In a triangle $ABC$, the points $A_1, B_1, C_1$ are symmetric to $A, B,C$ with respect to $B,C, A$, respectively. Given the points $A_1, B_1,C_1$ reconstruct the triangle $ABC$.

On the plane are given four distinct points $A,B,C,D$ on a line $g$ in this order, at the mutual distances $AB = a, BC = b, CD = c$. (a) Construct (if possible) a point $P$ outside line $g$ such that $\angle APB =\angle BPC =\angle CPD$. (b) Prove that such a point $P$ exists if and only if $ (a+b)(b+c) < 4ac$

Let be given an obtuse angle $AKS$ in the plane. Construct a triangle $ABC$ such that $S$ is the midpoint of $BC$ and $K$ is the intersection point of $BC$ with the bisector of $\angle BAC$.

Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane; but she can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment into $17$ equal parts. Laura drew a parallelogram $ABCD$ and decided to try out her magical ruler; with it, she found the midpoint $M$ of side $CD$, and she extended $CB$ beyond $B$ to point $N$ so that segments $CB$ and $BN$ were equal in length. Unfortunately, her mischievous little brother came along and erased everything on Laura's picture except for points $A, M$, and $N$. Using Laura's magical ruler, help her reconstruct the original parallelogram $ABCD$: write down the steps that she needs to follow and prove why this will lead to reconstructing the original parallelogram $ABCD$.

A $2 \times 2$ square was cut out of a sheet of grid paper. Using only a ruler without divisions and without going beyond the square, divide the diagonal of the square into $6$ equal parts.

Let $m, n \ge 2$ be integers. Carl is given $n$ marked points in the plane and wishes to mark their centroid.* He has no standard compass or straightedge. Instead, he has a device which, given marked points $A$ and $B$, marks the $m-1$ points that divide segment $\overline{AB}$ into $m$ congruent parts (but does not draw the segment). For which pairs $(m,n)$ can Carl necessarily accomplish his task, regardless of which $n$ points he is given? *Here, the [i]centroid[/i] of $n$ points with coordinates $(x_1, y_1), \dots, (x_n, y_n)$ is the point with coordinates $\left( \frac{x_1 + \dots + x_n}{n}, \frac{y_1 + \dots + y_n}{n}\right)$. [i]Proposed by Holden Mui and Carl Schildkraut[/i]

An acute-angled non-isosceles triangle $ABC$ is drawn, a circumscribed circle and its center $O$ are drawn. The midpoint of side $AB$ is also marked. Using only a ruler (no divisions), construct the triangle's orthocenter by drawing no more than $6$ lines. (Yu. Blinkov)

A triangle $A_1B_1C_1$ is a parallel projection of a triangle $ABC$ in space. The parallel projections $A_1H_1$ and $C_1L_1$ of the altitude $AH$ and the bisector $CL$ of $\vartriangle ABC$ respectively are drawn. Using a ruler and compass, construct a parallel projection of : (a) the orthocenter, (b) the incenter of $\vartriangle ABC$.

Given two concentric circles and a point $A$. Through point $A$, draw a secant such that its segment contained by the larger circle is divided by the smaller circle into three equal parts.

Given circle arc, whose center is an inaccessible point. $A$ is a point on this arc (see fig.). How to construct using compass and ruler without divisions, a tangent to given circle arc at point $A$ ? [img]https://1.bp.blogspot.com/-7oQBNJGLsVw/W6dYm4Xw7bI/AAAAAAAAJH8/sJ-rgAQZkW0kvlPOPwYiGjnOXGQZuDnRgCK4BGAYYCw/s1600/Yasinsky%2B2017%2BVIII-IX%2Bp3.png[/img]

Two different points $A,S$ are given in the plane. Furthermore, positive numbers $d,\omega$ are given, $\omega<180^\circ.$ Let $X$ be a point and $X'$ its image under the rotation by the angle $\omega$ (in counter-clockwise direction) with respect to the origin $S.$ Construct all points $X$ such that $XX'=d$ and $A$ is a point of the segment $XX'.$ Discuss conditions of solvability (in terms of $d,\omega,SA$).

The board features a triangle $ABC$, its center of the circle circumscribed is the point $O$, the midpoint of the side $BC$ is the point $F$, and also some point $K$ on side $AC$ (see fig.). Master knowing that $\angle BAC$ of this triangle is equal to the sharp angle $\alpha$ has separately drawn an angle equal to $\alpha$. After this teacher wiped the board, leaving only the points $O, F, K$ and the angle $\alpha$. Is it possible with a compass and a ruler to construct the triangle $ABC$ ? Justify the answer. (Grigory Filippovsky) [img]https://1.bp.blogspot.com/-RRPt8HbqW4I/XObthZFXyyI/AAAAAAAAKOo/zfHemPjUsI4XAfV_tcmKA6_al0i_gQ9iACK4BGAYYCw/s1600/Yasinsky%2B2019%2Bp6.png[/img]

On the base $BC$ of the isosceles triangle $ABC$ chose a point $D$ and in each of the triangles $ABD$ and $ACD$ inscribe a circle. Then everything was wiped, leaving only two circles. It is known from which side of their line of centers the apex $A$ is located . Use a compass and ruler to restore the triangle $ABC$ , if we know that : a) $AD$ is angle bisector, b) $AD$ is median.

Let $\ell$ be a given line, $A$ and $B$ given points of the plane. Choose a point $P$ on $\ell $ so that the longer of the segments $AP$, $BP$ is as short as possible. (If $AP = BP,$ either segment may be taken as the longer one.)

An infinite sequence $ \,x_{0},x_{1},x_{2},\ldots \,$ of real numbers is said to be [b]bounded[/b] if there is a constant $ \,C\,$ such that $ \, \vert x_{i} \vert \leq C\,$ for every $ \,i\geq 0$. Given any real number $ \,a > 1,\,$ construct a bounded infinite sequence $ x_{0},x_{1},x_{2},\ldots \,$ such that \[ \vert x_{i} \minus{} x_{j} \vert \vert i \minus{} j \vert^{a}\geq 1 \] for every pair of distinct nonnegative integers $ i, j$.

Given a point $P$ lying on a line $g,$ and given a circle $K.$ Construct a circle passing through the point $P$ and touching the circle $K$ and the line $g.$

The given points are $ A $ and $ B $ and the circle $ k $. Draw a circle passing through the points $ A $ and $ B $ and defining, at the intersection with the circle $ k $, a common chord of a given length $ d $.

Line segment $AM=d>0$ is given in the plane. Furthermore, a positive number $v$ is given. Construct a right triangle $ABC$ with hypotenuse $AB$, altitude to the hypotenuse of the length $v$ and the leg $BC$ being divided by $M$ in ration $MB/MC=2/3$. Discuss conditions of solvability in terms of $d, v$.

Let $A$, $X$ and $Y$ be three non-collinear points on the plane. Construct with a straightedge and compass a square $ABCD$ such that $X$ is on the line $BC$ and $Y$ is on the line $CD$.

Inscribe the equilateral triangle of the largest perimeter in a given semicircle.

Let $AC$ be the hypothenuse of a right-angled triangle $ABC$. The bisector $BD$ is given, and the midpoints $E$ and $F$ of the arcs $BD$ of the circumcircles of triangles $ADB$ and $CDB$ respectively are marked (the circles are erased). Construct the centers of these circles using only a ruler.

On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$.

Let $n$ be a positive integer. The Georgian folk dance team consists of $2n$ dancers, with $n$ males and $n$ females. Each dancer, both male and female, is assigned a number from 1 to $n$. During one of their dances, all the dancers line up in a single line. Their wish is that, for every integer $k$ from 1 to $n$, there are exactly $k$ dancers positioned between the $k$th numbered male and the $k$th numbered female. Prove the following statements: a) If $n \equiv 1 \text{ or } 2 \mod{4}$, then the dancers cannot fulfill their wish. b) If $n \equiv 0 \text{ or } 3 \mod{4}$, then the dancers can fulfill their wish. [i]Proposed by Giorgi Arabidze, Georgia[/i]

Construct the $ \triangle ABC $, with $ AC <BC $, if the circumcircle is known, and the points $ D, E, F $ in it, where they intersect, respectively, the altitude, the median and the angle bisector that they start from the vertex $ C $.

A circle $k$ is drawn using a given disc (e.g. a coin). A point $A$ is chosen on $k$. Using just the given disc, determine the point $B$ on $k$ so that $AB$ is a diameter of $k$. (You are allowed to choose an arbitrary point in one of the drawn circles, and using the given disc it is possible to construct either of the two circles that passes through the points at a distance that is smaller than the radius of the circle.)