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

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Found problems: 329

i) $ABCD$ is a rhombus. A tangent to the inscribed circle meets $AB, DA, BC, CD$ at $M, N, P, Q$ respectively. Find a relationship between $BM$ and $DN$. ii) $ABCD$ is a rhombus and $P$ a point inside. The circles through $P$ with centers $A, B, C, D$ meet the four sides $AB, BC, CD, DA$ in eight points. Find a property of the resulting octagon. Use it to construct a regular octagon. iii) Rotate the figure about the line $AC$ to form a solid. State a similar result.
Let $\mathbb{N}_0$ denote the set of non-negative integers. Determine all non-negative integers $k$ for which there exists a function $f: \mathbb{N}_0 \to \mathbb{N}_0$ such that $f(2024) = k$ and $f(f(n)) \leq f(n+1) - f(n)$ for all non-negative integers $n$.
The convex pentagon $ABCDE$ was drawn in the plane. $A_1$ was symmetric to $A$ with respect to $B$. $B_1$ was symmetric to $B$ with respect to $C$. $C_1$ was symmetric to $C$ with respect to $D$. $D_1$ was symmetric to $D$ with respect to $E$. $E_1$ was symmetric to $E$ with respect to $A$. How is it possible to restore the initial pentagon with the compasses and ruler, knowing $A_1,B_1,C_1,D_1,E_1$ points?
Restore a triangle by one of its vertices, the circumcenter and the Lemoine's point. [i](The Lemoine's point is the intersection point of the reflections of the medians in the correspondent angle bisectors)[/i]
Prove that infinitely many positive integers can be represented as $(a-1)/b + (b-1)/c + (c-1)/a$, where $a$, $b$ and $c$ are pairwise distinct positive integers greater than 1.
Triangle $ABC$ is inscribed into circle $k$. Points $A_1,B_1, C_1$ on its sides were marked, after this the triangle was erased. Prove that it can be restored uniquely if and only if $AA_1, BB_1$ and $CC_1$ concur.
Let \(AOB\) be a given angle less than \(180^{\circ}\) and let \(P\) be an interior point of the angular region determined by \(\angle AOB\). Show, with proof, how to construct, using only ruler and compass, a line segment \(CD\) passing through \(P\) such that \(C\) lies on the way \(OA\) and \(D\) lies on the ray \(OB\), and \(CP:PD=1:2\).
Let $I$ be the incenter of triangle $ABC$. $K_1$ and $K_2$ are the points on $BC$ and $AC$ respectively, at which the inscribed circle is tangent. Using a ruler and a compass, find the center of the inscribed circle for triangle $CK_1K_2$ in the minimal possible number of steps (each step is to draw a circle or a line). (Hryhorii Filippovskyi)
$AB$ and $CD$ are perpendicular diameters of a circle. $L$ is the tangent to the circle at $A$. $M$ is a variable point on the minor arc $AC$. The ray $BM, DM$ meet the line $L$ at $P$ and $Q$ respectively. Show that $AP\cdot AQ = AB\cdot PQ$. Show how to construct the point $M$ which gives$ BQ$ parallel to $DP$. If the lines $OP$ and $BQ$ meet at $N$ find the locus of $N$. The lines $BP$ and $BQ$ meet the tangent at $D$ at $P'$ and $Q'$ respectively. Find the relation between $P'$ and $Q$'. The lines $D$P and $DQ$ meet the line $BC$ at $P"$ and $Q"$ respectively. Find the relation between $P"$ and $Q"$.
Given a triangle $ABC$, $\angle C= 60^o$. Construct a point $P$ on the side $AC$ and a point $Q$ on side $BC$ such that $ABQP$ is a trapezoid whose diagonals make an angle of $60^o$ with each other.
We are given a ruler with two marks at a distance 1. With its help we can do all possible constructions as with a ruler with no measurements, including one more: If there is a line $l$ and point $A$ on $l$, then we can construct points $P_1,P_2\in l$ for which $AP_1=AP_2=1$. By using this ruler, construct a perpendicular from a given point to a given line.
Let's call a convex figure, the boundary of which consists of two segments and an arc of a circle, a mushroom-gon (see fig.). An arbitrary mushroom-gon is given. Use a compass and straightedge to draw a straight line dividing its area in half. [img]https://cdn.artofproblemsolving.com/attachments/d/e/e541a83a7bb31ba14b3637f82e6a6d1ea51e22.png[/img]
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.
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 $\vartriangle ABC$.
Let \( \omega \) be the circumcircle of triangle \( ABC \), where \( AB > AC \). Let \( N \) be the midpoint of arc \( \smile\!BAC \), and \( D \) a point on the circle \( \omega \) such that \( ND \perp AB \). Let \( I \) be the incenter of triangle \( ABC \). Reconstruct triangle \( ABC \), given the marked points \( A, D, \) and \( I \). Proposed by Oleksii Karlyuchenko and Hryhorii Filippovskyi
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.
Using only angle with angle $\frac{\pi}{7}$ and a ruler, constuct angle $\frac{\pi}{14}$
Consider a triangle $ABC$. For every point M $\in BC$ ,we define $N \in CA$ and $P \in AB$ such that $APMN$ is a parallelogram. Let $O$ be the intersection of $BN$ and $CP$. Find $M \in BC$ such that $\angle PMO=\angle OMN$
Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.
For some values of c, the equation $x^c + y^c = z^c$ can be illustrated geometrically. For example, the case $c = 2$ can be illustrated by a right-angled triangle. By this we mean that, x, y, z is a solution of the equation $x^2 + y^2 = z^2$ if and only if there exists a right-angled triangle with catheters $x$ and $y$ and hypotenuse $z$. In this problem we will look at the cases $c = -\frac{1}{2}$ and $c = - 1$. a) Let $x, y$ and $z$ be the radii of three circles intersecting each other and a line, as shown, in the figure. Show that, $x^{-\frac{1}{2}}+ y^{-\frac{1}{2}} = z^{-\frac{1}{2}}$ [img]https://cdn.artofproblemsolving.com/attachments/5/7/5315e33e1750a3a49ae11e1b5527311117ce70.png[/img] b) Draw a geometric figure that illustrates the case in a similar way, $c = - 1$. The figure must be able to be constructed with a compass and a ruler. Describe such a construction and prove that, in the figure, lines $x, y$ and $z$ satisfy $x^{-1}+ y^{-1} = z^{-1}$. (All positive solutions of this equation should be possible values for $x, y$, and $z$ on such a figure, but you don't have to prove that.)
Three points are given on the plane. With the help of compass and ruler construct a straight line in this plane, which will be equidistant from these three points. Explore how many solutions have this construction.
Construct a triangle given an angle, the side opposite the angle and the median to the other side (researching the number of solutions is not required).
Using only a compass and a ruler, reconstruct triangle $ABC$ given the following three points: point $M$ the intersection of its medians, point $I$ is the center of its inscribed circle and the point $Q_a$ is touch point of the inscribed circle to side $BC$.
Three rays with a common origin are drawn on the plane, dividing the plane into three angles. One point is marked inside each corner. Using one ruler, construct a triangle whose vertices lie on the given rays and whose sides contain the given points.
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.