Inside a right angle is given a point $A$. Construct an equilateral triangle, one of the vertices of which is point $A$, and two others lie on the sides of the angle (one on each side).

The plane has a circle $\omega$ and a point $A$ outside it. For any point $C$, the point $B$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$, then point $C$ also moves along a circle.

Let $P$ be an arbitrary interior point of the equilateral triangle $ABC$. From $P$ draw parallel to the sides: $A'_1A_1 \parallel AB$, $B' _1B_1 \parallel BC$ and $C'_1C_1 \parallel CA$. Prove that the sum of legths $| AC_1 | + | BA_1 | + | CB_1 |$ is independent of the choice of point $P$. [img]https://cdn.artofproblemsolving.com/attachments/5/a/15b06706c09e2458fb5938807b9f3833ffb62e.png[/img]

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 $.

Prove that for any integer $n\ge 6$ we can divide an equilateral triangle completely into $n$ smaller equilateral triangles.

Suppose that an interior point $O$ of a triangle $ABC$ is such that the angles $\angle BAO,\angle CBO, \angle ACO$ are all greater than or equal to $30^o$. Prove that the triangle $ABC$ is equilateral.

The plane is cut up into equilateral triangles by three families of parallel lines. Is it possible to find $4$ vertices of these triangles which form a square?

A black unit equilateral triangle is drawn on the plane. How can we place nine tiles, each a unit equilateral triangle, on the plane so that they do not overlap, and each tile covers at least one interior point of the black triangle? (Folklore)

On the side $BC$ of the equilateral triangle $ABC$, choose any point $D$, and on the line $AD$, take the point $E$ such that $| B A | = | BE |$. Prove that the size of the angle $AEC$ is of does not depend on the choice of point $D$, and find its size.

The shape of a military base is an equilateral triangle of side $10$ kilometers. Security constraints make cellular phone com­munication possible only within $2.5$ kilometers. Each of $17$ soldiers patrols the base randomly and tries to contact all others. Prove that at each moment at least two soldiers can communicate.

The vertices of an equilateral triangle lie on sides $ AB$, $CD$ and $EF$ of a regular hexagon $ABCDEF$. Prove that the triangle and the hexagon have a common centre. (N Sedrakyan, Yerevan )

Points $A, B$, and $C$ are on a line in this order. Points $D$ and $E$ lie on the same side of this line, in such a way that triangles $ABD$ and $BCE$ are equilateral. The segments $AE$ and $CD$ intersect in point $S$. Prove that $\angle ASD = 60^o$. [asy] unitsize(1.5 cm); pair A, B, C, D, E, S; A = (0,0); B = (1,0); C = (2.5,0); D = dir(60); E = B + 1.5*dir(60); S = extension(C,D,A,E); fill(A--B--D--cycle, gray(0.8)); fill(B--C--E--cycle, gray(0.8)); draw(interp(A,C,-0.1)--interp(A,C,1.1)); draw(A--D--B--E--C); draw(A--E); draw(C--D); draw(anglemark(D,S,A,5)); dot("$A$", A, dir(270)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$D$", D, N); dot("$E$", E, N); dot("$S$", S, N); [/asy]

Barack is an equilateral triangle and Michelle is a square. If Barack and Michelle each have perimeter $ 12$, find the area of the polygon with larger area.

A circle rolls along a side of an equilateral triangle. The radius of the circle is equal to the height of the triangle. Prove that the measure of the arc intercepted by the sides of the triangle on this circle is equal to $60^o$ at all times.

Let $ABC$ be an equilateral triangle, and let $P$ stand for an arbitrary point inside the triangle. Is it true that $| \angle PAB - \angle PAC| \ge | \angle PBC - \angle PCB|$ ?

In the equilateral triangle $ABC, M, N$ are the midpoints of the sides $AB, AC$, respectively. The line $MN$ intersects the circumcircle of $\vartriangle ABC$ at $K$ and $L$ and the lines $CK$ and $CL$ meet the line $AB$ at $P$ and $Q$, respectively. Prove that $PA^2 \cdot QB = QA^2 \cdot PB$.

Outside the parallelogram $ABCD$ on its sides $AB$ and $BC$ are constructed equilateral triangles $ABK$, and $BCM$. Prove that the triangle $KMD$ is equilateral.

Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.

On the sides $ BC $, $ CA $, $ AB $ of the triangle $ ABC $, the points $ D $, $ E $, $ F $ are chosen respectively, such that $$ \frac{|DB|}{|DC|} = \frac{|EC|}{|EA|} = \frac{|FA|}{|FB|}$$ Prove that if the triangle $ DEF $ is equilateral, then the triangle $ ABC $ is also equilateral.

Let $ABC$ be an equilateral triangle with centre $O$. A line through $C$ meets the circumcircle of triangle $AOB$ at points $D$ and $E$. Prove that points $A, O$ and the midpoints of segments $BD, BE$ are concyclic.

To the exterior of side $AB$ of square $ABCD$, we have drawn the regular triangle $ABE$. Point $A$ reflected on line $BE$ is $F$, and point $E$ reflected on line $BF$ is $G$. Let the perpendicular bisector of segment $FG$ meet segment $AD$ at $X$. Show that the circle centered at $X$ with radius $XA$ touches line$ FB$.

Let $A'$, $B'$ and $C'$ be points on the sides $BC$, $CA$ and $AB$, respectively, of an equilateral triangle $ABC$. If $AC' = 2C'B$, $BA' = 2A'C$ and $CB' = 2B'A$, prove that the lines $AA'$, $BB'$ and $CC'$ enclose a triangle whose area is $1/7$ that of $ABC$.

Outside triangle $ ABC $ equilateral triangles $ BMC $, $ CNA $, and $ APB $ are constructed. Prove that the centers $ S $, $ T $, $ U $ of these triangles form an equilateral triangle.

Prove that the sides of any equilateral triangle you can either increase everything or decrease everything by the same amount so that you get a right triangle.

Consider in the interior of an equilateral triangle $ABC$ points $D, E$ and $F$ such that$ D$ belongs to segment $BE$, $E$ belongs to segment $CF$ and$ F$ to segment $AD$. If $AD=BE = CF$ then $DEF$ is equilateral.