This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 329

In triangle $ABC$, points $D, E$, and $F$ are marked on sides $AC, BC$, and $AB$ respectively, so that $AD = AB$, $EC = DC$, $BF = BE$. After that, they erased everything except points $E, F$ and $D$. Reconstruct the triangle $ABC$ (no study required).
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 $ABCD$.
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 $s$ be the circumcircle of triangle $ABC, L$ and $W$ be common points of angle's $A$ bisector with side $BC$ and $s$ respectively, $O$ be the circumcenter of triangle $ACL$. Restore triangle $ABC$, if circle $s$ and points $W$ and $O$ are given. (D.Prokopenko)
For which positive integers $n$, one can find real numbers $x_1,x_2,\cdots ,x_n$ such that $$\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{\left(x_1+2x_2+\cdots+nx_n\right)^2}=\dfrac{27}{4n(n+1)(2n+1)}$$ and $i\leq x_i\leq 2i$ for all $i=1,2,\cdots ,n$ ?
Given a circle and two tangents to this circle. Draw a third tangent to the circle in such a way that its segment contained by the given tangents has the given length $ d $.
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 $ABC$.
Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.
Let $CH$ be the altitude of the triangle $ABC$ drawn on the board, in which $\angle C = 90^o$, $CA \ne CB$. The mathematics teacher drew the perpendicular bisectors of segments$ CA$ and $CB$, which cut the line CH at points $K$ and $M$, respectively, and then erased the drawing, leaving only the points $C, K$ and $M$ on the board. Restore triangle $ABC$, using only a compass and a ruler.
Let $AP$ and $BQ$ be the altitudes of acute-angled triangle $ABC$. Using a compass and a ruler, construct a point $M$ on side $AB$ such that $\angle AQM = \angle BPM$.
Using a compass and a ruler, construct a point $K$ inside an acute-angled triangle $ABC$ so that $\angle KBA = 2\angle KAB$ and $ \angle KBC = 2\angle KCB$.
Let $r>1$ be a rational number. Alice plays a solitaire game on a number line. Initially there is a red bead at $0$ and a blue bead at $1$. In a move, Alice chooses one of the beads and an integer $k \in \mathbb{Z}$. If the chosen bead is at $x$, and the other bead is at $y$, then the bead at $x$ is moved to the point $x'$ satisfying $x'-y=r^k(x-y)$. Find all $r$ for which Alice can move the red bead to $1$ in at most $2021$ moves.
There is a ruler and a "rusty" compass, with which you can construct a circle of radius $R$. The point $K$ is from the line $\ell$ at a distance greater than $R$. How to use this ruler and this compass to draw a line passing through the point $K$ and perpendicular to line $\ell$? (Misha Sidorenko, Katya Sidorenko, Rodion Osokin)
A trapezoid is given in which one base is twice as large as the other. Use one ruler (no divisions) to draw the midline of this trapezoid.
In triangle $ABC$, the bisector $CL$ was drawn. The incircles of triangles $CAL$ and $CBL$ touch $AB$ at points $M$ and $N$ respectively. Points $M$ and $N$ are marked on the picture, and then the whole picture except the points $A, L, M$, and $N$ is erased. Restore the triangle using a compass and a ruler. (V.Protasov)
Construct a triangle given a side, the radius of the inscribed circle, and the radius of the exscribed circle tangent to that side. (Research is not required.)
Consider a circle and a point $P$ outside the circle. The angle of given measure with vertex at $P$ subtends a diameter of the circle. Construct the circle’s diameter with ruler and compass.
Given a circle $ k $ and a point inside it $ H $. Inscribe a triangle in the circle such that this point $ H $ is the point of intersection of the triangle's altitudes.
Draw a circle that has a given radius $R$ and is tangent to a given line and a given circle. How many solutions does this problem have?
Construct a triangle knowing an angle, the ratio of the sides that form it and the radius of the inscribed circle.
We know the lengths of the $3$ altitudes of a triangle. Construct the triangle.
Construct a right triangle given the hypotenuse and the median drawn to the leg.
Given: point $ A $, line $ p $, and circle $ k $. Construct a triangle $ ABC $ with angles $ A = 60^\circ $, $ B = 90^\circ $, whose vertex $ B $ lies on line $ p $, and vertex $ C $ - on circle $ k $.
On the plane there is an image of a circle with a marked center. Let an arbitrary angle be drawn on this plane. Using one ruler, construct the bisector of this angle.
Inside the triangle $ ABC $ a point $ P $ is given; find a point $ Q $ on the perimeter of this triangle such that the broken line $ APQ $ divides the triangle into two parts with equal areas.