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: 296

Let $ABC$ be an equilateral triangle and let $D$ be an inner point of the side $BC$. A circle is tangent to $BC$ at $D$ and intersects the sides $AB$ and $AC$ in the inner points $M, N$ and $P, Q$ respectively. Prove that $|BD| + |AM| + |AN| = |CD| + |AP| + |AQ|$.
Let $ABCDEF$ be a convex hexagon such that $BCEF$ is a parallelogram and $ABF$ an equilateral triangle. Given that $BC = 1, AD = 3, CD+DE = 2$, compute the area of $ABCDEF$
Consider an equilateral triangle $\vartriangle ABC$ and points $D$ and $E$ on the sides $AB$ and $BC$csuch that $AE = CD$. Find the locus of intersection points of $AE$ with $CD$ as points $D$ and $E$ vary.
A triangle is cut into several (not less than two) triangles. One of them is isosceles (not equilateral), and all others are equilateral. Determine the angles of the original triangle.
Consider a unit square $ABCD$ and a (variable) equilateral triangle $XYZ$ such that $X, Z$ lie on rays $AB, DC,$ respectively, and $Y$ lies on segment $AD$. Compute the area of triangle $XYZ$ in terms of $x=AX$ and determine its maximum and minimum.
Let $\vartriangle ABC$ be an equilateral triangle with side $1$. $1011$ points $P_1$, $P_2$, $P_3$, $...$, $P_{1011}$ on the side $AC$ and $1011$ points $Q_1$, $Q_2$, $Q_3$, $...$ ,$ Q_{1011}$ on side AB (see figure) in such a way as to generate $2023$ triangles of equal area. Find the length of the segment $AP_{1011}$. [img]https://cdn.artofproblemsolving.com/attachments/f/6/fea495c16a0b626e0c3882df66d66011a1a3af.png[/img] PS. Harder version of [url=https://artofproblemsolving.com/community/c4h3323135p30741470]2023 Chile NMO L1 P3[/url]
Let $C$ be an interior point of line segment $AB$. Equilateral triangles $ADC$ and $CEB$ are constructed to the same side from $AB$. Find all points which can be the midpoint of the segment $DE$.
a) The maximal height $|AH|$ of the acute-angled triangle $ABC$ equals the median $|BM|$. Prove that the angle $ABC$ isn't greater than $60$ degrees. b) The height $|AH|$ of the acute-angled triangle $ABC$ equals the median $|BM|$ and bisectrix $|CD|$. Prove that the angle $ABC$ is equilateral.
Triangle $ABC$ is given. On its sides $AB$, $BC$ and $CA$, respectively, points $X, Y, Z$ are chosen so that $$AX : XB =BY : YC = CZ : ZA = 2:1.$$ It turned out that the triangle $XYZ$ is equilateral. Prove that the original triangle $ABC$ is also equilateral.
Given a regular triangle $ABC$, whose area is $1$, and the point $P$ on its circumscribed circle. Lines $AP, BP, CP$ intersect, respectively, lines $BC, CA, AB$ at points $A', B', C'$. Find the area of the triangle $A'B'C'$.
Two-mutually perpendicular lines $\ell$ and $m$ intersect each other at a point of the circumference of a circle, dividing it into three arcs. A point $M_i$ ($i = 1$,$2$,$3$) is taken on each arc so that the tangent line to the circumference at the point $M_i$ intersects $\ell$ and $m$ in two points at the same distance from $M_i$ (that is $M_i$ is the midpoint of the segment between them). Prove that the triangle $M_1M_2M_3$ is equilateral. (Przhevalsky)
On the sides $AB, \, \, BC, \, \, CA$ of the triangle $ABC$ the points ${{C} _ {1}}, \, \, {{A} _ { 1}},\, \, {{B} _ {1}}$ are selected respectively, that are different from the vertices. It turned out that $\Delta {{A} _ {1}} {{B} _ {1}} {{C} _ {1}}$ is equilateral, $\angle B{{C}_{1}}{{A}_{1}}=\angle {{C}_{1}}{{B}_{1}}A$ and $\angle B{{A}_{1}}{{C}_{1}}=\angle {{A}_{1}}{{B}_{1}}C$ . Is $ \Delta ABC$ equilateral?
Let $ABC$ be an equilateral triangle and let $D$ be an inner point of the side $BC$. A circle is tangent to $BC$ at $D$ and intersects the sides $AB$ and $AC$ in the inner points $M, N$ and $P, Q$ respectively. Prove that $|BD| + |AM| + |AN| = |CD| + |AP| + |AQ|$.
Let $A,B,C,D$, be four different points on a line $\ell$, so that $AB=BC=CD$. In one of the semiplanes determined by the line $\ell$, the points $P$ and $Q$ are chosen in such a way that the triangle $CPQ$ is equilateral with its vertices named clockwise. Let $M$ and $N$ be two points of the plane be such that the triangles $MAP$ and $NQD$ are equilateral (the vertices are also named clockwise). Find the angle $\angle MBN$.
Points on the sides $ BC $, $ CA $ and $ AB $ of the triangle $ ABC $ are respectively $ A_1 $, $ B_1 $ and $ C_1 $ such that $ AC_1: C_1B = BA_1: A_1C = CB_1: B_1A = 2: 1 $. Prove that if triangle $ A_1B_1C_1 $ is equilateral, then triangle $ ABC $ is also equilateral.
Find the locus of centers of regular triangles such that three given points $A, B, C$ lie respectively on three lines containing sides of the triangle.
(a) Given a point $O$ inside an equilateral triangle $\vartriangle ABC$. Line $OG$ connects $O$ with the center of mass $G$ of the triangle and intersects the sides of the triangle, or their extensions, at points $A', B', C'$ . Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G} = 3.$$ (b) Point $G$ is the center of the sphere inscribed in a regular tetrahedron $ABCD$. Straight line $OG$ connecting $G$ with a point $O$ inside the tetrahedron intersects the faces at points $A', B', C', D'$. Prove that $$\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G}+ \frac{D'O}{D'G} = 4.$$
In triangle $\vartriangle ABC$, we have marked points $A_1$ on side $BC, B_1$ on side $AC$, and $C_1$ on side $AB$ so that $AA_1$ is an altitude, $BB_1$ is a median, and $CC_1$ is an angle bisector. It is known that $\vartriangle A_1B_1C_1$ is equilateral. Prove that $\vartriangle ABC$ is equilateral too. (Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $BB_1$ we know that $B_1$ is the midpoint of side $AC$ in $\vartriangle ABC$.)
The points $P,Q,R$ and $A,B,C,D$ lie on a circle (clockwise) such that $\vartriangle PQR$ is equilateral and $ABCD$ is a square. The points $A$ and $P$ coincide. Prove that the symmetric of $B$ and $D$ wrt $PQ$ and $PR$ respectively lie on the sidelines of the symmetric square wrt $QR$.
On sides $BC,CA,AB$ of a triangle $ABC$ points $D,E,F$ respectively are chosen so that $AD,BE,CF$ have a common point, say $G$. Suppose that one can inscribe circles in the quadrilaterals $AFGE,BDGF,CEGD$ so that each two of them have a common point. Prove that triangle $ABC$ is equilateral.
An equilateral triangle $DCE$ is placed outside a square $ABCD$. The center of this triangle is denoted as $M$ and the intersection of the straight line $AC$ and $BE$ with $S$. Prove that the triangle $CMS$ is isosceles.
Let $ABC$ be a given triangle. Find the locus $\mathbf M$ of all vertices $Z$ such that triangle $XYZ$ is equilateral where $X$ is any point of segment $AB$ and $Y\neq X$ lies on ray $AC.$
Let $\vartriangle ABC$ be an equilateral triangle with side length $M$ such that points $E_1$ and $E_2$ lie on side $AB$, $F_1$ and $F_2$ lie on side $BC$, and $G1$ and $G2$ lie on side $AC$, such that $$m = \overline{AE_1} = \overline{BE_2} = \overline{BF_1} = \overline{CF_2} = \overline{CG_1} = \overline{AG_2}$$ and the area of polygon $E_1E_2F_1F_2G_1G_2$ equals the combined areas of $\vartriangle AE_1G_2$, $\vartriangle BF_1E_2$, and $\vartriangle CG_1F_2$. Find the ratio $\frac{m}{M}$. [img]https://cdn.artofproblemsolving.com/attachments/a/0/88b36c6550c42d913cdddd4486a3dde251327b.png[/img]
$ABC$ is a triangle with $\angle A = 90^\circ$, $\angle B = 60^\circ$. The points $A_1$, $B_1$, $C_1$ on $BC$, $CA$, $AB$ respectively are such that $A_1B_1C_1$ is equilateral and the perpendiculars (to $BC$ at $A_1$, to $CA$ at $B_1$ and to $AB$ at $C_1$) meet at a point $P$ inside the triangle. Find the ratios $PA_1:PB_1:PC_1$.
Each of the sides $ BC, CA, AB $ of the triangle $ ABC $ was divided into three equal parts and on the middle sections of these sides as bases, equilateral triangles were built outside the triangle $ ABC $, the third vertices of which were marked with the letters $ A', B' , C' $ respectively. In addition, points $ A'', B'', C'' $ were determined, symmetrical to $ A', B', C' $ respectively with respect to the lines $ BC, CA, AB $. Prove that the triangles $ A'B'C' $ and $ A''B''C'' $ are equilateral and have the same center of gravity as the triangle $ ABC $.