Rev. Acad. Canar. Cienc., XX (Núms. 1-2), 113-118 (2008) (publicado en septiembre de 2009)
A FORGOTTEN LITTLE CHAPTER ON ISOPERIMETRIC
INEQUALITIES: ON THE FRACTION OF A CONVEX
AND CLOSED PLANE AREA LYING OUTSIDE A CIRCLE
WITH WHICH IT SHARES A DIAMETER
José M. Pacheco
Departamento de Matemáticas. Universidad de Las Palmas de Gran Canaria
Abstract
Often sorne interesting or simply curious points are left out when developing a theory. lt seems that
one of them is the existence of an upper bound for the fraction of area of a convex and closed plane
area lying outside a circle with which it shares a diameter, a problem stemming from the theory of
isoperimetric inequalities. In this paper such a bound is constructed and shown to be attained for a
particular area. lt is also shown that convexity is a necessary condition in order to avoid the whole
area lying outside the circle.
Key Words: convex plane area, isoperimetric inequality.
AMS Classification numbers: 52AIO, 52A38.
lntroduction
Possibly one of the oldest extrema) problems is to find a set in Euclidean space with
given surface area and enclosing maxirnum volume. There exist considerable
differences in the rnathematical treatment of the cases n = 2 and general n, as shown in
the classical reference [2], where the general setting directly invites the reader to the
realrn of geometric measure theory. These problems pervade mathematical activity and
many mathernaticians have dealt with them to different depth degrees. As the
motivating exarnple for this paper, on reading the book Littlewood's Miscellany [l] one
finds in page 32 the following observation:
"An isoperimetrical problem: an area of (greatest) diameter not greater than 1 is at most
-t 1l" "
Littlewood's "greatest diameter" is now better known as the diameter d ofthe plane area,
defined as
d = sup{dist(X,Y)JX, Y E boundary},
and without loss of generality, we can suppose in Littlewood's remark the area to be
convex and bounded by a continuous closed curve. Indeed non-convexity would only
arnount to reducing the enclosed area, thus enhancing the inequality (see figure 1).
1 Postal address: Campus de Tafira Baja, 35017 LAS PALMAS, Spain. E-mail: pacheco@dma.ulpgc.es
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Figure 1: Nonconvexity implies area loss.
Littlewood's argument is the following (figure in p. 33 of[J]), see figure 2:
/
p
q
O ·
Q
Figure 2: A recreation of the figure in [1], p. 33.
1l ., ., 2 • ') where OP = p(B), OQ = p(B- - ) , and Qp- + OQ- = PQ ~ dwm - ~l . Therefore
2
The bound -tn is attained for the unit diameter circle, and the classical isoperimetric
problem. was to actually prove that the circle is the only plane area having this property.
A nice proofbased on Fourier expansion techniques can be found in [4], pp. 181-187,
anda standard proofis offered in [2], pp. 104 ff.
From an elementary viewpoint there is something counterintuitive in the geometrical
presentation of this inequality because, on a first and crude approximation, the layman
could make a very nalve remark: Draw an area with unit diameter, and then a circle
sharing a diameter with it. It "seems obvious" that the whole area is contained in the
circle, so the inequality would be an immediate one (see figure 3, left).
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/
Figure 3: The na'lve idea and the counterexample
Of course this is erroneous, as is readily shown by considering an isosceles triangle
KLM whose legs KL and KM are longer than the basis LM This is a convex area
whose diarneter equals one of the legs, say KL. Now, Jet a circle with KL as a diarneter
be drawn: There is sorne portion of triangle area in the neighbourhood of comer M that
lies outside the circle (see figure 3, right)
A natural question and its answer
After the above observations, a rather natural question irnrnediately arises: Is there any
upper bound to thefraction of area -of a given convex plane area- lying outside a circle
which shares with it a diameter?
Figure 4: lllustrating the procedure.
The classical references [2] and [3] do not rnention this topic, and a rather thorough
Internet search did not provide direct results, so an atternpt to fill this little gap will now
be rnade. In what follows the diarneter will be d = l, therefore the area is bounded
by -;\-n. To start, considera unit diarneter circle (see figure 4 to follow the discussion),
and Jet K and L be the endpoints of sorne diarneter thereof.
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We shall build a convex area sharing the diameter KL with the given circle, and Jet X be
any point in the plane. lt is evident that both conditions
dist(X,K) :51, dist(X,L) :51
must be satisfied in order that X be either an interior or a boundary point of the sought
area. Therefore the area is enclosed in the figure defined by two intersecting circle ares
centred at K and L and having unit radius. They define two points C and D outside the
circle and a curvilinear figure KDLC. Of coursedist(C,D) > 1, so KDLC cannot be a
candidate for solving our problem. Therefore we restrict our attention to vertex C (of
course, its symmetrical point D could be employed as well) and observe that no point in
the boundary of the sought area can be more than l apart from it. Although it seems
rather natural to draw the unit radius circle are centred at C and joining K and L to
obtain a curvilinear triangle KLC as a more appropriate candidate, it is clear that the
diameter KL <loes a better job, and we claim that the mixed triangle KLC mix is the
solution to our problem. lncidentally, the curvilinear triangle KLC is called "the
Reuleaux triangle" (see Appendix).
A measure ofhow much area líes outside the circle can now be defined: Simply, it is the
ratio between the area outside the circle and the total area just constructed.
area outside the circle μ = :5 1.
total area
It is an easy task to compute the total area ofthe mixed triangleKLCmix:
" ..Jj 81' - 6..J3
total area(KLCmix) = 3--4- = --2-4--
and the fraction outside the circle is:
exterw. r area( KLCm ix) = total area( KLC,,,ix ) - -" = -51-' - -6..J-3
Therefore, the following value is obtained:
μ = 51' - 6..J3 = 0.36
8Jr-6..J3
8 24
i.e. the maximum fraction of area lying outside the circle amounts to approximately
36% of the total area.
ln order to show optimality ofthis result, Jet us consider adding sorne area to KLCmix by
modifying the boundary curves. This cannot be done by changing the curved side CK
( or LC) into another convex curve joining both points for this would imply the existence
of sorne boundary point at a distance from L (respectively, K) larger than 1, thus
contradicting the fact that the sought figure must have unit diameter. Sorne area can be
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added below the diameter KL preserving both convexity and unit diameter, but in this
case the denominator in the definition of the measure would increase, thus reducing the
value ofμ.
The construction also shows that convexity is a necessary condition for the bound to be
a valid one. lt is enough to observe (see figure 5) that the area outside the circle is a
non-convex figure sharing the unit diameter KL with the circle, but 100% of it lies
outside the circle, (and indeed is less than -;\- Jr ).
Figure 5: Convexity is a necessary condition.
References
[1] Bollobás B (ed) (1997) Littlewood's miscellany, Cambridge University Press,
Cambridge UK.
[2] Eggleston H (1958) Convexity, Cambridge University Press, Cambridge UK.
[3] Mayer A (1935) Der Inhalt der Gleichdicke: Abschatzungen fur ebene Gleichdicke,
Mathematische Annalen, 110 (97-127).
[ 4] Nahin P (2006) Dr. Euler 's fabulous formula, Princeton University Press, New York.
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Appendix on the Reuleaux triangle
The Reuleaux triangle is a rather familiar curvilinear triangle (see figure A 1) obtained
by drawing three circle ares centred at the vertices of an equilateral triangle with a
radius equal to the side, it is indeed convex, and its area is the mínimum of al i possible
figures of constant width having the same diameter d, a result known as the BlaschkeLebesgue
Theorem. These figures share the common length 7l x d (a Theorem by
Barbier [3]), so the circle and the Reuleaux triangle are extrema! curves -in the sense of
enclosed area- with this property (see again [3]). Reuleaux triangles have been
employed for decorative purposes (see figure A2) and in technological applications,
such as the Wankel rotary engines. See also Chapter 7 in [2].
Figure A 1: The Reuleaux triangle
Figure A2: Reuleaux triangles in Camden Town, London (photograph by the author).
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