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] $\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5\qquad$

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]

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 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.

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

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is $ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $

16. Create a cube $C_1$ with edge length $1$. Take the centers of the faces and connect them to form an octahedron $O_1$. Take the centers of the octahedron’s faces and connect them to form a new cube $C_2$. Continue this process infinitely. Find the sum of all the surface areas of the cubes and octahedrons. 17. Let $p(x) = x^2 - x + 1$. Let $\alpha$ be a root of $p(p(p(p(x)))$. Find the value of $$(p(\alpha) - 1)p(\alpha)p(p(\alpha))p(p(p(\alpha))$$ 18. An $8$ by $8$ grid of numbers obeys the following pattern: 1) The first row and first column consist of all $1$s. 2) The entry in the $i$th row and $j$th column equals the sum of the numbers in the $(i - 1)$ by $(j - 1)$ sub-grid with row less than i and column less than $j$. What is the number in the 8th row and 8th column?

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)

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

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.

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?

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex? $ \textbf{(A)}\ \frac {5}{256} \qquad \textbf{(B)}\ \frac {21}{1024} \qquad \textbf{(C)}\ \frac {11}{512} \qquad \textbf{(D)}\ \frac {23}{1024} \qquad \textbf{(E)}\ \frac {3}{128}$

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$

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).

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.

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.

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$

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.

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.

We look at an octahedron, a regular octahedron, having painted one of the side surfaces red and the other seven surfaces blue. We throw the octahedron like a die. The surface that comes up is painted: if it is red it is painted blue and if it is blue it is painted red. Then we throw the octahedron again and paint it again according to the above rule. In total we throw the octahedron $10$ times. How many different octahedra can we get after finishing the $10$th time? [i]Two octahedra are different if they cannot be converted into each other by rotation.[/i]

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]

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

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.)

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 $\sphericalangle{EMZ}=\sphericalangle{EYF}$.