With a compass and a ruler, split a triangle into two smaller triangles with the same sum of squares of sides.

On the plane, a non-isosceles triangle is given, a circle circumscribed around it and the center of its inscribed circle are marked. Using only a ruler without tick marks and drawing no more than seven lines, construct the diameter of the circumcircle.

You are given a ruler with two parallel straight edges a distance $d$ apart. It may be used (1) to draw the line through two points, (2) given two points a distance $\ge d$ apart, to draw two parallel lines, one through each point, (3) to draw a line parallel to a given line, a distance d away. One can also (4) choose an arbitrary point in the plane, and (5) choose an arbitrary point on a line. Show how to construct : (A) the bisector of a given angle, and (B) the perpendicular to the midpoint of a given line segment.

The figure shows a triangle, a circle circumscribed around it and the center of its inscribed circle. Using only one ruler (one-sided, without divisions), construct the center of the circumscribed circle.

A circle is inscribed in the triangle $ ABC$ and it's center $I$ and the points of tangency $P, Q, R$ with the sides $BC$, $C A$ and $AB$ are marked, respectively. With a single ruler, build a point $K$ at which the circle passing through the vertices B and $C$ touches (internally) the inscribed circle.

A circle is drawn on the plane. How to use only a ruler to draw a perpendicular from a given point outside the circle to a given line passing through the center of this circle?

On a blank paper there were drawn two perpendicular axes $x$ and $y$ with the same scale. The graph of a function $y=f(x)$ was drawn in this coordinate system. Then the $y$ axis and all the scale marks on the $x$ axis were erased. Provide a way how to draw again the $y$ axis using pencil, ruler and compass: (a) $f(x)= 3^x$; (b) $f(x)= \log_a x$, where $a>1$ is an unknown number.

A trapezoid $ABCD$ with bases $BC = a$ and $AD = 2a$ is drawn on the plane. Using only with a ruler, construct a triangle whose area is equal to the area of the trapezoid. With the help of a ruler you can draw straight lines through two known points. (Rozhkova Maria)

With an unmarked ruler only, reconstruct the trapezoid $ABCD$ ($AD \parallel BC$) given the vertices $A$ and $B$, the intersection point $O$ of the diagonals of the trapezoid and the midpoint $M$ of the base $AD$. (Yukhim Rabinovych)

A $2\times 2$ square was cut from a squared sheet of paper. Using only a ruler without divisions and without going beyond the square, divide the diagonal of the square into $6$ equal parts.

Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane; but she can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment into $17$ equal parts. Laura drew a parallelogram $ABCD$ and decided to try out her magical ruler; with it, she found the midpoint $M$ of side $CD$, and she extended $CB$ beyond $B$ to point $N$ so that segments $CB$ and $BN$ were equal in length. Unfortunately, her mischievous little brother came along and erased everything on Laura's picture except for points $A, M$, and $N$. Using Laura's magical ruler, help her reconstruct the original parallelogram $ABCD$: write down the steps that she needs to follow and prove why this will lead to reconstructing the original parallelogram $ABCD$.

A $2 \times 2$ square was cut out of a sheet of grid paper. Using only a ruler without divisions and without going beyond the square, divide the diagonal of the square into $6$ equal parts.

An acute-angled non-isosceles triangle $ABC$ is drawn, a circumscribed circle and its center $O$ are drawn. The midpoint of side $AB$ is also marked. Using only a ruler (no divisions), construct the triangle's orthocenter by drawing no more than $6$ lines. (Yu. Blinkov)

On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$.

In a Cartesian coordinate system (with the same scale on the x and y axes)there is a graph of the exponential function $y=3^x$. Then the y-axis and all marks on the x-axis erased. Only the graph of the function and the x-axis remained without a scale and a mark of $0$. How can you restore the y-axis using a compass and ruler?

A $2 \times 2$ square was cut out of a sheet of grid paper. Using only a ruler without divisions and without going beyond the square, divide the diagonal of the square into $6$ equal parts.

Juku invented an apparatus that can divide any segment into three equal segments. How can you find the midpoint of any segment, using only the Juku made, a ruler and pencil?

On the plane there is an image of a circle with a marked center. Let an arbitrary angle be drawn on this plane. Using one ruler, construct the bisector of this angle.

Given three equidistant parallel lines. Express by points of the corresponding lines the values of the resistance, voltage and current in a conductor so as to obtain the voltage $V = I \cdot R$ by connecting with a ruler the points denoting the resistance $R$ and the current $I$. (Each point of each scale denotes only one number). [hide=similar wording]Three parallel straight lines are given at equal distances from each other. How to depict by points of the corresponding straight lines the values of resistance, voltage and the current in the conductor, so that, applying a ruler to to points depicting the values of resistance R and values of current I, obtain on the voltage scale a point depicting the value of voltage V = I R (point each scale represents one and only one number).[/hide]

On the plane is drawn a triangle $ABC$ and a circle $\omega$ passing through the vertex $C$, the midpoints of the sides $AC$ and $BC$ and the point of intersection of the medians of the triangle $ABC$. The point $K$ lies on the circle $\omega$ such that $\angle AKB=90^o$. Using only with a ruler, draw a tangent to the circle $\omega$ at point $K$.

Circle $\omega$ is inscribed in the $\vartriangle ABC$, with center $I$. Using only a ruler, divide segment $AI$ in half. (Grigory Filippovsky)