In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

Do there exist $6$ circles in the plane such that every circle passes through centers of exactly $3$ other circles? by Morteza Saghafian

There are four circles $k_1 , k_2 , k_3$ and $k_4$ of equal radius inside the triangle $ABC$. The circle $k_1$ touches the sides $AB, CA$ and the circle $k_4 $, $k_2$ touches the sides $AB,BC$ and $k_4$, and $k_3$ touches the sides $AC, BC$ and $k_4.$ Prove that the center of $k_4$ lies on the line connecting the incenter and circumcenter of $ABC.$

Let $C_1$ and $C_2$ be two circles with centers $O_1$ and $O_2$, respectively, intersecting at $A$ and $B$. Let $l_1$ be the line tangent to $C_1$ passing trough $A$, and $l_2$ the line tangent to $C_2$ passing through $B$. Suppose that $l_1$ and $l_2$ intersect at $P$ and $l_1$ intersects $C_2$ again at $Q$. Show that $PO_1B$ and $PO_2Q$ are similar triangles. [i]Proposed by Pablo Valeriano[/i]

In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

Three different circles of equal radii intersect in point $Q$. The circle $C$ touches all of them. Prove that $Q$ is the center of $C$.

Let $G$ be a finite group, and let $H_1, H_2 \subset G$ be two subgroups. Suppose that for any representation of $G$ on a finite-dimensional complex vector space $V$, one has that \[\text{dim} V^{H_1}=\text{dim} V^{H_2},\] where $V^{H_i}$ is the subspace of $H_i$-invariant vectors in $V$ ($i=1,2$). Prove that \[Z(G) \cap H_1=Z(G) \cap H_2.\] Here $Z(G)$ denotes the center of $G$.

Some squares of an $n \times n$ chessboard have been marked ($n \in N^*$). Prove that if the number of marked squares is at least $n\left(\sqrt{n} + \frac12\right)$, then there exists a rectangle whose vertices are centers of marked squares.