geometryquadrilateral
I was studying about cyclic quadrilaterals , and a thought came that are there infinite number of cyclic quadrilaterals having perimeter equal to its area or if they are finite and how many are there in total?
I just cannot understand how to proceed to solve this or how to count them ?
Also , If there is any way to analyse this using a program , that would also help .
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.
You will find that the optimal solution is for the opposite angles to add to $\pi$ and, thus, for the quadrilateral to be concyclic (inscribed in a circle).
Best Answer
Any cyclic quadrilateral (a quadrilateral that can be inscribed in a circle) can be scaled to have equal perimeter and area. Assume that in a unit circle an inscribed quadrilateral has area $A$ and perimeter $P$. Then scaling to a circle of radius $r$ gives a similar quadrilateral of area $r^2A$ and perimeter $rP$. Setting $r = P/A$ makes these equal.
The answer then comes down to noticing that there are infinitely many (dissimilar) cyclic quadrilaterals in a unit circle. Indeed we can fix one side of the quadrilateral to be a diameter and still obtain infinitely many dissimilar cyclic quadrilaterals by varying the length and position of the "opposite" side.
Added: There is no upper bound on the area (or perimeter) of a cyclic quadrilateral having equal area and perimeter. To see this we may restrict attention to rectangles, which are cyclic quadrilaterals, and note that "area equals perimeter" can be stated in terms of width $x \gt 0$ and height $y \gt 0$:
$$ xy = 2(x+y) $$
$$ (x-2)(y-2) = 4 $$
We recognize this as the equation of a hyperbola, having asymptotes $x=2,y=2$, with points on its upper branch in the first quadrant arbitrarily far from the origin. That is, take $x$ to be as big as we want, and solve for $y = 2 + 4/(x-2)$. Now the area and perimeter are equal, and this quantity exceeds $2x$.
On the other hand there is a minimum area (resp. perimeter). From the analysis above we should minimize $P^2/A$ among all cyclic quadrilaterals, which is tantamount to maximizing the area among cyclic quadrilaterals with specified perimeter. The isoperimetric inequality for quadrilaterals says that among quadrilaterals with the same perimeter, the one with the greatest area is regular, i.e. a square, which is of course cyclic. Thus $P^2/A \ge 16$, and we attain this lower bound by taking a square with side 4.
Also, among quadrilaterals with specified side lengths, the maximum area is that of a cyclic quadrilateral.