Given regular unit circle and a n-gon, circumscribed around this unit circle. I need to minimize area of n-gon.
Also i need to find the limit of n-gon area with $n\rightarrow \infty$.
Intuitively i understand that n-gon must be regular for minimizing it's area, but how formally can i show that?
Also by intuition i now that in limit area of n-gon converges to a circle's area, but can't prove it formally.
Note: i'm trying to solve it with inequalities, without using derivatives
On
Divide the circumscribed $n$-gon into $2n$ right triangles, each with one vertex at the center of the circle, one vertex at a vertex of the polygon, and the third vertex at the point of tangency of a side of the polygon to the circle.
If the central angle in triangle $k$ is $\theta_k$, we have $0<\theta_k<\pi,\ k=1,\dots,2n, \sum_{k=1}^{2n}\theta_k=2\pi,$ and the area of the polygon is $$A:= \frac12\sum_{k=1}^{2n}\tan\theta_k$$
The area of the regular circumscribed $n$-gon is $$\frac{2n}2\tan\frac{2\pi}{2n} =\frac{2n}2\tan\left(\frac1{2n}\sum_{k=1}^{2n}\theta_k\right)\leq \frac{2n}2\sum_{k=1}^{2n}\frac1{2n}\tan\theta_k=A,$$ where the inequality follows from the convexity of the tangent function on $[0,\pi/2).$
NOTE
I wasn't able to do this without appealing to convexity. Without using convexity I was able to show that if a minimum exists, it is the regular $n$-gon, but I couldn't show existence, since the domain isn't compact.
Although I believe this argument is valid (by the less than Hilbertian standard of rigour that seems to apply to arguments in synthetic Euclidean geometry - that's a topic for another question), it only proves that if an $n$-gon circumscribing a circle has minimal area, then it must be regular. It does not prove that a regular $n$-gon circumscribing a circle has minimal area. For a modified argument that does prove this, see under the horizontal line after Theorem 1. Although I haven't deleted the original argument (because I haven't found any errors in it), it can now be skipped in its entirety.
Old version
Lemma 1. In a triangle $LMN,$ let the bisector of the angle at $L$ meet $MN$ at $K.$ If $|LN| > |LM|,$ then $|KN| > |KM|.$
Proof. By Euclid I.18, the angle at $M$ is greater than the angle at $N,$ therefore $\angle LKM > \angle LKN.$ Construct $H$ on $LN$ such that $\angle LKH = \angle LKM.$ Then $\triangle LKH$ is congruent to $\triangle LKM$ (ASA), therefore: $$ \text{area } \triangle LKN > \text{ area } \triangle LKH = \text{ area } \triangle LKM. $$ But triangles $LKN$ and $LKM$ have the same height, therefore the base of the first, $|KN|,$ must be greater than the base of the second, $|KM|.$ $\ \square$
Lemma 2. Let $LHK$ be an isosceles triangle, with base $HK.$ Let $G$ on $HK$ and $N$ on $HK$ produced be such that $\angle HLG = \angle KLN.$ Then $|GN| > |HK|.$
Proof. Construct $M$ on $HK$ such that: $$ \angle KLM = \angle HLG = \angle KLN. $$ Then triangle $KLM$ is congruent to triangle $HLG$ (ASA), therefore $|KM| = |HG|.$ In the triangle $LKN,$ side $LN$ is opposite the obtuse angle $\angle LKN,$ therefore $|LN| > |LK| = |LH|.$ The segment $LG$ lies inside one of two right-angled triangles, with equal hypotenuses $LH$ and $LK,$ therefore $|LH| > |LG|.$ Because triangles $KLM$ and $HLG$ are congruent (already proved), $|LG| = |LM|.$ Putting all of this together: $$ |LN| > |LK| = |LH| > |LG| = |LM|. $$ Therefore Lemma 1 applies, giving $|KN| > |KM|,$ whence: $$ |GN| = |HN| - |HG| = |HN| - |KM| > |HN| - |KN| = |HK|, $$ as required. $\ \square$
Lemma 3. Let $\gamma$ be a circle. Let the tangents from an external point $Z$ meet $\gamma$ at $A, C.$ Let $B$ be a third point on $\gamma,$ restricted to lie on one of the two arcs determined by $A, C$. Let the tangent at $B$ meet the tangents $ZA, ZC$ at $X, Y,$ respectively. Then the length $|XY|$ of the line segment $XY$ is minimal when $B$ is midway between $A$ and $C.$
Proof. Let $O$ be the centre of $\gamma.$ Then $OX$ bisects $\angle AOB,$ and $OY$ bisects $\angle BOC,$ therefore $\angle XOY$ has the constant value $\frac12\angle AOC.$ The altitude of the triangle $XOY$ is also constant, being equal to the radius $r$ of $\gamma.$ Construct an isosceles triangle $LHK$ with altitude $r$ and apex angle $\frac12\angle AOC.$ This is congruent to the triangle $XOY$ when $B$ is at the midpoint of the arc $AC.$ For any other position of $B$ on the arc, the triangle $XOY$ is congruent to a triangle $GMN$ with the same apex $L,$ and base vertices $G,N$ satisfying the conditions of Lemma 2, because $\angle GLN = \angle HLK.$ The conclusion of Lemma 2 therefore applies, so we have $|GN| > |HK|.$ But the triangles were constructed so that $|HK| = |XY|$ when $B$ is at the midpoint of the arc, and $|GN| = |XY|$ when $B$ is in any position other than that. $\ \square$
Theorem 1. Let $\gamma$ be a circle. Let $n$ be an integer $\geqslant 3.$ Let $P$ be an $n$-gon circumscribed about $\gamma.$ Then the area of $P$ is minimal only if $P$ is regular.
Proof. The area of $P$ is $rs,$ where $r$ is the radius of $\gamma$ and $s$ is the semiperimeter of $P.$ Minimising the area of $P$ is equivalent to minimising $s$. Each vertex of $P$ contributes, to the sum $s,$ the common length of the two tangents to $\gamma$ from that vertex. Therefore, if all but one of the $n$ points of contact of $P$ with $\gamma$ are held constant, and the edge of $P$ that meets $\gamma$ at that one variable point is $XY,$ then the total contribution to $s$ made by the vertices $X$ and $Y$ is equal to $|XY|.$ If $P$ is not regular, then it has three successive points of contact $A, B, C$ such that the arcs $AB$ and $BC$ are unequal. Let $X, Y, Z$ be the points of intersection of the pairs of tangents at $(A, B),$ $(B,C),$ $(A, C)$ respectively. $X$ and $Y$ are successive vertices of $P.$ By Lemma 3, the length $|XY|$ is strictly greater than what it would be if $B$ were moved to the midpoint of the arc $AC$ upon which it lies. Therefore the semiperimeter of $P$ is not minimal, therefore neither is its area. $\ \square$
New version
Lemma 4. Let $B_1, B_2$ be points on an arc of a circle $\sigma$ between rays $l, l'$ from the centre $O.$ For $i = 1, 2,$ let the tangent to $\sigma$ at $B_i$ meet $l$ at $X_i,$ and $l'$ at $Y_i.$ If $\angle X_1OB_1 < \angle X_2OB_2,$ and if $\angle X_1OB_1 + \angle X_2OB_2$ is strictly less than the angle between $l$ and $l'$ - in particular, if $B_2$ lies at most halfway around the arc between $l$ and $l'$ - then $|X_1Y_1| > |X_2Y_2|.$
Proof. Construct the ray $l''$ through $O$ making an angle equal to $\angle X_1OB_1$ with $OB_2.$ By the second hypothesis, $l''$ lies between $l$ and $l'.$ For $i = 1, 2,$ let $l''$ intersect $X_iY_i$ at $W_i.$ Let $X_1Y_1$ and $X_2Y_2$ intersect at $V.$ Because $OB_1$ and $OB_2$ are reflections of each other in $OV,$ the tangents to $\sigma$ at $B_1$ and $B_2$ are reflections of each other. Because $l$ and $l''$ are also reflections of each other, the points of intersection $W_2, W_1$ are reflections of $X_1, X_2$ respectively; therefore the triangles $VW_2W_1$ and $VX_1X_2$ are reflections of each other. But $\triangle VW_2W_1$ is strictly contained in $\triangle VY_2Y_1.$ Therefore the area of $\triangle VX_1X_2$ is strictly less than that of $\triangle VY_2Y_1.$ Therefore the area of $\triangle OX_2Y_2$ is strictly less than that of $\triangle OX_1Y_1.$ But both triangles have altitudes equal to the radius of $\sigma.$ Therefore $|X_1Y_1| > |X_2Y_2|.$ $\ \square$
Lemma 5. Let $\gamma$ be a circle. Let the tangents from an external point $Z$ meet $\gamma$ at $A, C.$ Let $B$ be a point on the shorter arc $AC$. Let the tangent at $B$ meet the tangents $ZA, ZC$ at $X, Y,$ respectively. Then the length $|XY|$ of the segment $XY$ is least when $B$ is midway between $A$ and $C,$ and increases strictly as $B$ moves from the midpoint towards $A$ or $C.$
Proof. Let $O$ be the centre of $\gamma.$ Then $OX$ bisects $\angle AOB,$ and $OY$ bisects $\angle BOC,$ therefore $\angle XOY$ has the constant value $\frac12\angle AOC.$ The altitude of the triangle $XOY$ is also constant, being equal to the radius $r$ of $\gamma.$
Construct lines $l, l'$ meeting at angle $\frac12\angle AOC$ (i.e. at an angle whose value is copied from $\frac12\angle AOC$ in the present configuration) at the centre $O'$ of a circle of radius $r.$ Then the figure $OXBY$ is congruent to the similarly labelled figure in Lemma 4, which traverses the circular arc from $l$ to $l'$ as the present figure $OXBY$ traverses the arc $AC.$ Because both figures occupy the midpoint position at the same time (so to speak), the conclusion of Lemma 4 applies also to the present configuration. $\ \square$
Theorem 2. Let $\gamma$ be a circle. Let $n$ be an integer $\geqslant 3.$ Let $P$ be an $n$-gon circumscribed about $\gamma.$ Then the area of $P$ is minimal if and only if $P$ is regular.
Proof. The area of $P$ is $rs,$ where $r$ is the radius of $\gamma$ and $s$ is the semiperimeter of $P.$ Minimising the area of $P$ is equivalent to minimising $s$. Each vertex of $P$ contributes, to the sum $s,$ the common length of the two tangents to $\gamma$ from that vertex. Therefore, if all but one of the $n$ points of contact of $P$ with $\gamma$ are held constant, and the edge of $P$ that meets $\gamma$ at that one variable point is $XY,$ then the total contribution to $s$ made by the vertices $X$ and $Y$ is equal to $|XY|.$
If $P$ is not regular, then it has three successive points of contact $A, B, C$ such that the arcs $AB$ and $BC$ are unequal. Let $X, Y, Z$ be the points of intersection of the pairs of tangents at $(A, B),$ $(B,C),$ $(A, C)$ respectively. $X$ and $Y$ are successive vertices of $P.$ By Lemma 5, the length $|XY|$ is strictly greater than what it would be if $B$ were moved closer to the midpoint of the arc $AC$ upon which it lies.
Mark the $n$ arcs between the $n$ points of contact of $P$ with $\gamma$ with a cyclic sequence of $n$ symbols $+, 0, -,$ one per arc, according as the length of the arc is greater than, equal to, or less than $1/n$ of the circumference of $\gamma.$ If $P$ is not regular, there must occur at least one sequence consisting of a $+,$ followed by a possibly empty sequence of $0$s, followed by a $-.$ Choose one such sequence.
Iteratively, modify $P$ by reducing the length of the first arc in the sequence, which is marked with a $+,$ to $1/n$ of the circumference of $\gamma.$ If the next arc in the sequence was a $0,$ it becomes the $+$ for the next iteration, and the numbers of $+, 0, -$ symbols in the sequence remain unchanged. Eventually we come to a sequence $+-,$ which in the next step becomes either a $0+,$ a $00,$ or a $0-.$
Any such sequence of steps increases by either $1$ or $2$ the number of arcs of length $1/n$ of the circumference of $\gamma$; thus eventually $P$ becomes regular.
In any step in which a sequence $+0$ is replaced by $0+,$ the area of $P$ is unchanged. There is at least one step in which a sequence $+-$ is replaced by $0+,$ $00,$ or $0-.$ In each of those three cases, a point of contact $B$ has been moved closer to the midpoint of the arc $AC$ on which it lies. (In the case $00,$ $B$ becomes the midpoint.) By Lemma 5, therefore, the semiperimeter $s,$ and therefore also the area of $P,$ has been made strictly smaller.
Therefore, the area of the initial irregular circumscribing polygon $P$ is strictly greater than that of some regular circumscribing $n$-gon. All regular circumscribing $n$-gons are congruent, by rotation, therefore all have the same area. Therefore, every regular circumscribing $n$-gon has a strictly smaller area than every irregular circumscribing $n$-gon. $\ \square$
A shorter, almost equally low-tech trigonometric proof, avoiding calculus
For $i = 1, 2, \ldots, n,$ let $\theta_i$ be half the angle subtended at the centre of $\gamma$ by the arc between the $i^\text{th}$ and $(i+1)^\text{th}$ points of contact of $P$ with $\gamma.$ (Addition of indices $i$ is modulo $n,$ i.e. $i + n = i$ for all $i.$) Then: $$ s = r(\tan\theta_1 + \tan\theta_2 + \cdots + \tan\theta_n). $$
Let $M, m$ be indices such that $\tan\theta_M = \max\{\tan\theta_1, \tan\theta_2, \ldots, \tan\theta_n\}$ and $\tan\theta_m = \min\{\tan\theta_1, \tan\theta_2, \ldots, \tan\theta_n\}.$ Then $\tan\theta_M = \tan\theta_m$ if and only if $P$ is regular.
If $P$ is not regular, then $\theta_m < \pi/n < \theta_M.$ Define $\theta_M' = \pi/n$ and $\theta_m' = \theta_m + \theta_M - \pi/n.$ Then $\theta_M' + \theta_m' = \theta_M + \theta_m,$ and: $$ 0 < \theta_m < \min\{\theta_m', \theta_M'\} \leqslant \max\{\theta_m', \theta_M'\} < \theta_M < \pi/2. $$
By an argument along the lines of Michael Rozenberg's first proof in his answer to this question: $$ \tan\theta_M + \tan\theta_m > \tan\theta_M' + \tan\theta_m'. $$ So if we replace $\theta_M, \theta_m$ with $\theta_M', \theta_m',$ and set $\theta_i' = \theta_i$ for all $i \ne M, m$ (in essentially the same way as in the synthetic proof above, where we used Lemma 5, instead of this trigonometric inequality), we get a polygon $P'$ circumscribing $\gamma$ whose semiperimeter and therefore area are strictly less than those of $P.$
Because the number of values of $i$ such that $\theta_i' = \pi/n$ is at least one more than the number of values of $i$ such that $\theta_i = \pi/n,$ this iterative process (which is a little simpler than the one described before) terminates after at most $n - 1$ steps, yielding a regular circumscribing polygon with an area strictly less than that of the initial polygon $P.$ $\ \square$