$ABCD$ is a rectangle with $BC / AB = \sqrt2$. $ABEF$ is a congruent rectangle in a different plane. Find the angle $DAF$ such that the lines $CA$ and $BF$ are perpendicular. In this configuration, find two points on the line $CA$ and two points on the line $BF$ so that the four points form a regular tetrahedron.

A regular triangular pyramid $ABCD$ with the base side $AB=a$ and the lateral edge $AD=b$ is given. Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. A line $\alpha$ through $MN$ intersects the edges $AD$ and $BC$ at $P$ and $Q$, respectively. (a) Prove that $AP/AD=BQ/BC$. (b) Find the ratio $AP/AD$ which minimizes the area of $MQNP$.

$ABCD$ is a tetrahedron such that we can find a sphere $k(A,B,C)$ through $A, B, C$ which meets the plane $BCD$ in the circle diameter $BC$, meets the plane $ACD$ in the circle diameter $AC$, and meets the plane $ABD$ in the circle diameter $AB$. Show that there exist spheres $k(A,B,D)$, $k(B,C,D)$ and $k(C,A,D)$ with analogous properties.

The figure shows a glass prism which is partially filled with liquid. The surface of the prism consists of two isosceles right triangles, two squares with side length $10$ cm and a rectangle. The prism can lie in three different ways. If the prism lies as shown in figure $1$, the height of the liquid is $5$ cm. [img]https://cdn.artofproblemsolving.com/attachments/4/2/cda98a00f8586132fe519855df123534516b50.png[/img] a) What is the height of the liquid when it lies as shown in figure $2$? b) What is the height of the liquid when it lies as shown in figure$ 3$?

A paper cup has a base that is a circle with radius $r$, a top that is a circle with radius $2r$, and sides that connect the two circles with straight line segments as shown below. This cup has height $h$ and volume $V$. A second cup that is exactly the same shape as the first is held upright inside the first cup so that its base is a distance of $\tfrac{h}2$ from the base of the first cup. The volume of liquid that will t inside the first cup and outside the second cup can be written $\tfrac{m}{n}\cdot V$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] pair s = (10,1); draw(ellipse((0,0),4,1)^^ellipse((0,-6),2,.5)); fill((3,-6)--(-3,-6)--(0,-2.1)--cycle,white); draw((4,0)--(2,-6)^^(-4,0)--(-2,-6)); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-2.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); pair s = (10,-2); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-4.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); //darn :([/asy]

A $ 9\times9\times9$ cube is composed of twenty-seven $ 3\times3\times3$ cubes. The big cube is 'tunneled' as follows: First, the six $ 3\times3\times3$ cubes which make up the center of each face as well as the center of $ 3\times3\times3$ cube are removed. Second, each of the twenty remaining $ 3\times3\times3$ cubes is diminished in the same way. That is, the central facial unit cubes as well as each center cube are removed. [asy] import three; size(4.5cm); triple eye = (6, 9, 5); currentprojection = perspective(eye); real eps = 0.001; for(int i = 0; i < 3; ++i){ for(int j = 0; j < 3; ++j){ for(int k = 0; k < 3; ++k){ if(i == 1 && j == 1) continue; if(j == 1 && k == 1) continue; if(k == 1 && i == 1) continue; draw(shift(i, j, k) * scale(1 - eps, 1 - eps, 1 - eps) * unitcube, gray(0.9), nolight); draw(shift(i, j, k) * (X--(X + Y)--Y--(Y+Z)--Z--(Z + X)--cycle)); draw(shift(i, j, k) * (X + Y + Z--X + Y)); draw(shift(i, j, k) * (X + Y + Z--Y + Z)); draw(shift(i, j, k) * (X + Y + Z--Z + X)); } } } [/asy] The surface area of the final figure is $ \textbf{(A)}\ 384\qquad \textbf{(B)}\ 729\qquad \textbf{(C)}\ 864\qquad \textbf{(D)}\ 1024\qquad \textbf{(E)}\ 1056$

Each edge of a cube is increased by $50 \%$. The percent of increase of the surface area of the cube is: $ \textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750 $

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?

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$, and let $L$ and $N$ be points on the side $AC$. The point $K$ is between $M$ and $B$, and the point $L$ is between $N$ and $C$. If $\frac{BK}{KM}=\frac{CL}{LN}$, then prove that the orthocentres of the triangles $\triangle{ABC}$, $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

For what polynomials $P(n)$ with integer coefficients can a positive integer be assigned to every lattice point in $\mathbb{R}^3$ so that for every integer $n \ge 1$, the sum of the $n^3$ integers assigned to any $n \times n \times n$ grid of lattice points is divisible by $P(n)$? [i]Proposed by Andre Arslan[/i]

Does there exist a convex polyhedron which has a triangular section (by a plane not passing through the vertices) and each vertex of the polyhedron belonging to (a) no less than $ 5$ faces? (b) exactly $5$ faces? (G. Galperin)

Each face of a cube is painted either red or blue, each with probability $ 1/2$. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color? $ \textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac{5}{16} \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac{7}{16} \qquad \textbf{(E)}\ \frac12$

Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1-\sqrt[3]2+\sqrt[3]4)=0$. (A polynomial is called $\textit{monic}$ if its leading coefficient is $1$.)

A round pencil has length $ 8$ when unsharpened, and diameter $ \frac {1}{4}$. It is sharpened perfectly so that it remains $ 8$ inches long, with a $ 7$ inch section still cylindrical and the remaining $ 1$ inch giving a conical tip. What is its volume?

A right rectangular prism $P$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P$ cuts $P$ into two prisms, one of which is similar to $P,$ and both of which have nonzero volume. Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist?

A right circular cone has base radius 1 cm and slant height 3 cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back to $P$ is drawn (see diagram). What is the minimum distance from the vertex $V$ to this path? [asy] import graph; unitsize(1 cm); filldraw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(Circle((0,0),1)),gray(0.9),nullpen); draw(yscale(0.3)*(arc((0,0),1.5,0,180)),dashed); draw(yscale(0.3)*(arc((0,0),1.5,180,360))); draw((1.5,0)--(0,4)--(-1.5,0)); draw((0,0)--(1.5,0),Arrows); draw(((1.5,0) + (0.3,0.1))--((0,4) + (0.3,0.1)),Arrows); draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,0,180)),dashed); draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,180,360))); label("$V$", (0,4), N); label("1 cm", (0.75,-0.5), N); label("$P$", (-1.5,0), SW); label("3 cm", (1.7,2)); [/asy]

A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. [asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]

Consider positive numbers $h, s_1, s_2$, and a spatial triangle $\vartriangle ABC$. How many ways are there to select a point $D$ so that the height of tetrahedron $ABCD$ drawn from $D$ equals $h$, and the areas of faces $ACD$ and $BCD$ equal $s_1$ and $s_2$, respectively?

With $150$ white cubes of $1 \times 1 \times 1$ a prism of $6 \times 5 \times 5$ is assembled, its six faces are painted blue and then the prism is disassembled. Lucrecia must build a new prism, without holes, exclusively using cubes that have at least one blue face and so that the faces of Lucrecia's prism are all completely blue. Give the dimensions of the prism with the largest volume that Lucrecia can assemble.

A cube is constructed from $4$ white unit cubes and $4$ black unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.) $\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$

Given is a trihedral angle with edges $ SA $, $ SB $, $ SC $, all plane angles of which are acute, and the dihedral angle at edge $ SA $ is right. Prove that the section of this triangle with any plane perpendicular to any edge, at a point different from the vertex $ S $, is a right triangle.

Which of the following polynomials does not divide $x^{60} - 1$? $ \textbf{a)}\ x^2+x+1 \qquad\textbf{b)}\ x^4-1 \qquad\textbf{c)}\ x^5-1 \qquad\textbf{d)}\ x^{15}-1 \qquad\textbf{e)}\ \text{None of above} $

A regular tetrahedron of edge length $1$ is orthogonally projected onto a plane. Find the largest possible area of its image.

The base of a pyramid is an isosceles triangle with the vertex angle $\alpha$. The pyramid’s lateral edges are at angle $\phi$ to the base. Find the dihedral angle $\theta$ at the edge connecting the pyramid’s vertex to that of angle $\alpha$.

A right prism with height $h$ has bases that are regular hexagons with sides of length $12$. A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60^\circ$. Find $h^2$.