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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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Bulletin mathématique de la Société Roumaine des Sciences
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Societatea de Științe Matematice din România
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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}) that if ABC is an equilateral triangle and P a point of its plane, with the lengthsbar(PA), bar(PB), bar(PC)\overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{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
If ABC is an equilateral triangle and if P is a point on its plane, we have
( 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}}) .
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 polygonA_(0)A_(1)dotsA_(n-1)(n >= 3)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}(n \geqq 3)and let P be a point of its plane. The expression
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}}
Orrris a positive number, is a continuous function of P in the entire plane, except at the pointA_(0)A_{0}, where it is not defined.
We propose to study the minimum ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P}).
Without restricting the generality we can assume that the verticesA_(k)\mathrm{A}_{k}are represented in the plane by complex numbersi(2k pi)/(2)i \frac{2 k \pi}{2}
e quad,k=0,1,dots,n-1e \quad, k=0,1, \ldots, n-1and the variable point P by the complex numberrhoe^(i0),rho >= 0.0 <= theta <= 2pi\rho e^{i 0}, \rho \geqq 0,0 \leqq \theta \leqq 2 \pi. We then have
e quad,k=0,1,dots,n-1e \quad, k=0,1, \ldots, n-1and the variable point P by the complex numberrhoe^(i0),rho >= 0.0 <= theta <= 2pi\rho e^{i 0}, \rho \geqq 0,0 \leqq \theta \leqq 2 \pi. We then have
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 .
If P is the pointA_(0)^(')\mathrm{A}_{0}^{\prime}diametrically opposed toA_(0)\mathrm{A}_{0}on the circumscribed circle, we have
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 .
So we surely have
minE_(r)(P) <= alpha.\min \mathrm{E}_{r}(\mathrm{P}) \leqq \alpha .
But we have
|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,
SO
E_(r)(P) >= (n|rho-1|^(r))/((rho+1)^(r)).\mathrm{E}_{r}(\mathrm{P}) \geqq \frac{n|\rho-1|^{r}}{(\rho+1)^{r}} .
If we takerho > (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}}}, We haveE_(r)(P) > alpha\mathrm{E}_{r}(\mathrm{P})>\alpha.
It follows that the minimum ofE_(r)(P)E_{r}(P)is the same as in a closed circle with center origin and radius(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}}}. On the other hand
It follows that the minimum ofE_(r)(P)E_{r}(P)is the same as in a closed circle with center origin and radius(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}}}. On the other hand
{:[ limE_(r)(P)=+oo],[PrarrA_(0).]:}\begin{aligned}
& \lim \mathrm{E}_{r}(\mathrm{P})=+\infty \\
& \mathrm{P} \rightarrow \mathrm{~A}_{0} .
\end{aligned}
We can therefore affirm that:
The minimum ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P})is reached at least at one point of the plane. Note that such a point is necessarily distinct from the origin O since
The minimum ofE_(r)(P)\mathrm{E}_{r}(\mathrm{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.\mathrm{E}_{r}(\mathrm{O})=n .
-
- Let us consider the pointP_(m)\mathrm{P}_{m}represented by the complex numberi(0+(2m pi)/(n))i\left(0+\frac{2 m \pi}{n}\right)
wheremmis an integer. As a result of symmetry, we have
- Let us consider the pointP_(m)\mathrm{P}_{m}represented by the complex numberi(0+(2m pi)/(n))i\left(0+\frac{2 m \pi}{n}\right)
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}
Still as a result of symmetry, we see that it is sufficient to examine the expressionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})only for points P for which
(1)
(1)
pi-(pi )/(n) <= theta <= pi\pi-\frac{\pi}{n} \leqq \theta \leqq \pi
Let's fix such atheta\thetaand let's varyrho\rhofrom 0 to+oo+\infty.
If we pose
(2)
If we pose
(2)
t=(2rho)/(rho^(2)+1)t=\frac{2 \rho}{\rho^{2}+1}
We have
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}}
From (1) it follows thatcos theta < 0\cos \theta<0, SOE_(r)(P)\mathrm{E}_{r}(\mathrm{P})is a continuous function ofttin the closed interval[0,1][0,1]and is surely differentiable in the interval[0,1)[0,1)open to the right. We have, according to an easy calculation,
{:[(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}
and taking into account (1),
(dE_(r)(P))/(dt) < 0,quad0 <= t < 1.\frac{d \mathrm{E}_{r}(\mathrm{P})}{d t}<0, \quad 0 \leqq t<1 .
The functionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})ofttis therefore decreasing in the interval[0,1][0,1]. Its minimum is reached only fort=1t=1so, according to (2), only forrho=1\rho=1.
The minimum of expressionE_(r)(P)\mathrm{E}_{\mathrm{r}}(\mathrm{P})can only be reached on the circle circumscribed to the polygon.
3. - Let us therefore supposep=1p=1. The expressionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})becomes
3. - Let us therefore supposep=1p=1. The expressionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})becomes
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}
Or
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}
by posing
u=cotg((theta)/(2))u=\operatorname{cotg} \frac{\theta}{2}
It is a continuous function ofuuin the closed interval[0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right]and is surely indefinitely differentiable in the open interval(0,tg((pi)/(2n)))\left(0, \operatorname{tg} \frac{\pi}{2 n}\right). We have
{:[(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}
So we have
(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}
following thatr < 1,=1r<1,=1, resp.> 1>1.
The functionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})ofuuis therefore a continuous linear concave or convex function in the closed interval[0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right]. To go further let us examine the first derivative ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P})near the ends.
The functionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})ofuuis therefore a continuous linear concave or convex function in the closed interval[0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right]. To go further let us examine the first derivative ofE_(r)(P)\mathrm{E}_{r}(\mathrm{P})near the ends.
Ifnnis odd, all thecos((k pi)/(n))\cos \frac{k \pi}{n}are!=0\neq 0, SO
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}=0
We can therefore say in this case that the minimum is reached foru=0u=0ifr > 1r>1and foru=tg((pi)/(2n))u=\operatorname{tg} \frac{\pi}{2 n}ifr < 1r<1and only for these values.
Ifnnis even thecos((k pi)/(n))\cos \frac{k \pi}{n}are!=0\neq 0except fork=(n)/(2)k=\frac{n}{2}. So we have
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}
and consequently
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.
We immediately deduce that ifr > 1r>1the minimum is reached foru=0u=0. Ifr < 1r<1, the functionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})being concave we know that the minimum can only be reached for the ends
u=0,u=tg((pi)/(2n))u=0, u=\operatorname{tg} \frac{\pi}{2 n}
To decide, let us note that in this case (nnpeer)
{:[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}
The functionE_(r)(P)\mathrm{E}_{r}(\mathrm{P})is therefore increasing and its minimum is reached foru=0u=0.
It remains to examine the special caser=1r=1. We then have
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}
Fornneven the minimum is reached foru=0u=0. FornnoddE_(1)(P)\mathrm{E}_{1}(\mathrm{P})is constant over all points considered, so its minimum is reached at any point in the interval[0,tg((pi)/(2n))]\left[0, \operatorname{tg} \frac{\pi}{2 n}\right].
Now let us note that iftheta\thetabelieves inpi-(pi )/(n)\pi-\frac{\pi}{n}haspi,u=cotg((theta)/(2))\pi, u=\operatorname{cotg} \frac{\theta}{2}decreases bytg((pi)/(2n))\operatorname{tg} \frac{\pi}{2 n}to 0.
Now let us pose
(3)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.
(3)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.
We can then state the following theorem.
Theorem 1. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon, r a given positive number and P a point in the plane of the polygon, we have the inequality
Theorem 1. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is 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) >= lambda_(n)^((r)).\mathrm{E}_{r}(\mathrm{P}) \geqq \lambda_{n}^{(r)} .
Equality holds if and only if:
1^(0)n1^{0} \mathrm{n}being even andr > 0\mathrm{r}>0, or n being odd andr >= 1,P\mathrm{r} \geqq 1, \mathrm{P}coincides with the pointA_(0)^(')\mathrm{A}_{0}^{\prime}of the diametrically opposite circumscribed circlea^(`)A_(0)\grave{a} \mathrm{~A}_{0}.
2^(0)n2^{0} \mathrm{n}being odd and0 < r < 1,P0<\mathrm{r}<1, \mathrm{P}coincides with one of the opposite vertices toA_(0)\mathrm{A}_{0}, SOP=A_((n-1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}-1}{2}}OrP=A_((n+1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}+1}{2}}.
3^(0)n3^{0} \mathrm{n}being odd andr=1,P\mathrm{r}=1, \mathrm{P}coincides 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)\frac{\mathrm{A}_{\frac{\mathrm{n}-1}{2}}}{2}, \frac{\mathrm{~A}_{\frac{\mathrm{n}+1}{2}}}{2}.
1^(0)n1^{0} \mathrm{n}being even andr > 0\mathrm{r}>0, or n being odd andr >= 1,P\mathrm{r} \geqq 1, \mathrm{P}coincides with the pointA_(0)^(')\mathrm{A}_{0}^{\prime}of the diametrically opposite circumscribed circlea^(`)A_(0)\grave{a} \mathrm{~A}_{0}.
2^(0)n2^{0} \mathrm{n}being odd and0 < r < 1,P0<\mathrm{r}<1, \mathrm{P}coincides with one of the opposite vertices toA_(0)\mathrm{A}_{0}, SOP=A_((n-1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}-1}{2}}OrP=A_((n+1)/(2))\mathrm{P}=\mathrm{A}_{\frac{\mathrm{n}+1}{2}}.
3^(0)n3^{0} \mathrm{n}being odd andr=1,P\mathrm{r}=1, \mathrm{P}coincides 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)\frac{\mathrm{A}_{\frac{\mathrm{n}-1}{2}}}{2}, \frac{\mathrm{~A}_{\frac{\mathrm{n}+1}{2}}}{2}.
This is, of course, the circle circumscribed to the polygon.
4. - The previous result can be put into various forms. The expression
4. - The previous result can be put into various forms. The expression
{:(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*}
is the power averagerrdistancesbar(PA)_(0), bar(PA)_(1),dots, bar(PA)_(n-1)\overline{\mathrm{PA}}_{0}, \overline{\mathrm{PA}}_{1}, \ldots, \overline{\mathrm{PA}}_{n-1}. We know that we have the upper limitation
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)
Theorem 1 gives, for a regular polygon, a lower bound ofM_(r)(P)\mathrm{M}_{r}(\mathrm{P}). We have the
Theorem 2. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is 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( 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)
Or
C_(n)^((r))=((lambda_(n)^((r)))/(n))^((1)/(r))\mathrm{C}_{n}^{(r)}=\left(\frac{\lambda_{n}^{(r)}}{n}\right)^{\frac{1}{r}}
Note that
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<1
SOC_(n)^((r))\mathrm{C}_{n}^{(r)}Forn longrightarrow oon \longrightarrow \inftytends towards the average power valuerrof the functioncos x\cos xin the meantime[0,(pi)/(2)]\left[0, \frac{\pi}{2}\right].
To complete the previous results we will examine in a little more detail the case whererris an integer.
5. - Let firstr=1r=1. We then have
5. - Let firstr=1r=1. We then have
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}
The direct generalization of MD Pompeiu's theorem can be stated as follows
Theorem 3. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}is a regular polygon and P a point of its plane, we have
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)
OrC_(n)^((1))\mathrm{C}_{n}^{(1)}is equal to
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)
OrC_(n)^((1))\mathrm{C}_{n}^{(1)}is equal to
(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}}
depending on whether n is even or odd.
Note that fornnpeerC_(n)^((1))\mathrm{C}_{n}^{(1)}believes and tends towards(2)/(pi)\frac{2}{\pi}and fornnodd it decreases and tends towards(2)/(pi)\frac{2}{\pi}Forn longrightarrow oon \longrightarrow \infty, SO
Note that fornnpeerC_(n)^((1))\mathrm{C}_{n}^{(1)}believes and tends towards(2)/(pi)\frac{2}{\pi}and fornnodd it decreases and tends towards(2)/(pi)\frac{2}{\pi}Forn longrightarrow oon \longrightarrow \infty, SO
Consequence 1. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon with an odd number of sides and P a point on its plane, we have
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)
equality not being possible and(2)/(pi)\frac{2}{\pi}cannot be replaced by any other larger number.
Forn=4n=4we haveC_(n)^((1))=(sqrt2+1)/(4)\mathrm{C}_{n}^{(1)}=\frac{\sqrt{2}+1}{4}and we have
Theorem 4. If ABCD is a square and P a point on its plane, we have the inequality
Theorem 4. If ABCD is a square and P a point on its plane, we have the inequality
( 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}})
which is, moreover, equivalent to the four inequalities
{:[ 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}
MD Pompeiu has already found 2)C_(4)^((1)) > (1)/(2)\mathrm{C}_{4}^{(1)}>\frac{1}{2}.
6. - Let us now supposer=2m,mr=2 m, mbeing a natural number. Let us calculatelambda_(n)^((2m))\lambda_{n}^{(2 m)}. We have
6. - Let us now supposer=2m,mr=2 m, mbeing a natural number. Let us calculatelambda_(n)^((2m))\lambda_{n}^{(2 m)}. We have
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 x
and on the other hand
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}
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}
We deduce from this
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.
Or[alpha][\alpha]is the largest integer<= alpha\leqq \alpha.
Note that in this case the coefficientC_(n)^((2m))\mathrm{C}_{n}^{(2 m)}is independent of n forn > m\mathrm{n}>\mathrm{m}.
Note that in this case the coefficientC_(n)^((2m))\mathrm{C}_{n}^{(2 m)}is independent of n forn > m\mathrm{n}>\mathrm{m}.
Theorem 5. If m is a natural number,A_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}a regular polygon atn > m(n >= 3)\mathrm{n}>\mathrm{m}(\mathrm{n} \geqslant 3)sides and P a point on the plane of this polygon, we have
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)
The numberlambda_(n)^((r))\lambda_{n}^{(r)}can also be easily calculated ifr=2m+1r=2 m+1is an odd number. In this case we have
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}-1
But,
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) x
and on the other hand
[(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.
SO
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}
II.
- MD Pompeiu in his second cited work demonstrates that if the four lengthsa,b,c,da, b, c, dverify the inequality
a+b+c+d >= 2max(a,b,c,d)a+b+c+d \geqq 2 \max (a, 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 lengthsa,b,ca, b, csuch as
a+b+c >= 2max(a,b,c)a+b+c \geqq 2 \max (a, 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),dots,a_(n)\left.(\mathrm{n} \geqq 3)^{3}\right) \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{n}we can construct a closed polygonal line is that we have
(5)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).
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),dots,a_(n)\left.(\mathrm{n} \geqq 3)^{3}\right) \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{n}we can construct a closed polygonal line is that we have
(5)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).
Inequality (5) is equivalent tonninequalities
{:[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}
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 forn=3n=3. Let us assume the true forn( >= 3)n(\geq 3)and let's demonstrate it forn+1n+1lengths. Leta_(1),a_(2),dots,a_(n+1),n+1a_{1}, a_{2}, \ldots, a_{n+1}, n+1lengths such as
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)
Without restricting the generality we can assume
a_(1) <= a_(2) <= dots <= a_(n+1)a_{1} \leqq a_{2} \leqq \ldots \leqq a_{n+1}. We then have
(6)
a_(1) <= a_(2) <= dots <= a_(n+1)a_{1} \leqq a_{2} \leqq \ldots \leqq a_{n+1}. We then have
(6)
a_(1)+a_(2)+dots+a_(n) >= a_(n+1)a_{1}+a_{2}+\ldots+a_{n} \geqq a_{n+1}
Now let AB be a segment of lengtha_(n+1)a_{n+1}and let C be the point of this segment such that the segment CB has the lengtha_(1)a_{1}. The segment AC has the lengtha_(n+1)-a_(1)a_{n+1}-a_{1}. It is enough to demonstrate that with the lengthsa_(1),a_(2),dots,a_(n),a_(n+1)-a_(1)a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}-a_{1}we can construct a closed polygonal line. Indeed, we can then construct a closed polygonal lineAD...CA\mathrm{AD} . . . \mathrm{CA}with these lengths and the closed polygonal line AD ... CBA will be constructed with the lengthsa_(1),a_(2),dotsa_{1}, a_{2}, \ldots,a_(n+1)a_{n+1}.
We have
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)
It is therefore sufficient to demonstrate the inequalities
{:[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}
Now the first is none other than inequality (6). The second can be written
(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}
and is verified sincea_(2) >= a_(1),a_(n+1) >= a_(n)a_{2} \geqq a_{1}, a_{n+1} \geqq a_{n}. So we havea_(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)which demonstrates the property.
8. - Let us return to Theorem 5. Fornnodd we have
8. - Let us return to Theorem 5. Fornnodd we have
{:(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*}
equality being true only forn=3n=3. Fornnpair we have
{:(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*}
so we have the
Consequence 2. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon(n >= 3)(\mathrm{n} \geqq 3)and P a point of its plane, we have
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)
and we can state
Theorem 7. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon (n >= 3\mathrm{n} \geqq 3) and P a point of its plane, with the lengthsbar(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}we can always form a closed polygonal line.
Theorem 7. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon (n >= 3\mathrm{n} \geqq 3) and P a point of its plane, with the lengthsbar(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}we 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), forn >= 4n \geqq 4there surely exist non-regular polygons verifying Mr. Pompeiu's theorem.
9. - Ifrris a positive number, the unit of length being chosen, we can consider the lengthsbar(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}.
9. - Ifrris a positive number, the unit of length being chosen, we can consider the lengthsbar(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}.
We have
lim_(n rarr oo)lambda_(n)^((r))=+oo\lim _{n \rightarrow \infty} \lambda_{n}^{(r)}=+\infty
so there exists a natural numberN_(r)\mathrm{N}_{r}dependent onrr, such as
lambda_(n)^((r)) >= 2,quad" pour "n >= N_(r).\lambda_{n}^{(r)} \geqq 2, \quad \text { pour } n \geqq \mathrm{~N}_{r} .
Eitherr <= 1r \leqq 1. Taking into account (3), forn >= 3n \geqq 3, We have
{:[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}
therefore.
Theorem 8. IfA_(0)A_(0)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{0} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon (n >= 3\mathrm{n} \geqq 3), r a positive number<= 1\leqq 1and P a point on the plane of the polygon, with the lengthsbar(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}we can always form a closed polygonal line.
Theorem 8. IfA_(0)A_(0)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{0} \ldots \mathrm{~A}_{\mathrm{n}-1}is a regular polygon (n >= 3\mathrm{n} \geqq 3), r a positive number<= 1\leqq 1and P a point on the plane of the polygon, with the lengthsbar(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}we can always form a closed polygonal line.
We can also note that, for an >= 3n \geqq 3given,lambda_(n)^((r))\lambda_{n}^{(r)}is a decreasing function ofrr.
Theorem 7 therefore follows from inequalities (7), (8) and
lambda_(n)^((r)) >= lambda_(n)^((1)),r <= 1\lambda_{n}^{(r)} \geqq \lambda_{n}^{(1)}, r \leqq 1
Forr > 1r>1, we cannot takeN_(r)=3\mathrm{N}_{r}=3. The minimum ofN_(r)\mathrm{N}_{r}believes towards+oo+\inftyForr longrightarrow oor \longrightarrow \infty. Simple calculations show us that
forr <= 2r \leqq 2we can takeN_(r)=4\mathrm{N}_{r}=4. On the caser=2r=2we will come back later with more details.
forr <= 2r \leqq 2we can takeN_(r)=4\mathrm{N}_{r}=4. On the caser=2r=2we will come back later with more details.
III.
-
- Whenr=2r=2we can easily complete the results of § I.
Let us considernnpoints forming any polygonA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}and let P be any point, all located in the same plane. The case where the pointsA_(k)\mathrm{A}_{k}are not all distinct is not excluded.
The expression
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}}
is a function of P , continuous in the entire plane, except at the pointA_(i)\mathrm{A}_{i}.
We propose to study the minimum of this expression.
Let G be the center of gravity of the pointsA_(0),A_(1),dots,A_(n-1)\mathrm{A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}. We have
We propose to study the minimum of this expression.
Let G be the center of gravity of the pointsA_(0),A_(1),dots,A_(n-1)\mathrm{A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}. We have
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}}
with the notation (4).
If the vertexA_(i)A_{i}coincides withGGwe have
If the vertexA_(i)A_{i}coincides withGGwe have
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}}
and we see then that
minE_(2)^((i))(P)=n\min \mathrm{E}_{2}^{(i)}(\mathrm{P})=n
but this minimum is not reached by any point P of the plane, unless all the pointsA_(k)\mathrm{A}_{k}are confused.
IfA_(i)A_{i}does not coincide withGG, the minimum ofE_(2)^((i))(P)E_{2}^{(i)}(P)can only be reached for a point on the lineGA_(i)\mathrm{GA}_{i}. Indeed, on the circumference of a circle with centerA_(i)\mathrm{A}_{i}the minimum can only be reached at the points where this circumference intersects the lineGA_(i^(**))\mathrm{GA}_{i^{*}}
If P is on the rightGA_(i)\mathrm{GA}_{i}we can write
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}}
Orlambda\lambdais a real parameter. We have in particularP=lambdaG+(1-lambda)A_(i^('))\mathrm{P}=\lambda \mathrm{G}+(1-\lambda) \mathrm{A}_{i^{\prime}}
The minimum is reached only for
The minimum is reached only for
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}}
and we have
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})}
We see that in this case
minE_(2)^((i))(P) < n\min \mathrm{E}_{2}^{(i)}(\mathrm{P})<n
-
- Now let us consider the expressions
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})
Each has a minimum. The smallest of these minima,
{:(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]
is therefore determined and depends on the polygonA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}.
When the polygon varies, this minimum has a maximum. We propose to determine this maximum.
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 pointsA_(k)\mathrm{A}_{k}are confused the minimum (9) is equal tonn. In the following we exclude this case.
IfsspointsA_(k)(1 <= s <= n)\mathrm{A}_{k}(1 \leqq s \leqq n)are distinct from G and ifd,d^(')d, d^{\prime}are respectively the largest and smallest distances from G to these points, we have
(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}
and the number (9) is>= (d^('2))/(d^(2))*(ns)/(n+s)\geqq \frac{d^{\prime 2}}{d^{2}} \cdot \frac{n s}{n+s}, equality being possible and actually taking place only ifd=d^(')d=d^{\prime}. On the other hand
(ns)/(n+s) <= (n)/(2)\frac{n s}{n+s} \leqq \frac{n}{2}
equality being true only fors=ns=n. It follows that the maximum of (9) is(n)/(2)\frac{n}{2}and is achieved only if the distancesbar(GA)_(k)\overline{\mathrm{GA}}_{k}are all equal. We can therefore state the following property, a generalization of Theorem 5 form=1m=1.
Theorem 9. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a polygon inscribed in a circle whose center coincides with the center of gravity G of the verticesA_(k)\mathrm{A}_{\mathrm{k}}and if P is a point in the plane of this polygon, we have
{:(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*}
the equality being true only if P coincides with the symmetrical with respect to G of one of the pointsA_(k)\mathrm{A}_{\mathrm{k}}.
If the polygonA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}does 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 forn=3n=3and is a rectangle forn=4n=4.
12. - Inequality (10) also gives us the following property.
12. - Inequality (10) also gives us the following property.
Theorem 10. IfA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{\mathrm{n}-1}is a polygon withn >= 4n \geqq 4vertices inscribed in a circle whose center coincides with the center of gravity of the pointsA_(k)\mathrm{A}_{\mathrm{k}}and if P is a point in the plane of this polygon, with the lengthsbar(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}we can construct a closed polygonal line.
Forn=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 lengthsbar(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}we can always construct a closed polygonal line.
The property is not true for any other quadrilateral ABCD .
IV.
-
- We have considered so farnnpointA_(0),A_(1),dots,A_(n-1)\mathrm{A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}in 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 polygonA_(0)A_(1)dotsA_(n-1)\mathrm{A}_{0} \mathrm{~A}_{1} \ldots \mathrm{~A}_{n-1}a circumferenceGamma\boldsymbol{\Gamma}.
We then have the following theorem, limit of Theorem 2.
Theorem 12. If P is a point on the plane of the circumferenceGamma\GammaAndM_(r)(P)\mathrm{M}_{\mathrm{r}}(\mathrm{P})the average power valuer > 0\mathrm{r}>0distancesbar(PA)du\overline{\mathrm{PA}} d upoint P to a point A of 1 ', we have the inequality
Theorem 12. If P is a point on the plane of the circumferenceGamma\GammaAndM_(r)(P)\mathrm{M}_{\mathrm{r}}(\mathrm{P})the average power valuer > 0\mathrm{r}>0distancesbar(PA)du\overline{\mathrm{PA}} d upoint P to a point A of 1 ', we have the inequality
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}
OrP_(1)\mathrm{P}_{1}is the point ofGamma\Gammathe 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 caser=2r=2we can still ask ourselves a more complete problem.
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 caser=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)
Orf(t),g(t)f(t), g(t)are two continuous functions ofttin a closed interval[a,b][a, b].
Let P be a point on the plane andM_(2)(P)\mathrm{M}_{2}(\mathrm{P})the root mean square value of the distances from point P to the points on the curveGamma\Gamma. We then have,x,yx, ybeing the coordinates of P,
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 t
Let us study the minimum of the report
(M_(2)(P))/( bar(PA)_(0))\frac{\mathrm{M}_{2}(\mathrm{P})}{\overline{\mathrm{PA}}_{0}}
or better of the report
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}}
OrA_(0)\mathrm{A}_{0}is a given point ofGamma\Gamma.
If G is the center of gravity of the lineGamma\Gamma, we still have
If G is the center of gravity of the lineGamma\Gamma, we still have
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}}
We see, as in the previous §, that ifA_(0)\mathrm{A}_{0}coincides with G, we have
minE_(2)^((A_(0)))(P)=1\min E_{2}^{\left(\mathrm{A}_{0}\right)}(\mathrm{P})=1
and this minimum is not reached if, as we suppose, the lineGamma\mathbf{\Gamma}does not reduce to a point.
IfA_(0)\mathrm{A}_{0}does not coincide with G the minimum is equal to
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)}
So that
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]
is maximum it is necessary thatbar(GA)^(2)\overline{\mathrm{GA}}^{2}where A is a current point of I reduces to a constant. Indeed, ifddis. the maximum andd^(')d^{\prime}the minimum ofbar(GA)\overline{\mathrm{GA}}, we have
d^(') <= M_(2)(G) <= dd^{\prime} \leqq \mathrm{M}_{2}(\mathrm{G}) \leqq d
equality, not being possible, due to the continuity of functionsf(t)f(t),g(t)g(t), that ifbar(GA)_(0)\overline{\mathrm{GA}}_{0}is constant. The minimum (11) is therefore
>= (d^('2))/(2d^(2))\geq \frac{d^{\prime 2}}{2 d^{2}}
But(d^('2))/(2d^(2)) <= (1)/(2)\frac{d^{\prime 2}}{2 d^{2}} \leqq \frac{1}{2}, equality being possible only ifd^(')=dd^{\prime}=d. It follows that the maximum of (11) is equal to(1)/(2)\frac{1}{2}and this maximum is reached only ifbar(GA)\overline{\mathrm{G} A}is constant. For this, we can easily see that it is necessary and sufficient thatGamma\Gammabe a circumference and let P be on this circumference. We therefore finally have the
Theorem 13. IfGamma\Gammais a circumference and P a point on the plane of this circumference, we have
{:(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*}
P_(1)\mathrm{P}_{1}being the point ofGamma\Gammathe furthest from P.
IfGamma\Gammais a continuous curve different from a circumference there exists at least one point P of the plane for which inequality (12) is not true.
IfGamma\Gammais 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)}D. Pompeiu. "An identity between complex numbers and a theorem of elementary geometry." Bulletin of Mathematics and Physics, 6, 6-7 (1936).
- ^(2){ }^{2}) D. Pompeiu "Geometry and imaginaries: demonstration of some elementary theorems". Bulletin of Mathematics and Physics, 11,
(1941). - ^(3){ }^{3}) In reality the property is also true forn=2n=2, but in this case it is obvious.