# Area of a circle inscribed in a polygon

If a circle is inscribed in a polygon, show that,

$$\dfrac{\text{(Area of inscribed circle)}}{\text{(Perimeter of inscribed circle)}} = \dfrac{\text{(Area of Polygon)}}{\text{(Perimeter of Polygon)}}$$

For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}{2\tan\frac{\pi}{n}}$ with the opposite side (base of the triangle) of length $a$. This altitude height will also be the radius of the circle inscribed in it.

So,

$\text{Area of incribed circle} = \pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)^2$

$\text{Perimeter of incribed circle} = 2\pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)$

$\text{Area of polygon} = n \cdot \frac{1}{2}a\left(\frac{a}{2\tan\frac{\pi}{n}}\right)$

$\text{Perimeter of polygon} = n \cdot a$

• Wouldn't the answer be different if it wasn't a regular polygon? – Umashankar Sasikumar Jun 18 '17 at 5:08
• The polygon need not be regular. See my answer below. – Christian Blatter Jun 18 '17 at 10:30

The figure shows a portion of the polygon and its inscribed circle. It is sufficient to show:-

$\dfrac {[polygon (OAPB)]}{[sector (OAB)]} = \dfrac {AP + PB}{arc (AB)}$.

Note that, LHS $= \dfrac {R^2 \times \tan \theta}{0.5R^2(2 \theta)}= \dfrac {\tan \theta}{\theta}$ and RHS $= \dfrac {2R \times \tan \theta}{R(2\theta)} = \dfrac {\tan \theta}{\theta}$.

Let the next adjacent portion be OBQC and the next be OCRD …. Then,

$[Polygon(OAPB)] = \dfrac {\tan \theta}{\theta} \times [sector(OAB)]$ …..(1)

$[Polygon(OBQC)] = \dfrac {\tan \theta}{\theta} \times [sector (OBC)]$ …..(2)

:

After adding up all these equations up vertically, we have $[given(polygon)] = \dfrac {\tan \theta }{\theta} \times [inscribed(circle)]$.

That is, $\dfrac {[given(polygon)]}{ [inscribed(circle)]} = \dfrac {\tan \theta }{\theta}$.

The comparison of the lengths can be worked out in the similar fashion.

Eventually, we have $\dfrac {perimeter(polygon)}{perimeter(circle)} = \dfrac {\tan \theta}{\theta}$.

Result follows.

Remark:-

As pointed out by @expiTTp1z0, this only works for the case when the "central angle" is constantly equal to $2 \theta$ for each subdivision. (That is, the proof is valid only when the polygon is a regular one.) I should compare [Sector] : [Arc-length] and [OAPB] : [AP + PB] instead (like what C. Blatter did). Then, they both are equal to R/2.

• Does $x_1/y_1 = x_2/y_2$ and $a_1/b_1 = a_2/b_2$ imply $(x_1+a_1)/(y_1+b_1) = (x_2+a_2)/(y_2+b_2)$? If not, then the result does not seem to be true for any arbitrary polygon. – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 6:34
• @expiTTp1z0 See added. – Mick Jun 18 '17 at 8:14
• Why would you assume same $\theta$ for each portions (or sectors). If the polygon is irregular, those $\theta$'s could be unequal. Isn't it? – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 9:26
• @expiTTp1z0 Oh! I have totally neglected that. Maybe I should compare [Sector] : [Arc-length] and [OAPB] : [AP + PB] instead (like what C. Blatter did). Then, they both are equal to R/2. I will have my post deleted shortly because my post works only when the polygon is regular.. – Mick Jun 18 '17 at 11:19
• No need to delete the post I guess, because it helps to properly visualize the answer by @Christian. May be you can edit the post appropriately. – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 11:23

Consider the circular sector of central angle $\alpha$ between two successive points of tangency. One has $${{\rm area(circular\ sector)}\over{\rm length(circular\ arc)}}={{1\over 2}\alpha r^2\over \alpha r}={r\over2}\ ,$$ and for the corresponding part of the polygon (a kite) one has $${{\rm area(kite)}\over{\rm length(outer\ edges)}}={2\cdot{1\over 2}r\cdot r\tan{\alpha\over2}\over 2 r\tan{\alpha\over2}}={r\over2}$$ as well. Now, if $a_i={r\over2}\> b_i$ $(1\leq i\leq n)$ then $${\sum_i a_i\over\sum_i b_i}={\sum_i {r\over2} b_i\over\sum_i b_i}={r\over2}\ ,$$ and this is valid for the union of circular sectors as well as for the union of kites.

Divide the polygon into triangles by drawing a line segment from each vertex to the center of the circle. Each triangle includes one side of the polygon and a sector of the inscribed circle. Letting $r$ = the radius of the circle:

Area of sector = $(r/2)$×(Arc length of sector)

Area of triangle = $(r/2)$×(Length of included polygonal side)

Add up all the triangle and sector areas as above, and find that the area/perimeter ratio for both the polygon and the circle are $r/2$.