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László Fejes Tóth was a Hungarian mathematician who specialized in ...
In elementary geometry, a polygon is a plane figure that is bounded...
Toth's proof starts with a construction of the circumcircle $C$ of ...
Now for $P_n$ we can have vertices lying inside or outside $C$. ...
Since $P_n$ might expand outside of $C$ it is obvious that $P_n \ge...
MATHEMATICAL NOTES
EDITED BY E. F. BECKENBACIE,
University
of California
Material for this department should be sent directly to E. F. Beckenbach, University of
California,
Los
Angeles
24,
California.
NEW PROOF OF A MINIMUM PROPERTY OF THE REGULAR n-GON
L.
F. ToTH, Budapest, Hungary
J. Kurschik gives in his paper
Ober
dem Kreis
ein-
und urmgeschriebene
Vielecke*
among others
a
complete and entirely elementary geometrical proof
of the well known fact according to which the regular n-gon Pn has
a
minimal
area among all n-gons
P
circumscribed about a circle c. In this proof
P.
is
carried,
after
a
dismemberment
and
a suitable reassembly,
in n
-I
steps
into
Pn
so
that
the
area increases at
every
step.
In
this note we give an extremely simple proof,t which appears
to
be new,
showing immediately that if Pn is not regular, then P,
,,,
where
the area
is
denoted by the same symbol as the domain.
Consider the circle C circumscribed about P. We
show
that already
for
the
part Pn- C of
P.
lying in C we have
Pn-C
>
Pn.
We
have Pn
C=C-ns+(S1S2+S2s3+
**
+SnSI),
where
we
denote
by
S, S2,
S* *
,
sn
the circular sections of C cut off by the consecutive sides of
Pn,
and
by s the circular section of C cut off by a tangent to c. Hence
Pn*C
_ C - nS.
Then
P7 >!
Pn
C
>
C
-
ns
=Pn.
Equality holds
in
P.
r!
P.C
resp. in
PnC
?
Pn
only if no vertex
of
Pn
lies
in
the
outside
resp.
in
the
inside
of
C;
this
completes
the
proof.
BINOMIAL
COEFFICIENTS MODULO A PRIME
N. J. FINE,
University of
Pennsylvania
The
following
theorem,
although
given by
Lucas
in
his
Theorie
des
Nombres
(pp.
417-420),
does not
appear
to be as
widely
known as
it
deserves
to
be:
THEOREM 1.
Let p be
a prime,
and let
M
=
MO
+
MIP
+
M2p2
+
*
.
+
Mkpk
(O
<
Mr
<
P),
No
+ Np + N2p2+
**
+Nkpk
(O
5
Nr
<
P).
*
Mathematische Annalen 30 (1887), pp. 578-581.
t
As
P. Sz6sz
remarked [Bemerkung zu einer
Arbeit von K.
Ktirschik,
Matematikai
es
Fizikai Lapok
XLIV (1937), p. 167, note 3]
Kurschik's
proof is independent of the axiom
of
parallels.
This
advantage
is
preserved in the present
proof.
589
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Discussion

Toth's proof starts with a construction of the circumcircle $C$ of the regular n-gon $\overline{P}_n$ that circumscribes $c$. ![](http://i.imgur.com/Hcr6PS1.png?1) As can be seen in the image above, all the sides of $\overline{P}_n$ cut the same circular section $s$ of $C$ because each side is the same distance from the center $O$. As a consequence, the area of $\overline{P}_n=C-ns$. In elementary geometry, a polygon is a plane figure that is bounded by a finite chain of straight line segments closing in a loop. Here are a few examples of polygons. ![](https://upload.wikimedia.org/wikipedia/commons/1/1f/Assorted_polygons.svg) A regular polygon is a polygon where all angles are equal in measure and all sides have the same length. Regular polygons may be convex or star. ![](https://upload.wikimedia.org/wikipedia/commons/thumb/7/76/Regular_star_polygons.svg/512px-Regular_star_polygons.svg.png) In this proof regular polygon seems to denote a convex regular polygon. ($\overline{P_n} ≠ C - ns$ for a regular star $n$-gon) You are right Neven. László Fejes Tóth was a Hungarian mathematician who specialized in geometry. He proved that a lattice pattern is the most efficient way to pack centrally symmetric convex sets on the Euclidean plane. He also investigated the sphere packing problem. Tóth worked extensively with H.S.M. Coxeter and Paul Erdős, with whom he laid the foundations of discrete geometry. ![](https://upload.wikimedia.org/wikipedia/commons/a/a2/L%C3%A1szl%C3%B3_Fejes_T%C3%B3th-1958-TableTennis.png) @Ning no. Does this mean that any polygon can be constructed from a regular polygon by extending/shortening of one or more of its sides? Now for $P_n$ we can have vertices lying inside or outside $C$. ![](http://i.imgur.com/LBLaAnC.png?1) Since all sides of $P_n$ and $\overline{P}_n$ are tangents to $c$, each side of $P_n$ that cuts all the way through $C$ will cut from it a chunk of area $s$. For the vertices that occur in the ring between $c$ and $C$ we will extend both sides so that they cut all the way through $C$ as can be seen in the following image (the area of these chunks will be $s$ for the reason discussed above). We will also name the circular sections cut off by the consecutive sides of $P_n$ $s_1,s_2,...,s_n$. ![](http://i.imgur.com/Vw6aZ05.png?1) To calculate $P_n \cdot C$, the area of $P_n$ lying in $C$ we need to subtract from $C$ the areas cut by the sides of $P_n$. For the sides meeting at a vertex outside of $C$ each side cuts a full $s$ from $C$. For the sides meeting at a vertex in the ring between $c$ and $C$ we have 2 segments $s_i$ and $s_{i+1}$ (each with area $s$) which overlap. If we subtract and area $s$ for each of $s_i$ we would be taking away their intersection $s_is_{i+1}$ twice, and so we need to correct this by adding the area of the intersection $s_is_{i+1}$ for each vertex. Finally $$ P_n \cdot C = C-ns+(s_1s_2 + s_2s_3 + ... + s_ns_1) $$ Note that if the vertex lies exactly on $C$, $s_is_{i+1}=0$ for that vertex. Since $P_n$ might expand outside of $C$ it is obvious that $P_n \geq P_n \cdot C = C-ns+(s_1s_2 + s_2s_3 + ... + s_ns_1)$. Also since the areas of the intersections $s_is_{i+1}$ are not negative, $(s_1s_2 + s_2s_3 + ... + s_ns_1)$ is not negative as well and so $P_n \geq P_n \cdot C \geq C-ns = \overline{P}_n$. We finally proved that the regular n-gon $\overline{P}_n$ has a minimal area among all n-gons $P_n$! Note that if $P_n$ has any vertex outside $C$ we have $P_n > P_n \cdot C$ and if $P_n$ has a vertex in the ring between $c$ and $C$ we have $(s_1s_2 + s_2s_3 + ... + s_ns_1)>0$ and thus $$ P_n \cdot C > C-ns = \overline{P}_n $$ The only case when the two areas are equal $P_n=\overline{P}_n $ is when all the vertices of $P_n$ are on C, which means that $P_n$ is also a regular n-gon.