> jli
c 5jbjb!! bKKx'88Vd4e$ZZM$qp#r#r#r#r#r#r#,%R(## $ p# BTp# #T#) #T#5$0e$(#,( (T# z
z
Inversion of the "midpoints polygon" construction
by Branko Grnbaum
University of Washington, Box 354350, Seattle, WA 98195
email: grunbaum@math.washington.edu
Any given polygon leads naturally to the polygon that has vertices at midpoints of the sides of the original one. Although there is nothing particularly remarkable about this "midpoints polygon" construction, its repetition leads in the limit to surprising results. These were first investigated by Darboux [D], and later by Berlekamp et al. [B] and many others. The result is, roughly, that in most cases the limit is an affinelyregular polygon, that is, an affine image of a regular polygon. For the much more complicated precise results see, for example, Grnbaum [G] or Shephard [S].
Here we shall be interested in the inverse of the midpoints polygon construction. By this is meant the following.
Problem. Given a (cyclically understood) sequence of n points M1, M2, ... , Mn, the vertices of a polygon Q, find the ngon or ngons P = [V0, V1, ... , Vn1, Vn = V0] such that Mj is the midpoint of the edge [Vj1Vj] of P for all j = 1, 2, ... , n, or decide that there is no such ngon.
For n = 3 and n = 4 the problem has easy solutions, as indicated in Figure 1. For every triangle Q there is a unique solution, the triangle homothetic with ratio 2 with the triangle [M1M2M3], and with the same centroid. For quadrangles, a solution is not possible unless Q = [M1M2M3M4] is a parallelogram; in that case there are infinitely many solutions in fact, any point of the plane can be chosen as V0.
The source of these results is lost in the mists of the past. The first reference I have to additional development is the article of Lionnet [L], where the above problem was formulated; Lionnet added the restriction requiring that the inverse construction be
Figure 1. The unique solution of the inverse midpoint problem for a triangle, and two of the infinitely many solutions for a parallelogram.
carried out using only compasses but this is not relevant to the problem or its solution. Lionnet's problem was solved by Prouhet [P], and the same problem was solved, apparently independently, by Nagel [N]. Both authors first establish (or quote) the above results for n = 3, 4, and then proceed by imprecisely described constructions. Each of the authors gives for even n>4 a construction that involves an arbitrary segment centered at one of the points Mj, and then says that if the construction closes in a suitable way there are infinitely many solutions, otherwise there are none. This is not a satisfactory proof, since there are many arbitrary choices involved, and it is entirely possible that other choices would lead to different conclusions. For the odd case a similar construction is given, and it is asserted that it is always possible, and that it yields a unique polygon with the given midpoints of sides without any information about this polygon.
The two main shortcomings of these presentations are the absence of any indications of the dependence of the (unique) solution on the starting set of midpoints in the odd case, and of any characterization of the even ngons (with n > 4) for which the problem has a solution. As we shall see, all these questions can be resolved in a very simple and satisfactory manner by investigating what happens to the attempt of constructing the desired polygon without worrying (to begin with) about the outcome. In each case a basic step in the construction consists of taking a point V and finding its image M* under a 180 rotation about a chosen point M. Formally, we have V* = 2M V.
In order to formulate our results in a convenient way, we need some notation. As before, the starting polygon Q has vertices M1, M2, ... , Mn. An arbitrarily chosen point V0 is the start of a (finite or infinite) sequence P of points V0, V1, ... , Vj1, Vj ... obtained by successive application of the basic step to the appropriate point of Q. Specifically, Mj is the midpoint of the segment with endpoints Vj1, Vj; the subscripts of the Mj's are to be taken (mod n). When convenient and appropriate, we may consider P as a polygon or polygonal line.
From the expression for the basic step and the definition, it follows that
V1 = 2M1 V0
V2 = 2M2 (2M1 V0) = 2(M2 M1) + V0
V3 = 2M3 V2 = 2(M3 M2 + M1) V0
. . .
Vj = 2Mj Vj1 = 2(Mj Mj1 . . . M3 + M2 M1) + (1)jV0
Therefore,
Vj+n Vj = 2(Mn+j Mn+j1 + . . . + M3 M2 + M1) + (1)n+jV0
2((Mj Mj1 + . . . + M3 M2 + M1) + (1)jV0). (*)
Hence, for even n, we have
Vj+n Vj = 2(Mn+j Mn+j1 + . . . Mj+3 + Mj+2 Mj+1)
= 2(1)j ((Mn + . . . + M4 + M2) (Mn1 + . . . + M3 + M1))
= 2(1)j n (Ge Go)
where Ge and Go are the centroids of the even resp. odd labeled vertices of Q.
From this we conclude the following:
Theorem 1. For even n e" 4, a given Q and an arbitrary V0, all vectors Vj+n Vj have the same length, and alternate in direction. Hence, either Ge Go = 0 or Ge Go `" 0.
(i) If Ge Go = 0 then Vj+n Vj = 0, that is, the sequence of Vj's repeats after n steps. This means that P is an ngon, and an arbitrary V0 can serve as the starting vertex. This is illustrated in Figure 2.
(ii) If Ge Go `" 0 then for each j, the points Vj, Vj+n, Vj+2n, ..., form an infinite sequence of equidistant points, that is, an apeirogonal ray with starting point Vj. This is illustrated in Figure 3.
Corollary 2. The inverse midpoint construction leads for even n to an ngon if and only if the centroid of the evenlabeled points coincides with the centroid of the odd labeled points.
Hence for n = 4 we get back the result that the construction closes after one circuit around Q if and only if Q is a parallelogram. But already for each n e" 6 we find the apparently new result that the coincidence of the centroids is a necessary and sufficient condition for the closure of the construction after a single circuit around Q; this includes various moreorless symmetric polygons, but also some that have no particular symmetry (see Figure 2). In all other cases, that is, if Ge `" Go, repeated circuits around Q yield the families of apeirogonal rays described above shown in Fig. 3.
Figure 2. An illustration for n = 6 of the special case in which Ge = Go, and the inverse midpoint construction ends after one circuit around the ngon.
Figure 3. An illustration of the case n = 6. The edges of the starting polygon P and the lines carrying the points of the apeirogon rays and shown with heavy lines, the line determined by the centroids of the even and the odd vertices, and the path of the basic construction are shown with thin lines. To avoid clutter, the points on all lines except the first are indicated by numerals only.
In case Q is an ngon with odd n = 2k1, the arguments are similar but the conclusions are quite different. As before, we start with an arbitrary point V0, perform the sequence of basic steps. Then (*) yields
Vj+n Vj = 2(Mn+j Mn+j1 + . . . + M3 M2 + M1) + (1)n+jV0
2((Mj Mj1 + . . . + M3 M2 + M1) + (1)jV0)
= 2(Mn+j Mn+j1 + . . . Mj+3 + Mj+2 Mj+1) 2(1)jV0 .
Therefore
Vj+nVj = (1)j 2((Mn+. . .+M3+ M1) (Mn1 +. . .+ M4 + M2 + V0))
= (1)j 2k(Ge Go) = (1)j (Vn V0 ), (**)
where here Go is the centroid of the odd labeled vertices of Q, and Ge is the centroid of the set consisting of the even labeled vertices and the starting vertex V0. This implies:
Theorem 3. For odd n e" 3 all vectors Vj+n Vj have equal lengths and alternating directions. The case Vn = V0 (the special polygon) happens if and only if Ge Go, that is, if the starting point is
V0 = W0 = Mn Mn 1 + . . . + M3 M2 + M1.
On the other hand, if Vn `" V0 then
V2n V0 = V2n Vn + Vn V0
= (1)n (Vn V0 ) + (Vn V0 ) = 0.
This means that for odd n, regardless of the choice of V0, the sequence of Vj's ends after two circuits around Q, so P is a (2n)gon.
As is obvious both geometrically and calculationally, the midpoints Wj of the segments with endpoints V2n+j Vj are the vertices of the unique special ngon, regardless of the choice of V0.
The situation is illustrated in Figure 4 for n = 7.
Figure 4. An illustration of the construction for n = 7. The points Ge and Go, were calculated with respect to the starting point V0. From (**) it follows that the distance between V0 and V7 is n+1 = 8 times the distance between Ge and Go. The midpoints of the segments Vj+nVj are the vertices Wj of the exceptional polygon.
It is easy to confirm that the lengths of the sides Wj+1,Wj of the special polygon are uniquely determined by the centroids of even resp. odd labeled centroids of P different from Mj. This is illustrated for n = 7 in Figure 5 (and for n = 3 in Figure 1a).
References
[B] E. R. Berlekamp, E. N. Gilbert and F. W. Sinden, A polygon problem. Amer. Math. Monthly 72(1965), pp. 233241; reprinted in Selected Papers on Algebra, Mathematical Association of America, Washington, D.C., 1977.
[D] G. Darboux, Sur un problme de gometrie lmentaire. Bull. Sci. Math. (2)2(1878), pp. 298  304.
[G] B. Grnbaum, Lecture Notes on Modern Elementary Geometry. University of Washington, Spring 1997. Available at http://hdl.handle.net/1773/15592
Figure 5. The same situation as in Figure 4, but illustrating the fact that the vector Wj+1Wj is n1 = 6 times the vector Ge* Go*, where Ge* and Go* are the centroids of the even and odd labeled vertices of P other than Mj.
[L] Lionnet, Problme 63. Nouvelles annales de mathmatiques 1re sr. 2(1843), 416.
[N] Nagel, Ueber die Bestimmung der Vielecke durch die Halbierungspunkte ihrer Seiten. Archiv der Mathematik und Physik 53(1871), 378 381 + Figs. III, IV of Plate VII.
[P] A. Prouhet, Solution du Problme 63 (Tome II, p. 416). Nouvelles annales de mathmatiques 1re sr. 3(1844), 19 21.
[S] G. C. Shephard, Sequences of smoothed polygons. In: Discrete Geometry, In Honor of W. Kuperberg's 60th Birthday. A. Bezdek, ed. Dekker, New York 2003, pp. 407 430.
Page PAGE 3
GEOMBINATORICS 21(2012), 89 96
2}~pqa c k
<
?
@
A
:;CDEFGH۾ۦۦۦۦۦۦۦۦۦۦۦۦۦۦۦۦۦۦh<h1H*OJQJh<h15OJQJh1h<h16OJQJh<h1NHOJQJh<h1OJQJhrh1OJQJh<h15CJOJQJA2E}c
&*
,
.
tt$
88d,]8^8a$gd1$
88hd,x]8^8`ha$gd1$88hd,]8^8`ha$gd1$88d]8^8a$gd1$88]8^8a$$88d]8^8a$$88]8^8a$gd145#$*
+
,

.
t
u
:;wx9:RS[\EF78³h<h16OJQJh=
h1CJNHOJQJh=
h1CJOJQJjh^dh1OJQJUh^dh1OJQJjh^dh1OJQJUh<h1NHOJQJh<h1H*OJQJh<h1OJQJ6.
+U@L$88dx]8^8a$gd1$
88hd,]8^8`ha$gd1$
88hd,x]8^8`ha$gd1$
88d,]8^8a$gd1$
88d,]8^8a$gd1%(+,FGJK
)*./34<=>?MPTUh<h1H*OJQJh<h1NHOJQJh<h1H*OJQJh<h1OJQJRU[^bhstxy}~
"&)45679:=>HIMNTXcdhiuvxyh<h1CJOJQJhk&h1CJOJQJhk&h1H*OJQJh<h1CJ OJQJh<h1H*OJQJh<h1H*OJQJh<h1OJQJC,m07v#z#x$88]8^8a$gd1$
88hd,x]8^8`ha$gd1$
88hd,x]8^8`ha$gd1$88d,x]8^8a$gd1$88dx]8^8a$gd1$88d]8^8a$gd1
y{BDbhpr "@BJLbhpr68JL68>DJR7Cnoz{$!&!""""""6#8#h<h1H*OJQJhk&h1H*OJQJhk&h15OJQJh<h1NHOJQJh=
h1H*OJQJh<h1OJQJH8#v#x##$$$$$V$W$X$$$e%f%%z&{&&&&&&&&&&&&&&&&&&&'''''''' '!'%'&'.'/'0'1'2'D'G'K'״h<h1CJ OJQJh<h1H*OJQJh<h1H*OJQJjh1Uh=
h1CJNHOJQJhTKh1H*OJQJh=
h1CJOJQJ!jX
h=
h1CJOJQJUh<h1OJQJ8z#V$X$%&&5''''&()~~$88d]8^8a$gd1$88hdx]8^8`ha$gd1$88dx]8^8a$gd1$88hd]8^8`ha$gd1$
88d,]8^8a$gd1$88]8^8a$gd1K'Q'\'_'c'f'j'm'v'w'x'y'''''''''''''''''''''''''''''''''''''>(@(((`)b)l)n))))))))誟hk&h1OJQJhk&h15OJQJh<h1NHOJQJh=
h1H*OJQJh<h1CJ OJQJh<h1H*OJQJh<h1OJQJh<h1H*OJQJ>)d*f*n*p*****&+(+0+2+:+<+D+L+b+d+l+n+v+x+++++++++++++++++
,,,,,,,,,, ,$,%,f,g,n,o,z,{,,,,,,,h<h1NHOJQJhk&h1OJQJh<h1H*OJQJhk&h1H*OJQJhTKh1H*OJQJh=
h1H*OJQJh<h1H*OJQJh<h1OJQJ?)$+++,.,,t..//t$
,88d,x]8^8a$gd1$88x]8^8a$gd1$88hd]8^8`ha$gd1$88hdx]8^8`ha$gd1$88dx]8^8a$gd1$88hd]8^8`ha$gd1
, $%23qr../.@.A.Y.Z.a.b.h.i.................(/+/굩phTKh1CJNHOJQJhTKh1CJH*OJQJhTKh1CJOJQJhTKh1CJNHOJQJhTKh1H*OJQJhTKh1CJOJQJjYhTKh1OJQJUh<h1NHOJQJhk&h1H*OJQJh<h1OJQJhk&h1OJQJ,+//./////////08090F0Z0s0t00000001#1>1E1O1v1w1{111111зra!jrhTKh1CJOJQJUh)ddh1>*CJh)ddh1CJh=
h16NHOJQJh=
h16OJQJh=
h1NHOJQJh=
h1OJQJh*Jh1OJQJh=
h15NHOJQJh=
h15OJQJh<h1NHOJQJhk&h1H*OJQJh<h1OJQJ$/0>1112222334444445555$88hdx]8^8`ha$gd1$88x]8^8a$gd1$88x]8^8a$gd1$
88x]8^8a$gd111,2.2/21222P2Q2V2W2a2b2i2j222222223I3J3443474l4n4o4444444xjxjxbh1OJQJh=
h16NHOJQJh=
h16OJQJh*Jh1NHOJQJh=
h1OJQJh=
h1H*OJQJh*Jh1OJQJh<h1OJQJhTKh1CJH*OJQJhTKh1H*OJQJhk&h1CJH*OJQJhTKh1CJOJQJhTKh1CJOJQJ%44444444555h*Jh1OJQJhTKh1CJOJQJh1h1OJQJmHnHuh1OJQJjh1OJQJU
0
00P:p1/ =!"#$x%HH@ Rc(hh@dDdW7LN
C*AFigure 1BWW{aޮk:#D3D0T+t^k sSk3W{aޮk:#D V6xMlE8PQT2 z R6ֻc{{w3vlYzAT.u ƝC3zN7̬"
GsQ"}/ـ}!
$q3CjF,
/4:kKFIp6jr@P.Fg6j2Lm=>aop;zzjVwNqNA+e;+e7+<(zB`0o+p i_lP~^}ȫt=j>{e߄)cx(u?`{xXTL@}yv2O94D=SKnryi/@SWԚpīG,kUf"%eTcp:˦v{;eS#IMfkgO$;7]o8N{\T_+TyyF߫īM7\?_sC纐:GT;Kx=vz*
p{kNXm_DdK'N
C*AFigure 2B.qGxˤ) =0T'_YYgW;.qGxˤ) lJOxVߋU>7mNT
Kg[Ul\
[[UdM2$o"J"u">,HAjKjEܙILf>w23L溷?߽_F0J1D8
Ր.͞94Chǈ+]pg@Mb^g}\~7Ꜣb_ņg$hJa3$u(6_cN,
OIĦcPVoMX319y(bP;8?݊zkw,
/;bN@RA'0l9%8K{t; e\VڸpWf:Ks0Zw(H8+lR0Fvޗ)IRwy<`}}t;ζgNL5&Diߨܩ3٩19l_{;MgzjkmCZf#7'aoX>܈_ާQzXN;!w7~Zt;ɟޛ/֟w G5bog[yDDdd2SRN
C*AFigure 3B.ޘ/01ogQ~
0TvDu\kRP>.ޘ/01ogQc14xoDl@qCRJKҦ C%T!*&EPמ5][BHpB9ܸUčUJO@H 딵C=ػ̼w>tM
fNY;K~K]KP)]8ˊp21ΟE9GsFW=H95s.ʙ5Ǿ=ęo̹ɃԱ^]1ϱr$Jhb%G9Һߊs(G^Wߌu87y]5(G^83Du!3ӽDd`" ^
C:A"6pivot typical*BoF
sɯ@
j0T({fIƄoF
sɯ@
jx{lU)eۮBmK+` *QhKvwv;A"G0&2!0wP7"YPOvD/{`.ZFd
N+(AB*3q3i;PZoڥ5eۧKhS86,xOT<&GJ+GצaIm̵{J⯂!oKbu+Ybw{oWiMrK:p:KCwy8eWsBx@o^36j?@١^oNuzj_B'鿺YQ'?[vd/ſ#/Y bO˒k5D'Yп/ o"l2wjuvRKDdm'9:N
C*AFigure 4Bw߄<`PS0TKFA9jZ߄<`PFl xMpE@m6ɐB ACH$ "Ufwzw3 dA*jEI+,f\,N*9Y]42]לp jiBDITȲV8X)iA)flKYQA̋TiY$4e\ؖ,˧dɫE[E܂<ǥ,d}gRT33cDK>6zzI`vJPўe>G,9iBd",sTzhRrlWfIF]iy~]7oDb䕒:%6dWRWkթ옭Tc~wh:*ШuڹF
roJ}4";�emuz, xi7
{jC^T]x
T<; I50{8H F
:uP08cU*@O(.XOUQ كENVRJa8B㽩UYt;
fW[+2B̦Axc*N!FUC� 4D4Q
k6KP`/הax>TGNT$#W2qJ&~)\"ex.3g�?eICE Xp
5)܍YܧAGx
o馦C~
ڂ;>ܥbN
>Kf$,a
5qD7apP<4&q38(㚈07PS{xEyWx3Qw&aπnmR+_SoXKpڐBP~gk r/ue?eхhBI" o) b8pim au.qŗeonanw#!n'`y su96wލat amgzi {#jo:\\xgv{wbx7]yڿpf"gxϩ[71��d�d�q����������������'99���������������������������������n��� ������ ��c�*���a������������f�i�g�u�r�e� �5����������b���20!3)�k�����������0tc��g020!3)b����p����������!���xsuiˤipqzmiir@t)3i6ɶivcg58("388 y?qq4ic~{}p7wxlz �b ehKS2gN1_څ <
\īr
SܩQ+2^&(gnu/_
j#nCA
V؇(%>ۊHcTPYqX
j=jryE8S8`"$G95GMAuSPo<=3QFpލ: WEPixCXmRk.,]Auw2VtgkKaC}{~߭>4~3ڎYåqcYY=
[Jg'G%Q9ۻ%_..
>W+~.hw'ڡF{Ip\]o
CEke/@EET=zj=^=k^6B\R{_P
͵+\U=XmP ZoXqNת9պK(D@DNormalCJOJQJkH'mH sH tH DA@DDefault Paragraph FontZiZTable Normal :V4
l4a_H(k(No List4 @4Footer
!4@4Header
!:@:
Footnote TextCJNB@"N Body Text$pd,]pa$OJQJ0U@10W` Hyperlink>*B*@V@A@[=FollowedHyperlink>*B*'b2b z z z z z z z z*S
LTzB k$'1A2E}c&*,.+U@L,mwDTVzKba+W{M
B D !!""q##k$m$Q%R%S%T%%S&&w'x''''''0׀0׀0׀0׀0׀0ʀ0׀0׀0ʀ0׀0׀0׀0׀0ʀ0׀0ʀ0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0ʀ0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0׀0ʀ0׀0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ@0ʀ@0ʀ@0ʀ0ʀ0ʀ0Ѐ0ʀ0ʀ0ʀ0Ѐ0Ѐ0Ѐ0ʀ0ɀ0Ѐ0ɀ0Ѐ{2E}c&*,.+U@L,mwDTVzKba+W{M
B D ""q##k$m$Q%R%S%T%%S&&w'x''''''0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ@<`l@<`l
@<`lF02F02F02F0200ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ00ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ000@0@00ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ0ʀ@0@0@0@0@0ʀ@0ʀ@0@0@0@0ʀ@0ʀk$0ʀ0000ʀ00ɀ
Y0܂0ɀ
>0>{77:Uy8#K'),+/145!#$&')*,/0.
z#)/5"%(+.5 ':!867@60(
B
S ?'
OLE_LINK19
OLE_LINK20 OLE_LINK5 OLE_LINK6%%&&'&&'
_c*,
LPSUZ^ae
8:SUz=ADF!KORTY]`d#$&24FHvx57:<mo )!+!S!Y!k!m!""@"B"""##########$$$$L%N%a%i%n%w%x%%%%%%%%%%%%%%%%%%%%%%&&&
&&&& &l&n&o&w&&&&&&&&&''2'E'K'Q'W'x''24}~IY8>#:?25sv
+.or ######/$5$F$N$%%%%%%&<&&&x''::::::::::::::::::::::::::::::::::::::{&U808^8`0o(.^`.pLp^p`L.@@^@`.^`.L^`L.^`.^`.PLP^P`L.{ S%k''i.@
v
'`@```<@` `"`H@`(`*`X@UnknownGTimes New Roman5Symbol3Arial3Times9New York5Geneva"AhBfljGk E+xx>d'6`+#EConnected (n4) configurations exist for almost all n an updateGalya DimentBranko Grnbaum
Oh+'0( @L
lx
'HConnected (n4) configurations exist for almost all n an updateGalya DimentNormalBranko Grnbaum273Microsoft Word 11.6.6@A0@nR@66@}8
՜.+,D՜.+,\`hpx
'E'FConnected (n4) configurations exist for almost all n an updateTitle 8@_PID_HLINKS'A$HS*
Figure 1KS,
Figure 2JSv# Figure 3POV$6pivot typical*MS Figure 4LS1 Figure 5
!"#$%&'()*+,./013456789:;<=>?@ABCEFGHIJKLMNOPQRSTUVWXZ[\]^_`bcdefghkRoot Entry F;mData
2#1TableD)WordDocumentbSummaryInformation(YDocumentSummaryInformation8aCompObjX FMicrosoft Word DocumentNB6WWord.Document.8