Decomposition of convex figures into similar pieces
Discrete Comput Geom
Geometry Discrete & Computational
M. Laczkovich 0
0 Department of Analysis, E6tv6s Lor~nd University , Budapest, Mfizeum krt 6-8, Hungary 1088
If a convex plane figure P can be decomposed into finitely many nonoverlapping convex figures such that one of these pieces is similar to P, then P is a polygon. Also, if P can be decomposed into infinitely many nonoverlapping sets such that each of the pieces is similar to P, then P is a polygon. By a convex figure we mean a compact convex subset of the plane with nonempty interior. The following statement is mentioned, without proof, on p. 1 of [1].
-
I f a convex figure P is the union o f finitely many (but at least two) nonoverlapping
and congruent sets similar to P, then P is a polygon.
In this note we prove the following two generalizations of this result.
Theorem 1. Suppose that the convex figure P is the union o f finitely many (but at least
two) nonoverlapping convex figures such that one o f them is similar to P. Then P is a
polygon.
Theorem 2. Suppose that the convex figure P is the union o f infinitely many
nonoverlapping sets similar to P. Then P is a polygon.
W e use the following notation. The diameter, interior, closure, boundary, and
derived set (set of points of accumulation) of a set A are d e n o t e d by diam A, int A ,
cl A , OA, and A'. If A is convex, then we denote the set of extremal points of A by
E ( A ) . The isolated points of E ( A ) are called vertices, and the set of vertices of A is
d e n o t e d by V ( A ) . It is easy to see that p ~ V ( A ) if and only if p is the c o m m o n
endpoint of two nonparallel segments contained in OA. The angle of these segments
is d e n o t e d by aA(p). It is clear that V ( A ) is always countable and for every e > 0
the set {p ~ V(A): a.4(p) < 1r - e} is finite. Therefore, if V ( A ) is infinite and
P l , P2 . . . . is an e n u m e r a t i o n of V(A), then O~A(pn) ~ 7r.
W e denote the set ( E ( A ) ) ' by K ( A ) . Thus E ( A ) = V ( A ) u K(A). It is easy to
see that p ~ OA \ K ( A ) if and only if there is an open subarc I of 0A such that
p ~ I and I consists of at most two line segments. Clearly, K ( A ) = 0 if and only if
A is a polygon.
To prove T h e o r e m 1, suppose that P is a convex figure, P = U ~=1 Pi, where
P1 . . . . . Pn are nonoverlapping convex figures and P1 is similar to P. Then
n
OP1 n i n t P c U (P1 N
Pi)i=2
(
1
)
Indeed, every x E 3P l N int P is a p o i n t of accumulation of U in--2Pi. Thus x
P1 N P,' c P1 n Pi for at least one i _> 2. Since the sets P1 N Pi are line segments
(or points), it follows from (
1
) that OP1 n int P can be covered by finitely many
lines. Therefore the next lemma yields T h e o r e m 1.
L e m m a 1. Let P and P1 ~ P be similar convex figures such that OP1 n int P can be
covered by finitely many lines. Then P is a polygon.
Proof. First we note that E ( P 1) N int P is finite, as E ( P 0 intersects every line in
at most two points. Let 4, d e n o t e the similarity transformation mapping P onto P1.
W e show that 4,(K(P)) c K(P). Let p e K ( P ) and P l = 4,(P). Obviously, P l ~
K ( P 0 and hence Pl ~ E(P1)'. Since E(P~) n i n t P is finite, it follows that P l E
(E(P~) n OP)'. W e claim that Pl ~ K ( P ) . If this is not true, then p l E O P \ E ( P )
or p E V(P). In both cases there are two line segments, I and J, in c)P having P l as
a c o m m o n endpoint. Since p l ~ ( E ( P 1) n OP)', one of these line segment, say I,
contains at least two distinct elements q :/: p~ and r 4=P l of E ( P 0 n OP. Then Px,
q, and r are collinear points of OP1 and hence there is a line segment I ' c I N OP~
containing Pl, q, and r. However, in this case q and r cannot be b o t h extremal
points of P1, a contradiction. T h e r e f o r e Pl ~ K ( P ) as we stated.
Consequently, K ( P ) is a compact subset of OP that is m a p p e d into itself by the
contraction 4,. If P is not a polygon, then K ( P ) 4=~ and it follows that 4' has a fixed
point Po ~ K(P). Let S denote the angular domain between the half-tangents of P
at P0 and containing P (S is a half-plane if P has a tangent at P0)- Since
4 ' ( P ) = P1 c P, it is easy to see that ~ b ( S ) = S. This implies that either 4' is a
h o m o t h e t i c transformation or it is a homothetic transformation followed by a
reflection. In both cases, q, 4'2(q), and P0 are collinear for every point q.
Since Po ~ K ( P ) , we have P0 = q~(Po) ~ K(P1) = E(P1)'. Since E ( P 1) N int P is
finite, it follows that P0 E ( E ( P 0 n 0P)'. This implies, in particular, that E ( P x) n
0 P is infinite. Let ql ~ E(P1) n OP be arbitrary, and put q = 4 ' - 1 ( q l ) and q2 =
4'(q0. Then q ~ E ( P ) c OP and q2 E o~P1, since ql ~ OP. As we r e m a r k e d above,
the points q, q2, and P0 are collinear. Now we distinguish between two cases. If
q2 E 3P, then q, q2, and P0 are collinear points of OP and thus the segment I with
endpoints q and P0 belongs to OP. In this case qa = 4' 1(q2) @ ~b - l ( I ) ' Clear (...truncated)