geometric-constructiongeometry
I'm having trouble with this one:
Draw a parallelogram knowing the lengths of its sides and the angle between the diagonals.
Bonus points if the answer uses a translation, because that's where this exercise comes from.
I tried to find the length of the diagonals by using the cosine law, but I got a very complicated expression and I'm pretty sure there is a better way.
Given two quadrilaterals $ABCD$ and $EFGH$, with side lengths equal, (i.e. $d(A,B)=d(E,F)$, etc.) then $ABCD \cong EFGH$ if and only if the diagonals have equal length: $d(A,C)=d(E,G)$ and $d(B,D)=d(F,H)$, where $d$ is the distance between the points.
The forward implication is obvious. To see the converse, we first notice that the diagonal's length being equal gives us a few congruent triangles. Namely, $\triangle ABC \cong \triangle EFG$, $\triangle ACD \cong \triangle EGH$ from $d(A,C)=d(E,G)$, by the good old SSS property.
Using only the one diagonals being equal, we would have two possible quadrilaterals from these facts. They come from reflecting one of the triangles over the diagonal.
Next, we see that we are given the distance between $D$ and $B$. Therefore, we are forced in which of these two options must be our congruent quadrilateral. The only way that both quadrilaterals could be acceptable is if $B$ is the proper distance from $D$ in both. But this will force $B$ onto the diagonal between $A$ and $C$, which contradicts the definition of a polygon (giving you a triangle instead), and finishes the proof.
I hope this was helpful. I believe a slight variation of this will work if you know the length of one diagonal, say $d(A,C)$ and the angle $\angle BCD$ or $\angle BAD$. It would seem reasonable that the same will hold if you know area, the four sides and a diagonal also. The fact that you have two possibilities right away from one triangle gives you most of the information. Also, it is pretty clear that the reflection over one of the diagonals will always produce one convex quadrilateral and one not.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
Best Answer
Fix one side, say segment $AB$, then draw (one arc of) the circle $c$ which consists of those points from where $AB$ is seen in the given angle. So that the centre of the parallelogramma has to be on $c$. Then enlarge $c$ to its double from center $A$, getting arc of circle $c'$ which must contain a $3$rd vertex of the parallelogramma, so finally just draw the circle centered at $B$ with the length of the other side.