Some remarks on a theorem of M. D. Pompeiu

Tiberiu Popoviciu
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Tiberiu Popoviciu

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T. Popoviciu, Quelques remarques sur une théorème de M. D. Pompeiu, Bull. Math. de la Soc. Roum. des Sci., 43 (1941), pp. 27-44 (in French).

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1941 b -Popoviciu- Bull. Math. de la Soc. Roum. des Sci. – Quelques remarques sur une theoreme de M. Pompeiu

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Bulletin mathématique de la Société Roumaine des Sciences

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Societatea de Științe Matematice din România

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[MR0012412, Zbl 0060.34208]

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1941 b -Popoviciu- Bull. Math. of the Society. Rum. of Sci. - Some remarks on a theorem of M.
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TIBERIU POPOVICIU

SOME REMARKS ON A THEOREM OF MR. POMPEIU

SOME REMARKS ON A THEOREM OF MR. POMPEIU BY

TIBERIU POPOVICIU

MD Pompeiu demonstrated 1 1 ^(1){ }^{1}1) that if ABC is an equilateral triangle and P a point of its plane, with the lengths PA ― , PB ― , PC ― PA ¯ , PB ¯ , PC ¯ bar(PA), bar(PB), bar(PC)\overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{PC}}PA―,PB―,PC―we can always form a triangle.
We can state this result in the following form:
If ABC is an equilateral triangle and if P is a point on its plane, we have
PA ― + PB ― + PC ― 3 ≧ 2 3 max ( PA ― , PB ― , PC ― ) . PA ¯ + PB ¯ + PC ¯ 3 ≧ 2 3 max ( PA ¯ , PB ¯ , PC ¯ ) . ( bar(PA)+ bar(PB)+ bar(PC))/(3) >= (2)/(3)max( bar(PA), bar(PB), bar(PC)).\frac{\overline{\mathrm{PA}}+\overline{\mathrm{PB}}+\overline{\mathrm{PC}}}{3} \geqq \frac{2}{3} \max (\overline{\mathrm{PA}}, \overline{\mathrm{~PB}}, \overline{\mathrm{PC}}) .PA―+PB―+PC―3≧23max(PA―, PB―,PC―).
In the following we propose to generalize this property for a regular polygon with any number of sides. We will also make some other remarks. The reader will easily realize that these problems raise others that it would be interesting to examine more closely.

I.

    • Consider a regular polygon HAS 0 HAS 1 … HAS n − 1 ( n ≧ 3 ) HAS 0 HAS 1 … HAS n − 1 ( n ≧ 3 ) A_(0)A_(1)dotsA_(n-1)(n >= 3)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}(n \geqq 3)HAS0 HAS1… HASn−1(n≧3)and let P be a point of its plane. The expression
E r ( P ) = PA ― 0 r + PA ― 1 r + … + PA ― n − 1 r PA ― 0 r E r ( P ) = PA ¯ 0 r + PA ¯ 1 r + … + PA ¯ n − 1 r PA ¯ 0 r E_(r)(P)=( bar(PA)_(0)^(r)+ bar(PA)_(1)^(r)+dots+ bar(PA)_(n-1)^(r))/( bar(PA)_(0)^(r))\mathrm{E}_{r}(\mathrm{P})=\frac{\overline{\mathrm{PA}}_{0}^{r}+\overline{\mathrm{PA}}_{1}^{r}+\ldots+\overline{\mathrm{PA}}_{n-1}^{r}}{\overline{\mathrm{PA}}_{0}^{r}}Er(P)=PA―0r+PA―1r+…+PA―n−1rPA―0r
Or r r rrris a positive number, is a continuous function of P in the entire plane, except at the point HAS 0 HAS 0 A_(0)A_{0}HAS0, where it is not defined.
We propose to study the minimum of E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P).
Without restricting the generality we can assume that the vertices HAS k HAS k A_(k)\mathrm{A}_{k}HASkare represented in the plane by complex numbers i 2 k π 2 i 2 k π 2 i(2k pi)/(2)i \frac{2 k \pi}{2}i2kπ2
e , k = 0 , 1 , … , n − 1 e , k = 0 , 1 , … , n − 1 e quad,k=0,1,dots,n-1e \quad, k=0,1, \ldots, n-1e,k=0,1,…,n−1and the variable point P by the complex number ρ e i 0 , ρ ≧ 0 , 0 ≦ θ ≦ 2 π ρ e i 0 , ρ ≧ 0 , 0 ≦ θ ≦ 2 π rhoe^(i0),rho >= 0.0 <= theta <= 2pi\rho e^{i 0}, \rho \geqq 0,0 \leqq \theta \leqq 2 \piρei0,ρ≧0,0≦θ≦2π. We then have
PA ― k = ρ 2 + 1 − 2 ρ cos ⁡ ( θ − 2 k π n ) , k = 0 , 1 , … , n − 1 . PA ¯ k = ρ 2 + 1 − 2 ρ cos ⁡ θ − 2 k π n , k = 0 , 1 , … , n − 1 . bar(PA)_(k)=sqrt(rho^(2)+1-2rho cos(theta-(2k pi)/(n))),quad k=0.1,dots,n-1.\overline{\mathrm{PA}}_{k}=\sqrt{\rho^{2}+1-2 \rho \cos \left(\theta-\frac{2 k \pi}{n}\right)}, \quad k=0.1, \ldots, n-1 .PA―k=ρ2+1−2ρcos⁡(θ−2kπn),k=0,1,…,n−1.
If P is the point HAS 0 ′ HAS 0 ′ A_(0)^(')\mathrm{A}_{0}^{\prime}HAS0′diametrically opposed to HAS 0 HAS 0 A_(0)\mathrm{A}_{0}HAS0on the circumscribed circle, we have
E r ( HAS 0 ′ ) = ∑ k = 0 n − 1 | cos ⁡ k π n | r = α < n . E r HAS 0 ′ = ∑ k = 0 n − 1   cos ⁡ k π n r = α < n . E_(r)(A_(0)^('))=sum_(k=0)^(n-1)|cos((k pi)/(n))|^(r)=alpha < n.\mathrm{E}_{r}\left(\mathrm{~A}_{0}^{\prime}\right)=\sum_{k=0}^{n-1}\left|\cos \frac{k \pi}{n}\right|^{r}=\alpha<n .Er( HAS0′)=∑k=0n−1|cos⁡kπn|r=α<n.
So we surely have
min E r ( P ) ≦ α . min E r ( P ) ≦ α . minE_(r)(P) <= alpha.\min \mathrm{E}_{r}(\mathrm{P}) \leqq \alpha .minEr(P)≦α.
But we have
| ρ − 1 | ≦ PA k ≦ ρ + 1 , k = 0 , 1 , … , n − 1 , | ρ − 1 | ≦ PA k ≦ ρ + 1 , k = 0 , 1 , … , n − 1 , |rho-1| <= PA_(k) <= rho+1,quad k=0.1,dots,n-1,|\rho-1| \leqq \mathrm{PA}_{k} \leqq \rho+1, \quad k=0.1, \ldots, n-1,|ρ−1|≦PAk≦ρ+1,k=0,1,…,n−1,
SO
E r ( P ) ≧ n | ρ − 1 | r ( ρ + 1 ) r . E r ( P ) ≧ n | ρ − 1 | r ( ρ + 1 ) r . E_(r)(P) >= (n|rho-1|^(r))/((rho+1)^(r)).\mathrm{E}_{r}(\mathrm{P}) \geqq \frac{n|\rho-1|^{r}}{(\rho+1)^{r}} .Er(P)≧n|ρ−1|r(ρ+1)r.
If we take ρ > n 1 r + α 1 r n 1 r − α 1 r ρ > n 1 r + α 1 r n 1 r − α 1 r rho > (n^((1)/(r))+alpha^((1)/(r)))/(n^((1)/(r))-alpha^((1)/(r)))\rho>\frac{n^{\frac{1}{r}}+\alpha^{\frac{1}{r}}}{n^{\frac{1}{r}}-\alpha^{\frac{1}{r}}}ρ>n1r+α1rn1r−α1r, We have E r ( P ) > α E r ( P ) > α E_(r)(P) > alpha\mathrm{E}_{r}(\mathrm{P})>\alphaEr(P)>α.
It follows that the minimum of E r ( P ) E r ( P ) E_(r)(P)E_{r}(P)Er(P)is the same as in a closed circle with center origin and radius n 1 r + α 1 r n 1 r − α 1 r n 1 r + α 1 r n 1 r − α 1 r (n^((1)/(r))+alpha^((1)/(r)))/(n^((1)/(r))-alpha^((1)/(r)))\frac{n^{\frac{1}{r}}+\alpha^{\frac{1}{r}}}{n^{\frac{1}{r}}-\alpha^{\frac{1}{r}}}n1r+α1rn1r−α1r. On the other hand
lim E r ( P ) = + ∞ P → A 0 . lim E r ( P ) = + ∞ P → A 0 . {:[ limE_(r)(P)=+oo],[PrarrA_(0).]:}\begin{aligned} & \lim \mathrm{E}_{r}(\mathrm{P})=+\infty \\ & \mathrm{P} \rightarrow \mathrm{~A}_{0} . \end{aligned}limEr(P)=+∞P→ HAS0.
We can therefore affirm that:
The minimum of E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)is reached at least at one point of the plane. Note that such a point is necessarily distinct from the origin O since
E r ( O ) = n . E r ( O ) = n . E_(r)(O)=n.\mathrm{E}_{r}(\mathrm{O})=n .Er(O)=n.
    • Let us consider the point P m P m P_(m)\mathrm{P}_{m}Pmrepresented by the complex number i ( 0 + 2 m π n ) i 0 + 2 m π n i(0+(2m pi)/(n))i\left(0+\frac{2 m \pi}{n}\right)i(0+2mπn)
      where m m mmmis an integer. As a result of symmetry, we have
∑ k = 0 n − 1 P m A k r ― = ∑ k = 0 n − 1 PA ― k r ∑ k = 0 n − 1   P m A k r ¯ = ∑ k = 0 n − 1   PA ¯ k r sum_(k=0)^(n-1) bar(P_(m)A_(k)^(r))=sum_(k=0)^(n-1) bar(PA)_(k)^(r)\sum_{k=0}^{n-1} \overline{\mathrm{P}_{m} \mathrm{~A}_{k}^{r}}=\sum_{k=0}^{n-1} \overline{\mathrm{PA}}_{k}^{r}∑k=0n−1Pm HASkr―=∑k=0n−1PA―kr
Still as a result of symmetry, we see that it is sufficient to examine the expression E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)only for points P for which
(1)
π − π n ≦ θ ≦ π π − π n ≦ θ ≦ π pi-(pi )/(n) <= theta <= pi\pi-\frac{\pi}{n} \leqq \theta \leqq \piπ−πn≦θ≦π
Let's fix such a θ θ theta\thetaθand let's vary ρ ρ rho\rhoρfrom 0 to + ∞ + ∞ +oo+\infty+∞.
If we pose
(2)
t = 2 ρ ρ 2 + 1 t = 2 ρ ρ 2 + 1 t=(2rho)/(rho^(2)+1)t=\frac{2 \rho}{\rho^{2}+1}t=2ρρ2+1
We have
E r ( P ) = 1 ( 1 − t cos ⁡ θ ) r 2 ∑ k = 0 n − 1 [ 1 − t cos ⁡ ( θ − 2 k π n ) ] r 2 E r ( P ) = 1 ( 1 − t cos ⁡ θ ) r 2 ∑ k = 0 n − 1   1 − t cos ⁡ θ − 2 k π n r 2 E_(r)(P)=(1)/((1-t cos theta)^((r)/(2)))sum_(k=0)^(n-1)[1-t cos(theta-(2k pi)/(n))]^((r)/(2))\mathrm{E}_{r}(\mathrm{P})=\frac{1}{(1-t \cos \theta)^{\frac{r}{2}}} \sum_{k=0}^{n-1}\left[1-t \cos \left(\theta-\frac{2 k \pi}{n}\right)\right]^{\frac{r}{2}}Er(P)=1(1−tcos⁡θ)r2∑k=0n−1[1−tcos⁡(θ−2kπn)]r2
From (1) it follows that cos ⁡ θ < 0 cos ⁡ θ < 0 cos theta < 0\cos \theta<0cos⁡θ<0, SO E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)is a continuous function of t t tttin the closed interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]and is surely differentiable in the interval [ 0 , 1 ) [ 0 , 1 ) [0,1)[0,1)[0,1)open to the right. We have, according to an easy calculation,
d E r ( P ) d t = = − r ( 1 − t cos ⁡ θ ) r 2 + 1 ∑ k = 0 n − 1 sin ⁡ ( θ − k π n ) sin ⁡ k π n [ 1 − t cos ⁡ ( θ − 2 k π n ) ] r 2 − 1 d E r ( P ) d t = = − r ( 1 − t cos ⁡ θ ) r 2 + 1 ∑ k = 0 n − 1   sin ⁡ θ − k π n sin ⁡ k π n 1 − t cos ⁡ θ − 2 k π n r 2 − 1 {:[(dE_(r)(P))/(dt)=],[=-(r)/((1-t cos theta)^((r)/(2))+1)sum_(k=0)^(n-1)sin(theta-(k pi)/(n))sin((k pi)/(n)[1-t cos(theta-(2k pi)/(n))]^((r)/(2)-1))]:}\begin{gathered} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d t}= \\ =-\frac{r}{(1-t \cos \theta)^{\frac{r}{2}}+1} \sum_{k=0}^{n-1} \sin \left(\theta-\frac{k \pi}{n}\right) \sin \frac{k \pi}{n}\left[1-t \cos \left(\theta-\frac{2 k \pi}{n}\right)\right]^{\frac{r}{2}-1} \end{gathered}dEr(P)dt==−r(1−tcos⁡θ)r2+1∑k=0n−1sin⁡(θ−kπn)sin⁡kπn[1−tcos⁡(θ−2kπn)]r2−1
and taking into account (1),
d E r ( P ) d t < 0 , 0 ≦ t < 1 . d E r ( P ) d t < 0 , 0 ≦ t < 1 . (dE_(r)(P))/(dt) < 0,quad0 <= t < 1.\frac{d \mathrm{E}_{r}(\mathrm{P})}{d t}<0, \quad 0 \leqq t<1 .dEr(P)dt<0,0≦t<1.
The function E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)of t t tttis therefore decreasing in the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]. Its minimum is reached only for t = 1 t = 1 t=1t=1t=1so, according to (2), only for ρ = 1 ρ = 1 rho=1\rho=1ρ=1.
The minimum of expression E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{\mathrm{r}}(\mathrm{P})Er(P)can only be reached on the circle circumscribed to the polygon.
3. - Let us therefore suppose p = 1 p = 1 p=1p=1p=1. The expression E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)becomes
E r ( P ) = 1 sin r ⁡ θ 2 ∑ k = 0 n − 1 | sin ⁡ ( θ 2 − k π n ) | r E r ( P ) = 1 sin r ⁡ θ 2 ∑ k = 0 n − 1   sin ⁡ θ 2 − k π n r E_(r)(P)=(1)/(sin^(r)((theta)/(2)))sum_(k=0)^(n-1)|sin((theta)/(2)-(k pi)/(n))|^(r)\mathrm{E}_{r}(\mathrm{P})=\frac{1}{\sin ^{r} \frac{\theta}{2}} \sum_{k=0}^{n-1}\left|\sin \left(\frac{\theta}{2}-\frac{k \pi}{n}\right)\right|^{r}Er(P)=1sinr⁡θ2∑k=0n−1|sin⁡(θ2−kπn)|r
Or
E r ( P ) = ∑ k = 0 n − 1 | cos ⁡ k π n − u sin ⁡ k π n | r E r ( P ) = ∑ k = 0 n − 1   cos ⁡ k π n − u sin ⁡ k π n r E_(r)(P)=sum_(k=0)^(n-1)|cos((k pi)/(n))-u sin((k pi)/(n))|^(r)\mathrm{E}_{r}(\mathrm{P})=\sum_{k=0}^{n-1}\left|\cos \frac{k \pi}{n}-u \sin \frac{k \pi}{n}\right|^{r}Er(P)=∑k=0n−1|cos⁡kπn−usin⁡kπn|r
by posing
u = cotg θ 2 u = cotg θ 2 u=cotg((theta)/(2))u=\operatorname{cotg} \frac{\theta}{2}u=cotgθ2
It is a continuous function of u u uuuin the closed interval [ 0 , tg π 2 n ] 0 , tg π 2 n [0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right][0,tgπ2n]and is surely indefinitely differentiable in the open interval ( 0 , tg π 2 n ) 0 , tg π 2 n (0,tg((pi)/(2n)))\left(0, \operatorname{tg} \frac{\pi}{2 n}\right)(0,tgπ2n). We have
d E r ( P ) d u = − r ∑ k = 0 n − 1 sin ⁡ k π n | cos ⁡ k π n − u sin ⁡ k π n | r cos ⁡ k π n − u sin ⁡ k π n d 2 E r ( P ) d u 2 = r ( r − 1 ) ∑ k = 0 n − 1 sin 2 ⁡ k π n | cos ⁡ k π n − u sin ⁡ k π n | r − 2 d E r ( P ) d u = − r ∑ k = 0 n − 1   sin ⁡ k π n cos ⁡ k π n − u sin ⁡ k π n r cos ⁡ k π n − u sin ⁡ k π n d 2 E r ( P ) d u 2 = r ( r − 1 ) ∑ k = 0 n − 1   sin 2 ⁡ k π n cos ⁡ k π n − u sin ⁡ k π n r − 2 {:[(dE_(r)(P))/(du)=-rsum_(k=0)^(n-1)(sin((k pi)/(n)|cos((k pi)/(n))-u sin((k pi)/(n))|^(r)))/(cos((k pi)/(n))-u sin((k pi)/(n)))],[(d^(2)E_(r)(P))/(du^(2))=r(r-1)sum_(k=0)^(n-1)sin^(2)((k pi)/(n)|cos((k pi)/(n))-u sin((k pi)/(n))|^(r-2))]:}\begin{gathered} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d u}=-r \sum_{k=0}^{n-1} \frac{\sin \frac{k \pi}{n}\left|\cos \frac{k \pi}{n}-u \sin \frac{k \pi}{n}\right|^{r}}{\cos \frac{k \pi}{n}-u \sin \frac{k \pi}{n}} \\ \frac{d^{2} \mathrm{E}_{r}(\mathrm{P})}{d u^{2}}=r(r-1) \sum_{k=0}^{n-1} \sin ^{2} \frac{k \pi}{n}\left|\cos \frac{k \pi}{n}-u \sin \frac{k \pi}{n}\right|^{r-2} \end{gathered}dEr(P)du=−r∑k=0n−1sin⁡kπn|cos⁡kπn−usin⁡kπn|rcos⁡kπn−usin⁡kπnd2Er(P)du2=r(r−1)∑k=0n−1sin2⁡kπn|cos⁡kπn−usin⁡kπn|r−2
So we have
d 2 E r ( P ) d u 2 < 0 , = 0 resp. > 0 , 0 < u < tg π 2 n d 2 E r ( P ) d u 2 < 0 , = 0  resp.  > 0 , 0 < u < tg π 2 n (d^(2)E_(r)(P))/(du^(2)) < 0,=0" resp. " > 0,quad0 < u < tg((pi)/(2n))\frac{d^{2} \mathrm{E}_{r}(\mathrm{P})}{d u^{2}}<0,=0 \text { resp. }>0, \quad 0<u<\operatorname{tg} \frac{\pi}{2 n}d2Er(P)du2<0,=0 resp. >0,0<u<tgπ2n
following that r < 1 , = 1 r < 1 , = 1 r < 1,=1r<1,=1r<1,=1, resp. > 1 > 1 > 1>1>1.
The function E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)of u u uuuis therefore a continuous linear concave or convex function in the closed interval [ 0 , tg π 2 n ] 0 , tg π 2 n [0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right][0,tgπ2n]. To go further let us examine the first derivative of E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)near the ends.
If n n nnnis odd, all the cos ⁡ k π n cos ⁡ k π n cos((k pi)/(n))\cos \frac{k \pi}{n}cos⁡kπnare ≠ 0 ≠ 0 !=0\neq 0≠0, SO
lim u → + 0 d E r ( P ) d u = − r ∑ k = 0 n − 1 tg k π n | cos ⁡ k π n | r = 0 lim u → + 0   d E r ( P ) d u = − r ∑ k = 0 n − 1   tg k π n cos ⁡ k π n r = 0 lim_(u rarr+0)(dE_(r)(P))/(du)=-rsum_(k=0)^(n-1)tg((k pi)/(n)|cos((k pi)/(n))|^(r))=0\lim _{u \rightarrow+0} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d u}=-r \sum_{k=0}^{n-1} \operatorname{tg} \frac{k \pi}{n}\left|\cos \frac{k \pi}{n}\right|^{r}=0limu→+0dEr(P)du=−r∑k=0n−1tgkπn|cos⁡kπn|r=0
We can therefore say in this case that the minimum is reached for u = 0 u = 0 u=0u=0u=0if r > 1 r > 1 r > 1r>1r>1and for u = tg π 2 n u = tg π 2 n u=tg((pi)/(2n))u=\operatorname{tg} \frac{\pi}{2 n}u=tgπ2nif r < 1 r < 1 r < 1r<1r<1and only for these values.
If n n nnnis even the cos ⁡ k π n cos ⁡ k π n cos((k pi)/(n))\cos \frac{k \pi}{n}cos⁡kπnare ≠ 0 ≠ 0 !=0\neq 0≠0except for k = n 2 k = n 2 k=(n)/(2)k=\frac{n}{2}k=n2. So we have
lim u → + 0 d E r ( P ) d u = r lim u → + 0 u r − 1 lim u → + 0   d E r ( P ) d u = r lim u → + 0   u r − 1 lim_(u rarr+0)(dE_(r)(P))/(du)=rlim_(u rarr+0)u^(r-1)\lim _{u \rightarrow+0} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d u}=r \lim _{u \rightarrow+0} u^{r-1}limu→+0dEr(P)du=rlimu→+0ur−1
and consequently
lim u → + 0 d E r ( P ) d u = { 0 , pour r > 1 + ∞ , pour r < 1 lim u → + 0   d E r ( P ) d u = 0 ,       pour  r > 1 + ∞ ,       pour  r < 1 lim_(u rarr+0)(dE_(r)(P))/(du)={[0","," pour "r > 1],[+oo","," pour "r < 1]:}\lim _{u \rightarrow+0} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d u}=\left\{\begin{array}{lr} 0, & \text { pour } r>1 \\ +\infty, & \text { pour } r<1 \end{array}\right.limu→+0dEr(P)du={0, For r>1+∞, For r<1
We immediately deduce that if r > 1 r > 1 r > 1r>1r>1the minimum is reached for u = 0 u = 0 u=0u=0u=0. If r < 1 r < 1 r < 1r<1r<1, the function E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)being concave we know that the minimum can only be reached for the ends
u = 0 , u = tg π 2 n u = 0 , u = tg π 2 n u=0,u=tg((pi)/(2n))u=0, u=\operatorname{tg} \frac{\pi}{2 n}u=0,u=tgπ2n
To decide, let us note that in this case ( n n nnnpeer)
lim u → tg π 2 n − 0 d E r ( P ) d u = − r cos r − 1 ⁡ π 2 n ∑ k = 0 n − 1 sin ⁡ k π n cos ⁡ ( 2 k + 1 ) π 2 n | cos ⁡ ( 2 k + 1 ) π 2 n | r = = r sin ⁡ π n cos ⁡ π 2 n ∑ k = 0 n − 2 2 cos r ⁡ ( 2 k + 1 ) π 2 n > 0 lim u → tg π 2 n − 0   d E r ( P ) d u = − r cos r − 1 ⁡ π 2 n ∑ k = 0 n − 1   sin ⁡ k π n cos ⁡ ( 2 k + 1 ) π 2 n cos ⁡ ( 2 k + 1 ) π 2 n r = = r sin ⁡ π n cos ⁡ π 2 n ∑ k = 0 n − 2 2   cos r ⁡ ( 2 k + 1 ) π 2 n > 0 {:[lim_(u rarr tg((pi)/(2n))-0)(dE_(r)(P))/(du)=-(r)/(cos^(r-1)((pi)/(2n)))sum_(k=0)^(n-1)(sin((k pi)/(n)))/(cos(((2k+1)pi)/(2n)))|cos(((2k+1)pi)/(2n))|^(r)=],[=(r sin((pi )/(n)))/(cos((pi)/(2n)))sum_(k=0)^((n-2)/(2))cos^(r)(((2k+1)pi)/(2n)) > 0]:}\begin{aligned} \lim _{u \rightarrow \operatorname{tg} \frac{\pi}{2 n}-0} \frac{d \mathrm{E}_{r}(\mathrm{P})}{d u} & =-\frac{r}{\cos ^{r-1} \frac{\pi}{2 n}} \sum_{k=0}^{n-1} \frac{\sin \frac{k \pi}{n}}{\cos \frac{(2 k+1) \pi}{2 n}}\left|\cos \frac{(2 k+1) \pi}{2 n}\right|^{r}= \\ & =\frac{r \sin \frac{\pi}{n}}{\cos \frac{\pi}{2 n}} \sum_{k=0}^{\frac{n-2}{2}} \cos ^{r} \frac{(2 k+1) \pi}{2 n}>0 \end{aligned}limu→tgπ2n−0dEr(P)du=−rcosr−1⁡π2n∑k=0n−1sin⁡kπncos⁡(2k+1)π2n|cos⁡(2k+1)π2n|r==rsin⁡πncos⁡π2n∑k=0n−22cosr⁡(2k+1)π2n>0
The function E r ( P ) E r ( P ) E_(r)(P)\mathrm{E}_{r}(\mathrm{P})Er(P)is therefore increasing and its minimum is reached for u = 0 u = 0 u=0u=0u=0.
It remains to examine the special case r = 1 r = 1 r=1r=1r=1. We then have
E r ( P ) = { 1 tg π 2 n + u , pour n pair 1 sin ⁡ π 2 n , pour n impair. E r ( P ) = 1 tg π 2 n + u ,       pour  n  pair  1 sin ⁡ π 2 n ,       pour  n  impair.  E_(r)(P)={[(1)/(tg((pi)/(2n)))+u","," pour "n" pair "],[(1)/(sin((pi)/(2n)))","," pour "n" impair. "]:}\mathrm{E}_{r}(\mathrm{P})= \begin{cases}\frac{1}{\operatorname{tg} \frac{\pi}{2 n}}+u, & \text { pour } n \text { pair } \\ \frac{1}{\sin \frac{\pi}{2 n}}, & \text { pour } n \text { impair. }\end{cases}Er(P)={1tgπ2n+u, For n peer 1sin⁡π2n, For n odd. 
For n n nnneven the minimum is reached for u = 0 u = 0 u=0u=0u=0. For n n nnnodd E 1 ( P ) E 1 ( P ) E_(1)(P)\mathrm{E}_{1}(\mathrm{P})E1(P)is constant over all points considered, so its minimum is reached at any point in the interval [ 0 , tg π 2 n ] 0 , tg π 2 n [0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right][0,tgπ2n].
Now let us note that if θ θ theta\thetaθbelieves in π − π n π − π n pi-(pi )/(n)\pi-\frac{\pi}{n}π−πnhas π , u = cotg θ 2 π , u = cotg θ 2 pi,u=cotg((theta)/(2))\pi, u=\operatorname{cotg} \frac{\theta}{2}π,u=cotgθ2decreases by tg π 2 n tg π 2 n tg((pi)/(2n))\operatorname{tg} \frac{\pi}{2 n}tgπ2nto 0.
Now let us pose
(3) λ n ( r ) = { ∑ k = 0 n − 1 | cos ⁡ k π n | r , si n est pair r > 0 et si n est impair r ≧ 1 1 cos r ⁡ π 2 n ∑ k = 0 n − 1 | cos ⁡ ( 2 k + 1 ) π 2 n | r , si n est impair 0 < r < 1 . λ n ( r ) = ∑ k = 0 n − 1   cos ⁡ k π n r ,  si  n  est pair  r > 0  et si  n  est impair  r ≧ 1 1 cos r ⁡ π 2 n ∑ k = 0 n − 1   cos ⁡ ( 2 k + 1 ) π 2 n r ,  si  n  est impair  0 < r < 1 . quadlambda_(n)^((r))={[sum_(k=0)^(n-1)|cos((k pi)/(n))|^(r)","" si "n" est pair "r > 0" et si "n" est impair "r >= 1],[(1)/(cos^(r)((pi)/(2n)))sum_(k=0)^(n-1)|cos(((2k+1)pi)/(2n))|^(r)","" si "n" est impair "0 < r < 1.]:}\quad \lambda_{n}^{(r)}=\left\{\begin{array}{l}\sum_{k=0}^{n-1}\left|\cos \frac{k \pi}{n}\right|^{r}, \text { si } n \text { est pair } r>0 \text { et si } n \text { est impair } r \geqq 1 \\ \frac{1}{\cos ^{r} \frac{\pi}{2 n}} \sum_{k=0}^{n-1}\left|\cos \frac{(2 k+1) \pi}{2 n}\right|^{r}, \text { si } n \text { est impair } 0<r<1 .\end{array}\right.λn(r)={∑k=0n−1|cos⁡kπn|r, if n is even r>0 and if n is odd r≧11cosr⁡π2n∑k=0n−1|cos⁡(2k+1)π2n|r, if n is odd 0<r<1.
We can then state the following theorem.
Theorem 1. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a regular polygon, r a given positive number and P a point in the plane of the polygon, we have the inequality
E r ( P ) ≧ λ n ( r ) . E r ( P ) ≧ λ n ( r ) . E_(r)(P) >= lambda_(n)^((r)).\mathrm{E}_{r}(\mathrm{P}) \geqq \lambda_{n}^{(r)} .Er(P)≧λn(r).
Equality holds if and only if:
1 0 n 1 0 n 1^(0)n1^{0} \mathrm{n}10nbeing even and r > 0 r > 0 r > 0\mathrm{r}>0r>0, or n being odd and r ≧ 1 , P r ≧ 1 , P r >= 1,P\mathrm{r} \geqq 1, \mathrm{P}r≧1,Pcoincides with the point A 0 ′ A 0 ′ A_(0)^(')\mathrm{A}_{0}^{\prime}HAS0′of the diametrically opposite circumscribed circle a ` A 0 a ` A 0 a^(`)A_(0)\grave{a} \mathrm{~A}_{0}has` HAS0.
2 0 n 2 0 n 2^(0)n2^{0} \mathrm{n}20nbeing odd and 0 < r < 1 , P 0 < r < 1 , P 0 < r < 1,P0<\mathrm{r}<1, \mathrm{P}0<r<1,Pcoincides with one of the opposite vertices to A 0 A 0 A_(0)\mathrm{A}_{0}HAS0, SO P = A n − 1 2 P = A n − 1 2 P=A_((n-1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}-1}{2}}P=HASn−12Or P = A n + 1 2 P = A n + 1 2 P=A_((n+1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}+1}{2}}P=HASn+12.
3 0 n 3 0 n 3^(0)n3^{0} \mathrm{n}30nbeing odd and r = 1 , P r = 1 , P r=1,P\mathrm{r}=1, \mathrm{P}r=1,Pcoincides with one of the points of the arc of the circumscribed circle included between the vertices A n − 1 2 2 , A n + 1 2 2 A n − 1 2 2 , A n + 1 2 2 (A_((n-1)/(2)))/(2),(A_((n+1)/(2)))/(2)\frac{\mathrm{A}_{\frac{\mathrm{n}-1}{2}}}{2}, \frac{\mathrm{~A}_{\frac{\mathrm{n}+1}{2}}}{2}HASn−122, HASn+122.
This is, of course, the circle circumscribed to the polygon.
4. - The previous result can be put into various forms. The expression
(4) M r ( P ) = ( PA ― 0 r + PA ― 1 r + … + PA ― n − 1 r n ) 1 r (4) M r ( P ) = PA ¯ 0 r + PA ¯ 1 r + … + PA ¯ n − 1 r n 1 r {:(4)M_(r)(P)=(( bar(PA)_(0)^(r)+ bar(PA)_(1)^(r)+dots+ bar(PA)_(n-1)^(r))/(n))^((1)/(r)):}\begin{equation*} \mathrm{M}_{r}(\mathrm{P})=\left(\frac{\overline{\mathrm{PA}}_{0}^{r}+\overline{\mathrm{PA}}_{1}^{r}+\ldots+\overline{\mathrm{PA}}_{n-1}^{r}}{n}\right)^{\frac{1}{r}} \tag{4} \end{equation*}(4)Mr(P)=(PA―0r+PA―1r+…+PA―n−1rn)1r
is the power average r r rrrdistances PA ― 0 , PA ― 1 , … , PA ― n − 1 PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1)\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}PA―0,PA―1,…,PA―n−1. We know that we have the upper limitation
M r ( P ) ≦ max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) M r ( P ) ≦ max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 M_(r)(P) <= max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\mathrm{M}_{r}(\mathrm{P}) \leqq \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right)Mr(P)≦max(PA―0,PA―1,…,PA―n−1)
Theorem 1 gives, for a regular polygon, a lower bound of M r ( P ) M r ( P ) M_(r)(P)\mathrm{M}_{r}(\mathrm{P})Mr(P). We have the
Theorem 2. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a regular polygon, r a positive number and P a point in the plane of the polygon, we have
M r ( P ) ≧ C n ( r ) max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) M r ( P ) ≧ C n ( r ) max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 M_(r)(P) >= C_(n)^((r))max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\mathrm{M}_{r}(\mathrm{P}) \geqq \mathrm{C}_{n}^{(r)} \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right)Mr(P)≧Cn(r)max(PA―0,PA―1,…,PA―n−1)
Or
C n ( r ) = ( λ n ( r ) n ) 1 r C n ( r ) = λ n ( r ) n 1 r C_(n)^((r))=((lambda_(n)^((r)))/(n))^((1)/(r))\mathrm{C}_{n}^{(r)}=\left(\frac{\lambda_{n}^{(r)}}{n}\right)^{\frac{1}{r}}Cn(r)=(λn(r)n)1r
Note that
lim n → ∞ λ n ( r ) n = 1 π ∫ 0 π | cos ⁡ x | r d x = 2 π ∫ 0 π 2 cos r ⁡ x d x < 1 lim n → ∞   λ n ( r ) n = 1 π ∫ 0 π   | cos ⁡ x | r d x = 2 π ∫ 0 π 2   cos r ⁡ x d x < 1 lim_(n rarr oo)(lambda_(n)^((r)))/(n)=(1)/(pi)int_(0)^(pi)|cos x|^(r)dx=(2)/(pi)int_(0)^((pi)/(2))cos^(r)xdx < 1\lim _{n \rightarrow \infty} \frac{\lambda_{n}^{(r)}}{n}=\frac{1}{\pi} \int_{0}^{\pi}|\cos x|^{r} d x=\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos ^{r} x d x<1limn→∞λn(r)n=1π∫0π|cos⁡x|rdx=2π∫0π2cosr⁡xdx<1
SO C n ( r ) C n ( r ) C_(n)^((r))\mathrm{C}_{n}^{(r)}Cn(r)For n ⟶ ∞ n ⟶ ∞ n longrightarrow oon \longrightarrow \inftyn⟶∞tends towards the average power value r r rrrof the function cos ⁡ x cos ⁡ x cos x\cos xcos⁡xin the meantime [ 0 , π 2 ] 0 , π 2 [0,(pi)/(2)]\left[0, \frac{\pi}{2}\right][0,π2].
To complete the previous results we will examine in a little more detail the case where r r rrris an integer.
5. - Let first r = 1 r = 1 r=1r=1r=1. We then have
λ n ( 1 ) = { 1 tg π 2 n , n pair 1 sin ⁡ π 2 n , n impair λ n ( 1 ) = 1 tg π 2 n ,      n  pair  1 sin ⁡ π 2 n ,      n  impair  lambda_(n)^((1))={[(1)/(tg((pi)/(2n)))",",n" pair "],[(1)/(sin((pi)/(2n)))",",n" impair "]:}\lambda_{n}^{(1)}= \begin{cases}\frac{1}{\operatorname{tg} \frac{\pi}{2 n}}, & n \text { pair } \\ \frac{1}{\sin \frac{\pi}{2 n}}, & n \text { impair }\end{cases}λn(1)={1tgπ2n,n peer 1sin⁡π2n,n odd 
The direct generalization of MD Pompeiu's theorem can be stated as follows
Theorem 3. If 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}HAS0 HAS1… HASn−1is a regular polygon and P a point of its plane, we have
M 1 ( P ) = PA ― 0 + PA ― 1 + … + PA ― n − 1 n ≧ C n ( 1 ) max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) M 1 ( P ) = PA ¯ 0 + PA ¯ 1 + … + PA ¯ n − 1 n ≧ C n ( 1 ) max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 M_(1)(P)=( bar(PA)_(0)+ bar(PA)_(1)+dots+ bar(PA)_(n-1))/(n) >= C_(n)^((1))max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\mathrm{M}_{1}(\mathrm{P})=\frac{\overline{\mathrm{PA}}_{0}+\overline{\mathrm{PA}}_{1}+\ldots+\overline{\mathrm{PA}}_{n-1}}{n} \geqq \mathrm{C}_{n}^{(1)} \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right)M1(P)=PA―0+PA―1+…+PA―n−1n≧Cn(1)max(PA―0,PA―1,…,PA―n−1)
Or C n ( 1 ) C n ( 1 ) C_(n)^((1))\mathrm{C}_{n}^{(1)}Cn(1)is equal to
1 n ⋅ 1 tg π 2 n ou 1 n ⋅ 1 sin ⁡ π 2 n 1 n ⋅ 1 tg π 2 n  ou  1 n ⋅ 1 sin ⁡ π 2 n (1)/(n)*(1)/(tg((pi)/(2n)))" ou "(1)/(n)*(1)/(sin((pi)/(2n)))\frac{1}{n} \cdot \frac{1}{\operatorname{tg} \frac{\pi}{2 n}} \text { ou } \frac{1}{n} \cdot \frac{1}{\sin \frac{\pi}{2 n}}1n⋅1tgπ2n Or 1n⋅1sin⁡π2n
depending on whether n is even or odd.
Note that for n n nnnpeer C n ( 1 ) C n ( 1 ) C_(n)^((1))\mathrm{C}_{n}^{(1)}Cn(1)believes and tends towards 2 π 2 π (2)/(pi)\frac{2}{\pi}2πand for n n nnnodd it decreases and tends towards 2 π 2 π (2)/(pi)\frac{2}{\pi}2πFor n ⟶ ∞ n ⟶ ∞ n longrightarrow oon \longrightarrow \inftyn⟶∞, SO
Consequence 1. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a regular polygon with an odd number of sides and P a point on its plane, we have
M 1 ( P ) > 2 π max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) M 1 ( P ) > 2 π max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 M_(1)(P) > (2)/(pi)max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\mathrm{M}_{1}(\mathrm{P})>\frac{2}{\pi} \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right)M1(P)>2πmax(PA―0,PA―1,…,PA―n−1)
equality not being possible and 2 π 2 π (2)/(pi)\frac{2}{\pi}2πcannot be replaced by any other larger number.
For n = 4 n = 4 n=4n=4n=4we have C n ( 1 ) = 2 + 1 4 C n ( 1 ) = 2 + 1 4 C_(n)^((1))=(sqrt2+1)/(4)\mathrm{C}_{n}^{(1)}=\frac{\sqrt{2}+1}{4}Cn(1)=2+14and we have
Theorem 4. If ABCD is a square and P a point on its plane, we have the inequality
PA ― + PB ― + PC ― + PD ― 4 ≧ 2 + 1 4 max ( PA ― , PB ― , PC ― , PD ― ) PA ¯ + PB ¯ + PC ¯ + PD ¯ 4 ≧ 2 + 1 4 max ( PA ¯ , PB ¯ , PC ¯ , PD ¯ ) ( bar(PA)+ bar(PB)+ bar(PC)+ bar(PD))/(4) >= (sqrt2+1)/(4)max( bar(PA), bar(PB), bar(PC), bar(PD))\frac{\overline{\mathrm{PA}}+\overline{\mathrm{PB}}+\overline{\mathrm{PC}}+\overline{\mathrm{PD}}}{4} \geqq \frac{\sqrt{2}+1}{4} \max (\overline{\mathrm{PA}}, \overline{\mathrm{~PB}}, \overline{\mathrm{PC}}, \overline{\mathrm{PD}})PA―+PB―+PC―+PD―4≧2+14max(PA―, PB―,PC―,PD―)
which is, moreover, equivalent to the four inequalities
PB ― + PC ― + PD ― ≧ 2 PA ― , PC ― + PD ― + PA ― ≧ 2 PB ― PD ― + PA ― + PB ― ≧ 2 PC ― PA ― + PB ― + PC ― ≧ 2 PD ― . PB ¯ + PC ¯ + PD ¯ ≧ 2 PA ¯ ,      PC ¯ + PD ¯ + PA ¯ ≧ 2 PB ¯ PD ¯ + PA ¯ + PB ¯ ≧ 2 PC ¯      PA ¯ + PB ¯ + PC ¯ ≧ 2 PD ¯ . {:[ bar(PB)+ bar(PC)+ bar(PD) >= sqrt2 bar(PA)",", bar(PC)+ bar(PD)+ bar(PA) >= sqrt2 bar(PB)],[ bar(PD)+ bar(PA)+ bar(PB) >= sqrt2 bar(PC), bar(PA)+ bar(PB)+ bar(PC) >= sqrt2 bar(PD).]:}\begin{array}{ll} \overline{\mathrm{PB}}+\overline{\mathrm{PC}}+\overline{\mathrm{PD}} \geqq \sqrt{2} \overline{\mathrm{PA}}, & \overline{\mathrm{PC}}+\overline{\mathrm{PD}}+\overline{\mathrm{PA}} \geqq \sqrt{2} \overline{\mathrm{~PB}} \\ \overline{\mathrm{PD}}+\overline{\mathrm{PA}}+\overline{\mathrm{PB}} \geqq \sqrt{2} \overline{\mathrm{PC}} & \overline{\mathrm{PA}}+\overline{\mathrm{PB}}+\overline{\mathrm{PC}} \geqq \sqrt{2} \overline{\mathrm{PD}} . \end{array}PB―+PC―+PD―≧2PA―,PC―+PD―+PA―≧2 PB―PD―+PA―+PB―≧2PC―PA―+PB―+PC―≧2PD―.
MD Pompeiu has already found 2) C 4 ( 1 ) > 1 2 C 4 ( 1 ) > 1 2 C_(4)^((1)) > (1)/(2)\mathrm{C}_{4}^{(1)}>\frac{1}{2}C4(1)>12.
6. - Let us now suppose r = 2 m , m r = 2 m , m r=2m,mr=2 m, mr=2m,mbeing a natural number. Let us calculate λ n ( 2 m ) λ n ( 2 m ) lambda_(n)^((2m))\lambda_{n}^{(2 m)}λn(2m). We have
cos 2 m ⁡ x = 1 2 2 m ( 2 m m ) + 2 ∑ j = 1 m 1 2 2 m ( 2 m m + j ) cos ⁡ 2 j x cos 2 m ⁡ x = 1 2 2 m ( 2 m m ) + 2 ∑ j = 1 m   1 2 2 m ( 2 m m + j ) cos ⁡ 2 j x cos^(2m)x=(1)/(2^(2m))((2m)/(m))+2sum_(j=1)^(m)(1)/(2^(2m))((2m)/(m+j))cos 2jx\cos ^{2 m} x=\frac{1}{2^{2 m}}\binom{2 m}{m}+2 \sum_{j=1}^{m} \frac{1}{2^{2 m}}\binom{2 m}{m+j} \cos 2 j xcos2m⁡x=122m(2mm)+2∑I=1m122m(2mm+I)cos⁡2Ix
and on the other hand
∑ k = 0 n − 1 cos ⁡ 2 j k n = { 0 , si j ≡≡ 0 ( mod n ) n , si j ≡ 0 ( mod n ) . ∑ k = 0 n − 1   cos ⁡ 2 j k n = 0 ,       si  j ≡≡ 0 ( mod n ) n ,       si  j ≡ 0 ( mod n ) . sum_(k=0)^(n-1)cos 2j(k)/(n)={[0","," si "j≡≡0(mod n)],[n","," si "j-=0(mod n).]:}\sum_{k=0}^{n-1} \cos 2 j \frac{k}{n}= \begin{cases}0, & \text { si } j \equiv \equiv 0(\bmod n) \\ n, & \text { si } j \equiv 0(\bmod n) .\end{cases}∑k=0n−1cos⁡2Ikn={0, if I≡≡0(modn)n, if I≡0(modn).
We deduce from this
λ n ( 2 m ) = { n 2 2 m ( 2 m m ) , si m < n n 2 2 m [ ( 2 m m ) + 2 ∑ j = 1 [ m n ] ( 2 m m + j n ) ] , si m ≧ n λ n ( 2 m ) = n 2 2 m ( 2 m m ) ,  si  m < n n 2 2 m ( 2 m m ) + 2 ∑ j = 1 m n   ( 2 m m + j n ) ,  si  m ≧ n lambda_(n)^((2m))={[(n)/(2^(2m))((2m)/(m))","quad" si "m < n],[(n)/(2^(2m))[((2m)/(m))+2sum_(j=1)^([(m)/(n)])((2m)/(m+jn))]","quad" si "m >= n]:}\lambda_{n}^{(2 m)}=\left\{\begin{array}{l} \frac{n}{2^{2 m}}\binom{2 m}{m}, \quad \text { si } m<n \\ \frac{n}{2^{2 m}}\left[\binom{2 m}{m}+2 \sum_{j=1}^{\left[\frac{m}{n}\right]}\binom{2 m}{m+j n}\right], \quad \text { si } m \geqq n \end{array}\right.λn(2m)={n22m(2mm), if m<nn22m[(2mm)+2∑I=1[mn](2mm+In)], if m≧n
Or [ α ] [ α ] [alpha][\alpha][α]is the largest integer ≦ α ≦ α <= alpha\leqq \alpha≦α.
Note that in this case the coefficient C n ( 2 m ) C n ( 2 m ) C_(n)^((2m))\mathrm{C}_{n}^{(2 m)}Cn(2m)is independent of n for n > m n > m n > m\mathrm{n}>\mathrm{m}n>m.
Theorem 5. If m is a natural number, 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1a regular polygon at n > m ( n ⩾ 3 ) n > m ( n ⩾ 3 ) n > m(n >= 3)\mathrm{n}>\mathrm{m}(\mathrm{n} \geqslant 3)n>m(n⩾3)sides and P a point on the plane of this polygon, we have
M 2 m ( P ) ≧ 1 2 ( 2 m m ) 1 2 m max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) M 2 m ( P ) ≧ 1 2 ( 2 m m ) 1 2 m max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 M_(2m)(P) >= (1)/(2)((2m)/(m))^((1)/(2m))max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\mathrm{M}_{2 m}(\mathrm{P}) \geqq \frac{1}{2}\binom{2 m}{m}^{\frac{1}{2 m}} \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right)M2m(P)≧12(2mm)12mmax(PA―0,PA―1,…,PA―n−1)
The number λ n ( r ) λ n ( r ) lambda_(n)^((r))\lambda_{n}^{(r)}λn(r)can also be easily calculated if r = 2 m + 1 r = 2 m + 1 r=2m+1r=2 m+1r=2m+1is an odd number. In this case we have
λ n ( 2 m + 1 ) = 2 ∑ k = 0 [ n − 1 2 ] cos 2 m + 1 ⁡ k π n − 1 λ n ( 2 m + 1 ) = 2 ∑ k = 0 n − 1 2   cos 2 m + 1 ⁡ k π n − 1 lambda_(n)^((2m+1))=2sum_(k=0)^([(n-1)/(2)])cos^(2m+1)((k pi)/(n))-1\lambda_{n}^{(2 m+1)}=2 \sum_{k=0}^{\left[\frac{n-1}{2}\right]} \cos ^{2 m+1} \frac{k \pi}{n}-1λn(2m+1)=2∑k=0[n−12]cos2m+1⁡kπn−1
But,
cos 2 m + 1 ⁡ x = 1 2 2 m ∑ j = 1 m + 1 ( 2 m + 1 m + j ) cos ⁡ ( 2 j − 1 ) x cos 2 m + 1 ⁡ x = 1 2 2 m ∑ j = 1 m + 1   ( 2 m + 1 m + j ) cos ⁡ ( 2 j − 1 ) x cos^(2m+1)x=(1)/(2^(2m))sum_(j=1)^(m+1)((2m+1)/(m+j))cos(2j-1)x\cos ^{2 m+1} x=\frac{1}{2^{2 m}} \sum_{j=1}^{m+1}\binom{2 m+1}{m+j} \cos (2 j-1) xcos2m+1⁡x=122m∑I=1m+1(2m+1m+I)cos⁡(2I−1)x
and on the other hand
[ n − 1 2 ] cos ⁡ ( 2 j − 1 ) k π n = { 1 2 + 1 2 ⋅ ( − 1 ) j + 1 tg ( 2 j − 1 ) π 2 n , n pair 1 2 + 1 2 ⋅ ( − 1 ) j + 1 sin ⁡ ( 2 j − 1 ) π 2 n , n impair n − 1 2 cos ⁡ ( 2 j − 1 ) k π n = 1 2 + 1 2 ⋅ ( − 1 ) j + 1 tg ( 2 j − 1 ) π 2 n , n  pair  1 2 + 1 2 ⋅ ( − 1 ) j + 1 sin ⁡ ( 2 j − 1 ) π 2 n , n  impair  [(n-1)/(2)]cos(2j-1)(k pi)/(n)={[(1)/(2)+(1)/(2)*((-1)^(j+1))/(tg(((2j-1)pi)/(2n)))","quad n" pair "],[(1)/(2)+(1)/(2)*((-1)^(j+1))/(sin(((2j-1)pi)/(2n)))","quad n" impair "]:}\left[\frac{n-1}{2}\right] \cos (2 j-1) \frac{k \pi}{n}=\left\{\begin{array}{l} \frac{1}{2}+\frac{1}{2} \cdot \frac{(-1)^{j+1}}{\operatorname{tg} \frac{(2 j-1) \pi}{2 n}}, \quad n \text { pair } \\ \frac{1}{2}+\frac{1}{2} \cdot \frac{(-1)^{j+1}}{\sin \frac{(2 j-1) \pi}{2 n}}, \quad n \text { impair } \end{array}\right.[n−12]cos⁡(2I−1)kπn={12+12⋅(−1)I+1tg(2I−1)π2n,n peer 12+12⋅(−1)I+1sin⁡(2I−1)π2n,n odd 
SO
λ n ( 2 m + 1 ) = { 1 2 2 m ∑ j = 1 m + 1 ( − 1 ) j + 1 ( 2 m + 1 m + j ) tg ( 2 j − 1 ) π 2 n , n pair 1 2 2 m ∑ j = 1 m + 1 ( − 1 ) j + 1 ( 2 m + 1 m + j ) sin ⁡ ( 2 j − 1 ) π 2 n , n impair. λ n ( 2 m + 1 ) = 1 2 2 m ∑ j = 1 m + 1   ( − 1 ) j + 1 ( 2 m + 1 m + j ) tg ( 2 j − 1 ) π 2 n ,      n  pair  1 2 2 m ∑ j = 1 m + 1   ( − 1 ) j + 1 ( 2 m + 1 m + j ) sin ⁡ ( 2 j − 1 ) π 2 n ,      n  impair.  lambda_(n)^((2m+1))={[(1)/(2^(2m))sum_(j=1)^(m+1)((-1)^(j+1)((2m+1)/(m+j)))/(tg(((2j-1)pi)/(2n)))",",n" pair "],[(1)/(2^(2m))sum_(j=1)^(m+1)((-1)^(j+1)((2m+1)/(m+j)))/(sin(((2j-1)pi)/(2n)))",",n" impair. "]:}\lambda_{n}^{(2 m+1)}= \begin{cases}\frac{1}{2^{2 m}} \sum_{j=1}^{m+1} \frac{(-1)^{j+1}\binom{2 m+1}{m+j}}{\operatorname{tg} \frac{(2 j-1) \pi}{2 n}}, & n \text { pair } \\ \frac{1}{2^{2 m}} \sum_{j=1}^{m+1} \frac{(-1)^{j+1}\binom{2 m+1}{m+j}}{\sin \frac{(2 j-1) \pi}{2 n}}, & n \text { impair. }\end{cases}λn(2m+1)={122m∑I=1m+1(−1)I+1(2m+1m+I)tg(2I−1)π2n,n peer 122m∑I=1m+1(−1)I+1(2m+1m+I)sin⁡(2I−1)π2n,n odd. 

II.

  1. MD Pompeiu in his second cited work demonstrates that if the four lengths a , b , c , d a , b , c , d a,b,c,da, b, c, dhas,b,c,dverify the inequality
a + b + c + d ≧ 2 max ( a , b , c , d ) a + b + c + d ≧ 2 max ( a , b , c , d ) a+b+c+d >= 2max(a,b,c,d)a+b+c+d \geqq 2 \max (a, b, c, d)has+b+c+d≧2max(has,b,c,d)
we can construct with these lengths a closed polygonal contour. This is the extension of the well-known property that with three lengths a , b , c a , b , c a,b,ca, b, chas,b,csuch as
a + b + c ≧ 2 max ( a , b , c ) a + b + c ≧ 2 max ( a , b , c ) a+b+c >= 2max(a,b,c)a+b+c \geqq 2 \max (a, b, c)has+b+c≧2max(has,b,c)
we can form a triangle.
The generalization of this property is as follows.
Theorem 6. The necessary and sufficient condition for that with n lengths ( n ≧ 3 ) 3 ) a 1 , a 2 , … , a n ( n ≧ 3 ) 3 a 1 , a 2 , … , a n {:(n >= 3)^(3))a_(1),a_(2),dots,a_(n)\left.(\mathrm{n} \geqq 3)^{3}\right) \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{n}(n≧3)3)has1,has2,…,hasnwe can construct a closed polygonal line is that we have
(5) a 1 + a 2 + … + a n ≧ 2 max ( a 1 , a 2 , … , a n ) a 1 + a 2 + … + a n ≧ 2 max a 1 , a 2 , … , a n quada_(1)+a_(2)+dots+a_(n) >= 2max(a_(1),a_(2),dots,a_(n))\quad a_{1}+a_{2}+\ldots+a_{n} \geqq 2 \max \left(a_{1}, a_{2}, \ldots, a_{n}\right)has1+has2+…+hasn≧2max(has1,has2,…,hasn).
Inequality (5) is equivalent to n n nnninequalities
a 1 + a 2 + … + a i − 1 + a i + 1 + … + a n ≧ a i i = 1 , 2 , … , n a 1 + a 2 + … + a i − 1 + a i + 1 + … + a n ≧ a i i = 1 , 2 , … , n {:[a_(1)+a_(2)+dots+a_(i-1)+a_(i+1)+dots+a_(n) >= a_(i)],[i=1","2","dots","n]:}\begin{gathered} a_{1}+a_{2}+\ldots+a_{i-1}+a_{i+1}+\ldots+a_{n} \geqq a_{i} \\ i=1,2, \ldots, n \end{gathered}has1+has2+…+hasi−1+hasi+1+…+hasn≧hasii=1,2,…,n
The condition is necessary since a broken line is at least as long as the segment joining its ends.
It remains to be demonstrated that the condition is also sufficient. We will demonstrate this by complete induction.
The property is true and well known for n = 3 n = 3 n=3n=3n=3. Let us assume the true for n ( ≥ 3 ) n ( ≥ 3 ) n( >= 3)n(\geq 3)n(≥3)and let's demonstrate it for n + 1 n + 1 n+1n+1n+1lengths. Let a 1 , a 2 , … , a n + 1 , n + 1 a 1 , a 2 , … , a n + 1 , n + 1 a_(1),a_(2),dots,a_(n+1),n+1a_{1}, a_{2}, \ldots, a_{n+1}, n+1has1,has2,…,hasn+1,n+1lengths such as
a 1 + a 2 + … + a n + 1 ≧ 2 max ( a 1 , a 2 , … , a n + 1 ) a 1 + a 2 + … + a n + 1 ≧ 2 max a 1 , a 2 , … , a n + 1 a_(1)+a_(2)+dots+a_(n+1) >= 2max(a_(1),a_(2),dots,a_(n+1))a_{1}+a_{2}+\ldots+a_{n+1} \geqq 2 \max \left(a_{1}, a_{2}, \ldots, a_{n+1}\right)has1+has2+…+hasn+1≧2max(has1,has2,…,hasn+1)
Without restricting the generality we can assume
a 1 ≦ a 2 ≦ … ≦ a n + 1 a 1 ≦ a 2 ≦ … ≦ a n + 1 a_(1) <= a_(2) <= dots <= a_(n+1)a_{1} \leqq a_{2} \leqq \ldots \leqq a_{n+1}has1≦has2≦…≦hasn+1. We then have
(6)
a 1 + a 2 + … + a n ≧ a n + 1 a 1 + a 2 + … + a n ≧ a n + 1 a_(1)+a_(2)+dots+a_(n) >= a_(n+1)a_{1}+a_{2}+\ldots+a_{n} \geqq a_{n+1}has1+has2+…+hasn≧hasn+1
Now let AB be a segment of length a n + 1 a n + 1 a_(n+1)a_{n+1}hasn+1and let C be the point of this segment such that the segment CB has the length a 1 a 1 a_(1)a_{1}has1. The segment AC has the length a n + 1 − a 1 a n + 1 − a 1 a_(n+1)-a_(1)a_{n+1}-a_{1}hasn+1−has1. It is enough to demonstrate that with the lengths a 1 , a 2 , … , a n , a n + 1 − a 1 a 1 , a 2 , … , a n , a n + 1 − a 1 a_(1),a_(2),dots,a_(n),a_(n+1)-a_(1)a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}-a_{1}has1,has2,…,hasn,hasn+1−has1we can construct a closed polygonal line. Indeed, we can then construct a closed polygonal line AD . . . CA AD . . . CA AD...CA\mathrm{AD} . . . \mathrm{CA}AD...THATwith these lengths and the closed polygonal line AD ... CBA will be constructed with the lengths a 1 , a 2 , … a 1 , a 2 , … a_(1),a_(2),dotsa_{1}, a_{2}, \ldotshas1,has2,…, a n + 1 a n + 1 a_(n+1)a_{n+1}hasn+1.
We have
max ( a 2 , a 3 , … , a n , a n + 1 − a 1 ) = max ( a n , a n + 1 − a 1 ) max a 2 , a 3 , … , a n , a n + 1 − a 1 = max a n , a n + 1 − a 1 max(a_(2),a_(3),dots,a_(n),a_(n+1)-a_(1))=max(a_(n),a_(n+1)-a_(1))\max \left(a_{2}, a_{3}, \ldots, a_{n}, a_{n+1}-a_{1}\right)=\max \left(a_{n}, a_{n+1}-a_{1}\right)max(has2,has3,…,hasn,hasn+1−has1)=max(hasn,hasn+1−has1)
It is therefore sufficient to demonstrate the inequalities
a 2 + a 3 + … + a n ≧ a n + 1 − a 1 a 2 + a 3 + … + a n + 1 + ( a n + 1 − a 1 ) ≧ a n . a 2 + a 3 + … + a n ≧ a n + 1 − a 1 a 2 + a 3 + … + a n + 1 + a n + 1 − a 1 ≧ a n . {:[a_(2)+a_(3)+dots+a_(n) >= a_(n+1)-a_(1)],[a_(2)+a_(3)+dots+a_(n+1)+(a_(n+1)-a_(1)) >= a_(n).]:}\begin{aligned} & a_{2}+a_{3}+\ldots+a_{n} \geqq a_{n+1}-a_{1} \\ & a_{2}+a_{3}+\ldots+a_{n+1}+\left(a_{n+1}-a_{1}\right) \geqq a_{n} . \end{aligned}has2+has3+…+hasn≧hasn+1−has1has2+has3+…+hasn+1+(hasn+1−has1)≧hasn.
Now the first is none other than inequality (6). The second can be written
( a 2 − a 1 ) + a 3 + … + a n − 1 + a n + 1 ≧ a n a 2 − a 1 + a 3 + … + a n − 1 + a n + 1 ≧ a n (a_(2)-a_(1))+a_(3)+dots+a_(n-1)+a_(n+1) >= a_(n)\left(a_{2}-a_{1}\right)+a_{3}+\ldots+a_{n-1}+a_{n+1} \geqq a_{n}(has2−has1)+has3+…+hasn−1+hasn+1≧hasn
and is verified since a 2 ≧ a 1 , a n + 1 ≧ a n a 2 ≧ a 1 , a n + 1 ≧ a n a_(2) >= a_(1),a_(n+1) >= a_(n)a_{2} \geqq a_{1}, a_{n+1} \geqq a_{n}has2≧has1,hasn+1≧hasn. So we have a 2 + a 3 + … + a n + ( a n + 1 − a 1 ) ≧ 2 max ( a 2 , a 3 , … , a n , a n + 1 − a 1 ) a 2 + a 3 + … + a n + a n + 1 − a 1 ≧ 2 max a 2 , a 3 , … , a n , a n + 1 − a 1 a_(2)+a_(3)+dots+a_(n)+(a_(n+1)-a_(1)) >= 2max(a_(2),a_(3),dots,a_(n),a_(n+1)-a_(1))a_{2}+a_{3}+\ldots+a_{n}+\left(a_{n+1}-a_{1}\right) \geqq 2 \max \left(a_{2}, a_{3}, \ldots, a_{n}, a_{n+1}-a_{1}\right)has2+has3+…+hasn+(hasn+1−has1)≧2max(has2,has3,…,hasn,hasn+1−has1)which demonstrates the property.
8. - Let us return to Theorem 5. For n n nnnodd we have
(7) n C n ( 1 ) = 1 sin ⁡ π 2 n ≧ 1 sin ⁡ π 6 = 2 (7) n C n ( 1 ) = 1 sin ⁡ π 2 n ≧ 1 sin ⁡ π 6 = 2 {:(7)nC_(n)^((1))=(1)/(sin((pi)/(2n))) >= (1)/(sin((pi)/(6)))=2:}\begin{equation*} n \mathrm{C}_{n}^{(1)}=\frac{1}{\sin \frac{\pi}{2 n}} \geqq \frac{1}{\sin \frac{\pi}{6}}=2 \tag{7} \end{equation*}(7)nCn(1)=1sin⁡π2n≧1sin⁡π6=2
equality being true only for n = 3 n = 3 n=3n=3n=3. For n n nnnpair we have
(8) n C n ( 1 ) = 1 tg π 2 n ≧ 1 tg π 8 = V 2 ― + 1 > 2 (8) n C n ( 1 ) = 1 tg π 2 n ≧ 1 tg π 8 = V 2 ¯ + 1 > 2 {:(8)nC_(n)^((1))=(1)/(tg((pi)/(2n))) >= (1)/(tg((pi)/(8)))=V bar(2)+1 > 2:}\begin{equation*} n C_{n}^{(1)}=\frac{1}{\operatorname{tg} \frac{\pi}{2 n}} \geqq \frac{1}{\operatorname{tg} \frac{\pi}{8}}=V \overline{2}+1>2 \tag{8} \end{equation*}(8)nCn(1)=1tgπ2n≧1tgπ8=V2―+1>2
so we have the
Consequence 2. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a regular polygon ( n ≧ 3 ) ( n ≧ 3 ) (n >= 3)(\mathrm{n} \geqq 3)(n≧3)and P a point of its plane, we have
PA ― 0 + PA ― 1 + … + PA ― n − 1 ≧ 2 max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) PA ¯ 0 + PA ¯ 1 + … + PA ¯ n − 1 ≧ 2 max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 bar(PA)_(0)+ bar(PA)_(1)+dots+ bar(PA)_(n-1) >= 2max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1))\overline{\mathrm{PA}}_{0}+\overline{\mathrm{PA}}_{1}+\ldots+\overline{\mathrm{PA}}_{n-1} \geqq 2 \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{\mathrm{n}-1}\right)PA―0+PA―1+…+PA―n−1≧2max(PA―0,PA―1,…,PA―n−1)
and we can state
Theorem 7. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a regular polygon ( n ≧ 3 n ≧ 3 n >= 3\mathrm{n} \geqq 3n≧3) and P a point of its plane, with the lengths PA ― 0 , PA ― 1 , … , PA ― n − 1 PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1)\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{\mathrm{n}-1}PA―0,PA―1,…,PA―n−1we can always form a closed polygonal line.
This is MD Pompeiu's theorem for any regular polygon.
Note that, by virtue of inequalities (7) and (8), for n ≧ 4 n ≧ 4 n >= 4n \geqq 4n≧4there surely exist non-regular polygons verifying Mr. Pompeiu's theorem.
9. - If r r rrris a positive number, the unit of length being chosen, we can consider the lengths PA ― 0 r , PA ― 1 r , … , PA ― n − 1 r PA ¯ 0 r , PA ¯ 1 r , … , PA ¯ n − 1 r bar(PA)_(0)^(r), bar(PA)_(1)^(r),dots, bar(PA)_(n-1)^(r)\overline{\mathrm{PA}}_{0}^{r}, \overline{\mathrm{PA}}_{1}^{r}, \ldots, \overline{\mathrm{PA}}_{n-1}^{r}PA―0r,PA―1r,…,PA―n−1r.
We have
lim n → ∞ λ n ( r ) = + ∞ lim n → ∞   λ n ( r ) = + ∞ lim_(n rarr oo)lambda_(n)^((r))=+oo\lim _{n \rightarrow \infty} \lambda_{n}^{(r)}=+\inftylimn→∞λn(r)=+∞
so there exists a natural number N r N r N_(r)\mathrm{N}_{r}Nrdependent on r r rrr, such as
λ n ( r ) ≧ 2 , pour n ≧ N r . λ n ( r ) ≧ 2 ,  pour  n ≧ N r . lambda_(n)^((r)) >= 2,quad" pour "n >= N_(r).\lambda_{n}^{(r)} \geqq 2, \quad \text { pour } n \geqq \mathrm{~N}_{r} .λn(r)≧2, For n≧ Nr.
Either r ≦ 1 r ≦ 1 r <= 1r \leqq 1r≦1. Taking into account (3), for n ≧ 3 n ≧ 3 n >= 3n \geqq 3n≧3, We have
λ n ( r ) ≧ 1 + 2 | cos ⁡ π n | r ≧ 1 + 2 2 r ≧ 2 , n pair λ n ( r ) ≧ 1 cos r ⁡ π 2 n 2 | cos ⁡ π 2 n | r = 2 , n impair λ n ( r ) ≧ 1 + 2 cos ⁡ π n r ≧ 1 + 2 2 r ≧ 2 , n  pair  λ n ( r ) ≧ 1 cos r ⁡ π 2 n 2 cos ⁡ π 2 n r = 2 , n  impair  {:[lambda_(n)^((r)) >= 1+2|cos((pi )/(n))|^(r) >= 1+(2)/(2^(r)) >= 2","quad n" pair "],[lambda_(n)^((r)) >= (1)/(cos^(r)((pi)/(2n)))2|cos((pi)/(2n))|^(r)=2","quad n" impair "]:}\begin{aligned} & \lambda_{n}^{(r)} \geqq 1+2\left|\cos \frac{\pi}{n}\right|^{r} \geqq 1+\frac{2}{2^{r}} \geqq 2, \quad n \text { pair } \\ & \lambda_{n}^{(r)} \geqq \frac{1}{\cos ^{r} \frac{\pi}{2 n}} 2\left|\cos \frac{\pi}{2 n}\right|^{r}=2, \quad n \text { impair } \end{aligned}λn(r)≧1+2|cos⁡πn|r≧1+22r≧2,n peer λn(r)≧1cosr⁡π2n2|cos⁡π2n|r=2,n odd 
therefore.
Theorem 8. If A 0 A 0 … A n − 1 A 0 A 0 … A n − 1 A_(0)A_(0)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{0} \ldots \mathrm{~A}_{\mathrm{n}-1}HAS0 HAS0… HASn−1is a regular polygon ( n ≧ 3 n ≧ 3 n >= 3\mathrm{n} \geqq 3n≧3), r a positive number ≦ 1 ≦ 1 <= 1\leqq 1≦1and P a point on the plane of the polygon, with the lengths PA ― 0 r , PA ― 1 r , … , PA ― n − 1 r PA ¯ 0 r , PA ¯ 1 r , … , PA ¯ n − 1 r bar(PA)_(0)^(r), bar(PA)_(1)^(r),dots, bar(PA)_(n-1)^(r)\overline{\mathrm{PA}}_{0}^{r}, \overline{\mathrm{PA}}_{1}^{r}, \ldots, \overline{\mathrm{PA}}_{n-1}^{r}PA―0r,PA―1r,…,PA―n−1rwe can always form a closed polygonal line.
We can also note that, for a n ≧ 3 n ≧ 3 n >= 3n \geqq 3n≧3given, λ n ( r ) λ n ( r ) lambda_(n)^((r))\lambda_{n}^{(r)}λn(r)is a decreasing function of r r rrr.
Theorem 7 therefore follows from inequalities (7), (8) and
λ n ( r ) ≧ λ n ( 1 ) , r ≦ 1 λ n ( r ) ≧ λ n ( 1 ) , r ≦ 1 lambda_(n)^((r)) >= lambda_(n)^((1)),r <= 1\lambda_{n}^{(r)} \geqq \lambda_{n}^{(1)}, r \leqq 1λn(r)≧λn(1),r≦1
For r > 1 r > 1 r > 1r>1r>1, we cannot take N r = 3 N r = 3 N_(r)=3\mathrm{N}_{r}=3Nr=3. The minimum of N r N r N_(r)\mathrm{N}_{r}Nrbelieves towards + ∞ + ∞ +oo+\infty+∞For r ⟶ ∞ r ⟶ ∞ r longrightarrow oor \longrightarrow \inftyr⟶∞. Simple calculations show us that
for r ≦ 2 r ≦ 2 r <= 2r \leqq 2r≦2we can take N r = 4 N r = 4 N_(r)=4\mathrm{N}_{r}=4Nr=4. On the case r = 2 r = 2 r=2r=2r=2we will come back later with more details.

III.

    • When r = 2 r = 2 r=2r=2r=2we can easily complete the results of § I.
Let us consider n n nnnpoints forming any 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}HAS0 HAS1… HASn−1and let P be any point, all located in the same plane. The case where the points A k A k A_(k)\mathrm{A}_{k}HASkare not all distinct is not excluded.
The expression
E 2 ( i ) ( P ) = PA ― 0 2 + PA ― 1 2 + … + PA ― n − 1 2 PA ― i 2 E 2 ( i ) ( P ) = PA ¯ 0 2 + PA ¯ 1 2 + … + PA ¯ n − 1 2 PA ¯ i 2 E_(2)^((i))(P)=( bar(PA)_(0)^(2)+ bar(PA)_(1)^(2)+dots+ bar(PA)_(n-1)^(2))/( bar(PA)_(i)^(2))\mathrm{E}_{2}^{(i)}(\mathrm{P})=\frac{\overline{\mathrm{PA}}_{0}^{2}+\overline{\mathrm{PA}}_{1}^{2}+\ldots+\overline{\mathrm{PA}}_{n-1}^{2}}{\overline{\mathrm{PA}}_{i}^{2}}E2(i)(P)=PA―02+PA―12+…+PA―n−12PA―i2
is a function of P , continuous in the entire plane, except at the point A i A i A_(i)\mathrm{A}_{i}HASi.
We propose to study the minimum of this expression.
Let G be the center of gravity of the points A 0 , A 1 , … , A n − 1 A 0 , A 1 , … , A n − 1 A_(0),A_(1),dots,A_(n-1)\mathrm{A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}HAS0, HAS1,…, HASn−1. We have
E 2 ( i ) ( P ) = n GP ― 2 + ∑ k = 0 n − 1 GA ― k 2 PA ― i 2 = n GP ― 2 + M 2 2 ( G ) PA ― i 2 E 2 ( i ) ( P ) = n GP ¯ 2 + ∑ k = 0 n − 1   GA ¯ k 2 PA ¯ i 2 = n GP ¯ 2 + M 2 2 ( G ) PA ¯ i 2 E_(2)^((i))(P)=(n bar(GP)^(2)+sum_(k=0)^(n-1) bar(GA)_(k)^(2))/( bar(PA)_(i)^(2))=n( bar(GP)^(2)+M_(2)^(2)(G))/( bar(PA)_(i)^(2))\mathrm{E}_{2}^{(i)}(\mathrm{P})=\frac{n \overline{\mathrm{GP}}^{2}+\sum_{k=0}^{n-1} \overline{\mathrm{GA}}_{k}^{2}}{\overline{\mathrm{PA}}_{i}^{2}}=n \frac{\overline{\mathrm{GP}}^{2}+\mathrm{M}_{2}^{2}(\mathrm{G})}{\overline{\mathrm{PA}}_{i}^{2}}E2(i)(P)=nGP―2+∑k=0n−1GA―k2PA―i2=nGP―2+M22(G)PA―i2
with the notation (4).
If the vertex A i A i A_(i)A_{i}HASicoincides with G G GGGwe have
E 2 ( i ) ( P ) = n + ∑ k = 0 n − 1 GA ― k 2 PA ― i 2 E 2 ( i ) ( P ) = n + ∑ k = 0 n − 1   GA ¯ k 2 PA ¯ i 2 E_(2)^((i))(P)=n+(sum_(k=0)^(n-1) bar(GA)_(k)^(2))/( bar(PA)_(i)^(2))\mathrm{E}_{2}^{(i)}(\mathrm{P})=n+\frac{\sum_{k=0}^{n-1} \overline{\mathrm{GA}}_{k}^{2}}{\overline{\mathrm{PA}}_{i}^{2}}E2(i)(P)=n+∑k=0n−1GA―k2PA―i2
and we see then that
min E 2 ( i ) ( P ) = n min E 2 ( i ) ( P ) = n minE_(2)^((i))(P)=n\min \mathrm{E}_{2}^{(i)}(\mathrm{P})=nminE2(i)(P)=n
but this minimum is not reached by any point P of the plane, unless all the points A k A k A_(k)\mathrm{A}_{k}HASkare confused.
If A i A i A_(i)A_{i}HASidoes not coincide with G G GGG, the minimum of E 2 ( i ) ( P ) E 2 ( i ) ( P ) E_(2)^((i))(P)E_{2}^{(i)}(P)E2(i)(P)can only be reached for a point on the line GA i GA i GA_(i)\mathrm{GA}_{i}GAi. Indeed, on the circumference of a circle with center A i A i A_(i)\mathrm{A}_{i}HASithe minimum can only be reached at the points where this circumference intersects the line GA i ∗ GA i ∗ GA_(i^(**))\mathrm{GA}_{i^{*}}GAi∗
If P is on the right GA i GA i GA_(i)\mathrm{GA}_{i}GAiwe can write
E 2 ( i ) ( P ) = n ( 1 − λ ) 2 GA ― i 2 + M 2 2 ( G ) λ 2 GA ― i 2 E 2 ( i ) ( P ) = n ( 1 − λ ) 2 GA ¯ i 2 + M 2 2 ( G ) λ 2 GA ¯ i 2 E_(2)^((i))(P)=n((1-lambda)^(2) bar(GA)_(i)^(2)+M_(2)^(2)(G))/(lambda^(2) bar(GA)_(i)^(2))\mathrm{E}_{2}^{(i)}(\mathrm{P})=n \frac{(1-\lambda)^{2} \overline{\mathrm{GA}}_{i}^{2}+\mathrm{M}_{2}^{2}(\mathrm{G})}{\lambda^{2} \overline{\mathrm{GA}}_{i}^{2}}E2(i)(P)=n(1−λ)2GA―i2+M22(G)λ2GA―i2
Or λ λ lambda\lambdaλis a real parameter. We have in particular P = λ G + ( 1 − λ ) A i ′ P = λ G + ( 1 − λ ) A i ′ P=lambdaG+(1-lambda)A_(i^('))\mathrm{P}=\lambda \mathrm{G}+(1-\lambda) \mathrm{A}_{i^{\prime}}P=λG+(1−λ)HASi′
The minimum is reached only for
λ = GA ― i 2 + M 2 2 ( G ) GA ― i 2 λ = GA ¯ i 2 + M 2 2 ( G ) GA ¯ i 2 lambda=( bar(GA)_(i)^(2)+M_(2)^(2)(G))/( bar(GA)_(i)^(2))\lambda=\frac{\overline{\mathrm{GA}}_{i}^{2}+\mathrm{M}_{2}^{2}(\mathrm{G})}{\overline{\mathrm{GA}}_{i}^{2}}λ=GA―i2+M22(G)GA―i2
and we have
min E 2 ( i ) ( P ) = n M 2 2 ( G ) GA i 2 + M 2 2 ( G ) min E 2 ( i ) ( P ) = n M 2 2 ( G ) GA i 2 + M 2 2 ( G ) minE_(2)^((i))(P)=n(M_(2)^(2)(G))/(GA_(i)^(2)+M_(2)^(2)(G))\min \mathrm{E}_{2}^{(i)}(\mathrm{P})=n \frac{\mathrm{M}_{2}^{2}(\mathrm{G})}{\mathrm{GA}_{i}^{2}+\mathrm{M}_{2}^{2}(\mathrm{G})}minE2(i)(P)=nM22(G)GAi2+M22(G)
We see that in this case
min E 2 ( i ) ( P ) < n min E 2 ( i ) ( P ) < n minE_(2)^((i))(P) < n\min \mathrm{E}_{2}^{(i)}(\mathrm{P})<nminE2(i)(P)<n
    • Now let us consider the expressions
E 2 ( 0 ) ( P ) , E 2 ( 1 ) ( P ) , … , E 2 ( n − 1 ) ( P ) E 2 ( 0 ) ( P ) , E 2 ( 1 ) ( P ) , … , E 2 ( n − 1 ) ( P ) E_(2)^((0))(P),E_(2)^((1))(P),dots,E_(2)^((n-1))(P)\mathrm{E}_{2}^{(0)}(\mathrm{P}), \mathrm{E}_{2}^{(1)}(\mathrm{P}), \ldots, \mathrm{E}_{2}^{(n-1)}(\mathrm{P})E2(0)(P),E2(1)(P),…,E2(n−1)(P)
Each has a minimum. The smallest of these minima,
(9) i = 0 , 1 , … , n − 1 [ min ( P ) E 2 ( i ) ( P ) ] (9) i = 0 , 1 , … , n − 1 min ( P )   E 2 ( i ) ( P ) {:(9)i=0","1","dots","n-1[min_((P))E_(2)^((i))(P)]:}i=0,1, \ldots, n-1\left[\begin{array}{l} \min _{(\mathrm{P})} \mathrm{E}_{2}^{(i)}(\mathrm{P}) \tag{9}\\ \end{array}\right](9)i=0,1,…,n−1[min(P)E2(i)(P)]
is therefore determined and depends on the 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}HAS0 HAS1… HASn−1.
When the polygon varies, this minimum has a maximum. We propose to determine this maximum.
The results of the previous No. show us that if all the points A k A k A_(k)\mathrm{A}_{k}HASkare confused the minimum (9) is equal to n n nnn. In the following we exclude this case.
If s s ssspoints A k ( 1 ≦ s ≦ n ) A k ( 1 ≦ s ≦ n ) A_(k)(1 <= s <= n)\mathrm{A}_{k}(1 \leqq s \leqq n)HASk(1≦s≦n)are distinct from G and if d , d ′ d , d ′ d,d^(')d, d^{\prime}d,d′are respectively the largest and smallest distances from G to these points, we have
s d ′ 2 n ≦ M 2 2 ( G ) ≦ s d 2 n s d ′ 2 n ≦ M 2 2 ( G ) ≦ s d 2 n (sd^('2))/(n) <= M_(2)^(2)(G) <= (sd^(2))/(n)\frac{s d^{\prime 2}}{n} \leqq \mathrm{M}_{2}^{2}(\mathrm{G}) \leqq \frac{s d^{2}}{n}sd′2n≦M22(G)≦sd2n
and the number (9) is ≧ d ′ 2 d 2 ⋅ n s n + s ≧ d ′ 2 d 2 ⋅ n s n + s >= (d^('2))/(d^(2))*(ns)/(n+s)\geqq \frac{d^{\prime 2}}{d^{2}} \cdot \frac{n s}{n+s}≧d′2d2⋅nsn+s, equality being possible and actually taking place only if d = d ′ d = d ′ d=d^(')d=d^{\prime}d=d′. On the other hand
n s n + s ≦ n 2 n s n + s ≦ n 2 (ns)/(n+s) <= (n)/(2)\frac{n s}{n+s} \leqq \frac{n}{2}nsn+s≦n2
equality being true only for s = n s = n s=ns=ns=n. It follows that the maximum of (9) is n 2 n 2 (n)/(2)\frac{n}{2}n2and is achieved only if the distances GA ― k GA ¯ k bar(GA)_(k)\overline{\mathrm{GA}}_{k}GA―kare all equal. We can therefore state the following property, a generalization of Theorem 5 for m = 1 m = 1 m=1m=1m=1.
Theorem 9. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a polygon inscribed in a circle whose center coincides with the center of gravity G of the vertices A k A k A_(k)\mathrm{A}_{\mathrm{k}}HASkand if P is a point in the plane of this polygon, we have
(10) M 2 ( P ) ≧ 1 2 max ( PA ― 0 , PA ― 1 , … , PA ― n − 1 ) (10) M 2 ( P ) ≧ 1 2 max PA ¯ 0 , PA ¯ 1 , … , PA ¯ n − 1 {:(10)M_(2)(P) >= (1)/(sqrt2)max( bar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1)):}\begin{equation*} \mathrm{M}_{2}(\mathrm{P}) \geqq \frac{1}{\sqrt{2}} \max \left(\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}\right) \tag{10} \end{equation*}(10)M2(P)≧12max(PA―0,PA―1,…,PA―n−1)
the equality being true only if P coincides with the symmetrical with respect to G of one of the points A k A k A_(k)\mathrm{A}_{\mathrm{k}}HASk.
If the 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}HAS0 HAS1… HASn−1does not verify the previous condition there exists at least one point P for which inequality (10) is not verified.
The "maximizing" polygon is an equilateral triangle for n = 3 n = 3 n=3n=3n=3and is a rectangle for n = 4 n = 4 n=4n=4n=4.
12. - Inequality (10) also gives us the following property.
Theorem 10. If 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}_{\mathrm{n}-1}HAS0 HAS1… HASn−1is a polygon with n ≧ 4 n ≧ 4 n >= 4n \geqq 4n≧4vertices inscribed in a circle whose center coincides with the center of gravity of the points A k A k A_(k)\mathrm{A}_{\mathrm{k}}HASkand if P is a point in the plane of this polygon, with the lengths PA ― 0 2 , PA ― 1 2 , … , PA ― n − 1 2 PA ¯ 0 2 , PA ¯ 1 2 , … , PA ¯ n − 1 2 bar(PA)_(0)^(2), bar(PA)_(1)^(2),dots, bar(PA)_(n-1)^(2)\overline{\mathrm{PA}}_{0}^{2}, \overline{\mathrm{PA}}_{1}^{2}, \ldots, \overline{\mathrm{PA}}_{n-1}^{2}PA―02,PA―12,…,PA―n−12we can construct a closed polygonal line.
For n = 4 n = 4 n=4n=4n=4we have, in particular, the characteristic property of the rectangle expressed by the
Theorem 11. If ABCD is a rectangle and P a point on its plane, with the lengths PA ― 2 , PB ― 2 , PC ― 2 , PD ― 2 PA ¯ 2 , PB ¯ 2 , PC ¯ 2 , PD ¯ 2 bar(PA)^(2), bar(PB)^(2), bar(PC)^(2), bar(PD)^(2)\overline{\mathrm{PA}}^{2}, \overline{\mathrm{~PB}}^{2}, \overline{\mathrm{PC}}^{2}, \overline{\mathrm{PD}}^{2}PA―2, PB―2,PC―2,PD―2we can always construct a closed polygonal line.
The property is not true for any other quadrilateral ABCD .

IV.

    • We have considered so far n n nnnpoint A 0 , A 1 , … , A n − 1 A 0 , A 1 , … , A n − 1 A_(0),A_(1),dots,A_(n-1)\mathrm{A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}HAS0, HAS1,…, HASn−1in a plane and the point P varying in this plane. We can assume, more generally, that these points are in a space with any number of dimensions. In particular, the problem treated in § III is immediately resolved in the space with any number of dimensions. Theorem 9 generalizes immediately and it is not even necessary to insist on this question here.
Now consider, instead of 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}HAS0 HAS1… HASn−1a circumference Γ Γ Gamma\boldsymbol{\Gamma}Γ.
We then have the following theorem, limit of Theorem 2.
Theorem 12. If P is a point on the plane of the circumference Γ Γ Gamma\GammaΓAnd M r ( P ) M r ( P ) M_(r)(P)\mathrm{M}_{\mathrm{r}}(\mathrm{P})Mr(P)the average power value r > 0 r > 0 r > 0\mathrm{r}>0r>0distances PA ― d u PA ¯ d u bar(PA)du\overline{\mathrm{PA}} d uPA―dupoint P to a point A of 1 ', we have the inequality
M r ( P ) ≧ ( 2 π ∫ 0 π 2 cos r ⁡ x d x ) 1 r PP ― 1 M r ( P ) ≧ 2 π ∫ 0 π 2   cos r ⁡ x d x 1 r PP ¯ 1 M_(r)(P) >= ((2)/(pi)int_(0)^((pi)/(2))cos^(r)xdx)^((1)/(r)) bar(PP)_(1)\mathrm{M}_{r}(\mathrm{P}) \geqq\left(\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos ^{r} x d x\right)^{\frac{1}{r}} \overline{\mathrm{PP}}_{1}Mr(P)≧(2π∫0π2cosr⁡xdx)1rPP―1
Or P 1 P 1 P_(1)\mathrm{P}_{1}P1is the point of Γ Γ Gamma\GammaΓthe furthest from P .
The equality holds only if P is on the circumference r .
We see immediately that the passage to the limit is perfectly justified. It is easy, moreover, to demonstrate the property directly.
14. - In the case r = 2 r = 2 r=2r=2r=2we can still ask ourselves a more complete problem.
Let us consider in the plane a continuous curve I. More precisely a curve represented parametrically by
x = f ( t ) , y = g ( t ) x = f ( t ) , y = g ( t ) x=f(t),y=g(t)x=f(t), y=g(t)x=f(t),y=g(t)
Or f ( t ) , g ( t ) f ( t ) , g ( t ) f(t),g(t)f(t), g(t)f(t),g(t)are two continuous functions of t t tttin a closed interval [ a , b ] [ a , b ] [a,b][a, b][has,b].
Let P be a point on the plane and M 2 ( P ) M 2 ( P ) M_(2)(P)\mathrm{M}_{2}(\mathrm{P})M2(P)the root mean square value of the distances from point P to the points on the curve Γ Γ Gamma\GammaΓ. We then have, x , y x , y x,yx, yx,ybeing the coordinates of P,
M 2 2 ( P ) = 1 b − a ∫ a b { [ x − f ( t ) ] 2 + [ y − g ( t ) ] 2 } d t M 2 2 ( P ) = 1 b − a ∫ a b   [ x − f ( t ) ] 2 + [ y − g ( t ) ] 2 d t M_(2)^(2)(P)=(1)/(b-a)int_(a)^(b){[x-f(t)]^(2)+[y-g(t)]^(2)}dt\mathrm{M}_{2}^{2}(\mathrm{P})=\frac{1}{b-a} \int_{a}^{b}\left\{[x-f(t)]^{2}+[y-g(t)]^{2}\right\} d tM22(P)=1b−has∫hasb{[x−f(t)]2+[y−g(t)]2}dt
Let us study the minimum of the report
M 2 ( P ) PA ― 0 M 2 ( P ) PA ¯ 0 (M_(2)(P))/( bar(PA)_(0))\frac{\mathrm{M}_{2}(\mathrm{P})}{\overline{\mathrm{PA}}_{0}}M2(P)PA―0
or better of the report
E 2 ( A 0 ) ( P ) = M 2 2 ( P ) PA ― 0 2 E 2 A 0 ( P ) = M 2 2 ( P ) PA ¯ 0 2 E_(2)^((A_(0)))(P)=(M_(2)^(2)(P))/( bar(PA)_(0)^(2))\mathrm{E}_{2}^{\left(\mathrm{A}_{0}\right)}(\mathrm{P})=\frac{\mathrm{M}_{2}^{2}(\mathrm{P})}{\overline{\mathrm{PA}}_{0}^{2}}E2(HAS0)(P)=M22(P)PA―02
Or A 0 A 0 A_(0)\mathrm{A}_{0}HAS0is a given point of Γ Γ Gamma\GammaΓ.
If G is the center of gravity of the line Γ Γ Gamma\GammaΓ, we still have
E 2 ( A 0 ) ( P ) = GP ― 2 + M 2 2 ( G ) PA ― 0 2 E 2 A 0 ( P ) = GP ¯ 2 + M 2 2 ( G ) PA ¯ 0 2 E_(2)^((A_(0)))(P)=( bar(GP)^(2)+M_(2)^(2)(G))/( bar(PA)_(0)^(2))\mathrm{E}_{2}^{\left(\mathrm{A}_{0}\right)}(\mathrm{P})=\frac{\overline{\mathrm{GP}}^{2}+\mathrm{M}_{2}^{2}(\mathrm{G})}{\overline{\mathrm{PA}}_{0}^{2}}E2(HAS0)(P)=GP―2+M22(G)PA―02
We see, as in the previous §, that if A 0 A 0 A_(0)\mathrm{A}_{0}HAS0coincides with G, we have
min E 2 ( A 0 ) ( P ) = 1 min E 2 A 0 ( P ) = 1 minE_(2)^((A_(0)))(P)=1\min E_{2}^{\left(\mathrm{A}_{0}\right)}(\mathrm{P})=1minE2(HAS0)(P)=1
and this minimum is not reached if, as we suppose, the line Γ Γ Gamma\mathbf{\Gamma}Γdoes not reduce to a point.
If A 0 A 0 A_(0)\mathrm{A}_{0}HAS0does not coincide with G the minimum is equal to
min E 2 ( A 0 ) ( P ) = M 2 2 ( G ) GA 0 2 + M 2 2 ( G ) min E 2 A 0 ( P ) = M 2 2 ( G ) GA 0 2 + M 2 2 ( G ) minE_(2)^((A_(0)))(P)=(M_(2)^(2)(G))/(GA_(0)^(2)+M_(2)^(2)(G))\min E_{2}^{\left(A_{0}\right)}(P)=\frac{M_{2}^{2}(G)}{\mathrm{GA}_{0}^{2}+M_{2}^{2}(G)}minE2(HAS0)(P)=M22(G)GA02+M22(G)
So that
min ( A 0 ) [ min ( P ) E 2 ( A 0 ) ( P ) ] min A 0   min ( P )   E 2 A 0 ( P ) min_((A_(0)))[min_((P))E_(2)^((A_(0)))(P)]\min _{\left(\mathrm{A}_{0}\right)}\left[\begin{array}{c} \min _{(\mathrm{P})} \mathrm{E}_{2}^{\left(\mathrm{A}_{0}\right)}(\mathrm{P}) \\ \end{array}\right]min(HAS0)[min(P)E2(HAS0)(P)]
is maximum it is necessary that GA ― 2 GA ¯ 2 bar(GA)^(2)\overline{\mathrm{GA}}^{2}GA―2where A is a current point of I reduces to a constant. Indeed, if d d dddis. the maximum and d ′ d ′ d^(')d^{\prime}d′the minimum of GA ― GA ¯ bar(GA)\overline{\mathrm{GA}}GA―, we have
d ′ ≦ M 2 ( G ) ≦ d d ′ ≦ M 2 ( G ) ≦ d d^(') <= M_(2)(G) <= dd^{\prime} \leqq \mathrm{M}_{2}(\mathrm{G}) \leqq dd′≦M2(G)≦d
equality, not being possible, due to the continuity of functions f ( t ) f ( t ) f(t)f(t)f(t), g ( t ) g ( t ) g(t)g(t)g(t), that if GA ― 0 GA ¯ 0 bar(GA)_(0)\overline{\mathrm{GA}}_{0}GA―0is constant. The minimum (11) is therefore
≥ d ′ 2 2 d 2 ≥ d ′ 2 2 d 2 >= (d^('2))/(2d^(2))\geq \frac{d^{\prime 2}}{2 d^{2}}≥d′22d2
But d ′ 2 2 d 2 ≦ 1 2 d ′ 2 2 d 2 ≦ 1 2 (d^('2))/(2d^(2)) <= (1)/(2)\frac{d^{\prime 2}}{2 d^{2}} \leqq \frac{1}{2}d′22d2≦12, equality being possible only if d ′ = d d ′ = d d^(')=dd^{\prime}=dd′=d. It follows that the maximum of (11) is equal to 1 2 1 2 (1)/(2)\frac{1}{2}12and this maximum is reached only if G A ― G A ¯ bar(GA)\overline{\mathrm{G} A}GHAS―is constant. For this, we can easily see that it is necessary and sufficient that Γ Γ Gamma\GammaΓbe a circumference and let P be on this circumference. We therefore finally have the
Theorem 13. If Γ Γ Gamma\GammaΓis a circumference and P a point on the plane of this circumference, we have
(12) M 2 ( P ) ≧ 1 V 2 PP 1 ― (12) M 2 ( P ) ≧ 1 V 2 PP 1 ¯ {:(12)M_(2)(P) >= (1)/(V2) bar(PP_(1)):}\begin{equation*} \mathrm{M}_{2}(\mathrm{P}) \geqq \frac{1}{V 2}{\overline{\mathrm{PP}_{1}}} \tag{12} \end{equation*}(12)M2(P)≧1V2PP1―
P 1 P 1 P_(1)\mathrm{P}_{1}P1being the point of Γ Γ Gamma\GammaΓthe furthest from P.
If Γ Γ Gamma\GammaΓis a continuous curve different from a circumference there exists at least one point P of the plane for which inequality (12) is not true.
Bucuresti, December 11, 1941.
  1. 1 ) 1 ) ^(1)){ }^{1)}1)D. Pompeiu. "An identity between complex numbers and a theorem of elementary geometry." Bulletin of Mathematics and Physics, 6, 6-7 (1936).
  2. 2 2 ^(2){ }^{2}2) D. Pompeiu "Geometry and imaginaries: demonstration of some elementary theorems". Bulletin of Mathematics and Physics, 11,
    (1941).
  3. 3 3 ^(3){ }^{3}3) In reality the property is also true for n = 2 n = 2 n=2n=2n=2, but in this case it is obvious.
1941

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