A unit cube has vertices $P_1, P_2, P_3, P_4, P_1', P_2', P_3'$, and $P_4'$. Vertices $P_2, P_3$, and $P_4$ are adjacent to $P_1$, and for $1\leq i\leq 4$, vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2, P_1P_3, P_1P_4, P_1'P_2', P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length? [asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label("$P_1$", P[1], dir(P[1])); label("$P_2$", P[2], dir(P[2])); label("$P_3$", P[3], dir(-45)); label("$P_4$", P[4], dir(P[4])); label("$P'_1$", Pp[1], dir(Pp[1])); label("$P'_2$", Pp[2], dir(Pp[2])); label("$P'_3$", Pp[3], dir(-100)); label("$P'_4$", Pp[4], dir(Pp[4])); [/asy] $ \textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2} $

Points $STABCD$ in space form a convex octahedron with faces $SAB,SBC,SCD,SDA,TAB,TBC,TCD,TDA$ such that there exists a sphere that is tangent to all of its edges. Prove that $A,B,C,D$ lie in one plane.

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]

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 $m+n.$

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).

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.

You have three colors $\{\text{red}, \text{blue}, \text{green}\}$ with which you can color the faces of a regular octahedron (8 triangle sided polyhedron, which is two square based pyramids stuck together at their base), but you must do so in a way that avoids coloring adjacent pieces with the same color. How many different coloring schemes are possible? (Two coloring schemes are considered equivalent if one can be rotated to fit the other.)

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]

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 $AB$.

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.)

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)

$E_1$ and $E_2$ are parallel planes and their distance is $p$. (a) How long is the seitenkante of the regular octahedron such that a side lies in $E_1$ and another in $E_2$? (b) $E$ is a plane between $E_1$ and $E_2$, parallel to $E_1$ and $E_2$, so that its distances from $E_1$ and $E_2$ are in ratio $1:2$ Draw the intersection figure of $E$ and the octahedron for $P=4\sqrt{\frac32}$ cm and justifies, why the that figure must look in such a way

A convex octahedron in Cartesian space contains the origin in its interior. Two of its vertices are on the $x$-axis, two are on the $y$-axis, and two are on the $z$-axis. One triangular face $F$ has side lengths $\sqrt{17}$, $\sqrt{37}$, $\sqrt{52}$. A second triangular face $F_0$ has side lengths $\sqrt{13}$, $\sqrt{29}$, $\sqrt{34}$. What is the minimum possible volume of the octahedron?

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.

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

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

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 $mn=$________.

A regular octahedron has side length $ 1$. A plane parallel to two of its opposite faces cuts the octahedron into the two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $ \frac {a\sqrt {b}}{c}$, where $ a$, $ b$, and $ c$ are positive integers, $ a$ and $ c$ are relatively prime, and $ b$ is not divisible by the square of any prime. What is $ a \plus{} b \plus{} c$? $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14$

Two regular tetrahedrons $A$ and $B$ are made with the 8 vertices of a unit cube. (this way is unique) What's the volume of $A\cup B$?

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]

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.

The centers of the faces of a cube form a regular octahedron of volume $V$. Through each vertex of the cube we may take the plane perpendicular to the long diagonal from the vertex. These planes also form a regular octahedron. Show that its volume is $27V$.

The centers of the faces of the right rectangular prism shown below are joined to create an octahedron, What is the volume of the octahedron? [asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy] $\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15 $

Let $ABCA'B'C'$ be a centrosymmetric octahedron (vertices $A$ and $A'$, $B$ and $B'$, $C$ and $C'$ are opposite) such that the sums of four planar angles equal $240^o$ for each vertex. The Torricelli points $T_1$ and $T_2$ of triangles $ABC$ and $A'BC$ are marked. Prove that the distances from $T_1$ and $T_2$ to $BC$ are equal.

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.