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

Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that: [i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$ [i]b.)[/i] $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$ [i]c.)[/i] $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$
The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points $A (1, 2)$ and $B (3,1)$. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point $O (0, 0)$? Point A on the board located above and to the left of point $B$.
Two different points $A, M$ are given in a plane, $AM = d > 0$. Let a number $v > 0$ be given. Construct a rhombus $ABCD$ with the height of length $v$ and $M$ being a midpoint of $BC$. Discuss conditions of solvability and determine number of solutions. Can the resulting quadrilateral $ABCD$ be a square?
Given a segment $ AB $ and a line $ m $ parallel to this segment. Find the midpoint of the segment $ AB $ using only a ruler, i.e. drawing only straight lines.
On the plane, a non-isosceles triangle is given, a circle circumscribed around it and the center of its inscribed circle are marked. Using only a ruler without tick marks and drawing no more than seven lines, construct the diameter of the circumcircle.
In the plane there is a quadrilateral $ ABCD $ and a point $ M $. Construct a parallelogram with center $ M $ and its vertices lying on the lines $ AB $, $ BC $, $ CD $, $ DA $.
Show how to construct a line segment length $(a^4 + b^4)^{1/4}$ given segments lengths $a$ and $b$.
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 $d,\omega,SA$).
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]
A pair of positive integers $(x, y)$ is [i] good [/i] if they satisfy $\text{rad}(x) = \text{rad}(y)$ and they do not divide each-other. Given coprime positive integers $a$ and $b$, show that there exist infinitely many $n$ for which there exists a positive integer $m$ such that $(a^n + bm, b^n + am)$ is [i] good[/i]. (Here, $\text{rad}(x)$ denotes the product of $x$'s prime divisors, as usual.)
Betal marks $2021$ points on the plane such that no three are collinear, and draws all possible segments joining these. He then chooses any $1011$ of these segments, and marks their midpoints. Finally, he chooses a segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint using [b]only[/b] a straightedge. Can Vikram always complete this challenge? [i]Note.[/i] A straightedge is an infinitely long ruler without markings, which can only be used to draw the line joining any two given distinct points. [i]Proposed by Prithwijit De and Sutanay Bhattacharya[/i]
Show how to construct (by a ruler and a compass) a right-angled triangle, given its inradius and circumradius.
A circle is circumscribed around an isosceles triangle $ABC$ with base $BC$. The bisector of the angle $C$ and the bisector of the angles $A$ intersect the circle at the points $E$ and $D$, respectively, and the segment $DE$ intersects the sides $BC$ and $AB$ at the points $P$ and $Q$, respectively. Reconstruct $\vartriangle ABC$ given points $D, P, Q$, if it is known in which half-plane relative to the line $DQ$ lies the vertex $A$. (Maria Rozhkova)
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$
Restore the isosceles triangle $ABC$ ($AB = AC$) if the common points $I, M, H$ of bisectors, medians and altitudes respectively are given.
Given two sides of the triangle. Construct that triangle, if medians to those sides are orthogonal.
In triangle $ABC$, touching points $A', B'$ of the incircle with $BC, AC$ and common point $G$ of segments $AA'$ and $BB'$ were marked. After this the triangle was erased. Restore it by the ruler and the compass.
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 $.
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.
Let $ABC$ be a scalene triangle. Given the center $I$ of the inscribe circle and the points $K_1$, $K_2$ and $K_3$ where the inscribed circle is tangent to the sides $BC$, $AC$ and $AB$. Using only a ruler, construct the center of the circumscribed circle of triangle $ABC$. (Hryhorii Filippovskyi)
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]
Two circles, one inside the other, are given in the plane. Construct a point $O$, inside the inner circle, such that if a ray from $O$ cuts the circles at $A$ and $B$ respectively, then the ratio $OA/OB$ is constant. (Folklore)
Given the lengths of two sides of a triangle and that of the bisector of the angle between these sides, construct the triangle.
On the plane is drawn a triangle $ABC$ and a circle $\omega$ passing through the vertex $C$, the midpoints of the sides $AC$ and $BC$ and the point of intersection of the medians of the triangle $ABC$. The point $K$ lies on the circle $\omega$ such that $\angle AKB=90^o$. Using only with a ruler, draw a tangent to the circle $\omega$ at point $K$.