No polygon has the same area as the difference between its inscribed and circumscribed circles

Question

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.

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Edit:
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.

Bonus:
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!

Answer 1

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 $A=\frac{1}{4}\sqrt{(a+b)^2(a-b+2c)(b-a+2c)}$.

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.

Answer 2

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.