No polygon has the same area as the difference between its inscribed and circumscribed circles. The inscribed circles must touch every side and the circumscribed circle must touch each vertice. I have proved this for some simple cases but failed to prove it generally. Or is there any counter-proof? Please help.

dbx proved that it does not hold for some irregular polygons. A big round of applause for him on cracking that tough nut? So some new questions to ponder about:
Are there a finite number of irregular polygons who disobey this hypothesis?
Are there a finite number of irregular polygons who obey this hypothesis?
Could anyone give any more examples of such polygons who do not obey this hypothesis.
Also thanks to Ross and anderstood who proved this does hold for all regular polygons.

I have expanded on this idea: There is no such polygon whose perimeter is equal to the difference between the circumferences of its circumscribed and inscribed circle .
I may also continue this onto the third dimension if I get conclusive results for the above post. All the best!

  • $\begingroup$ do you mean regular polygons? not all polygons even have inscribed/circumstribed circles. $\endgroup$ Dec 25, 2017 at 16:05
  • $\begingroup$ Sounds too easy for a regular polygon. Either a triangle will have the difference area be too small, or there will be an n-gon will an area too large and an n+1 gon with the area too small. The proof would just be to find that $n$ through a simple search. $\endgroup$
    – DanielV
    Dec 25, 2017 at 16:07
  • $\begingroup$ I agree, but I'm not sure it's well-posed otherwise. $\endgroup$ Dec 25, 2017 at 16:07
  • 2
    $\begingroup$ Sounds like he defined it well enough...as long as the outer circle touches all vertices and the inner circle touches all edges then it is a valid construction to consider. Unless there is some kind of weird ambiguous case of like zero length sides. I think it is expected to assume convex polygons only. $\endgroup$
    – DanielV
    Dec 25, 2017 at 16:09
  • 3
    $\begingroup$ Do you have a reason to believe this statement is true? Just by virtue of the vast number of liberties in the construction of the polygon, it seems unlikely. $\endgroup$
    – DanielV
    Dec 25, 2017 at 17:27

2 Answers 2


Here is a proof that a counterexample exists.

Given a polygon, call its area $A$. Let $A_R$ be the area of the circumcircle, $A_r$ the area of the inscribed circle, and $A_\Delta$ be the difference $A_R - A_r$. We want to find a polygon such that $A=A_\Delta$. I will show that there is such a quadrilateral, specifically a trapezoid.

First consider the unit square, with area $A=1$. Its incircle has area $\pi/4$ and its circumcircle has area $\pi/2$, thus $A_\Delta=\pi/4 < 1 = A$. Now elongate one side, to create an isosceles trapezoid (see fig). The area of this trapezoid is $A=\frac{1}{4}\sqrt{(a+b)^2(a-b+2c)(b-a+2c)}$.

isosceles trapezoid ABCD

Every isosceles trapezoid has circumscribed circle, and furthermore, its area is given by: $$ A_R=\pi c^2 \frac{ab+c^2}{4c^2-(a-b)^2} $$

Now we can restrict the values $a,b,c$ to ensure there is an inscribed circle; in this case we need $a+b=2c$. We can also assume $b=1$, simplifying $A$ considerably: $$ A = \frac{1}{4}\sqrt{4c^2 \cdot 2a \cdot 2b} = c\sqrt{a} = \frac{1}{2}(a+1)\sqrt{a} $$

Now that an inscribed circle is guaranteed, we can find its area: $$ A_r=\pi\frac{a}{4} $$

Using $b=1$, we thus have: $$ A_\Delta = \pi \left( c^2 \frac{a+c^2}{4c^2-(a-1)^2} - \frac{a}{4} \right) = \pi \left( \frac{(a+1)^2}{4}\cdot\frac{a+(a+1)^2/4}{(a+1)^2-(a-1)^2} - \frac{a}{4} \right) $$ $$ = \pi \left( \frac{(a+1)^2}{4} \cdot \frac{a + (a+1)^2/4}{4a} - \frac{a}{4} \right)$$

It's admittedly a bit messy, but we can use the intermediate value theorem. Instead of looking for an $a$ that satisfies $A=A_\Delta$, we only need to find one with $A<A_\Delta$, since for the unit square we had $A > A_\Delta$. Choose $a=2$. Then $A_\Delta\approx 2.18$ and $A\approx 2.12$, i.e. $A<A_\Delta$.

Since the isosceles trapezoid is a continuous deformation of the square, the intermediate value theorem applies and there must be some value of $a$ between $1$ an $2$ with $A=A_\Delta$. The conjecture is false.

  • $\begingroup$ Does the inscribed circle touch all edges of the polygon? $\endgroup$ Dec 27, 2017 at 13:55
  • $\begingroup$ Yes, that's the reason for requiring $a+b=2c$ $\endgroup$ Dec 27, 2017 at 19:36

With regular polygons the claim is true. Let $R$ be the radius of the circumscribed circle, $r$ the radius of the inscribed circle, and $n$ the number of sides. We have $r=R\cos \frac {2\pi}n$ The area of the outer circle is $\pi R^2$ and the inner circle is $\pi R^2 \cos^2 \frac {2\pi}n$ so the difference is $\pi R^2 \sin^2 \frac {2 \pi}n$. The area of the polygon is $nR^2 \sin \frac {2\pi}n \cos \frac {2\pi}n=\frac n2 R^2 \sin \frac {4\pi}n$ The second is almost the area of the outer circle, while the first is smaller by a factor $(\frac {2\pi}n)^2$. The transition happens between $n=5$ and $n=6$.

Using the link from Blue in a comment, it appears the claim is false. We saw that for a regular hexagon the difference between the circles is smaller than the regular hexagon. Wikipedia states that for bicentric hexagons if $r$ is the inradius, $R$ the outradius, and $x$ the distance between the centers $$3(R^2-x^2)^4=4r^2(R^2+x^2)((R^2-x^2)^2+16r^4x^2R^2$$ As $x$ increases $r$ decreases increasing the difference of areas of the circles. The area of the hexagon looks like it is decreasing as well, so there will be some point the equality obtains.

  • $\begingroup$ Any ideas on irregular polygons? Anyways thanks for the proof for regular polygons. $\endgroup$ Dec 25, 2017 at 17:22
  • 1
    $\begingroup$ I think with regular polygons you might also be able to just say that (by scaling) all of the vertices are on algebraically closed points, so the area of the polygon is algebraically closed. And the radii of the circles is algebraically closed, so the circle difference is not algebraically closed (by virtue of multiplying by $\pi$). $\endgroup$
    – DanielV
    Dec 25, 2017 at 17:26
  • $\begingroup$ I did some sketches and it seems hard to get both inscribed and circumscribed circles around an irregular polygon, but don’t know if you can $\endgroup$ Dec 25, 2017 at 17:36
  • 1
    $\begingroup$ @RossMillikan: It's certainly possible to inscribe and circumscribe an irregular polygon. These things are called "bicentric polygons". $\endgroup$
    – Blue
    Dec 26, 2017 at 7:36
  • $\begingroup$ @Blue thanks for the help. It would be nice if anyone could find the exact values for computing. However I always thought that the inscribed circle and the circumscribed circle were concentric for regular polygons. Infact : "Every regular polygon is bicentric.[2] In a regular polygon, the incircle and the circumcircle are concentric—that is, they share a common center, which is also the center of the regular polygon, so the distance between the incenter and circumcenter is always zero" is from the link above. Does bicentric mean two distinct centres or just two centres? Please help. $\endgroup$ Dec 26, 2017 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.