Found problems: 296
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 $AQ = BP$.
$\vartriangle ABC$ is an equilateral triangle. $L, M$ and $N$ are points on $BC, CA$ and $AB$ respectively. Prove that $MA \cdot AN + NB \cdot BL + LC \cdot CM < BC^2$.
Given an equilateral triangle with sidelength $k$ cm. With lines parallel to it's sides, we split it into $k^2$ small equilateral triangles with sidelength $1$ cm. This way, a triangular grid is created. In every small triangle of sidelength $1$ cm, we place exactly one integer from $1$ to $k^2$ (included), such that there are no such triangles having the same numbers. With vertices the points of the grid, regular hexagons are defined of sidelengths $1$ cm. We shall name as [i]value [/i] of the hexagon, the sum of the numbers that lie on the $6$ small equilateral triangles that the hexagon consists of . Find (in terms of the integer $k>4$) the maximum and the minimum value of the sum of the values of all hexagons .
Find all quadrilaterals which can be covered (without overlappings) with squares with side $ 1$ and equilateral triangles with side $ 1$.
(Emanuele Tron)
In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$.
[img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]
Let $ABC$ be an equilateral triangle of area $1998$ cm$^2$. Points $K, L, M$ divide the segments $[AB], [BC] ,[CA]$, respectively, in the ratio $3:4$ . Line $AL$ intersects the lines $CK$ and $BM$ respectively at the points $P$ and $Q$, and the line $BM$ intersects the line $CK$ at point $R$. Find the area of the triangle $PQR$.
For what positive integers $n$ is it possible to tile an equilateral triangle of side $n$ with trapezoids each of which has sides $1, 1, 1, 2$?
(NB Vassiliev)
What is the largest area of a regular hexagon that can be drawn inside the equilateral triangle of side $3$?
A. $3\sqrt7$ B. $\frac{3 \sqrt3}{2}$ C. $2\sqrt5$ D. $\frac{3\sqrt3}{8}$ E. $3\sqrt5$
Let $ABC$ be a triangle.
Points $P$ and $Q$ lie on side $BC$ and satisfy $|BP| =|PQ| = |QC| = \frac13 |BC|$.
Points $R$ and $S$ lie on side $CA$ and satisfy $|CR| =|RS| = |SA| = 1 3 |CA|$.
Finally, points $T$ and $U$ lie on side $AB$ and satisfy $|AT| = |TU| = |UB| =\frac13 |AB|$.
Points $P, Q,R, S, T$ and $U$ turn out to lie on a common circle.
Prove that $ABC$ is an equilateral triangle.
Let $O$ be the center of the regular triangle $ABC$. Find the set of all points $M$ such that any line containing the point $M$ intersects one of the segments $AB, OC$.
In an acute triangle, the distance from the midpoint of any side to the opposite vertex is equal to the sum of the distances from it to sides of the triangle. Prove that this triangle is equilateral.
Given an equilateral triangle $ABC$ and a point $P$ so that the distances $P$ to $A$ and to $C$ are not farther than the distances $P$ to $B$. Prove that $PB = PA + PC$ if and only if $P$ lies on the circumcircle of $\vartriangle ABC$.
In the plane there is a square $ Q $ and a point $ P $. The point $ K $ runs through the perimeter of the square $ Q $. Find the locus of the vertex $ M $ of the equilateral triangle $ KPM $.
An arbitrary point $M$ inside an equilateral triangle $ABC$ was connected to vertices. Prove that on each side the triangle can be selected one point at a time so that the distances between them would be equal to $AM, BM, CM$.
Equilateral triangles $ABE$ and $BCF$ are erected externally onthe sidess $AB$ and $BC$ of a parallelogram $ABCD$. Prove that $\vartriangle DEF$ is equilateral.
Paint the maximum number of vertices of the cube red so that you cannot select three of the red vertices that form an equilateral triangle.
Determine a necessary and sufficient condition for the affixes of three complex numbers $z_1$ , $z_2$ and $z_3$ are the vertices of an equilateral triangle.
Let $M,N, P$ be the midpoints of the sides $BC,CA,AB$ of the triangle $ABC$, respectively, and let $G$ be the centroid of the triangle. Prove that if $BMGP$ is cyclic and $2BN = \sqrt3 AB$ , then triangle $ABC$ is equilateral.
Points $A,B,C$ with $AB = BC$ are given on a circle with radius $r$, and $D$ is a point inside the circle such that the triangle $BCD$ is equilateral. The line $AD$ meets the circle again at $E$. Show that $DE = r$.
Let $ACE$ be a triangle with $\angle ECA=60^{\circ}, \angle AEC=90^{\circ}$. Let $B$ and $D$ be points on the sides $AC$ and $CE$ respectively such that the $\triangle BCD$ is equilateral. Now suppose $BD \cap AE=F$. Find $\angle EAC+\angle EFD$.
Let $[ABC]$ be an equilateral triangle and $P$ be a point on $AC$ such that $\overline{PC}= 7$. The straight line that passes through $P$ and is perpendicular to $AC$, intersects $CB$ at point $M$ and intersects $AB$ at point $Q$. The midpoint $N$ of $[MQ]$ is such that $\overline{BN} = 14$. Determine the side of the triangle $[ABC]$.
A sequence whose first term is positive has the property that any given term is the area of an equilateral triangle whose perimeter is the preceding term. If the first three terms form an arithmetic progression, determine all possible values of the first term.
A triangle has side lengths $a, b, c$ and angles $A, B, C$ as usual (with $b$ opposite $B$ etc). Show that if $$a(1 - 2 \cos A) + b(1 - 2 \cos B) + c(1 - 2 \cos C) = 0$$ then the triangle is equilateral.
Let $t$ be a line passing through the vertex $A$ of the equilateral $ABC$, parallel to the side $BC$. On the side $AC$ arbitrarily mark the point $D$. Bisector of the angle $ABD$ intersects the line $t$at the point $E$. Prove that $BD=CD+AE$.
On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ are constructed the equilateral triangles $ABE$ and $BCF,$ so that the points $D$ and $E$ are on the same side of the line $AB$, and $F$ and $D$ on different sides of the line $BC$. If the points $E,D$ and $F$ are collinear, then prove that $ABCD$ is rhombus.