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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 .

Mod
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2 Answers2

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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.

hardmath
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  • then can the value of area attain any value or we can put contraint on it ? – Mod Jan 04 '14 at 22:41
  • Also , what if they have integer sides only ? – Mod Jan 04 '14 at 22:46
  • These are interesting questions, but significantly different from the original Question. I'll tackle the first variant (bounding the area of a cyclic quadrilateral whose perimeter equals its area), but the integer-only sides aspect seems to deserve a separate post. – hardmath Jan 04 '14 at 23:19
  • So should I put the integer sides question separately ? – Mod Jan 05 '14 at 05:52
  • @Mod: You might want to give some thought before posting such a Question. Certainly there exist cyclic quadrilaterals with integer sides that have equal area and perimeter, and it is not hard to identify the rectangles of this type. – hardmath Jan 05 '14 at 13:10
  • Ok, will ponder upon it. – Mod Jan 05 '14 at 18:50
  • @Mod: I'm encouraging you to post about cyclic quadrilaterals with integer side lengths having area = perimeter. I just want you to give a good statement: the minimum context needed for a precise Question, followed by some brief comments about what you've tried and any partial conclusions or difficulties encountered. – hardmath Jan 06 '14 at 17:28
  • @Mod: I've posted the essence of your problem about integer side length cases here. – hardmath Jan 09 '14 at 02:44
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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

Edwin Gray
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  • Then why are triangles with equal area and perimeter finite? There are only 5 such triangles. –  Jul 06 '23 at 20:51