[b]6.1[/b] Three shooters - Anilov, Borisov and Vorobiev - made $6$ each shots at one target and scored equal points. It is known that Anilov scored $43$ points in the first three shots, and Borisov scored $43$ points with the first shot knocked out 3 points. How many points did each shooter score per shot? if there was one hit in 50, two in 25, three in 20, three in 10, two in 5, in 3 - two, in 2 - two, in 1 - three? [img]https://cdn.artofproblemsolving.com/attachments/a/1/4abb71f7bccc0b9d2e22066ec17c31ef139d6a.png[/img] [b]6.2 / 7.4 [/b]Prove that a $10 \times 10$ chessboard cannot be covered with $ 25$ figures like [img]https://cdn.artofproblemsolving.com/attachments/0/4/89aafe1194628332ec13ad1c713bb35cbefff7.png[/img]. [b]6.3[/b] The squares of a chessboard contain natural numbers such that each is equal to the arithmetic mean of its neighbors. Sum of numbers standing in the corners of the board is $16$. Find the number standing on the field $e2$. [b]6.4 [/b] There is a table $ 100 \times 100$. What is the smallest number of letters which can be arranged in its cells so that no two are identical the letters weren't next to each other? [b]6.5[/b] The pioneer detachment is lined up in a rectangle. In each rank the tallest is noted, and from these pioneers the most short. In each row, the lowest one is noted, and from them is selected the tallest. Which of these two pioneers is taller? (This means that the two pioneers indicated are the highest of the low and the lowest of tall - must be different) [b]6.6[/b] Find the product of three numbers whose sum is equal to the sum of their squares, equal to the sum of their cubes and equal to $1$. PS. You should use hide for answers.Collected [url=https://artofproblemsolving.com/community/c3983461_1964_leningrad_math_olympiad]here[/url].

The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$. [b]proposed by L. Emelyanov, S. Berlov[/b]

a magical $3\times3$ square is a $3\times3$ matrix containing all number from 1 to 9, and of which the sum of every row, every column, every diagonal, are all equal. Determine all magical $3\times3$ square

$f: \mathbb R^{n}\longrightarrow\mathbb R^{n}$ is a bijective map, that Image of every $n-1$-dimensional affine space is a $n-1$-dimensional affine space. 1) Prove that Image of every line is a line. 2) Prove that $f$ is an affine map. (i.e. $f=goh$ that $g$ is a translation and $h$ is a linear map.)

Let $ ABC$ be a triangle, and $ D$ the midpoint of the side $ BC$. On the sides $ AB$ and $ AC$ we consider the points $ M$ and $ N$, respectively, both different from the midpoints of the sides, such that \[ AM^2\plus{}AN^2 \equal{}BM^2 \plus{} CN^2 \textrm{ and } \angle MDN \equal{} \angle BAC.\] Prove that $ \angle BAC \equal{} 90^\circ$.

Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral.

What is the smallest positive integer $t$ such that there exist integers $x_{1},x_{2}, \cdots, x_{t}$ with \[{x_{1}}^{3}+{x_{2}}^{3}+\cdots+{x_{t}}^{3}=2002^{2002}\;\;?\]

Be $R$ a positive real number. If $R, 1, R+\frac12$ are triangle sides, call $\theta$ the angle between $R$ and $R+\frac12$ (in rad). Prove $2R\theta$ is between $1$ and $\pi$.

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is: $ \textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ \textbf{(B)}\ \frac{1}{2} \text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2} \text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4} \text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}$

Suppose medians $m_a$ and $m_b$ of a triangle are orthogonal. Prove that: a.) Using medians of that triangle it is possible to construct a rectangular triangle. b.) The following inequality: \[5(a^2+b^2-c^2) \geq 8ab,\] is valid, where $a,b$ and $c$ are side length of the given triangle.

Let $ABCD$ be a trapezium $AB // CD,$ $M$ and $N$ are fixed points on $AB,$ $P$ is a variable point on $CD$. $E = DN \cap AP$, $F = DN \cap MC$, $G = MC \cap PB$, $DP = \lambda \cdot CD$. Find the value of $\lambda$ for which the area of quadrilateral $PEFG$ is maximum.

Let $\omega$ and $O$ be respectively the circumcircle and the circumcenter of a triangle $ABC$. The line $AO$ intersects $\omega$ second time at $A'$. $M_B$ and $M_C$ are the midpoints of $AC$ and $AB$, respectively. The lines $A'M_B$ and $A'M_C$ intersect $\omega$ secondly at points $B'$ and $C$, and also intersect $BC$ at points $D_B$ and $D_C$, respectively. The circumcircles of $CD_BB'$ and $BD_CC'$ intersect at points $P$ and $Q$. Prove that $O$, $P$, $Q$ are collinear. [i] (М. Германсков)[/i] [hide=Thanks]Thanks to the user Vlados021 for translating the problem.[/hide]

Let $ABC$ be a triangle with two angles $B,C$ not having the same measure, $I$ be its incircle, $(O)$ its circumcircle. Circle $(O_b)$ touches $BA,BC$ and is internally tangent to $(O)$ at $B_1$. Circle $(O_c)$ touches $CA,CB$ and is internally tangent to $(O)$ at $C_1$. Let $S$ be the intersection of $BC$ and $B_1C_1$. Prove that $\angle AIS = 90^o$. Nguyễn Minh Hà

In acute triangle ABC AB>AC. H is the orthocenter. M is midpoint of BC and AD is the symmedian line. Prove that if $\angle ADH= \angle MAH$, EF bisects segment AD. [img]https://s2.loli.net/2022/08/02/t9xzTV8IEv1qQRm.jpg[/img]

Does there exist a convex pentagon $A_1A_2A_3A_4A_5$ and a point $X$ inside it such that $XA_i=A_{i+2}A_{i+3}$ for all $i=1,...,5$ (all indices are considered modulo $5$) ? I. Voronovich

Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$ [i]Proposed by Michael Ren[/i]

Let a semicircle $AB$ be given and let $X$ be an inner point of the arc. Consider a point $Y$ on ray $XA$ such that $XY=XB$. Find the locus of all points $Y$ when $X$ moves on the arc $AB$ (excluding the endpoints).

A polygon $A$ that doesn't intersect itself and has perimeter $p$ is called [b]Rotund[/b] if for each two points $x,y$ on the sides of this polygon which their distance on the plane is less than $1$ their distance on the polygon is at most $\frac{p}{4}$. (Distance on the polygon is the length of smaller path between two points on the polygon) Now we shall prove that we can fit a circle with radius $\frac{1}{4}$ in any rotund polygon. The mathematicians of two planets earth and Tarator have two different approaches to prove the statement. In both approaches by "inner chord" we mean a segment with both endpoints on the polygon, and "diagonal" is an inner chord with vertices of the polygon as the endpoints. [b]Earth approach: Maximal Chord[/b] We know the fact that for every polygon, there exists an inner chord $xy$ with a length of at most 1 such that for any inner chord $x'y'$ with length of at most 1 the distance on the polygon of $x,y$ is more than the distance on the polygon of $x',y'$. This chord is called the [b]maximal chord[/b]. On the rotund polygon $A_0$ there's two different situations for maximal chord: (a) Prove that if the length of the maximal chord is exactly $1$, then a semicircle with diameter maximal chord fits completely inside $A_0$, so we can fit a circle with radius $\frac{1}{4}$ in $A_0$. (b) Prove that if the length of the maximal chord is less than one we still can fit a circle with radius $\frac{1}{4}$ in $A_0$. [b]Tarator approach: Triangulation[/b] Statement 1: For any polygon that the length of all sides is less than one and no circle with radius $\frac{1}{4}$ fits completely inside it, there exists a triangulation of it using diagonals such that no diagonal with length more than $1$ appears in the triangulation. Statement 2: For any polygon that no circle with radius $\frac{1}{4}$ fits completely inside it, can be divided into triangles that their sides are inner chords with length of at most 1. The mathematicians of planet Tarator proved that if the second statement is true, for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. (c) Prove that if the second statement is true, then for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. They found out that if the first statement is true then the second statement is also true, so they put a bounty of a doogh on proving the first statement. A young earth mathematician named J.N., found a counterexample for statement 1, thus receiving the bounty. (d) Find a 1392-gon that is counterexample for statement 1. But the Tarators are not disappointed and they are still trying to prove the second statement. (e) (Extra points) Prove or disprove the second statement. Time allowed for this problem was 150 minutes.

Let $ABCD$ be a unit square. For any interior points $M,N$ such that the line $MN$ does not contain a vertex of the square, we denote by $s(M,N)$ the least area of the triangles having their vertices in the set of points $\{ A,B,C,D,M,N\}$. Find the least number $k$ such that $s(M,N)\le k$, for all points $M,N$. [i]Dinu Șerbănescu[/i]

Let $M$ be a point inside quadrilateral $ABCD$ such that $ABMD$ is parallelogram. If $\angle CBM = \angle CDM$ prove that $\angle ACD = \angle BCM$

Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let$ K$ be the intersection point of the extension of side $AB$ (from $A$) with the bisector of the $\angle ACD$. If $MA \cdot MC + MA \cdot CD = MB\cdot MD$ , prove that $\angle BKC = \angle CDB$.

Three disks of diameter $d$ are touching a sphere in their centers. Besides, every disk touches the other two disks. How to choose the radius $R$ of the sphere in order that axis of the whole figure has an angle of $60^\circ$ with the line connecting the center of the sphere with the point of the disks which is at the largest distance from the axis ? (The axis of the figure is the line having the property that rotation of the figure of $120^\circ$ around that line brings the figure in the initial position. Disks are all on one side of the plane, passing through the center of the sphere and orthogonal to the axis).

In the cartesian plane, consider the curves $x^2+y^2=r^2$ and $(xy)^2=1$. Call $F_r$ the convex polygon with vertices the points of intersection of these 2 curves. (if they exist) (a) Find the area of the polygon as a function of $r$. (b) For which values of $r$ do we have a regular polygon?

Let $O$ be the circumcenter of triangle $ABC$. Choose a point $X$ on the circumcircle $\odot (ABC)$ such that $OX\parallel BC$. Assume that $\odot(AXO)$ intersects $AB, AC$ at $E, F$, respectively, and $OE, OF$ intersect $BC$ at $P, Q$, respectively. Furthermore, assume that $\odot(XP Q)$ and $\odot (ABC)$ intersect at $R$. Prove that $OR$ and $\odot (XP Q)$ are tangent to each other. (ltf0501)

Let $A,B$ and $C$ be three points on a circle of radius $1$. (a) Show that the area of the triangle $ABC$ equals \[\frac12(\sin(2\angle ABC)+\sin(2\angle BCA)+\sin(2\angle CAB))\] (b) Suppose that the magnitude of $\angle ABC$ is fixed. Then show that the area of the triangle $ABC$ is maximized when $\angle BCA=\angle CAB$ (c) Hence or otherwise, show that the area of the triangle $ABC$ is maximum when the triangle is equilateral.