PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 45, Number 2, August 1974
EXHAUSTINGAN AREA WITH DISCS
P. ERDOS AND D. J. NEWMAN
ABSTRACT.
The
may be covered
the unit
square
on how much
It is true,
as our given
is,
Rin)
so.
remarked
disparate
this situation
cise
result
then
give
of a given
matters
estimates
area
by choosing
(upper
and lower)
square
can have
by n discs.
not trivial,
by the removal
« disjoint
that
how much
We simplify
that
the unit
of open disjoint
discs
al-
in its inte-
how fast?
discs
of the areas
interior
one must leave
to S, the unit square.
Rin)-—* 0 and our job is to estimate
The straightforward
enormously
and
to be the minimum
when one removes
already
is, just
discs?
area,
perhaps
exhausted
Our question
So define
question
disjoint
of it may be so covered
although
most all its area
rior.
general
by interior
estimates
(e~c
somewhat
give
°ß n and
by showing
upper
n~cn
and lower
just
how it does
bounds
respectively)
that the truth
behind
We have
which
are
and we try to repair
is around
n~
. The pre-
is
Theorem.
Proof.
c.«~
>/?(«)>
We turn first
c7n~
, c,
to the lower
bound
and
c2 positive
and need
constants.
the following
elemen-
tary lemma.
Lemma
the smaller
1. Consider
region
has
a disc
height
cut by a chord of length
h and the larger
one has
c and suppose
height
that
H, We then
have
(a) The area
of the smaller
(b) The area
of the larger
To prove
which
(a) consider
is tangent'to
the slope
is bounded
the ellipse
of the chord
else
of the circle
Received
region
is bounded
which
by in/4)ch.
by in/2)H
has the chord
the minor arc at its midpoint.
meet at the endpoints
they meet nowhere
region
(4 points
is smaller
by the editors
and doubly
October
as major axis
The ellipse
at the tangency
are all they can share).
than
.
oo, the slope
and the circle
point
and so the
5, 1973.
(1970). Primary 52A45, 50B20.
Key words
area
Covering,
and so
At the endpoints
of the ellipse,
AMS(MOS)subject classifications
and phrases,
and
exhaustion,
discs.
Copyright © 1974, American Mathematical Society
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305
306
P. ERDÖS AND D. J. NEWMAN
entire
circular
half ellipse
segment
must stay
is exactly
in/4)ch,
As for (b) simply
construct
chord
and radius
tains
H. This
the larger
Now let
within
the half ellipse.
however,
with center
to our original
region.
Its area
is exactly
in/2)H
the strip,
of height
W, along
any collection
ters
If D is any one of them then part
E.
of disjoint
discs
at midpoint
circle
F denote
S, and consider
outside
of the
and (a) is proved.
the semicircle
is tangent
The area
of the
and hence
, however,
and (b) follows.
the bottom
in S which
con-
of our square
have their
(a) of the lemma
cen-
insures
that Area (D O E) < in/4)W • (chord of D on the boundary of E). These chords
are disjoint,
Next
however,
look at a disc,
our lemma applies
If we have
can cover
« disjoint
discs
in E is bounded
area
bound is established
at each
of S but this
in S, then,
by (77/4)^
area
is bounded by (n/4)W • 1.
inside
the total
amount
+ « ■ (n/2)W
is at least
E. Here part (b) of
O F) < in/2)W
so that
, choosing
.
of area
there
they
remains
in
W = (4 — n)/4rrn
((4 - 77) /3277) ■ n~
and our lower
with C2 = i4 - rr) /32/7 K .0074.
As for our upper
(It is not quite
that Area(D'
of (1 - n/4)W - nin/2)W
that the uncovered
moving,
area's
D , in S with its center
and we conclude
E an uncovered
shows
and so the sum of these
bound
turn,
we remove
the largest
trivial
that this
is known
« discs
disc
which
process
by the simple
is contained
first
(almost
noticed
of re-
in the residual
even exhausts
and was perhaps
device
set.
all) the area
by A. Beck.)
We need
the following
Lemma
2. Suppose
cave
arcs
area
by a and perimeter
therein
(its
(arcs
that a region
whose
interior
every
chord
has
by I. Suppose
lies
its boundary
is disjoint
in the region
that
composed
from the region)
a disc
of radius
of 3 conand denote
its
r is inscribed
and its circumference
meets
all 3 arcs).
Then r • l>2a.
The simple
y is the boundary,
the center
proof
is based
on the integral
~t is the radius
of our disc,
vector
n is the unit
formula
a = l/2 C r -nds
from the origin
outward
normal
and
0, which
where
we take as
ds represents
arc
length.
If we focus
support
serve
that,
first
of all that
at any point
so that
case,
on any one of our concave
from 0 then the two points
r • « = distance
then,
for any point
of A , the line
from
arcs
of contact
A, and draw the two lines
define
a subarc,
in A — A , ~r • « < 0. Secondly
of support,
L, separates
0 to L < distance
from
r • « < r and so we have
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of
A . We obwe note
the origin
from A
0 to A = r. In either
307
EXHAUSTING AN AREA WITH DISCS
a=—
t • nds < — I ds = —
2 Jy
-2 Jy
2
as required.
This
shows.
is clearly
a best
In our application,
circles
calculations
is then
would
allow
Now let us return
number
then,
that
k V —
to be removed).
(1) r.L
V
rather
k
over
, where,
and area
being
e.g.
than
tangent
that
only
the biggest
set
triangles"
discs.
and perimeter
v ir,
Summing
>2A
perimeter
k discs
the areas
for all
"circular
inscribed
will be mutually
r • I
«
. The
and so we omit the details.
the residual
a
/
n~
however,
of removing
call
total
us to obtain
(2& + 2) of disjoint
> 2a
arcs
of the triangle
It can be shown
to our strategy
will be one of their
and
the three
at the &th stage
disc
r.l
as the example
possible.
are very complicated
It is clear,
result
however,
and improvement
2. 2.8a and this
have
possible
disc
is composed
possible.
of a finite
and that the next chosen
Our lemma applies!
of these
the radius
circular
Indeed
triangles
of the next,
if we
then we
or ik + l)th,
disc
v gives
of course,
of the residual
L
and
A
are, respectively,
set
(the set remaining
after
the
the first
have been removed).
It is also
clear,
from the very definition
of our strategy,
that
(2)^-l>V
(3) Lfe_j = Lk-2mk_x,
The trick
is to notice
and
that these
entail
the nonincrease
of the quantity
yJr~kLk+ 2Ak/\]r~^. We have, namely, by (3) and (4),
(V^m.,
+ 2A,_1/v^71)
- (VTkLk+ 2Afe/v^)
= (V^-i/^)(Vr—rkLk-2Ak)
and by (2) the first
in turn,
> 0 by (1).
factor
is > 0 while
The montonicity
the second
is thereby
is > r L
established
- 2A
which
and in particular
we have
"KLn + 2VVT-< V^L0 + 2A0 / vr0 = 4V2.
Again,
quality
by (2) and
(3), we obtain
Ln > « . 2t77
so that
becomes
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the above
ine-
is,
P. ERDÖS AND D. J. NEWMAN
308
If, finally,
we apply
the arithmetic-geometric
inequality
to the left-hand
side
we find that it is
> 4[(r3/277«)(r~1/2A
/3)3]1/4 = 4(3z7«A 723)1/4/3
72
72
72
—
Combining
gives
A
< 3/\¡4nn
and our result
is established
^
with
C. =
3/\[4~n & 1.29.
HUNGARIANACADEMY OF SCIENCES, BUDAPEST, HUNGARY
DEPARTMENT OF MATHEMATICS, YESHIVA UNIVERSITY, NEW YORK, NEW YORK
10033
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