The numbers $p,q>0$ are given. Construct a rectangle $ABCD$ with $AE=p,AF=q$ where $E,F$ are midpoints of $BC,CD,$ respectively. Discuss conditions of solvability.

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.

Let's call a convex figure, the boundary of which consists of two segments and an arc of a circle, a mushroom-gon (see fig.). An arbitrary mushroom-gon is given. Use a compass and straightedge to draw a straight line dividing its area in half. [img]https://cdn.artofproblemsolving.com/attachments/d/e/e541a83a7bb31ba14b3637f82e6a6d1ea51e22.png[/img]

Given three points that are not on the same straight line. Three circles pass through each pair of the points so that the tangents to the circles at their intersection points are perpendicular to each other. Construct the circles.

Two different points $A,S$ are given in the plane. Furthermore, positive numbers $d,\omega$ are given, $\omega<180^\circ.$ Let $X$ be a point and $X'$ its image under the rotation by the angle $\omega$ (in counter-clockwise direction) with respect to the origin $S.$ Construct all points $X$ such that $XX'=d$ and $A$ is a point of the segment $XX'.$ Discuss conditions of solvability (in terms of $d,\omega,SA$).

Construct a convex polyhedron of equal “bricks” shown in Figure. [img]https://cdn.artofproblemsolving.com/attachments/6/6/75681a90478f978665b6874d0c0c9441ea3bd2.gif[/img]

Given an angle less than $180^o$, and a point $M$ outside the angle. Draw a line through $M$ so that the triangle, whose vertices are the vertex of the angle and the intersection points of its legs with the line drawn, has a given perimeter.

Given a circle, its diameter $[AB]$ and a point $C$ on it. Construct (with the help of compasses and ruler) two points $X$ and $Y$, that are symmetric with respect to $(AB)$ line, such that $(YC)$ is orthogonal to $(XA)$.

Albatross and Frankinfueter both own a circle. Frankinfueter also has stolen from Prof. Trugweg a ruler. Before that, Trugweg had two points with a distance of $1$ drawn his (infinitely large) board. For a natural number $n$, let A $(n)$ be the number of the construction steps that Albatross needs at least to create two points with a distance of $n$ to construct. Similarly, Frankinfueter needs at least $F(n)$ steps for this. How big can $\frac{A (n)}{F (n)}$ become? There are only the following three construction steps: a) Mark an intersection of two straight lines, two circles or a straight line with one circle. b) Pierce at a marked point $P$ and draw a circle around $P$ through one marked point . c) Draw a straight line through two marked points (this implies possession of a ruler ahead!).

The convex pentagon $ABCDE$ was drawn in the plane. $A_1$ was symmetric to $A$ with respect to $B$. $B_1$ was symmetric to $B$ with respect to $C$. $C_1$ was symmetric to $C$ with respect to $D$. $D_1$ was symmetric to $D$ with respect to $E$. $E_1$ was symmetric to $E$ with respect to $A$. How is it possible to restore the initial pentagon with the compasses and ruler, knowing $A_1,B_1,C_1,D_1,E_1$ points?

The median, bisector, and height, all originate at the same vertex of a triangle. Given the intersection points of the median, bisector, and height with the circumscribed circle, construct the triangle.

Two given circles $k_1$ and $k_2$ intersect at points $P$ and $Q$. Construct a segment $AB$ through $P$ with the endpoints at $k_1$ and $k_2$ for which $AP \cdot PB$ is maximal.

Consider a circle and a point $P$ outside the circle. The angle of given measure with vertex at $P$ subtends a diameter of the circle. Construct the circle’s diameter with ruler and compass.

For some values of c, the equation $x^c + y^c = z^c$ can be illustrated geometrically. For example, the case $c = 2$ can be illustrated by a right-angled triangle. By this we mean that, x, y, z is a solution of the equation $x^2 + y^2 = z^2$ if and only if there exists a right-angled triangle with catheters $x$ and $y$ and hypotenuse $z$. In this problem we will look at the cases $c = -\frac{1}{2}$ and $c = - 1$. a) Let $x, y$ and $z$ be the radii of three circles intersecting each other and a line, as shown, in the figure. Show that, $x^{-\frac{1}{2}}+ y^{-\frac{1}{2}} = z^{-\frac{1}{2}}$ [img]https://cdn.artofproblemsolving.com/attachments/5/7/5315e33e1750a3a49ae11e1b5527311117ce70.png[/img] b) Draw a geometric figure that illustrates the case in a similar way, $c = - 1$. The figure must be able to be constructed with a compass and a ruler. Describe such a construction and prove that, in the figure, lines $x, y$ and $z$ satisfy $x^{-1}+ y^{-1} = z^{-1}$. (All positive solutions of this equation should be possible values for $x, y$, and $z$ on such a figure, but you don't have to prove that.)

Given a circle $ k $ and a point inside it $ H $. Inscribe a triangle in the circle such that this point $ H $ is the point of intersection of the triangle's altitudes.

Triangle $ABC$ in the plane $\Pi$ is called [i]good [/i] if it has the following property: For any point $D$ in space outside the plane $\Pi$, it is possible to construct a triangle with sides of lengths $CD,BD,AD$. Find all good triangles

Construct the quadrilateral $ ABCD $ given the lengths of the sides $ AB $ and $ CD $ and the angles of the quadrilateral.

Two points $K$ and $L$ are given on a circle. Construct a triangle $ABC$ so that its vertex $C$ and the intersection points of its medians $AK$ and $BL$ both lie on the circle, $K$ and $L$ being the midpoints of its sides $BC$ and $AC$.

Given an acute angle $APX$ in the plane, construct a square $ABCD$ such that $P$ lies on the side $BC$ and ray $PX$ meets $CD$ in a point $Q$ such that $AP$ bisects the angle $BAQ$.

Three points $K, L$ and $M$ are given in the plane. It is known that they are the midpoints of three successive sides of an erased quadrilateral and that these three sides have the same length. Reconstruct the quadrilateral.

Construct a right triangle, given two medians drawn to its legs.

Divide a segment in halves using a right triangle. (With a right triangle one can draw straight lines and erect perpendiculars but cannot draw perpendiculars.)

Given points $M$ and $N$, the bases of heights $AM$ and $BN$ of $\vartriangle ABC$ and the line to which the side $AB$ belongs. Construct $\vartriangle ABC$.

Given the base, height and the difference between the angles at the base of a triangle, construct the triangle.

Consider points $A, B, C$. Draw a line through $A$ so that the sum of distances from $B$ and $C$ to this line is equal to the length of a given segment.