Cyclic quadrilateral with equal area and perimeter 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 .
 A: 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.
A: By Brahmagupta's formula,16A^2 = (p-2a)(p-2b)(p-2c)(p-2d) and 16p^2 =16(a+b+c+d)^2
let x = b+c+d-a, y = a+c+d-b, z = a+b+d-c, u = a+b+c-d. Then x+y+z+u =2p. Since
x=p-2a, y=p-2b, z=p-2c, u =p-2d, 16A^2 = xyzu. Then 16A^2 = 16p^2 leads to the Diophantine equation:  4(x+y+z+u)^2 = xyzu. There are some obvious answers such as x=y=z=u=8. Then 4(32)^2 = 4096 = 8^4. Then 2a= p-x =16-8,a=4;so does b,c, and d. So we have a 4X4 square. Another solution is x=12,y=6,z=12,u=6 leading to a
3x6 rectangle. Other solutions should be computer available.  Ed Gray
