Starting from a pyramid $T_0$ whose edges are all of length $2019$, we construct the Figure $T_1$ when considering the triangles formed by the midpoints of the edges of each face of $T_0$, building in each of these new pyramid triangles with faces identical to base. Then the bases of these new pyramids are removed. Figure $T_2$ is constructed by applying the same process from $T_1$ on each triangular face resulting from $T_1$, and so on for $T_3, T_4, ...$ Let $D_0= \max \{d(x,y)\}$, where $x$ and $y$ are vertices of $T_0$ and $d(x,y)$ is the distance between $x$ and $y$. Then we define $D_{n + 1} = \max \{d (x, y) |d (x, y) \notin \{D_0, D_1,...,D_n\}$, where $x, y$ are vertices of $T_{n+1}$. Find the value of $D_n$ for all $n$.

In regular $n$-regular pyramid, the range value of dihedral angle of two adjacent sides is $\text{(A)}\left(\frac{n-2}{n}\pi,\pi\right)\qquad\text{(B)}\left(\frac{n-1}{n}\pi,\pi\right)\qquad\text{(C)}\left(0,\frac{\pi}{2}\right)\qquad\text{(D)}\left(\frac{n-2}{n}\pi,\frac{n-1}{n}\pi\right)$

A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. [asy] // dragon96, replacing // [img]http://i.imgur.com/08FbQs.png[/img] size(140); defaultpen(linewidth(.7)); real alpha=10, x=-0.12, y=0.025, r=1/sqrt(3); path hex=rotate(alpha)*polygon(6); pair A = shift(x,y)*(r*dir(330+alpha)), B = shift(x,y)*(r*dir(90+alpha)), C = shift(x,y)*(r*dir(210+alpha)); pair X = (-A.x, -A.y), Y = (-B.x, -B.y), Z = (-C.x, -C.y); int i; pair[] H; for(i=0; i<6; i=i+1) { H[i] = dir(alpha+60*i);} fill(X--Y--Z--cycle, rgb(204,255,255)); fill(H[5]--Y--Z--H[0]--cycle^^H[2]--H[3]--X--cycle, rgb(203,153,255)); fill(H[1]--Z--X--H[2]--cycle^^H[4]--H[5]--Y--cycle, rgb(255,203,153)); fill(H[3]--X--Y--H[4]--cycle^^H[0]--H[1]--Z--cycle, rgb(153,203,255)); draw(hex^^X--Y--Z--cycle); draw(H[1]--B--H[2]^^H[3]--C--H[4]^^H[5]--A--H[0]^^A--B--C--cycle, linewidth(0.6)+linetype("5 5")); draw(H[0]--Z--H[1]^^H[2]--X--H[3]^^H[4]--Y--H[5]);[/asy]

Consider a (right) square pyramid $ABCDV$ with the apex $V$ and the base (square) $ABCD$. Denote $d=AB/2$ and $\varphi$ the dihedral angle between planes $VAD$ and $ABC$. (1) Consider a line $XY$ connecting the skew lines $VA$ and $BC$, where $X$ lies on line $VA$ and $Y$ lies on line $BC$. Describe a construction of line $XY$ such that the segment $XY$ is of the smallest possible length. Compute the length of segment $XY$ in terms of $d,\varphi$. (2) Compute the distance $v$ between points $V$ and $X$ in terms of $d,\varphi.$

Given the pyramid $ABCD$. Let $O$ be the midpoint of edge $AC$. Given that $DO$ is the height of the pyramid, $AB=BC=2DO$ and the angle $ABC$ is right. Cut this pyramid into $8$ equal and similar to it pyramids.

Let $SABCDE$ be a pyramid whose base $ABCDE$ is a regular pentagon and whose other faces are acute triangles. The altitudes from $S$ to the base sides dissect them into ten triangles, colored red and blue alternatingly. Prove that the sum of the squared areas of the red triangles is equal to the sum of the squared areas of the blue triangles.

A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $ W$? [asy]import three; size(200); defaultpen(linewidth(.8pt)+fontsize(10pt)); currentprojection=oblique; path3 p1=(0,2,2)--(0,2,0)--(2,2,0)--(2,2,2)--(0,2,2)--(0,0,2)--(2,0,2)--(2,2,2); path3 p2=(2,2,0)--(2,0,0)--(2,0,2); path3 p3=(0,0,2)--(0,2,1)--(2,2,1)--(2,0,2); path3 p4=(2,2,1)--(2,0,0); pen finedashed=linetype("4 4"); draw(p1^^p2^^p3^^p4); draw(shift((4,0,0))*p1); draw(shift((4,0,0))*p2); draw(shift((4,0,0))*p3); draw(shift((4,0,0))*p4); draw((4,0,2)--(5,2,2)--(6,0,2),finedashed); draw((5,2,2)--(5,2,0)--(6,0,0),finedashed); label("$W$",(3,0,2)); draw((2.7,.3,2)--(2.1,1.9,2),linewidth(.6pt)); draw((3.4,.3,2)--(5.9,1.9,2),linewidth(.6pt)); label("Figure 1",(1,-0.5,2)); label("Figure 2",(5,-0.5,2));[/asy]$ \textbf{(A)}\ \frac {1}{12}\qquad \textbf{(B)}\ \frac {1}{9}\qquad \textbf{(C)}\ \frac {1}{8}\qquad \textbf{(D)}\ \frac {1}{6}\qquad \textbf{(E)}\ \frac {1}{4}$

Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O$, and that the ratio of the volume of $O$ to that of $C$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find $m+n$.

A sphere is inscribed in the tetrahedron whose vertices are $A=(6,0,0), B=(0,4,0), C=(0,0,2),$ and $D=(0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

In convex hexagonal pyramid 11 edges are equal to 1.Find all possible values of 12th edge.

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?

Square pyramid $ABCDE$ has base $ABCD,$ which measures $3$ cm on a side, and altitude $\overline{AE}$ perpendicular to the base$,$ which measures $6$ cm. Point $P$ lies on $\overline{BE},$ one third of the way from $B$ to $E;$ point $Q$ lies on $\overline{DE},$ one third of the way from $D$ to $E;$ and point $R$ lies on $\overline{CE},$ two thirds of the way from $C$ to $E.$ What is the area, in square centimeters, of $\triangle PQR?$ $\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$

A triangular pyramid $ O.ABC$ with base $ ABC$ has the property that the lengths of the altitudes from $ A$, $ B$ and $ C$ are not less than $ \frac{OB \plus{}OC}{2}$, $ \frac{OC \plus{} OA}{2}$ and $ \frac{OA \plus{} OB}{2}$, respectively. Given that the area of $ ABC$ is $ S$, calculate the volume of the pyramid.

Let $VWXYZ$ be a square pyramid with vertex $V$ with height $1$, and with the unit square as its base. Let $STANFURD$ be a cube, such that face $FURD$ lies in the same plane as and shares the same center as square face $WXYZ$. Furthermore, all sides of $FURD$ are parallel to the sides of $WXY Z$. Cube $STANFURD$ has side length $s$ such that the volume that lies inside the cube but outside the square pyramid is equal to the volume that lies inside the square pyramid but outside the cube. What is the value of $s$?

Let $VABC$ be a regular triangular pyramid with base $ABC$, of center $O$. Points $I$ and $H$ are the center of the inscribed circle, respectively the orthocenter $\vartriangle VBC$. Knowing that $AH = 3 OI$, determine the measure of the angle between the lateral edge of the pyramid and the plane of the base.

Consider the pyramid $ VABCD, $ where $ V $ is the top and $ ABCD $ is a rectangular base. If $ \angle BVD = \angle AVC, $ then prove that the triangles $ VAC $ and $ VBD $ share the same perimeter and area.

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? [asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy] $\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. [asy] // dragon96, replacing // [img]http://i.imgur.com/08FbQs.png[/img] size(140); defaultpen(linewidth(.7)); real alpha=10, x=-0.12, y=0.025, r=1/sqrt(3); path hex=rotate(alpha)*polygon(6); pair A = shift(x,y)*(r*dir(330+alpha)), B = shift(x,y)*(r*dir(90+alpha)), C = shift(x,y)*(r*dir(210+alpha)); pair X = (-A.x, -A.y), Y = (-B.x, -B.y), Z = (-C.x, -C.y); int i; pair[] H; for(i=0; i<6; i=i+1) { H[i] = dir(alpha+60*i);} fill(X--Y--Z--cycle, rgb(204,255,255)); fill(H[5]--Y--Z--H[0]--cycle^^H[2]--H[3]--X--cycle, rgb(203,153,255)); fill(H[1]--Z--X--H[2]--cycle^^H[4]--H[5]--Y--cycle, rgb(255,203,153)); fill(H[3]--X--Y--H[4]--cycle^^H[0]--H[1]--Z--cycle, rgb(153,203,255)); draw(hex^^X--Y--Z--cycle); draw(H[1]--B--H[2]^^H[3]--C--H[4]^^H[5]--A--H[0]^^A--B--C--cycle, linewidth(0.6)+linetype("5 5")); draw(H[0]--Z--H[1]^^H[2]--X--H[3]^^H[4]--Y--H[5]);[/asy]

A square piece of paper has sides of length $ 100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance $ \sqrt {17}$ from the corner, and they meet on the diagonal at an angle of $ 60^\circ$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $ \sqrt [n]{m}$, where $ m$ and $ n$ are positive integers, $ m < 1000$, and $ m$ is not divisible by the $ n$th power of any prime. Find $ m \plus{} n$. [asy]import math; unitsize(5mm); defaultpen(fontsize(9pt)+Helvetica()+linewidth(0.7)); pair O=(0,0); pair A=(0,sqrt(17)); pair B=(sqrt(17),0); pair C=shift(sqrt(17),0)*(sqrt(34)*dir(75)); pair D=(xpart(C),8); pair E=(8,ypart(C)); draw(O--(0,8)); draw(O--(8,0)); draw(O--C); draw(A--C--B); draw(D--C--E); label("$\sqrt{17}$",(0,2),W); label("$\sqrt{17}$",(2,0),S); label("cut",midpoint(A--C),NNW); label("cut",midpoint(B--C),ESE); label("fold",midpoint(C--D),W); label("fold",midpoint(C--E),S); label("$30^\circ$",shift(-0.6,-0.6)*C,WSW); label("$30^\circ$",shift(-1.2,-1.2)*C,SSE);[/asy]

A triangular pyramid $ABCD$ is given. Prove that $\frac Rr > \frac ah$, where $R$ is the radius of the circumscribed sphere, $r$ is the radius of the inscribed sphere, $a$ is the length of the longest edge, $h$ is the length of the shortest altitude (from a vertex to the opposite face).

Are there convex $n$-gonal ($n \ge 4$) and triangular pyramids such that the four trihedral angles of the $n$-gonal pyramid are equal trihedral angles of a triangular pyramid? [hide=original wording] Существуют ли выпуклая n-угольная (n>= 4) и треугольная пирамиды такие, что четыре трехгранных угла n-угольной пирамиды равны трехгранным углам треугольной пирамиды?[/hide]

A pyramid with a square base is cut by a plane that is parallel to its base and is $ 2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid? $ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 2 \plus{} \sqrt{2}\qquad \textbf{(C)}\ 1 \plus{} 2\sqrt{2}\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 4 \plus{} 2\sqrt{2}$

The base of a right pyramid is a quadrilateral whose sides are each of length $a$. The planar angles at the vertex of the pyramid are equal to the angles between the lateral edges and the base. Find the volume of the pyramid.

The sum of the base-$ 10$ logarithms of the divisors of $ 10^n$ is $ 792$. What is $ n$? $ \textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

In $\triangle ABC$, $\angle C=90^{\circ},\angle B=30^{\circ}, AC=2$. $M$ is the midpoint of $AB$. Fold up $\triangle ACM$ along $CM$, satisfying that $|AB|=2\sqrt2$. The volume of triangular pyramid $A-BCM$ is________.