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172
Articole şi Note Matematice
Bibliografie
[1] D. Andrica, Asupra unor şiruri care au mulţimile punctelor limită intervale, Gazeta Matematică nr. 11/1979, pp. 404-406.
[2] V. Nicula, Asupra unor şiruri care au limita ∞, Gazeta Matematică
nr. 3/1995, pp. 97-101.
[3] G. Pólya, G. Szegö, Problems and theorems in analysis. Vol. l, SpringerVerlag, Berlin, 1998.
[4] D. Tănasie, Asupra unor clase de şiruri, Gazeta Matematică nr. 6/1982,
pp. 207-209.
[5] A. Vernescu, Asupra unor subserii ale seriei armonice, Gazeta Matematică nr. 2/2006, pp. 59-62.
A SIMPLE PROOF OF THE ERDÖS-MORDELL
INEQUALITY FOR POLYGONS
IN N-DIMENSIONAL SPACES
Marian Dincă1)
Abstract. The purpose of this article is a new proof of the generalization
of the Erdös-Mordell inequality
Keywords: Erdös-Mordell inequality
MSC : 51M16
We start with the following:
Theorem 1. For every xk ≥ 0, 1 ≤ k ≤ n, xn+1 = x1 and αk ∈ (0, π)
n
αk = π, the following inequality holds:
with
k=1
n
n
k=1
k=1
π 2 xk ≥
xk xk+1 cos αk .
cos
n
(1)
Proof. Let us consider, in the plane with the origin O(0; 0), the points
Pk (ak ; bk ), 1 ≤ k ≤ n such that:
(Pk OPk+1 ) = αk ,
and denote
n−1
1 ≤ k ≤ n − 1,
αk = π − αn . Denote:
k=1
xk = OPk =
1) Profesor, Şcoala nr. 1, Pantelimon, Ilfov.
a2k + b2k .
173
M. Dincă, Proof of the Erdös-Mordell inequality
By using the cosine theorem in the triangle Pk OPk+1 , 1 ≤ k ≤ n − 1
and in triangle P1 OPn , we obtain:
OPk 2 + OPk+1 2 − Pk Pk+1 2
=
2
1 2
ak + b2k + a2k+1 + b2k+1 − (ak − ak+1 )2 − (bk − bk+1 )2 = ak ak+1 +bk bk+1 ,
=
2
for k = 1, 2, . . . , n − 1 and
n−1 n−1
αk = −x1 xn cos
αk =
x1 xn cos αn = x1 xn cos π −
xk xk+1 cos αk =
k=1
=−
OP1
2
+ OPn
2
Then:
2
= cos
π
n
n
n
k=1
a2k + b2k −
k=1
n−2
1
k=1
(k+1)π
2 sin kπ
n sin
n
n−2
= −a1 an − b1 bn .
π 2 xk −
xk xk+1 cos αk =
n
k=1
=
− P1 Pn
n
cos
k=1
2
(ak ak+1 + bk bk+1 ) + a1 an + b1 bn =
k=1
kπ
π
(k + 1)π
ak − sin
ak+1 + sin an
sin
n
n
n
1
n−1
2
(k + 1)π
kπ
π
sin
bk − sin
bk+1 + sin bn
n
n
n
+
2
. (∗)
(k + 1)π
kπ
sin
2 sin
n
n
Let A1 , A2 , ..., An be the vertices of a convex n-gon and let P be an
internal point. Let Rk = P Ak and let rk be the distance from P to√the side
Ak Ak+1 . We consider An+1 = A1 . From the inequality (1) for xk = Rk , we
obtain:
n
n n
n
π
Rk ≥
Rk Rk+1 cos αk ≥
wk ≥
rk ,
cos
n
+
k=1
k=1
k=1
k=1
k=1
where wk is the length of the segment of the angle bisector Ak P Ak+1 . In
consequence we proved the well-known Erdös-Mordell inequality:
n
n
k=1
k=1
π
Rk ≥
rk .
cos
n
Now we prove the identity:
cos
n
n−1
k=1
k=1
π 2
ak −
ak ak+1 + a1 an =
n
(2)
174
=
Articole şi Note Matematice
n−2
1
(k + 1)π
kπ
π
sin
ak − sin
ak+1 + sin an
n
n
n
2
(3)
(k + 1)π
kπ
sin
2 sin
n
n
by comparing the coefficients of a2k and ak ak+1 . For example, the coefficient
of a2k is:
k=1
(k − 1)π
kπ
π
(k + 1)π
sin
2 sin
cos
n
n
n
n = cos π
+
=
kπ
kπ
kπ
n
2 sin
2 sin
2 sin
n
n
n
sin
for k = 2, 3, . . . , n − 2 and the coefficient of a21 is:
2π
n = cos π ,
π
n
2 sin
n
sin
the coefficient of a2n−1 is:
2π
(n − 2)π
sin
n
n = cos π
=
π
(n − 1)π
n
2 sin
2 sin
n
n
sin
and the coeficient of a2n is:
π
π
n−2
sin kπ
(k + 1)π
n
n
=
cot
− cot
=
(k + 1)π
kπ
2
n
n
k=1 2 sin
k=1
sin
n
n
n−2
sin2
π n cot π − cot (n − 1)π = cos π .
2
n
n
n
sin
=
The coefficient of ak ak+1 is:
kπ
(k + 1)π
sin
n
n = −1
(k + 1)π
kπ
sin
2 sin
n
n
−2 sin
for k = 1, 2, . . . , n − 2 and the coefficient of an−1 an is:
π
π
(n − 2)π
− sin
sin
n
n
n
=
π = −1.
(n − 1)π
(n − 2)π
sin
sin
2 sin
n
n
n
−2 sin
175
M. Dincă, Proof of the Erdös-Mordell inequality
The coefficient of a1 an is:
π
2π
sin
n
n =1
π
2π
2 sin sin
n
n
and the coeficient of ak an for k = 2, 3, . . . , n − 2 is:
2 sin
π
π
(k + 1)π
kπ
n−3
sin
sin
2 sin
n
n −
n
n
=
(k + 1)π
(k + 1)π
kπ
kπ
k=2 2 sin
k=1 2 sin
sin
sin
n
n
n
n
π
π
n−2
n−3
sin
sin
n
n
= 0.
−
=
kπ
(k + 1)π
k=2 sin
k=1 sin
n
n
Similarly we can prove the identity for variabile bk :
n−2
2 sin
cos
n−2
1
n
n−1
k=1
k=1
π 2 bk −
bk bk+1 + b1 bn =
n
(k + 1)π
kπ
π
sin
bk − sin
bk+1 + sin bn
n
n
n
2
. (4)
(k + 1)π
kπ
k=1 2 sin
sin
n
n
Now the identity (∗) follows by summing the relations (3) and (4).
We prove now the following generalization.
Theorem 2. Suppose x = (x1, x2 , . . . , xN ) and y = (y1 , y2 , . . . , yN ) are
vectors in RN , endowed with the inner product
=
x, y = x y cos θ =
N
xk yk ,
k=1
where θ is the angle between the vectors x and y, while
x = x21 + x22 + ... + x2N
is the corresponding norm. Then the following inequality of Erdös-Mordell
type holds:
n
n
n
π
Xk 2 ≥
Xk , Xk+1 =
Xk Xk+1 cos θk ,
cos
n
k=1
k=1
k=1
where Xk = (x1k , x2k , . . . , xN k ), k = 1, 2, ..., n and
n
k=1
θk = π.
When P is an exterior point, we have the following form of the ErdösMordell inequality for convex polygons:
176
Articole şi Note Matematice
Theorem 3. Let A1 A2 ...An be a convex polygon, P be an exterior
point and Ak P Ak+1 = αk , k = 1, ..., n. Let max(Ai P Aj ) = α, where
n
αk = 2α < 2π and
i = j ∈ {1, 2, ...n, } and
k=1
rk ≤ wk ≤
Then:
Rk Rk+1 cos
αk
.
2
n
n
n
n
k=1
k=1
k=1
k=1
α
π
cos
Rk −
rk ≥ cos
Rk −
rk ≥
n
n
n
n
n
n
π
αk
π
Rk −
wk ≥ cos
Rk −
Rk Rk+1 cos .
≥ cos
n
n
2
k=1
k=1
k=1
k=1
Theorem 4. Let A1 , A2 , ..., Ap be p arbitrary points in the N-dimensional euclidian space RN , M an arbitrary point in the same space and
Ak M Ak+1 = αk , k = 1, ..., p. Let max(Ai M Aj ) = α, i = j ∈ {1, 2, ...p}
p
αk = 2α < 2π. We consider Ap+1 = A1 . Then we obtain the following
and
k=1
inequality:
cos
p
p
p
p
π
α
Rk −
rk ≥ cos
Rk −
rk ≥
p
p
k=1
≥ cos
π
p
p
k=1
Rk −
p
k=1
k=1
wk ≥ cos
k=1
π
p
p
Rk −
k=1
p
k=1
k=1
Rk Rk+1 cos
αk
≥ 0,
2
where Rk is the norm of Ak M , rk is the distance from M to the segment
Ak Ak+1 , k ∈ {1, 2, ..., p} and wk is the length of the segment of the angle
bisector Ak P Ak+1 .
References
[1] M. Dincă, Generalizarea inegalităţii Erdös-Mordell, G. M.-B nr.7-8
(1998) 269-273.
[2] M. Dincă, Generalization of the inequality of Erdös-Mordell, Octogon
Math. Magazine, 1 (1999), 143-147, (in romanian).
[3] M. Dincă, A refinement of Erdös-Mordell’s inequality, Octogon Math.
Magazine, 10 (2002), 77-86 (in romanian).
[4] H. G. Eggleston, A triangle inequality, Math. Gaz., 42 (1958), 54-55.
[5] P. Erdös, Problem 3740, Amer. Math. Monthly, 42 (1937), 396.
[6] F. Faruk, Abi-Khuzam, A trigonometric inequality and its geometric
applications, Mathematical Inequalities & Application, vol. 3 (2003),
437-442.
[7] A. Florian, Zu einem Satz von P. Erdös, Elem. Math. 13 (1958), 55-58.
L. Ţurea, Aproximări raţionale. Numere Farey. Fracţii continue (II)
177
[8] S. Gueron, I. Shafrir, A weighted Erdös-Mordell inequality for polygon,
American Mathematical Monthly, 112 (2005), 257-263.
[9] D. K. Kazarinoff, A simple proof of the Erdös-Mordell inequality for
triangles, Michigan Math. J., 4 (1957), 97-98.
[10] H. C. Lenhard, Verallgemeinerung und Verschärfung der Erdös-Mordellschen Ungleichung für Polygone, Arh. Math. vol. XII (1961), 311-314.
[11] D. S. Mitrinović, J. E. Pečarić, On the Erdös-Mordell’s inequality for a
Polygon, J. College Arts Sci, Chiba Univ., B-19 (1986), 3-6.
[12] L. J. Mordell, D. F. Barrow, Solution of Problem 3740, American Mathematical Monthly, 44 (1937), 252-254.
[13] N. Ozeki, On Paul Erdös’s inequality for the triangle, J. College Arts
Sci., Chiba Univ., 2 (1957), 247-250.
[14] G. R. Veldkamp, The Erdös-Mordell inequality, Nieuw Tijdshr. Wisk. 45
(1957/58), 193-196.
[15] S. Wu, L. Debnath, Generalization of the Wolstenholme cyclic inequality
and its application, Computers & Mathematics with Application, vol. 53,
Issue 1, (2007), 104-114.
APROXIMĂRI RAŢIONALE. NUMERE FAREY.
FRACŢII CONTINUE (II)1)
de Lucian Ţurea2)
Abstract. This is the second part of an article about various approximations of real numbers using certain classes of rationals.
Keywords: rational approximation, Liouville numbers, Farey numbers,
continuous fractions, Kronecker density theorem.
MSC : 11J68, 11J70.
4. Cele mai bune aproximări. Fracţii continue.
La sfârşitul paragrafului anterior, după ce am studiat problema existenţei unor aproximări cât mai bune, am pus şi ı̂ntrebarea naturală despre cum
arată aceste aproximări. Pentru cazul lui ϕ am ı̂ntâlnit şi nişte coincidenţe
interesante. Un alt exemplu ı̂n care este interesant de studiat forma fracţiilor
22
este o aproximare foarte bună a acestuia (eroare
este π. De exemplu,
7
de ordinul lui 10−3 ). Cel mai ,,simplu” număr raţional care ı̂l aproximează
333
179
. Alte aproximări foarte bune ale acestuia sunt
pe π mai bine este
57
106
355
−5
−7
(eroare de ordinul lui 10 ). Un alt lucru
(eroare de ordinul lui 10 ) şi
113
interesant este că, pentru 106 < q < 113, nicio fracţie nu ı̂l aproximează pe
1) Continuare din G. M. - B nr. 3/2009.
2) Student, Facultatea de Matematică şi Informatică, Universitatea din Bucureşti.