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T. Popoviciu, Asupra poligoanelor regulate, Pozitiva, 2 (1941), pp. 92-97 (in Romanian).
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[MR0017921]
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On regular polygons
- Let's consider a regular polygonA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}ofnnsides in a plane. Let P be a point in the plane and
{:[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}
M_(1)(P)M_{1}(P)is the arithmetic mean andM_(2)(P)M_{2}(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) \leqq M_{?}(P)
and equality is only possible ifbar(PA)= bar(PA_(1))=dots= bar(PA_(n-1))\overline{\mathrm{PA}}=\overline{\mathrm{PA}_{1}}=\ldots=\overline{\mathrm{PA}_{n-1}}This actually happens if and only ifPPis the center of gravity of the polygon. In other words, the ratio
{:(1)E(P)=(M_(1)(P))/(M_(1)(P)):}\begin{equation*}
E(P)=\frac{M_{1}(P)}{M_{1}(P)} \tag{1}
\end{equation*}
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 numberse^(i(2k pi)/(n)),k=0,1,dots n-1e^{i \frac{2 k \pi}{n}}, k=0,1, \ldots n-1. The floating pointPPwill be represented by the complex numberrhoe^(i theta),rho\rho e^{i \theta}, \rhobeing the module andtheta\thetathe argument,rho >= 0,0 <= theta <= 2pi\rho \geqq 0,0 \leqq \theta \leqq 2 \pi. We then have
{:[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}
- 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)\mathrm{E}(\mathrm{P})is a continuous function in the entire plane and never vanishes. Based on the above observations, ifP_(1)P_{1}is a different point of origin,
so we have
E(P)\mathrm{E}(\mathrm{P})is a continuous function in the entire plane and never vanishes. Based on the above observations, ifP_(1)P_{1}is a different point of origin,
so we have
0 < E(P_(1))=0 < 10<E\left(P_{1}\right)=0<1
(010) quad minE(P) <= a < 1\quad \min \mathrm{E}(\mathrm{P}) \leqq a<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+1, so
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+1, so
E(P) >= (|p-1|)/(p+1)\mathrm{E}(\mathrm{P}) \geqq \frac{|p-1|}{p+1}
If we takerho > (1+alpha)/(1-alpha)\rho>\frac{1+\alpha}{1-\alpha}, we have
E(P) > alpha.E(P)>\alpha .
It follows that the minimum ofE(P)E(P)in the whole plan it is the samecuc uhis minimumE(P)\mathrm{E}(\mathrm{P})in the circle with center at the origin and radius(1+alpha)/(1-alpha)\frac{1+\alpha}{1-\alpha}This latter circle is a closed and bounded domain so, according to a theorem of Weierstruss, its minimumE(P)\mathrm{E}(\mathrm{P})is reached. This minimum cannot be zero becauseE(P)\mathrm{E}(\mathrm{P})does not cancel, so
min E(P) > 0.\min E(P)>0 .
We can also observe that, due to symmetry, the minimum is reached in at leastnnpoints forming a regular polygon with the center at the origin.
3. Let's now move on to the actual determination of the minimum.
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 thattheta\thetais fixed andrho\rhovaries from 0 to+oo+\inftyLet's put
t=(2p)/(p^(2)+1)t=\frac{2 p}{p^{2}+1}
Ifrho\rhovaries from 0 to+oo+\infty, ttfirst increases from 0 to its maximum of 1 which it reaches forrho=1\rho=1, and then decreases towards 0. We have
{:(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*}
E(P) is then a continuous function ofttin the closed range[0,1][0,1]and is indefinitely differentiable in the semiclosed interval[0,1)[0,1)We have
(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}}
so
(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)
It is then known thatE(P)\mathrm{E}(\mathrm{P})is a concave function ofttin the closed range[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, fort=0t=0ort=1t=1Fort=0t=0HAVEE(P)=1\mathrm{E}(\mathrm{P})=1so the minimum is only reached fort=1t=1and then we havep=1p=1.
The minimum of expression (1) can therefore only be reached on the circumscribed circle of the polygon. Letrho=1\rho=1and let's vary per.theta\thetaIt is sufficient to consider, due to symmetry, the case0 <= theta <= (2pi)/(n)0 \leqq \theta \leqq \frac{2 \pi}{n}. We then have
(3)
(3)
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|=
(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]
However, we have
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
so
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
min E(P)=(sqrt2)/(n)cotg((pi)/(2n))\min E(P)=\frac{\sqrt{2}}{n} \operatorname{cotg} \frac{\pi}{2 n}
which is reached only fortheta=0\theta=0andtheta=(2)/(n)\theta=\frac{2}{n}We can thus
state the following property
Theorem 1. IfA_(i)A_(I)dotsA_(n-1)\mathrm{A}_{\mathrm{i}} \mathrm{A}_{\mathrm{I}} \ldots \mathrm{A}_{n-1}is 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((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})
or
{: 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)
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 meanM_(1)(P)\mathrm{M}_{1}(\mathrm{P}), average powerrrof the distances of P to the vertices of the polygon, that is
state the following property
Theorem 1. IfA_(i)A_(I)dotsA_(n-1)\mathrm{A}_{\mathrm{i}} \mathrm{A}_{\mathrm{I}} \ldots \mathrm{A}_{n-1}is 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((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})
or
{: 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)
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 meanM_(1)(P)\mathrm{M}_{1}(\mathrm{P}), average powerrrof the distances of P to the vertices of the polygon, that is
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} .
If0 < r < 20<r<2we still haveM_(r)(P) <= M_(2)(P)\mathrm{M}_{r}(\mathrm{P}) \leqq \mathrm{M}_{2}(\mathrm{P})and ifr > 2r>2HAVEM_(r)(P) >= M_(2)(P)\mathrm{M}_{r}(\mathrm{P}) \geqq \mathrm{M}_{2}(\mathrm{P}), equality cannot occur, in both cases, unlessPPis the center of the polygon. Let
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})}
If0 < r < 2,E_(r)(P)0<r<2, \mathrm{E}_{r}(\mathrm{P})has reached a minimum, and ifr > 2r>2a maximum reached. In the present case, expression (2) becomes
and
and
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}}
(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}
which is< 0<0 resp. > 0>0as0 < r < 20<r<2resp.r > 2r>2. It is deduced as above that the minimum or maximum ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P})it can only be touched on the circle circumscribed around the polygon
which is< 0<0 resp. > 0>0as0 < r < 20<r<2resp.r > 2r>2. It is deduced as above that the minimum or maximum ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P})it can only be touched on the circle circumscribed around the polygon
Expression (3) now becomes
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]
This expression is a continuous function oftheta\thetain the interval[0,(2pi)/(n)]\left[0, \frac{2 \pi}{n}\right]We have, in the open interval(0,(2pi)/(n))\left(0, \frac{2 \pi}{n}\right)
(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.
and it is seen that this derivative is always negative ifr <= 1r \leqq 1. One can, however, state the following generalization of Theorem 1.
(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.
and it is seen that this derivative is always negative ifr <= 1r \leqq 1. One can, however, state the following generalization of Theorem 1.
Theorem 2. IfA_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}is a regular polygon ofnnswill,rra positive number<= 1\leqq 1andPPa point in the plane of the polygon, we have
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})
equality is not true unlessPPcoincides with one of the vertices of the polygon.
5. Butr > 1r>1the problem is more complicated. We will examine here only the case whenrris a positive and even integer,r=2mr=2 mIt all comes down to evaluating the amount.
5. Butr > 1r>1the problem is more complicated. We will examine here only the case whenrris a positive and even integer,r=2mr=2 mIt all comes down to evaluating the amount.
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]
Let's remember the formula
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 \alpha
However, we have
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.ă
so
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.ă
so
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 \theta
[ lambda\lambda] meaning the largest integer<= lambda\leqq \lambdaEspecially
ifm < nm<n, we have
ifm < nm<n, we have
S_(m)=(n)/(2^(2m))((2(m))/((m)))S_{m}=\frac{n}{2^{2 m}}\binom{2 \mathrm{~m}}{\mathrm{~m}}
and the following can be stated
Theorem 3. If m is an integer (> 1>1) andA_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}a regular polygon ofn >= mn \geq msides andPPa point in the plane of the polygon, we have
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})
equality is not true unlessPPis on the circle circumscribed around the polygon.
If we havem > nm>n, it can be seen from its structureS_(m)S_{m}that the maximum is reached fortheta=0\theta=0andtheta=(2pi)/(n)\theta=\frac{2 \pi}{n}ifnnit is even fortheta=(pi )/(n)\theta=\frac{\pi}{n}ifnnis odd, so
Theorem 4. IfA_(0)A_(1)dotsA_(n-1)A_{0} A_{1} \ldots A_{n-1}is a regular polygon ofn( >= 3)n(\geqq 3)sides andmma positive integer >in, whatever the pointPPfrom the polygon plane we have
(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}}
equality is not true unless
1^(0)n1^{0} nbeing even,PPcoincides with one of the vertices of the polygon.
2^(0)n2^{0} nbeing odd,PPcoincides 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 ifrris 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
1^(0)n1^{0} nbeing even,PPcoincides with one of the vertices of the polygon.
2^(0)n2^{0} nbeing odd,PPcoincides 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 ifrris 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))\frac{\mathrm{M}_{p}(\mathrm{P})}{\mathrm{M}_{q}(\mathrm{P})}
where0 < 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.