On the regular polygons

Tiberiu Popoviciu
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T. Popoviciu

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T. Popoviciu, Asupra poligoanelor regulate, Pozitiva, 2 (1941), pp. 92-97 (in Romanian).

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1941 a -Popoviciu- Pozitiva – Asupra poligoanelor regulate

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1941 -Popoviciu- Pozitiva - On regular polygons
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On regular polygons

by Tiberiu Popoviciu

  1. Let's consider a regular polygon A 0 A 1 … A n − 1 A 0 A 1 … A n − 1 A_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}A0 A1… An−1of n n nnnsides in a plane. Let P be a point in the plane and
M 1 ( P ) = PA 0 ― + PA 1 ― + ⋯ + PA n − 1 n , M 2 ( P ) = PA ― 0 2 + P A 1 ― + ⋯ + PA n − 1 2 n . M 1 ( P ) = PA 0 ¯ + PA 1 ¯ + ⋯ + PA n − 1 n , M 2 ( P ) = PA ¯ 0 2 + P A 1 ¯ + ⋯ + PA n − 1 2 n . {:[M_(1)(P)=( bar(PA_(0))+ bar(PA_(1))+cdots+PA_(n-1))/(n)","],[M_(2)(P)=sqrt(( bar(PA)_(0)^(2)+ bar(PA_(1))+cdots+PA_(n-1)^(2))/(n)).]:}\begin{gathered} \mathrm{M}_{1}(\mathrm{P})=\frac{\overline{\mathrm{PA}_{0}}+\overline{\mathrm{PA}_{1}}+\cdots+\mathrm{PA}_{n-1}}{n}, \\ \mathrm{M}_{2}(\mathrm{P})=\sqrt{\frac{\overline{\mathrm{PA}}_{0}^{2}+\overline{\mathrm{P} A_{1}}+\cdots+\mathrm{PA}_{n-1}^{2}}{n}} . \end{gathered}M1(P)=PA0―+PA1―+⋯+PAn−1n,M2(P)=PA―02+PA1―+⋯+PAn−12n.
M 1 ( P ) M 1 ( P ) M_(1)(P)M_{1}(P)M1(P)is the arithmetic mean and M 2 ( P ) M 2 ( P ) M_(2)(P)M_{2}(P)M2(P)the square root of the distances of P from the vertices of the polygon. We always have the inequality
M 1 ( P ) ≦ M ? ( P ) M 1 ( P ) ≦ M ? ( P ) M_(1)(P) <= M_(?)(P)M_{1}(P) \leqq M_{?}(P)M1(P)≦M?(P)
and equality is only possible if PA ― = PA 1 ― = … = PA n − 1 ― PA ¯ = PA 1 ¯ = … = PA n − 1 ¯ bar(PA)= bar(PA_(1))=dots= bar(PA_(n-1))\overline{\mathrm{PA}}=\overline{\mathrm{PA}_{1}}=\ldots=\overline{\mathrm{PA}_{n-1}}PA―=PA1―=…=PAn−1―This actually happens if and only if P P PPPis the center of gravity of the polygon. In other words, the ratio
(1) E ( P ) = M 1 ( P ) M 1 ( P ) (1) E ( P ) = M 1 ( P ) M 1 ( P ) {:(1)E(P)=(M_(1)(P))/(M_(1)(P)):}\begin{equation*} E(P)=\frac{M_{1}(P)}{M_{1}(P)} \tag{1} \end{equation*}(1)AND(P)=M1(P)M1(P)
has a maximum equal to 1 and this maximum is reached if and only if P is the center of the polygon.
In the following we propose to determine the minimum of the ratio (1). It is sufficient to assume that the center of the polygon is the origin and that the vertices are represented by complex numbers e i 2 k π n , k = 0 , 1 , … n − 1 e i 2 k π n , k = 0 , 1 , … n − 1 e^(i(2k pi)/(n)),k=0,1,dots n-1e^{i \frac{2 k \pi}{n}}, k=0,1, \ldots n-1andi2kpn,k=0,1,…n−1. The floating point P P PPPwill be represented by the complex number ρ e i θ , ρ ρ e i θ , ρ rhoe^(i theta),rho\rho e^{i \theta}, \rhorandii,rbeing the module and θ θ theta\thetaithe argument, ρ ≧ 0 , 0 ≦ θ ≦ 2 π ρ ≧ 0 , 0 ≦ θ ≦ 2 π rho >= 0,0 <= theta <= 2pi\rho \geqq 0,0 \leqq \theta \leqq 2 \pir≧0,0≦i≦2p. We then have
PA k = ρ 2 + 1 − 2 ρ cos ⁡ ( θ − 2 k π n ) , k = 0 , 1 , … , n − 1 M 3 ( P ) = ρ 2 + 1 PA k = ρ 2 + 1 − 2 ρ cos ⁡ θ − 2 k π n , k = 0 , 1 , … , n − 1 M 3 ( P ) = ρ 2 + 1 {:[PA_(k)=sqrt(rho^(2)+1-2rho cos(theta-(2k pi)/(n)))","k=0","1","dots","n-1],[M_(3)(P)=sqrt(rho^(2)+1)]:}\begin{gathered} \mathrm{PA}_{k}=\sqrt{\rho^{2}+1-2 \rho \cos \left(\theta-\frac{2 k \pi}{n}\right)}, k=0,1, \ldots, n-1 \\ \mathrm{M}_{3}(\mathrm{P})=\sqrt{\rho^{2}+1} \end{gathered}PAk=r2+1−2rcos⁡(i−2kpn),k=0,1,…,n−1M3(P)=r2+1
  1. We can directly prove the following
Lemma. The ratio (1) has a positive minimum reached for at least one point P in the plane.
E ( P ) E ( P ) E(P)\mathrm{E}(\mathrm{P})AND(P)is a continuous function in the entire plane and never vanishes. Based on the above observations, if P 1 P 1 P_(1)P_{1}P1is a different point of origin,
so we have
0 < E ( P 1 ) = 0 < 1 0 < E P 1 = 0 < 1 0 < E(P_(1))=0 < 10<E\left(P_{1}\right)=0<10<AND(P1)=0<1
(010) min E ( P ) ≦ a < 1 min E ( P ) ≦ a < 1 quad minE(P) <= a < 1\quad \min \mathrm{E}(\mathrm{P}) \leqq a<1minAND(P)≦a<1.
Dar PA k ― ≧ | ρ − 1 | , M 2 ( P ) ≦ ρ + 1 Dar PA k ¯ ≧ | ρ − 1 | , M 2 ( P ) ≦ ρ + 1 Dar bar(PA_(k)) >= |rho-1|,M_(2)(P) <= rho+1\operatorname{Dar} \overline{\mathrm{PA}_{k}} \geqq|\rho-1|, \mathrm{M}_{2}(\mathrm{P}) \leqq \rho+1ButPAk―≧|r−1|,M2(P)≦r+1, so
E ( P ) ≧ | p − 1 | p + 1 E ( P ) ≧ | p − 1 | p + 1 E(P) >= (|p-1|)/(p+1)\mathrm{E}(\mathrm{P}) \geqq \frac{|p-1|}{p+1}AND(P)≧|p−1|p+1
If we take ρ > 1 + α 1 − α ρ > 1 + α 1 − α rho > (1+alpha)/(1-alpha)\rho>\frac{1+\alpha}{1-\alpha}r>1+a1−a, we have
E ( P ) > α . E ( P ) > α . E(P) > alpha.E(P)>\alpha .AND(P)>a.
It follows that the minimum of E ( P ) E ( P ) E(P)E(P)AND(P)in the whole plan it is the same c u c u cuc ucinhis minimum E ( P ) E ( P ) E(P)\mathrm{E}(\mathrm{P})AND(P)in the circle with center at the origin and radius 1 + α 1 − α 1 + α 1 − α (1+alpha)/(1-alpha)\frac{1+\alpha}{1-\alpha}1+a1−aThis latter circle is a closed and bounded domain so, according to a theorem of Weierstruss, its minimum E ( P ) E ( P ) E(P)\mathrm{E}(\mathrm{P})AND(P)is reached. This minimum cannot be zero because E ( P ) E ( P ) E(P)\mathrm{E}(\mathrm{P})AND(P)does not cancel, so
min E ( P ) > 0 . min E ( P ) > 0 . min E(P) > 0.\min E(P)>0 .minAND(P)>0.
We can also observe that, due to symmetry, the minimum is reached in at least n n nnnpoints forming a regular polygon with the center at the origin.
3. Let's now move on to the actual determination of the minimum.
Suppose that P moves on a half-line starting from the origin, in other words that θ θ theta\thetaiis fixed and ρ ρ rho\rhorvaries from 0 to + ∞ + ∞ +oo+\infty+∞Let's put
t = 2 p p 2 + 1 t = 2 p p 2 + 1 t=(2p)/(p^(2)+1)t=\frac{2 p}{p^{2}+1}t=2pp2+1
If ρ ρ rho\rhorvaries from 0 to + ∞ + ∞ +oo+\infty+∞, t t tttfirst increases from 0 to its maximum of 1 which it reaches for ρ = 1 ρ = 1 rho=1\rho=1r=1, and then decreases towards 0. We have
(2) E ( P ) = 1 n ∑ k = 1 n − 1 1 − t cos ⁡ ( θ − 2 k π n ) (2) E ( P ) = 1 n ∑ k = 1 n − 1   1 − t cos ⁡ θ − 2 k π n {:(2)E(P)=(1)/(n)sum_(k=1)^(n-1)sqrt(1-t cos(theta-(2k pi)/(n))):}\begin{equation*} \mathrm{E}(\mathrm{P})=\frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-t \cos \left(\theta-\frac{2 k \pi}{n}\right)} \tag{2} \end{equation*}(2)AND(P)=1n∑k=1n−11−tcos⁡(i−2kpn)
E(P) is then a continuous function of t t tttin the closed range [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]and is indefinitely differentiable in the semiclosed interval [ 0 , 1 ) [ 0 , 1 ) [0,1)[0,1)[0,1)We have
d 2 E ( P ) d t 2 = − 1 4 n ∑ k = 0 n − 1 cos 2 ⁡ ( θ − 2 k π n ) [ 1 − t cos ⁡ ( θ − 2 k π n ) ] 3 d 2 E ( P ) d t 2 = − 1 4 n ∑ k = 0 n − 1   cos 2 ⁡ θ − 2 k π n 1 − t cos ⁡ θ − 2 k π n 3 (d^(2)E(P))/(dt^(2))=-(1)/(4n)sum_(k=0)^(n-1)(cos^(2)(theta-(2k pi)/(n)))/([sqrt(1-t cos(theta-(2k pi)/(n)))]^(3))\frac{d^{2} \mathrm{E}(\mathrm{P})}{d t^{2}}=-\frac{1}{4 n} \sum_{k=0}^{n-1} \frac{\cos ^{2}\left(\theta-\frac{2 k \pi}{n}\right)}{\left[\sqrt{1-t \cos \left(\theta-\frac{2 k \pi}{n}\right)}\right]^{3}}d2AND(P)dt2=−14n∑k=0n−1cos2⁡(i−2kpn)[1−tcos⁡(i−2kpn)]3
so
d 2 E ( P ) d t 2 < 0 , in intervalul [ 0 , 1 ) d 2 E ( P ) d t 2 < 0 ,  in intervalul  [ 0 , 1 ) (d^(2)E(P))/(dt^(2)) < 0,quad" in intervalul "[0,1)\frac{d^{2} \mathrm{E}(\mathrm{P})}{d t^{2}}<0, \quad \text { in intervalul }[0,1)d2AND(P)dt2<0, in the interval [0,1)
It is then known that E ( P ) E ( P ) E(P)\mathrm{E}(\mathrm{P})AND(P)is a concave function of t t tttin the closed range [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]This function does not reduce to a constant, so its minimum can only be reached for the extremes of the interval, that is, for t = 0 t = 0 t=0t=0t=0or t = 1 t = 1 t=1t=1t=1For t = 0 t = 0 t=0t=0t=0HAVE E ( P ) = 1 E ( P ) = 1 E(P)=1\mathrm{E}(\mathrm{P})=1AND(P)=1so the minimum is only reached for t = 1 t = 1 t=1t=1t=1and then we have p = 1 p = 1 p=1p=1p=1.
The minimum of expression (1) can therefore only be reached on the circumscribed circle of the polygon. Let ρ = 1 ρ = 1 rho=1\rho=1r=1and let's vary per. θ θ theta\thetaiIt is sufficient to consider, due to symmetry, the case 0 ≦ θ ≦ 2 π n 0 ≦ θ ≦ 2 π n 0 <= theta <= (2pi)/(n)0 \leqq \theta \leqq \frac{2 \pi}{n}0≦i≦2pn. We then have
(3)
E ( P ) = 2 n ∑ k = 0 n − 1 | sin ⁡ ( θ 2 − k π n ) | = E ( P ) = 2 n ∑ k = 0 n − 1   sin ⁡ θ 2 − k π n = E(P)=(sqrt2)/(n)sum_(k=0)^(n-1)|sin((theta)/(2)-(k pi)/(n))|=\mathrm{E}(\mathrm{P})=\frac{\sqrt{2}}{n} \sum_{k=0}^{n-1}\left|\sin \left(\frac{\theta}{2}-\frac{k \pi}{n}\right)\right|=AND(P)=2n∑k=0n−1|sin⁡(i2−kpn)|=
2 n [ sin ⁡ θ 2 + ∑ k = 1 n sin ⁡ ( k π n − θ 2 ) ] 2 n sin ⁡ θ 2 + ∑ k = 1 n   sin ⁡ k π n − θ 2 (sqrt2)/(n)[sin((theta)/(2))+sum_(k=1)^(n)sin((k pi)/(n)-(theta)/(2))]\frac{\sqrt{2}}{n}\left[\sin \frac{\theta}{2}+\sum_{k=1}^{n} \sin \left(\frac{k \pi}{n}-\frac{\theta}{2}\right)\right]2n[sin⁡i2+∑k=1nsin⁡(kpn−i2)]
However, we have
∑ k = 1 n − 1 sin ⁡ k π n = cotg π 2 n , ∑ k = 1 n − 1 cos ⁡ k π n = 0 ∑ k = 1 n − 1   sin ⁡ k π n = cotg π 2 n , ∑ k = 1 n − 1   cos ⁡ k π n = 0 sum_(k=1)^(n-1)sin((k pi)/(n))=cotg((pi)/(2n)),quadsum_(k=1)^(n-1)cos((k pi)/(n))=0\sum_{k=1}^{n-1} \sin \frac{k \pi}{n}=\operatorname{cotg} \frac{\pi}{2 n}, \quad \sum_{k=1}^{n-1} \cos \frac{k \pi}{n}=0∑k=1n−1sin⁡kpn=boxp2n,∑k=1n−1cos⁡kpn=0
so
E ( P ) = 2 n − [ sin ⁡ θ 2 + cotg π 2 n cos ⁡ θ 2 ] E ( P ) = 2 n − sin ⁡ θ 2 + cotg π 2 n cos ⁡ θ 2 E(P)=(sqrt2)/(n)-[sin((theta)/(2))+cotg((pi)/(2n))cos((theta)/(2))]\mathrm{E}(\mathrm{P})=\frac{\sqrt{2}}{n}-\left[\sin \frac{\theta}{2}+\operatorname{cotg} \frac{\pi}{2 n} \cos \frac{\theta}{2}\right]AND(P)=2n−[sin⁡i2+boxp2ncos⁡i2]
and
min E ( P ) = 2 n cotg π 2 n min E ( P ) = 2 n cotg π 2 n min E(P)=(sqrt2)/(n)cotg((pi)/(2n))\min E(P)=\frac{\sqrt{2}}{n} \operatorname{cotg} \frac{\pi}{2 n}minAND(P)=2nboxp2n
which is reached only for θ = 0 θ = 0 theta=0\theta=0i=0and θ = 2 n θ = 2 n theta=(2)/(n)\theta=\frac{2}{n}i=2nWe can thus
state the following property
Theorem 1. If A i A I … A n − 1 A i A I … A n − 1 A_(i)A_(I)dotsA_(n-1)\mathrm{A}_{\mathrm{i}} \mathrm{A}_{\mathrm{I}} \ldots \mathrm{A}_{n-1}AiAI…An−1is a regular polygon of n sides and P a point in the plane of the polygon, we have
M 1 ( P ) ≧ V ′ 2 n cotg π 2 n M 2 ( P ) M 1 ( P ) ≧ V ′ 2 n cotg π 2 n M 2 ( P ) M_(1)(P) >= (V^(')2)/(n)cotg((pi)/(2n)M_(2))(P)\mathrm{M}_{1}(\mathrm{P}) \geqq \frac{\mathrm{V}^{\prime} 2}{n} \operatorname{cotg} \frac{\pi}{2 n} \mathrm{M}_{2}(\mathrm{P})M1(P)≧In′2nboxp2nM2(P)
or
PA ― 0 + PA ― 1 + … + PA ― n − 1 ) 2 ≧ 2 cotg 2 π 2 n PA ― 0 2 + PA ― 1 2 + … + PA ― n − 1 ) PA ¯ 0 + PA ¯ 1 + … + PA ¯ n − 1 2 ≧ 2 cotg 2 π 2 n PA ¯ 0 2 + PA ¯ 1 2 + … + PA ¯ n − 1 {: bar(PA)_(0)+ bar(PA)_(1)+dots+ bar(PA)_(n-1))^(2) >= 2cotg^(2)(pi)/(2n) bar(PA)_(0)^(2)+ bar(PA)_(1)^(2)+dots+ bar(PA)_(n-1))\left.\left.\overline{\mathrm{PA}}_{0}+\overline{\mathrm{PA}}_{1}+\ldots+\overline{\mathrm{PA}}_{n-1}\right)^{2} \geqq 2 \operatorname{cotg}^{2} \frac{\pi}{2 n} \overline{\mathrm{PA}}_{0}{ }^{2}+\overline{\mathrm{PA}}_{1}{ }^{2}+\ldots+\overline{\mathrm{PA}}_{n-1}\right)PA―0+PA―1+…+PA―n−1)2≧2box2p2nPA―02+PA―12+…+PA―n−1)
the equality being true only if P coincides with one of the vertices of the polygon.
4. Some of the previous results can be easily generalized. Let us consider, instead of the arithmetic mean M 1 ( P ) M 1 ( P ) M_(1)(P)\mathrm{M}_{1}(\mathrm{P})M1(P), average power r r rrrof the distances of P to the vertices of the polygon, that is
M r ( P ) = [ PA 0 ― r + PA 1 r + … + PA ― n − r n ] 1 r . M r ( P ) = PA 0 ¯ r + PA 1 r + … + PA ¯ n − r n 1 r . M_(r)(P)=[( bar(PA_(0))^(r)+PA_(1)r+dots+ bar(PA)_(n-)r)/(n)](1)/(r).\mathrm{M}_{r}(\mathrm{P})=\left[\frac{{\overline{\mathrm{PA}_{0}}}^{r}+\mathrm{PA}_{1} r+\ldots+\overline{\mathrm{PA}}_{n-} r}{n}\right] \frac{1}{r} .Mr(P)=[PA0―r+PA1r+…+PA―n−rn]1r.
If 0 < r < 2 0 < r < 2 0 < r < 20<r<20<r<2we still have M r ( P ) ≦ M 2 ( P ) M r ( P ) ≦ M 2 ( P ) M_(r)(P) <= M_(2)(P)\mathrm{M}_{r}(\mathrm{P}) \leqq \mathrm{M}_{2}(\mathrm{P})Mr(P)≦M2(P)and if r > 2 r > 2 r > 2r>2r>2HAVE M r ( P ) ≧ M 2 ( P ) M r ( P ) ≧ M 2 ( P ) M_(r)(P) >= M_(2)(P)\mathrm{M}_{r}(\mathrm{P}) \geqq \mathrm{M}_{2}(\mathrm{P})Mr(P)≧M2(P), equality cannot occur, in both cases, unless P P PPPis the center of the polygon. Let
E r ( P ) = M r ( P ) M i ( P ) E r ( P ) = M r ( P ) M i ( P ) E_(r)(P)=(M_(r)(P))/(M_(i)(P))\mathrm{E}_{r}(\mathrm{P})=\frac{\mathrm{M}_{r}(\mathrm{P})}{\mathrm{M}_{i}(\mathrm{P})}ANDr(P)=Mr(P)Mi(P)
If 0 < r < 2 , E r ( P ) 0 < r < 2 , E r ( P ) 0 < r < 2,E_(r)(P)0<r<2, \mathrm{E}_{r}(\mathrm{P})0<r<2,ANDr(P)has reached a minimum, and if r > 2 r > 2 r > 2r>2r>2a maximum reached. In the present case, expression (2) becomes
and
E r r ( P ) = 1 n ∑ k = 0 n − 1 [ 1 − t cos ⁡ ( 0 − 2 k π n ) ] r 2 E r r ( P ) = 1 n ∑ k = 0 n − 1   1 − t cos ⁡ 0 − 2 k π n r 2 E_(r)^(r)(P)=(1)/(n)sum_(k=0)^(n-1)[1-t cos(0-(2k pi)/(n))]^((r)/(2))\mathrm{E}_{r}^{r}(\mathrm{P})=\frac{1}{n} \sum_{k=0}^{n-1}\left[1-t \cos \left(0-\frac{2 k \pi}{n}\right)\right]^{\frac{r}{2}}ANDrr(P)=1n∑k=0n−1[1−tcos⁡(0−2kpn)]r2
d 2 E r r ( P ) d t 2 = r ( r − 2 ) 4 ∑ k = 0 n − 1 cos 2 ⁡ ( θ − 2 k π n ) [ 1 − t cos ⁡ ( θ − 2 k π n ) ] r 2 − 2 d 2 E r r ( P ) d t 2 = r ( r − 2 ) 4 ∑ k = 0 n − 1   cos 2 ⁡ θ − 2 k π n 1 − t cos ⁡ θ − 2 k π n r 2 − 2 (d^(2)E_(r)^(r)(P))/(dt^(2))=(r(r-2))/(4)sum_(k=0)^(n-1)cos^(2)(theta-(2k pi)/(n))[1-t cos(theta-(2k pi)/(n))]^((r)/(2)-2)\frac{d^{2} \mathrm{E}_{r}^{r}(\mathrm{P})}{d t^{2}}=\frac{r(r-2)}{4} \sum_{k=0}^{n-1} \cos ^{2}\left(\theta-\frac{2 k \pi}{n}\right)\left[1-t \cos \left(\theta-\frac{2 k \pi}{n}\right)\right]^{\frac{r}{2}-2}d2ANDrr(P)dt2=r(r−2)4∑k=0n−1cos2⁡(i−2kpn)[1−tcos⁡(i−2kpn)]r2−2
which is < 0 < 0 < 0<0<0 resp. > 0 > 0 > 0>0>0as 0 < r < 2 0 < r < 2 0 < r < 20<r<20<r<2resp. r > 2 r > 2 r > 2r>2r>2. It is deduced as above that the minimum or maximum of E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})ANDr(P)it can only be touched on the circle circumscribed around the polygon
Expression (3) now becomes
E r r ( P ) = ( 2 ) r n [ sin r ⁡ θ 2 + ∑ k = 1 n − 1 sin r ⁡ ( k π n − θ 2 ) ] E r r ( P ) = ( 2 ) r n sin r ⁡ θ 2 + ∑ k = 1 n − 1   sin r ⁡ k π n − θ 2 E_(r)^(r)(P)=((sqrt2)^(r))/(n)[sin^(r)((theta)/(2))+sum_(k=1)^(n-1)sin^(r)((k pi)/(n)-(theta)/(2))]\mathrm{E}_{r}^{r}(\mathrm{P})=\frac{(\sqrt{2})^{r}}{n}\left[\sin ^{r} \frac{\theta}{2}+\sum_{k=1}^{n-1} \sin ^{r}\left(\frac{k \pi}{n}-\frac{\theta}{2}\right)\right]ANDrr(P)=(2)rn[sinr⁡i2+∑k=1n−1sinr⁡(kpn−i2)]
This expression is a continuous function of θ θ theta\thetaiin the interval [ 0 , 2 π n ] 0 , 2 π n [0,(2pi)/(n)]\left[0, \frac{2 \pi}{n}\right][0,2pn]We have, in the open interval ( 0 , 2 π n ) 0 , 2 π n (0,(2pi)/(n))\left(0, \frac{2 \pi}{n}\right)(0,2pn)
d 2 E r r ( P ) d t 2 = − r ( 2 ) r n − 1 4 n ∑ k = 0 1 | sin r − 2 ⁡ ( k π n − θ 2 ) | [ 1 − r cos 2 ⁡ ( k π n − θ 2 ] d 2 E r r ( P ) d t 2 = − r ( 2 ) r n − 1 4 n ∑ k = 0 1   sin r − 2 ⁡ k π n − θ 2 1 − r cos 2 ⁡ k π n − θ 2 (d^(2)E_(r)^(r)(P))/(dt^(2))=-(r(sqrt2)^(rn-1))/(4n)sum_(k=0)^(1)|sin^(r-2)((k pi)/(n)-(theta)/(2))|[1-rcos^(2)((k pi)/(n)-(theta)/(2)]:}\frac{d^{2} \mathrm{E}_{r}^{r}(\mathrm{P})}{d t^{2}}=-\frac{r(\sqrt{2})^{r n-1}}{4 n} \sum_{k=0}^{1}\left|\sin ^{r-2}\left(\frac{k \pi}{n}-\frac{\theta}{2}\right)\right|\left[1-r \cos ^{2}\left(\frac{k \pi}{n}-\frac{\theta}{2}\right]\right.d2ANDrr(P)dt2=−r(2)rn−14n∑k=01|sinr−2⁡(kpn−i2)|[1−rcos2⁡(kpn−i2]
and it is seen that this derivative is always negative if r ≦ 1 r ≦ 1 r <= 1r \leqq 1r≦1. One can, however, state the following generalization of Theorem 1.
Theorem 2. If A 0 A 1 … A n − 1 A 0 A 1 … A n − 1 A_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}A0A1…An−1is a regular polygon of n n nnnswill, r r rrra positive number ≦ 1 ≦ 1 <= 1\leqq 1≦1and P P PPPa point in the plane of the polygon, we have
M r ( P ) ≧ 2 n [ ∑ k = 1 n − 1 sin r ⁡ k π n ] r ¯ M 2 ( P ) M r ( P ) ≧ 2 n ∑ k = 1 n − 1   sin r ⁡ k π n r ¯ M 2 ( P ) M_(r)(P) >= (sqrt2)/(sqrtn)[sum_(k=1)^(n-1)sin^(r)((k pi)/(n))] bar(r)M_(2)(P)\mathrm{M}_{r}(\mathrm{P}) \geqq \frac{\sqrt{2}}{\sqrt{n}}\left[\sum_{k=1}^{n-1} \sin ^{r} \frac{k \pi}{\mathrm{n}}\right] \bar{r} \mathrm{M}_{2}(\mathrm{P})Mr(P)≧2n[∑k=1n−1sinr⁡kpn]r¯M2(P)
equality is not true unless P P PPPcoincides with one of the vertices of the polygon.
5. But r > 1 r > 1 r > 1r>1r>1the problem is more complicated. We will examine here only the case when r r rrris a positive and even integer, r = 2 m r = 2 m r=2mr=2 mr=2mIt all comes down to evaluating the amount.
S m = ∑ k = 0 n − 1 sin 2 m ⁡ [ k π n − θ 2 ] S m = ∑ k = 0 n − 1   sin 2 m ⁡ k π n − θ 2 S_(m)=sum_(k=0)^(n-1)sin^(2m)[(k pi)/(n)-(theta)/(2)]\mathrm{S}_{m}=\sum_{k=0}^{n-1} \sin ^{2 m}\left[\frac{k \pi}{n}-\frac{\theta}{2}\right]Sm=∑k=0n−1sin2m⁡[kpn−i2]
Let's remember the formula
sin 2 m ⁡ α = 1 2 2 m ( 2 m m ) + 2 ∑ s = 1 m ( − 1 ) s m m + s 2 m ) 2 2 m cos ⁡ 2 s α sin 2 m ⁡ α = 1 2 2 m ( 2 m m ) + 2 ∑ s = 1 m   ( − 1 ) s m m + s 2 m 2 2 m cos ⁡ 2 s α sin^(2m)alpha=(1)/(2^(2m))((2m)/(m))+2sum_(s=1)^(m)((-1)^(s)_(m)_(m+s)^(2m)))/(2^(2m))cos 2s alpha\sin ^{2 m} \alpha=\frac{1}{2^{2 m}}\binom{2 m}{m}+2 \sum_{s=1}^{m} \frac{\left.(-1)^{s}{ }_{m}{ }_{m+s}^{2 m}\right)}{2^{2 m}} \cos 2 s \alphasin2m⁡a=122m(2mm)+2∑s=1m(−1)smm+s2m)22mcos⁡2sa
However, we have
∑ k = 0 n − 1 cos ⁡ 2 s [ k π n − θ 2 ] = { 0 , dacă s ≢ 0 ( mod n ) ∑ k = 0 n − 1   cos ⁡ 2 s k π n − θ 2 = 0 ,       dacă       s ≢ 0 ( mod n ) sum_(k=0)^(n-1)cos 2s[(k pi)/(n)-(theta)/(2)]={[0","," dacă ",s≢0](modn):}\sum_{k=0}^{n-1} \cos 2 s\left[\frac{k \pi}{n}-\frac{\theta}{2}\right]=\left\{\begin{array}{lll}0, & \text { dacă } & s \not \equiv 0\end{array}(\bmod \mathrm{n})\right.∑k=0n−1cos⁡2s[kpn−i2]={0, if s≢0(againstn)
so
S m = n 2 2 m ( 2 m m ) + 2 n ∑ j = 1 ( m n ) ( − 1 ) j n ( 2 m m + j n ) 2 2 m cos ⁡ j n θ S m = n 2 2 m ( 2 m m ) + 2 n ∑ j = 1 m n   ( − 1 ) j n ( 2 m m + j n ) 2 2 m cos ⁡ j n θ S_(m)=(n)/(2^(2m))((2(m))/((m)))+2nsum_(j=1)^(((m)/(n)))((-1)^(jn)((2m)/(m+jn)))/(2^(2m))cos jn theta\mathrm{S}_{m}=\frac{\mathrm{n}}{2^{2 m}}\binom{2 \mathrm{~m}}{\mathrm{~m}}+2 n \sum_{j=1}^{\left(\frac{m}{n}\right)} \frac{(-1)^{j n}\binom{2 m}{m+j n}}{2^{2 m}} \cos j n \thetaSm=n22m(2 m m)+2n∑j=1(mn)(−1)jn(2mm+jn)22mcos⁡jni
[ λ λ lambda\lambdal] meaning the largest integer ≦ λ ≦ λ <= lambda\leqq \lambda≦lEspecially
if m < n m < n m < nm<nm<n, we have
S m = n 2 2 m ( 2 m m ) S m = n 2 2 m ( 2 m m ) S_(m)=(n)/(2^(2m))((2(m))/((m)))S_{m}=\frac{n}{2^{2 m}}\binom{2 \mathrm{~m}}{\mathrm{~m}}Sm=n22m(2 m m)
and the following can be stated
Theorem 3. If m is an integer ( > 1 > 1 > 1>1>1) and A 0 A 1 … A n − 1 A 0 A 1 … A n − 1 A_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}A0A1…An−1a regular polygon of n ≥ m n ≥ m n >= mn \geq mn≥msides and P P PPPa point in the plane of the polygon, we have
M 2 m ( P ) ≦ 1 2 ( 2 m m ) 1 2 m M 2 ( P ) M 2 m ( P ) ≦ 1 2 ( 2 m m ) 1 2 m M 2 ( P ) M_(2m)(P) <= (1)/(sqrt2)((2m)/(m))^((1)/(2m))M_(2)(P)\mathrm{M}_{2 m}(\mathrm{P}) \leqq \frac{1}{\sqrt{2}}\binom{2 m}{m}^{\frac{1}{2 m}} \mathrm{M}_{2}(\mathrm{P})M2m(P)≦12(2mm)12mM2(P)
equality is not true unless P P PPPis on the circle circumscribed around the polygon.
If we have m > n m > n m > nm>nm>n, it can be seen from its structure S m S m S_(m)S_{m}Smthat the maximum is reached for θ = 0 θ = 0 theta=0\theta=0i=0and θ = 2 π n θ = 2 π n theta=(2pi)/(n)\theta=\frac{2 \pi}{n}i=2pnif n n nnnit is even for θ = π n θ = π n theta=(pi )/(n)\theta=\frac{\pi}{n}i=pnif n n nnnis odd, so
Theorem 4. If A 0 A 1 … A n − 1 A 0 A 1 … A n − 1 A_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}A0A1…An−1is a regular polygon of n ( ≧ 3 ) n ( ≧ 3 ) n( >= 3)n(\geqq 3)n(≧3)sides and m m mmma positive integer >in, whatever the point P P PPPfrom the polygon plane we have
M 2 m ( P ) M 2 ( P ) ≦ 1 2 [ ( 2 m m ) + 2 ∑ j = 1 [ m n ] ( 2 m m + j n ) ] 1 2 m M 2 m ( P ) M 2 ( P ) ≦ 1 2 ( 2 m m ) + 2 ∑ j = 1 m n   ( 2 m m + j n ) 1 2 m (M_(2m)(P))/(M_(2)(P)) <= (1)/(sqrt2)[((2m)/(m))+2sum_(j=1)^([(m)/(n)])((2m)/(m+jn))]^((1)/(2m))\frac{\mathrm{M}_{2 m}(\mathrm{P})}{\mathrm{M}_{2}(\mathrm{P})} \leqq \frac{1}{\sqrt{2}}\left[\binom{2 m}{m}+2 \sum_{j=1}^{\left[\frac{m}{n}\right]}\binom{2 m}{m+j n}\right]^{\frac{1}{2 m}}M2m(P)M2(P)≦12[(2mm)+2∑j=1[mn](2mm+jn)]12m
equality is not true unless
1 0 n 1 0 n 1^(0)n1^{0} n10nbeing even, P P PPPcoincides with one of the vertices of the polygon.
2 0 n 2 0 n 2^(0)n2^{0} n20nbeing odd, P P PPPcoincides with the middle of a circumscribed arc limited by two consecutive vertices of the polygon.
6. To conclude, I draw the readers' attention to the fact that several of the previous results would be interesting to complete. First of all, it should be seen, in the special case examined by us, what happens if r r rrris any positive number greater than 1. We should then examine the case when instead of a regular polygon we take an arbitrary polygon. It can easily be seen that in general (for an arbitrary polygon) the ratio
M p ( P ) M q ( P ) M p ( P ) M q ( P ) (M_(p)(P))/(M_(q)(P))\frac{\mathrm{M}_{p}(\mathrm{P})}{\mathrm{M}_{q}(\mathrm{P})}Mp(P)Mq(P)
where 0 < p < q 0 < p < q 0 < p < q0<p<q0<p<q, has a minimum reached. This minimum depends on the polygon considered and has a maximum when this polygon varies. It is very likely that this "maximum minimorum" is reached, in particular, for a regular polygon.
Bucharest, November 29, 1941.
1941

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