Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.

Fencing Ferdinand wants to fence three rectangular areas. there are fences in three types, with $4$ amount of fences of each type. You will notice that there is always at least as much area it manages to enclose a total of three by enclosing three square areas (i.e., each area fencing elements of the same size to enclose it) as if it were three different, rectangular would encircle an area (i.e., use two different elements for each of the three areas). Why is this is so? When does it not matter how he fences the rectangles, in terms of the sum of the areas?

A rectangular sheet of paper is folded so that one corner lies on top of the corner diagonally opposite. The resulting shape is a pentagon whose area is $20\%$ one-sheet thick, and $80\%$ two-sheets-thick. Determine the ratio of the two sides of the original sheet of paper.

Given a regular tetrahedron $ABCD$ with center $O$, find $\sin \angle AOB$.

The points $A_1,B_1,C_1$ lie on the sides $BC,CA$ and $AB$ of the triangle $ABC$ respectively. Suppose that $AB_1-AC_1=CA_1-CB_1=BC_1-BA_1$. Let $O_A,O_B$ and $O_C$ be the circumcentres of triangles $AB_1C_1,A_1BC_1$ and $A_1B_1C$ respectively. Prove that the incentre of triangle $O_AO_BO_C$ is the incentre of triangle $ABC$ too.

Three equal regular tetrahedrons have the common center. Is it possible that all faces of the polyhedron that forms their intersection are equal?

The circles $\mathcal{K}_1,\mathcal{K}_2$ and $\mathcal{K}_3$ are pairwise externally tangent to each other; the point of tangency betwwen $\mathcal{K}_1$ and $\mathcal{K}_2$ is $T$. One of the external common tangents of $\mathcal{K}_1$ and $\mathcal{K}_2$ meets $\mathcal{K}_3$ at points $P$ and $Q$. Prove that the internal common tangent of $\mathcal{K}_1$ and $\mathcal{K}_2$ bisects the arc $PQ$ of $\mathcal{K}_3$ which is closer to $T$.

Let $ ABCD$ be a quadrilateral inscribed in a circle. The diagonal $ BD$ bisects $ AC$. If $ AB = 10$, $ AD = 12$ and $ DC = 11$, find $ BC$.

Let $ABC$ be an isosceles triangle with $AB=AC$ . Let also $\omega$ be a circle of center $K$ tangent to the line $AC$ at $C$ which intersects the segment $BC$ again at $H$ . Prove that $HK \bot AB $.

Let $L_i,\ i=1,2,3$, be line segments on the sides of an equilateral triangle, one segment on each side, with lengths $l_i,\ i=1,2,3$. By $L_i^{\ast}$ we denote the segment of length $l_i$ with its midpoint on the midpoint of the corresponding side of the triangle. Let $M(L)$ be the set of points in the plane whose orthogonal projections on the sides of the triangle are in $L_1,L_2$, and $L_3$, respectively; $M(L^{\ast})$ is defined correspondingly. Prove that if $l_1\ge l_2+l_3$, we have that the area of $M(L)$ is less than or equal to the area of $M(L^{\ast})$.

Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} \plus{} r_{j} \plus{} r_{k}$.

Consider the isosceles right triangle$ ABC, \angle A = 90^o$, and point $D \in (AB)$ such that $AD = \frac13 AB$. In the half-plane determined by the line $AB$ and point $C$ , consider a point $E$ such that $\angle BDE = 60^o$ and $\angle DBE = 75^o$. Lines $BC$ and $DE$ intersect at point $G$, and the line passing through point $G$ parallel to the line $AC$ intersects the line $BE$ at point $H$. Prove that the triangle $CEH$ is equilateral.

Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ and radius $10$. Let sides $AB$, $BC$, $CD$, and $DA$ have midpoints $M, N, P$, and $Q$, respectively. If $MP = NQ$ and $OM + OP = 16$, then what is the area of triangle $\vartriangle OAB$?

Several pupils wrote a solution of a math problem on the blackboard on the break. When the teacher came in, a pupil was just clearing the blackboard, so the teacher could only observe that there was a rectangle with the sides of integer lenghts and a diagonal of lenght $ 2002$. Then the teacher pointed out that there was a computation error in pupils' solution. Why did he conclude that?

[u]Round 1[/u] [b]p1.[/b] Ash is running around town catching Pokémon. Each day, he may add $3, 4$, or $5$ Pokémon to his collection, but he can never add the same number of Pokémon on two consecutive days. What is the smallest number of days it could take for him to collect exactly $100$ Pokémon? [b]p2.[/b] Jack and Jill have ten buckets. One bucket can hold up to $1$ gallon of water, another can hold up to $2$ gallons, and so on, with the largest able to hold up to $10$ gallons. The ten buckets are arranged in a line as shown below. Jack and Jill can pour some amount of water into each bucket, but no bucket can have less water than the one to its left. Is it possible that together, the ten buckets can hold 36 gallons of water? [img]https://cdn.artofproblemsolving.com/attachments/f/8/0b6524bebe8fe859fe7b1bc887ac786106fc17.png[/img] [b]p3.[/b] There are $2023$ knights and liars standing in a row. Knights always tell the truth and liars always lie. Each of them says, “the number of liars to the left of me is greater than the number of knights to the right.” How many liars are there? [b]p4.[/b] Camila has a deck of $101$ cards numbered $1, 2, ..., 101$. She starts with $50$ random cards in her hand and the rest on a table with the numbers visible. In an exchange, she replaces all $50$ cards in her hand with her choice of $50$ of the $51$ cards from the table. Show that Camila can make at most 50 exchanges and end up with cards $1, 2, ..., 50$. [img]https://cdn.artofproblemsolving.com/attachments/0/6/c89e65118764f3b593da45264bfd0d89e95067.png[/img] [b]p5.[/b] There are $101$ pirates on a pirate ship: the captain and $100$ crew. Each pirate, including the captain, starts with $1$ gold coin. The captain makes proposals for redistributing the coins, and the crew vote on these proposals. The captain does not vote. For every proposal, each crew member greedily votes “yes” if he gains coins as a result of the proposal, “no” if he loses coins, and passes otherwise. If strictly more crew members vote “yes” than “no,” the proposal takes effect. The captain can make any number of proposals, one after the other. What is the largest number of coins the captain can accumulate? [u]Round 2[/u] [b]p6.[/b] The town of Lumenville has $100$ houses and is preparing for the math festival. The Tesla wiring company will lay lengths of power wire in straight lines between the houses so that power flows between any two houses, possibly by passing through other houses. The Edison lighting company will hang strings of lights in straight lines between pairs of houses so that each house is connected by a string to exactly one other. Show that however the houses are arranged, the Edison company can always hang their strings of lights so that the total length of the strings is no more than the total length of the power wires the Tesla company used. [img]https://cdn.artofproblemsolving.com/attachments/9/2/763de9f4138b4dc552247e9316175036c649b6.png[/img] [b]p7.[/b] You are given a sequence of $16$ digits. Is it always possible to select one or more digits in a row, so that multiplying them results in a square number? [img]https://cdn.artofproblemsolving.com/attachments/d/1/f4fcda2e1e6d4a1f3a56cd1a04029dffcd3529.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Inside the acute triangle $ABC$ the point $P{}$ in on height $AA_1{}$. Lines $BP{}$ and $CP{}$ intersect the sides $AC{}$ and $AB{}$, respectively, in points $B_1{}$ and $C_1{}$. Prove that: [b]a)[/b] $AA_1{}$ is the bisector of the angle $B_1A_1C_1;$ [b]b)[/b] if the lines $BC$ and $B_1C_1$ are concurrent, then the position of theri intersection does not depend on $P.$

Inside the acute-angled triangle $ ABC $ we consider the point $ P $ and its projections $ L, M, N $ to the sides $ BC, CA, AB $, respectively. Determine the point $ P $ for which the sum $ |BL|^2 + |CM|^2 + |AN|^2 $ is the smallest.

Points $ A_{1}$, $ B_{1}$, $ C_{1}$ are chosen on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ respectively. The circumcircles of triangles $ AB_{1}C_{1}$, $ BC_{1}A_{1}$, $ CA_{1}B_{1}$ intersect the circumcircle of triangle $ ABC$ again at points $ A_{2}$, $ B_{2}$, $ C_{2}$ respectively ($ A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$). Points $ A_{3}$, $ B_{3}$, $ C_{3}$ are symmetric to $ A_{1}$, $ B_{1}$, $ C_{1}$ with respect to the midpoints of the sides $ BC$, $ CA$, $ AB$ respectively. Prove that the triangles $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar.

Triangle $\vartriangle ABC$ is isosceles with $AB = AC$ and $\angle ABC = 2\angle BAC$. Compute $\frac{AB}{BC}$ .

Consider $2002$ segments on a plane, such that their lengths are the same. Prove that there exists such a straight line $r$ such that the sum of the lengths of the projections of the $2002$ segments about $r$ is less than $\frac{2}{3}$.

A plastic ball with a radius of 45 mm has a circular hole made in it. The hole is made to fit a ball with a radius of 35 mm, in such a way that the distance between their centers is 60 mm. Calculate the radius of the hole.

In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$. The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$ M. Didin

Let $ABCD$ be the cyclic quadrilateral. Suppose that there exists some line $l$ parallel to $BD$ which is tangent to the inscribed circles of triangles $ABC, CDA$. Show that $l$ passes through the incenter of $BCD$ or through the incenter of $DAB$. [i](Proposed by Fedir Yudin)[/i]

Determine the least real number $a > 1$ such that for any point $P$ in the interior of a square $ABCD$, the ratio of the areas of some two triangle $PAB, PBC, PCD, PDA$ lies in the interval $[1/a, a]$.

Is it possible to locate in a rectangle of $5$ cm by $ 8$ cm, $51$ circles of diameter $ 1$ cm, so that they don't overlap? Could it be possible for more than $40$ circles ?