$

A net for a polyhedron is cut along an edge to give two [b]pieces[/b]. For example, we may cut a cube net along the red edge to form two pieces as shown. [asy] size(5.5cm); draw((1,0)--(1,4)--(2,4)--(2,0)--cycle); draw((1,1)--(2,1)); draw((1,2)--(2,2)); draw((1,3)--(2,3)); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((2,1)--(2,2),red+linewidth(1.5)); draw((3.5,2)--(5,2)); filldraw((4.25,2.2)--(5,2)--(4.25,1.8)--cycle,black); draw((6,1.5)--(10,1.5)--(10,2.5)--(6,2.5)--cycle); draw((7,1.5)--(7,2.5)); draw((8,1.5)--(8,2.5)); draw((9,1.5)--(9,2.5)); draw((7,2.5)--(7,3.5)--(8,3.5)--(8,2.5)--cycle); draw((11,1.5)--(11,2.5)--(12,2.5)--(12,1.5)--cycle); [/asy] Are there two distinct polyhedra for which this process may result in the same two pairs of pieces? If you think the answer is no, prove that no pair of polyhedra can result in the same two pairs of pieces. If you think the answer is yes, provide an example; a clear example will suffice as a proof.

For a regular hexahedron and a regular octahedron, all their faces are regular triangles, whose lengths of each side are $a$. Their inradius are $r_1,r_2$. $\frac{r_1}{r_2}=\frac{m}{n}, \gcd(m,n)=1$. Then $mn=$________.

Can the regular octahedron be inscribed into regular dodecahedron in such way that all vertices of octahedron be the vertices of dodecahedron? (B.Frenkin)

Can a regular octahedron be inscribed in a cube in such a way that all vertices of the octahedron are on cube's edges? (4)

Consider a regular octahedron $ABCDEF$ with lower vertex $E$, upper vertex $F$, middle cross-section $ABCD$, midpoint $M$ and circumscribed sphere $k$. Further, let $X$ be an arbitrary point inside the face $ABF$. Let the line $EX$ intersect $k$ in $E$ and $Z$, and the plane $ABCD$ in $Y$. Show that $\sphericalangle{EMZ}=\sphericalangle{EYF}$.