A hemisphere is inscribed in a cone so that its base lies on the base of the cone. The ratio of the area of the entire surface of the cone to the area of the hemisphere (without the base) is $\frac{18}5$. Compute the angle at the vertex of the cone.

Let $ABC$ be a triangle with $AB = 4024$, $AC = 4024$, and $BC=2012$. The reflection of line $AC$ over line $AB$ meets the circumcircle of $\triangle{ABC}$ at a point $D\ne A$. Find the length of segment $CD$. [i]Ray Li.[/i]

Three distinct points with integer coordinates lie in the plane on a circle of radius $r>0$. Show that two of these points are separated by a distance of at least $r^{1/3}$.

Three rods of lengths $1396, 1439$, and $2018$ millimeters have been hinged from one tip on the ground. What is the smallest value for the radius of the circle passing through the other three tips of the rods in millimeters?

Given the segments $AB$ and $CD$ not necessarily on the same plane. Point $X$ is the midpoint of the segment $AB$, and the point $Y$ is the midpoint of $CD$. Given that point $X$ is not on line $CD$, and that point $Y$ is not on line $AB$, prove that $2 | XY | \le | AD | + | BC |$. When is equality achieved?

The non-isosceles triangle $ABC$ is inscribed in the circle ω. The tangent to this circle at the point $C$ intersects the line $AB$ at the point $D$. Let the bisector of the angle $CDB$ intersect the segments $AC$ and $BC$ at the points $K$ and $L$, respectively. On the side $AB$, the point $M$ is taken such that $AK / BL = AM / BM$. Let the perpendiculars from the point $M$ to the lines $KL$ and $DC$ intersect the lines $AC$ and $DC$ at the points $P$ and $Q$, respectively. Prove that the angle $CQP$ is half of the angle $ACB$.

Points $W$, $X$, $Y,$ and $Z$ are chosen inside a regular octagon so that four congruent rhombuses are formed, as shown in the diagram below. If the side length of the octagon is $1$, compute the area of quadrilateral $WXY Z$. [img]https://cdn.artofproblemsolving.com/attachments/9/6/bb12385cbd9fd802b3f3960b5e449268be45d4.png[/img]

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots,5$ is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

Prove that for any positive integer $n$, the number of odd integers among the binomial coefficients $\binom nh \ ( 0 \leq h \leq n)$ is a power of 2.

Let $\vartriangle A_1B_1C_1$ and $l_1, m_1, n_1$ be the trisectors closest to $A_1B_1$, $B_1C_1$, $C_1A_1$ of the angles $A_1, B_1, C_1$ respectively. Let $A_2 = l_1 \cap n_1$, $B_2 = m_1 \cap l_1$, $C_2 = n_1 \cap m_1$. So on we create triangles $\vartriangle A_nB_nC_n$ . If $\vartriangle A_1B_1C_1$ is equilateral prove that exists $n \in N$, such that all the sides of $\vartriangle A_nB_nC_n$ are parallel to the sides of $\vartriangle A_1B_1C_1$.

Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Given a concyclic quadrilateral $ABCD$ with circumcenter $O$. Let $E$ be the intersection of $AD$ and $BC$, while $F$ be the intersection of $AC$ and $BD$. A circle $w$ are tangent to $BD$ and $AC$ such that $F$ is the orthocenter of $\triangle QEP$ where $PQ$ is a diameter of $w$. Prove that $EO$ passes through the center of $w$.

Prove that $$AB + PQ + QR + RP \le AP + AQ + AR + BP + BQ + BR$$ where $A, B, P, Q$ and $R $ are any five points in a plane.

Prove that if a set $X\subset S^n$ takes up more than half a Riemannian volume of a unit sphere $S^n$, then the set of all possible geodesic segments length less than $\pi$ with endpoints in the set $X$ covers the entire sphere. Geodetic on sphere $S^n$ is a curve lying on some circle of intersection of the sphere $S^n\subset R^{n+1}$ two-dimensional linear subspace $L \subset R^{n+1}$

In equilateral triangle $ABC$, the midpoint of $\overline{BC}$ is $M$. If the circumcircle of triangle $MAB$ has area $36\pi$, then find the perimeter of the triangle. [i]Proposed by Isabella Grabski [/i]

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).

Show that a triangle whose side lengths are prime numbers cannot have integer area.

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0,$ and the hunter's starting point, $B_0$ are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}.$ In the $n^{\text{th}}$ round of the game, three things occur in order: [list=i] [*]The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly $1.$ [*]A tracking device reports a point $P_n$ to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between $P_n$ and $A_n$ is at most $1.$ [*]The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly $1.$ [/list] Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds, she can ensure that the distance between her and the rabbit is at most $100?$ [i]Proposed by Gerhard Woeginger, Austria[/i]

Given the segment $ PQ$ and a circle . A chord $AB$ moves around the circle, equal to $PQ$. Let $T$ be the intersection point of the perpendicular bisectors of the segments $AP$ and $BQ$. Prove that all points of $T$ thus obtained lie on one line.

Let $JHIZ$ be a rectangle, and let $A$ and $C$ be points on sides $ZI$ and $ZJ,$ respectively. The perpendicular from $A$ to $CH$ intersects line $HI$ in $X$ and the perpendicular from $C$ to $AH$ intersects line $HJ$ in $Y.$ Prove that $X,$ $Y,$ and $Z$ are collinear (lie on the same line).

In a convex quadrilateral $ ABCD $ with area $ S $, each side was divided into 3 equal parts and segments were drawn connecting the appropriate points of division of the opposite sides in such a way that the quadrilateral was divided into 9 quadrilaterals. Prove that the sum of the areas of the following three quadrilaterals resulting from the division: the one containing the vertex $ A $, the middle one and the one containing the vertex $ C $ is equal to $ \frac{S}{3} $.

In acute-angled triangle $ABC$ ($AC > AB$), point $H$ is the orthocenter and point $M$ is the midpoint of the segment $BC$. The median $AM$ intersects the circumcircle of triangle $ABC$ at $X$. The line $CH$ intersects the perpendicular bisector of $BC$ at $E$ and the circumcircle of the triangle $ABC$ again at $F$. Point $J$ lies on circle $\omega$, passing through $X, E,$ and $F$, such that $BCHJ$ is a trapezoid ($CB \parallel HJ$). Prove that $JB$ and $EM$ meet on $\omega$. [i]Proposed by Alireza Dadgarnia[/i]

The square was cut into acute -angled triangles. Prove that there are at least eight of them.

Let $ABCD$ be a parallelogram. The line $\Delta$ meets the lines $AB, BC, CD$ and $DA$ at $M, N, P$ and $Q,$ respectively. Let $R$ be the intersection point of the lines $AB,DN$ and let $S$ be intersection point of the lines $AD, BP.$ Prove that $RS \parallel \Delta.$ [asy] import graph; size(400); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); pen qqzzcc = rgb(0,0.6,0.8); pen wwwwff = rgb(0.4,0.4,1); draw((2,2)--(6,2),qqzzcc+linewidth(1.6pt)); draw((6,2)--(4,0),qqzzcc+linewidth(1.6pt)); draw((-1.95,(+12-2*-1.95)/2)--(12.24,(+12-2*12.24)/2),qqzzcc+linewidth(1.6pt)); draw((-1.95,(-0+3*-1.95)/3)--(12.24,(-0+3*12.24)/3),qqzzcc+linewidth(1.6pt)); draw((-1.95,(-0-0*-1.95)/6)--(12.24,(-0-0*12.24)/6),qqzzcc+linewidth(1.6pt)); draw((4,0)--(4,4),wwwwff+linewidth(1.2pt)+linetype("3pt 3pt")); draw((2,2)--(8.14,0),wwwwff+linewidth(1.2pt)+linetype("3pt 3pt")); draw((-1.95,(+32.56-4*-1.95)/4.14)--(12.24,(+32.56-4*12.24)/4.14),qqzzcc+linewidth(1.6pt)); dot((0,0),ds); label("$A$", (0,-0.3),NE*lsf); dot((4,0),ds); label("$B$", (4.02,-0.33),NE*lsf); dot((2,2),ds); label("$D$", (1.81,2.07),NE*lsf); dot((6,2),ds); label("$C$", (6.16,2.08),NE*lsf); dot((3,3),ds); label("$Q$", (2.97,3.22),NE*lsf); dot((5,1),ds); label("$N$", (4.99,1.19),NE*lsf); label("$\Delta$", (1.7,3.76),NE*lsf); dot((6,0),ds); label("$M$", (5.9,-0.33),NE*lsf); dot((4,2),ds); label("$P$", (4.02,2.08),NE*lsf); dot((4,4),ds); label("$S$", (3.94,4.12),NE*lsf); dot((8.14,0),ds); label("$E$", (8.2,0.09),NE*lsf); clip((-1.95,-6.96)--(-1.95,4.99)--(12.24,4.99)--(12.24,-6.96)--cycle); [/asy]