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

All rectangles that can be inscribed in an isosceles triangle with two of their vertices on the triangle’s base have the same perimeter. Construct the triangle.

Construct a convex quadrilateral given the lengths of all its sides and the length of the segment between the midpoints of its diagonals. (Folklore)

$M$ is the midpoint of the side $BC$ of a triangle $ABC$. Construct a line $\ell$ intersecting the triangle and parallel to $BC$ such that the segment of $\ell$ between the sides $AB$ and $AC$ is the hypotenuse of a right-angled triangle with $M$ being its third vertex. (Folklore)

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

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

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 $\vartriangle ABC$.

Construct a circle equidistant from four points on a plane. How many solutions are there?

Given an isosceles trapezoid $ABCD$ and a point $P$. Prove that a quadrilateral can be constructed from segments $PA, PB, PC, PD$. Note: It is allowed that the vertices of a quadrilateral lie not only not only on the sides of the trapezoid, but also on their extensions.

The center of the circle circumscribing $\vartriangle ABC$ is mirrored through each side of the triangle and three points are obtained: $O_1, O_2, O_3$. Reconstruct $\vartriangle ABC$ from $O_1, O_2, O_3$ if everything else is erased.

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 $d,\omega,SA$).

Line segment $AM=d>0$ is given in the plane. Furthermore, a positive number $v$ is given. Construct a right triangle $ABC$ with hypotenuse $AB$, altitude to the hypotenuse of the length $v$ and the leg $BC$ being divided by $M$ in ration $MB/MC=2/3$. Discuss conditions of solvability in terms of $d, v$.

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!).

Given the lengths of two sides of a triangle and that of the bisector of the angle between these sides, construct the triangle.

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.

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 two non-intersecting chords $AB$ and $CD$ of a circle $k$ and a length $a <CD$. Determine a point $X$ on $k$ with the following property: If lines $XA$ and $XB$ intersect $CD$ at points $P$ and $Q$ respectively, then $PQ = a$. Show how to construct all such points $X$ and prove that the obtained points indeed have the desired property.

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.

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.

Let $AKL$ be a triangle such that $\angle ALK > 90^o +\angle LAK$. Construct an isosceles trapezoid $ABCD$ with $AB \parallel CD$ such that $K$ lies on the side $BC, L$ on the diagonal $AC$ and the lines $AK$ and $BL$ intersect at the circumcenter of the trapezoid.

Given two points $A$ and $B$ and a circle, find a point $X$ on the circle so that points $C$ and $D$ at which lines $AX$ and $BX$ intersect the circle are the endpoints of the chord $CD$ parallel to a given line $MN$.

Given a triangle $PQX$ in the plane, with $PQ = 3, PX = 2.6$ and $QX = 3.8$. Construct a right-angled triangle $ABC$ such that the incircle of $\vartriangle ABC$ touches $AB$ at $P$ and $BC$ at $Q$, and point $X$ lies on the line $AC$.

Consider a half-plane with the boundary line $p$ and two points $M,N$ in it such that the distances $Mp$ and $Np$ are different. Construct a trapezoid $MNPQ$ with area $MN^2$ such that $P,Q\in p.$ Discuss conditions of solvability.

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.