Let $ ABC$ be a triangle, and let $ X$, $ Y$, $ Z$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Denote by $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ the reflections of these points $ X$, $ Y$, $ Z$ in the midpoints of the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \left\vert XYZ\right\vert \equal{}\left\vert X^{\prime}Y^{\prime}Z^{\prime}\right\vert$.

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?

Given is a segment $AB$. Three points $X, Y, Z$ are picked in the space so that $ABX$ is an equilateral triangle and $ABYZ$ is a square. Prove that the orthocenters of all triangles $XYZ$ obtained in this way belong to a fixed circle. [i]Alexandr Matveev[/i]

Jamie counted the number of edges of a cube, Jimmy counted the number of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 26$

Determine the volume of the body obtained by cutting the ball of radius $R$ by the trihedron with vertex in the center of that ball, it its dihedral angles are $\alpha, \beta, \gamma.$

Let $ ABCDA’B’C’D’ $ a right parallelepiped and $ M,N $ the feet of the perpendiculars of $ BD $ through $ A’, $ respectively, $ C’. $ We know that $ AB=\sqrt 2, BC=\sqrt 3, AA’=\sqrt 2. $ [b]a)[/b] Prove that $ A’M\perp C’N. $ [b]b)[/b] Calculate the dihedral angle between the plane formed by $ A’MC $ and the plane formed by $ ANC’. $

(a) Given a point $O$ inside an equilateral triangle $\vartriangle ABC$. Line $OG$ connects $O$ with the center of mass $G$ of the triangle and intersects the sides of the triangle, or their extensions, at points $A', B', C'$ . Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G} = 3.$$ (b) Point $G$ is the center of the sphere inscribed in a regular tetrahedron $ABCD$. Straight line $OG$ connecting $G$ with a point $O$ inside the tetrahedron intersects the faces at points $A', B', C', D'$. Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G}+ \frac{D'O}{D'G} = 4.$$

Let $ABCDE$ be a right quadrangular pyramid with vertex $E$ and height $EO$. Point $S$ divides this height in the ratio $ES: SO=m$. In which ratio does the plane $(ABC)$ divide the lateral area of the pyramid.

We call a tetrahedron divisor of a parallelepiped if the parallelepiped can be divided into $6$ copies of that tetrahedron. Does there exist a parallelepiped that it has at least two different divisor tetrahedrons?

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points?

Triangles $\vartriangle ABC$ and $\vartriangle A_1B_1C_1$ lie on different planes. Line $AB$ intersects line $A_1B_1$, line $BC$ intersects line $B_1C_1$ and line $CA$ intersects line $C_1A_1$. Prove that either the three lines $AA_1, BB_1, CC_1$ meet at one point or that they are all parallel.

Prove that in each polyhedron there exist two faces with the same number of edges.

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.

Prove that the sum of the distances from any point in space from the vertices of a cube with edge $a$ is not less than $4\sqrt3 a$.

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