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CruxPublished by the Canadian Mathematical Society.
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Journal title history:
➢ The first 32 issues, from Vol. 1, No. 1 (March 1975) to Vol. 4, No.2 (February 1978) were published under the name EUREKA.
➢ Issues from Vol. 4, No. 3 (March 1978) to Vol. 22, No.8 (December 1996) were published under the name Crux Mathematicorum.
➢ Issues from Vol 23., No. 1 (February 1997) to Vol. 37, No. 8 (December 2011) were published under the name Crux Mathematicorum with Mathematical Mayhem.
➢ Issues since Vol. 38, No. 1 (January 2012) are published under the name Crux Mathematicorum.
Mat he m
at ico ru m
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ISSN 0705-0348
C R U X M A T H E M A T I C O R U M
Vol. 6, No. 6
June - July 1980
Sponsored by Carleton-Ottawa Mathematics Association Mathematique d'Ottawa-Carleton
Publie par le College Algonquin
The assistance of the publisher and the support of the Canadian Mathematical Olympiad Committee, the Carleton University Mathematics Department, the Ottawa Valley Education Liaison Council, and the University of Ottawa Mathematics Department are gratefully acknowledged.
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CRUX MATHEMATICORUM is a problem-solving journal at the senior secondary and university undergraduate levels for those who practise or teach mathematics. Its purpose is primarily educational, but it serves also those who read it for professional, cultural, or recreational reasons.
It is published monthly (except July and August). The yearly subscription rate for ten issues is $10.00. Back issues: $1.00 each. Bound volumes with index: Vols. 1&2 (combined), $10.00; Vols. 3,4,5, $10.00 each. Cheques and money orders, payable to CRUX MATHEMATICORVM (in US funds from outside Canada), should be sent to the managing editor.
All communications about the content of the maqazine (articles, problems, solutions, etc.) should be sent to the editor. All changes of address and inquiries about subscriptions and h&ck Issues should be sent to the managing editor.
Editor: Leo Sauve, Architecture Department, Algcr.cuin College, 281 Echo Drive, Ottawa, Ontario, K1S 1N3.
Managing Editor: F.G.B. Maskell, Mathematics Department, Algonquin College, 200 Lees Ave., Ottawa, Ontario, K1S OC5.
Typist-compositor: Gorettie C.F. Luk.
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CONTENTS
Propriety ggornGtriques et approximation des fonctions: II . . J.G. Dhombres 166
A Pandiagonal Fifth-Order Prime Magic Square Allan Wm. Johnson Jr. 175
Two Polyhedral Problems Howard Eves 176
The Olympiad Corner: 16 Murray S. Klamkin 176
Solutions to "Two Polyhedral Problems" . . . Howard Eves 182
A Mathematics Calendar 182
Problems - Probl^mes 183
Solutions 1 Q5
- 165 -
- 166 -
PROPRIETES GEOMETRIQUES ET APPROXIMATION DES FONCTIONS: II
J.G. DHOMBRES
Les rgsultats acquis dans un article precedent [1980: 132] sont utilises
pour de"montrer un glSgant re*sultat d'analyse harmonique.
3. Un exemple de meilleure approximation.
Nous allons nous placer sur 1'ensemble E des fonctions dgfinies sur Taxe r£el, a" valeurs complexes, peYiodiques de peYiode 2TT, continues ou possSdant un
nombre fini de discontinuity de premiere esp£ce sur Vintervalle d'une peYiode.
Sur cet espace E9 nous construisons une norme intSgrale en posant
Introduisons tout de suite un outil dont nous aurons besoin sur E9 a* savoir
le coefficient de Fourier o (f) d'ordre n d'une fonction / de E. Par definition,
on a pour tout entier relatif n
%(f) = 2? / f(t)e dU
•'-IT
Soit alors A un sous-ensemble non vide et propre de 1'ensemble des entiers
relatifs. On prend pour F(A) le sous-ensemble des f de E telles que o if)=o pour tout n n'appartenant pas Si A. On laisse au lecteur le soin de montrer que F(A) est un sous-espace ferae" propre de E. Notre propos est d'approximer un
element quelconque de E par F(A). Nous allons un peu specialiser cette etude en
remarquant que si f prend des valeurs replies et c (/) = 0 alors o_ (f) - o ggalement. Soit alors A un sous-ensemble de 1'ensemble des entiers positifs ou nuls et
soit S(A) la reunion de A et de -A. On dispose du theor£me suivant (cf. [3])»
surprenant au premier abord:
THEOREME 5. Soit A un sous-ensemble des entiers positifs ou nuts. Pour
tout f de E it existe une unique meilleure approximation par les elements de F(S(h))
si et seulement si A est une progression arithmetique infinie de raison impaire et
debutant & 0.
A) On commence par prouver que la condition donnSe au Th£or£me 5 est nScessaire.
On procSde en quatre Stapes. Dans les trois premieYes Stapes, A peut designer un
sous-ensemble propre non vide de 1'ensemble des entiers relatifs.
- 167 -
Demonstration.
l&re etape: Caraete'risation de oertaines formes lineaires sur E.
Pour pouvoir utiliser le Theor§me 4, il convient d'abord de caracteYiser les
formes lineaires complexes sur E muni de sa norme integrale, et qui satisfont
1'ggalite* (1) du Th£or£me 4. Soit g une fonction deTinie sur R et 3 valeurs complexes, pe>iodique et de periode 2% que nous supposerons born£e. L'application
de*finie sur E9 3 valeurs complexes, donn§e par 1'expression
I r+7T
L (f) = ~~/ f(t)g(t)dt9
est l i n ^ a i r e d&s que fg est intggrable sur C-TT,+7T] pour tout / de E. En
outre on dispose de la majorat!on
1 r + T r
HV^I I * W / l f ( t ) l \ffM\dt * (sup|^(t) |) | |f | | . 0 * J-Tt ten
III est done c l a i r que
" L J I = ,.S?P ' V f ) l * s u Pl^ ( t ) i = Halloo-
feE
De fait, on a m§me \\L || = ||̂|| . Nous nous contenterons de dgmontrer ce point dans le cas oQ g est continue et prend des valeurs replies. II existe alors
un point x0 de r-7r»+Tr] oQ |#(a?0)| = sup|#(t)| d'apr&s la proprigte" bien connue teR
des fonctions continues. Done, pour tout e > o , on peut trouver un n > o tel que
pour tout x satisfaisant \x -x\ < r\ on ait \g(x)\ z \g(xQ)\(l - e).
Puisque, pour une fonction periodique h et de peYiode 2TT, on a
+TT - +7T+iC0
h(t)dt = / h{t)dt9 •J -7T+.7!0 -i\+x0
on peut toujours supposer que a!0€l-TT,+7rr et prendre ri assez p e t i t de sorte que
]ar0-n,ff0+n[ c r-fT,+-nl. On d ^ f i n i t alors une fonct ion / sur C-TT,4TT], que 1 fon
prolonge par pe r iod i c i t y 2TT, en posant
]f(t) = o s i t {lx0-r)9x0+r\l,
-, s 2 IT
et / continue et l i n§a i r e par morceaux.
On calcule que | | / | | =1 et en outre on minore L if) selon
- 168 -
\L(f)\ = | F I/ f(t)g(t)dt\
> \g(xQ)\(±-e) = sup \g(t)\(l - e ) .
Comme cette inggalite" reste vraie pour tout e>o, on a IMnggalite* inverse chercM5e | | ^ | | > \\g\\. D'oQ Ton de^Iuit | | ^ | | = H^L-
Ce que nous venons de voir montre que toute fonction g de borne supeYieure ggale 3 l et 2Tr-peYiodique (telle que fg soit intggrable) fournit une forme linSaire L sur E satisfaisant la relation \\L || =1. On constate que Lg -Lg imp! i que gx =#2 , comme il est bien facile de le voir du mo ins d£s que gx et gz
sont continues. Nous admettrons1 la r^ciproque, a savoir que toute forme linSaire L sur E qui veYifie I'ggalite* (1) du Th£or£me 4, soit | |L| | =1 , provient ainsi dfune fonction g dont le module a une borne supeYieure ggale 3 l sur 1faxe rgel, 2iT~pe>iodique (et telle que fg soit intggrable pour tout / de E). Cela revient a* dire L-L .
9 2&me etape: Determination de g. Rappelons que nous voulons gtablir une condition n£cessaire. Soit f un
Pigment de E et Pf sa meilleure approximation par F(A). Ici, pour commencer,
A est un sous-ensemble propre non vide de 1'ensemble des entiers relatifs. D!apr£s
le Thgor£me 4, on dispose done pour toute fonction f de E dfune fonction g telle que
sup \g(t)\ = 1, (1) teE
Lgif) =\\f-Vf\\* (2) et
L (h) = o pour tout h de F(A). (3)
Re^crivons (2) sous une forme integrale en tenant compte de (3);
/+TT /-+TT
^ f(t)g(t)dt =J^ (f(t)-Pf(t))g(t)dt
ce qui s'eYrit encore ' _7r f™ \f(t)-Pf(t)\dt;
(2 bis)
/ l\f(t)-Pf(t)\ - (f(t)-Pf(t))g(t)ldt = 0.
Mais on a \f(t) - P f ( t ) g ( t ) \ < \f(t) - P f ( t ) \ en tous points t de C-TT,+7T1. Si g
Cette reciproque exige la notion d'integrale de Lebesgue. On pourra consulter C*+ II et C53 entre autres multiples r^ferenaes.
- 169 -
gtait continue, nous savons bien que ces deux relations impliquent
\f(t)-Pf(t)\ = (f(t)-Pf(t))g(t). (4)
Nous adlmettrons que cette relation reste vraie en general (quitte gventuellement
3 modifier g quelque peu). Cette dernigre ggalite" (<*) determine g en tous points oQ f-Pf ne s?annule pas.
Montrons maintenant que Pf est une meilleure approximation de f dans F(A) si
et seulement si I'on a une fonotion g satisfaisant (I), (4) et satisfaisant
Sgalement la relation
a (g) = 0 pour tout n de -A. (5)
II suffit de montrer que (5) gquivaut SI (3). D'une part si Ton a (3), ~int
}ue e appartient 3
si Ton a (5), on a encore
puisque e'tn appartient 3 F(A) si n appartient 3 -A, on a bian (5). R£ciproquement
/ +7T
g(t)h(t)dt = 0
pour toute somme finie de la forme
&(*> =• E c eikt (6) keh
oD les c« sont complexes. Nous admettrons (th€or§me de Fe\jer) que toute h de F(A) sfobtient comme limite, au sens de la norme intggrale de fonctions du type precedent
(6). On en de"duit (3) par passage 3 la limite.
3&me Stape: Invariance de la propriety par translation,
A de"signe encore un sous-ensemble non vide et propre de Tensemble des entiers
relatifs tel que tout f de E ait une unique meilleure approximation Pf par F(A). Soit n0 un entier relatif quelconque et d^signons par A0 le sous-ensemble
n0 +A. Montrons que tout f de E poss&de une meilleure approximation par F(A0).
En effet, soit fcE. Par hypoth&se, il existe une unique approximation dans F(A)
de f{t)e~%n<it\ Gr§ce au Thgor£me 4, on dispose done de g telle que
/sup|#(t)| = l, UR
\f{t)e~in^ -P(nt)e-in^)\ = (nt^™** -P(nt)e~in«%(t),
c (g) = 0 pour tout n de -A. iyi t ~~iyi t
Par suite, fQ(t) = e ° P(f(t)e ° ) appartient 3 F(A0) puisque, pour tout h de F, on a
- 170 -
o(Ht)ein'>t) = c „ (Ht)). n n-n0
Posons #0(t) = e~tn° git). II vient les trois relations
|
sup|^0(t)| = 1,
tcR
\f(t)'f0(t)\ = {f(t)-f9(t))g9(t),
Q (g0) = 0 pour tout n de -A0.
GrSce au Thgore"me M- une nouvelle fois, on constate que .f0eF(A0) est une
meilleure approximation de / par F(A0). L'unicite* d'une telle meilleure approx-
imatfon provient de ce que le passage de A 5 A0 est reversible par une operation
de meme nature.
La propriety gtablie a la 3$me etape se d^montre simplement grSce Si 1'utili-
sation de 1'exponentielle complexe. C'est uniquement pour cette raison de simpli
city que nous nous sommes imposes de prendre le corps des complexes C comme corps
de base des espaces vectoriels considers au lieu de se restreindre fi R.
n£me Stape: A dgsigne maintenant un sous-ensemble non vide et propre de
1'ensemble des entiers positifs ou nuls. Nous supposons que tout / de E a une unique meilleure approximation par F(S(A)). Rappelons que #(A) = Au (-A).
(a) Commengons par supposer que o n'appartient pas a" A. Soit feE et f(t)>o pour tout t de R. On a ngcessairement Pf=o car on peut prendre g(t)~l et on a ainsi une fonction g satisfaisant (l), (4), et enfin (5).
Soit alors n un element de A. ConsideYons les deux fonctions fi(t)=l et
f2(t) = l-sin nt. Ces deux fonctions sont positives et f2 - fx est un Element non
nul de F ( S ( A ) ) . Par sutte Pft =Pf 2=o et on dispose alors des inggalites suivantes
contradictoires puisquMl y a unicite" supposed de la meilleure approximation:
11/, II = I IA-P/JI < 11/,-(/,-/2>ll = ll/JI et
ll/JI = llf2-p/2ll < l l / 2 - ( / * - / , ) II = 11/, II. Par suite 0 appartient toujours 3 1'ensemble A.
(b) Supposons maintenant que nY et n2 soient deux elements de A. Montrons
que si £(n1+n2) est un nombre entier, alors (n1+n2)/2 et \n1 -n2\/2 appartiennent
a A.
En utilisant £(A), il suffit de montrer que si n[ et nx2 sont dans ${A) et si
(rcj + n2)/2 est un entier relatif, alors k-(n[ +np/2 appartient 8 5(A). On
- 171 -
utilise alors l'ensemble A0=£(A)-fc qui, dfapr£s la 3$me etape, possSde la m§me
propriete d'unicite que 5(A). Selon la m§me demarche qu'en (a), on aboutit &
une contradiction si o^A0, done si k£Sih).
(c) Soient X et X' deux elements successifs et distincts de 1'ensemble A,
c'est-cl-dire X < X \ X,X' e A et il nfy a aucun element de A entre X et X'.
Si X' -X est un nombre entier pair, le point milieu (X + X')/2 est un entier
qui appartient 5 A grSce 3 (b). Done il y a contradiction. On doit done toujours
avoir un intervalle de longueur impaire entre deux entiers successifs de A.
(d) Montrons que A est un ensemble infini. Par Tabsurde, supposons que
tout X de A satisfasse o<X<n. ConsideYons alors la fonction fit) definie sur
lfaxe reel par
fit) = +1 si sin nt > o,
= o si sin nt = 0,
= -1 si sin nt < 0.
Cette fonction est peYiodique integrable et de peYiode 2-n/n. Par suite,
supposons que k soit un entier relatif, non multiple de n. On a c, if) =o puisque
= ( 2? / f( )e )e
Soit ( l - e2iklT/n)okif) = o et e2 ^ 7 T / n * i par hypothese. Done c fc(/) = o. En outre
c 0 ( / ) = o puisque / est une fonction impaire. On deduit finalement que <?^(f) = c ^ ( / ) = o
pour tout fc de A. Naturellement
s u p | f ( t ) | = l . tei?
GrSce 131 la relation fit)fit) = |/(t)| on constate que o est 1'unique meilleure approx
imation de / par F(sih)) puisqu'on peut appliquer la 2£me etape avec git) ~ fit) et
P/=o.
Considerons maintenant la relation g(t)(fit) -1) = \fit) -1| oQ git) = fit) si
fit) * o et git) = +1 si fit) = o. II est facile de noter que e^Cg) ~G_J<g) ~ ° P°ur
tout k de A et
sup|g(t)| = +1. tcR
Done, encore grSce 3 la 2§me etape, on note que o est Tunique meilleure approxi
mation de fit) -1 par F(siK)).
- 172 -
La contradiction provient maintenant, comme Tors de 1'etape (a), de ce que
fit) - (fit) -1) = l est un element de F(S(A)) .
(e) Enfin, terminons la demonstration de la necessite du TheorSme 5 avec
les notations de (c) notamment.
Ici OeA. Supposons que A soit le premier element non nul de A, On a,
grSce 5 (c), A = 2fc+ 1. Soit A' le premier element de A a droite de A, lequel
existe grSce 5 (d). On a, grSce a" (c), A' = A + 2k% +1. Par suite A1 = 2(k t k1) + 2
et done k + k' +1 doit appartenir 5 A et ne peut que co'incider avec A, ce qui implique V -k. Puisque A est infini, on conclut de proche en proche que A est
une progression arithmetique infinie de raison 2& + 1 debutant en o.
B) Passons maintenant a" la demonstration de la suffisance de la condition
donnee au Theor£me 5. Soit done Alfensemble des nombres entiers de la forme
(2k + l)n oQ n parcourt les entiers positifs ou nuls et oO k est un nombre entier fixe, positif ou nul. Soit alors feE. Nous cherchons PfeF(S(K)) telle que Pf soit 1'unique meilleure approximation de / par F(S(A)). Comme 5 la deuxiGme
etape, nous admettrons (theor^me de Fe\jer) que tout element g de F(S(A)) est limi-
te au sens de la norme integrale de combinaisons lineaires finies dfexponentielles
de la forme e1"™ oQ meS(K). Ces exponentielles possedent toutes 2-n/(2k + l) comme
periode, ainsi en est-il de tout g de F(S(A)). Calculons alors \\f-Pf || selon
~f \f(t)-Pf(t)\dt^ £ / \f(t)-Pf(t)\dt J0 m=0J 2wn/(2k+l)
m=2fc /-27r/(2^+l) , 2TT .
! f2ir/(2k+l) im^k. .) ,
Nous pouvons essayer de determiner, pour chaque t f ixe de [o,2ir/(2k+l) l , un
element Pf(t) tel que
m~2k
£ l/w(*)-W*>l m=0
soit minimal.
II sfagit d!un probl§me de geometrie plane: a* 2fc+i points donnes s09z l,...9s ,
du plan, associer un unique point 2 de ce plan tel que
m=2k
mW m
- 173 -
soit minimal.
SI Ton peut rgsoudre ce probl§me ggome*trique, et si Z depend continQment des 2fc+l points, il est alors bien clair que toute / de E poss§de une unique meilleure approximation par F(£(A)), ce qui terminerait la demonstration du
ThSorfcme 5.
'.'existence du point Z nfest gu£re difficile 5 gtablir puisque la fonction
deTinie sur C, le corps des complexes, par
2k z - D«-»J =^(3)
m=0
est continue: on a m§me lfin£galite" dgduite de I'inggalite* triangulaire
|/(3)-/(a')| < (2fe + l)|a-g,|-
En outre lim f(z) = +°°.
12 (•+•+«>
Par suite la borne infeYi eure
inf f{z) zeC
est figale a la borne infgrieure inf f(z)
pour un i? assez grand. Mais comme le disque fermg de centre 0 et de rayon R est
compact, la fonction continue / atteint son minimum en au moins un point Z0 de
ce disque
inf /(a) = /(Z 0). zcC
Uunicite* du point Z exige une analyse un peu plus GtoffGe. Appelons K le
sous-ensemble non vide des points de c oQ / atteint son minimum. C'est gyidemment
un sous-ensemble ferine* et borne* de C9 done compact. Plus inteYessante est la
propr16te* de convexity de K: Si z et Z1 sont dans K9 AZ + (l - A)Z* est encore dans K et ce pour tout A
satisfaisant o < A < i . Ge*ometriquements le segment joignant z & z} appartient
tout entier 5 K. En effet
2k /(AZ+(1-A)Z') = y |A(Z-s ) + (l-A)(Z» -a )|.
m=o m m
Soit
- 171 -
2k n f{\Z+(l ~X)Z*) < If \z-z | + ( 1 - X ) 7 \Z% -z |
772=0 171=0
< Xf(Z) + (l-X)f(Zt)
<: (X + l-X)f(Z0) = /(Z0).
LMnegalite stricte ne pouvant avoir lieu, toutes les inegalites precedentes deviennent des egalites. En particulier pour tout m variant entre o et 2k,
| X ( Z - a w ) + ( l - X ) ( Z ' - a ) | = | X ( Z - s ) | + | ( 1 - X ) ( Z « - a ) | . tff m m m '
Par suite (Z - z ) et (Z* -a ) sont sur une meme demi-droite3 o' est-ci-dive m m qufil existe des nombves a 3 B 3 positifs ou nuts et
Soit
a ( Z - a ) = B (Z* - a ) at?e<2 a +B > 0. 77? m wi m mm
a Z - M 1 = (ex - B )a . m m m m m
SMI existe un indice pour lequel a = B on deduit aussitOt Z = Z'. On peut alors
supposer que a *B pour tout m = o,l9... ,2k, auquel cas tous les points z doivent
§tre situes sur la droite passant par Z et Z', et on est ramene* & un problSme rectiligne analogue. On note alors que z - z et Z' -a ont m§me signe c'est-3-dire quTil existe des e (=+l)tels que
m \Z~*J = EJZ~ZJ e t lZ' ~BJ = Em<Z' ~ZJ-1 TTT m m l m m m
D'oD 2k 2k 2k 2k
/ (Z 0 ) = V | Z - a J = V e m (Z-a ) = [ e ( Z ' - a ) = f | z ' - a w | = / (Z 0 ) m% m rfeom m m=0m m m=0
e t , par Vega l i t e cent ra le , 2k
(z -z*)y E = o.
w=0
On note que £ „ e est toujours different de o. Par suite Z = Z'. 777=0 777
La methode utilisee conduit evidemment a un resultat plus precis que celui
dont nous avons reellement besoin. Pour lfetude de tels minimums, et quelques
generalisations, cf. [6],
Montrons enfin que le point z depend continQment des points a (m - o,l,... , 2 k ) . On peut raisonner par Vabsurde. Soit done z^ une suite de points et lima^n^=a .
Soit z } le point assurant le minimum de
- 175 -
2k Jn)t
m=Q
et Z le point assurant le minimum de
2k I l*-« I.
m% m
Les Z'n restent dans un sous-ensemble borne\ Si done 1fon suppose que Z ne
converge pas vers z, alors il existe une sous-suite, encore not£e Z pour
simplifier les notations, et un point z1 (Z**z) tel que l|mz ( n ) = z'.
Dfis lors, pour tout s de C, on dispose de 1'inggalite* au sens large
mM) <- i M)\ m~0
En passant S la limite en n, puisque / est continue,
jr(z«) * £ |*-« j . m=0
Cette inggalite" contredit l'unicite* de z et termine la demonstration.
REVERENCES
(Pour les references [l]-[4], voir [1980: 140].)
5. J.P. Kahane, "Projection m^trique dans Lx(T)9n Comptes Rendus Acad. So.
Paris, Se> A , 276 (1973) 621-623; "Best Approximation in Ll(T)9" Bull. Amev. Math.
Soc, 1974, pp. 788-804.
6. J.G. Dhombres, "Et pan dans le mille," Bulletin de 1'Association des
Professeurs de Math£matiques, avril 1976.
U.E.R. de Mathematiques, Universite de Nantes, 2 chemin de la Houssiniere, BP 1044, 44037 Nantes Cedex, France.
* * *
A PANDIAQONAL FIFTH-ORDER PRIME MAGIC SQUARE
This pandiagonal fifth-order magic square is composed of distinct primes. The magic constant is 395, the smallest possible for a magic square of this type.
ALLAN WM, JOHNSON JR., Washington, D.C.
1 11
113
1167
7
97
43
31
71
23
227
83
137
103
67
5
127
41
17
197
33
131
73
37
101
53
- 176 -
TWO POLYHEDRAL PROBLEMS
The following two problems show that one must be wary inducing results by
intuition or by analogy.
Problem 1,
Of two regular polygons inscribed in the same circle, that with the greater
number of sides has the greater area. Is it true that of two regular polyhedra
inscribed in the same sphere, that with the greater number of faces has the
greater volume?
Problem 2.
Of two regular polygons circumscribed about the same circle, that with the
greater number of sides has the lesser area. Is it true that of two regular
polyhedra circumscribed about the same sphere, that with the greater number of
faces has the lesser volume?
HOWARD EVES, University of Maine.
Solutions to these problems appear on page 182 in this issue, ft ft ft
THE OLYMPIAD CORNER; 16
MURRAY S. KLAMKIN
We present this month the questions asked at the Ninth U.S.A. Mathematical
Olympiad on May 6, 1980 and, through the courtesy of Willie S.M. Yong of South
Wales, those asked at the Sixteenth British Mathematical Olympiad on March 13,
1980. Solutions to the British Olympiad problems will appear here in the next
issue. For the U.S.A. Olympiad, the solutions (along with those of the Twenty-
second International Mathematical Olympiad, which has yet to take place) will
appear later this year in a booklet compiled by Samuel L. Greitzer and available
(at 50<t per copy) from
Dr. Walter E. Mientka, Executive Director, MAA Committee on H.S. Contests, 917 Oldfather Hall, University of Nebraska, Lincoln, Nebraska 68588.
- 177 -
THE NINTH U.S.A. MATHEMATICAL OLYMPIAD
May 6, 1980 - 3* hours
1, A two-pan balance 1s inaccurate since its balance arms are of different
lengths and its pans are of different weights. Three objects of different
weights A, B9 and c are each weighed separately. When placed on the left-hand
pan, they are balanced by weights Al9 B19 c19 respectively. When A and B are
placed on the right-hand pan, they are balanced by A2 and B29 respectively. Deter
mine the true weight of C in terms of A1 9 Bx , C1, A29 and B2.
2, Determine the maximum number of different three-term arithmetic progres
sions which can be chosen from a sequence of n real numbers ax <a2 < ... <a .
3, Let F = a^sin (rA) +z/rsin (r£) fs^sin (rC),
where x9 y9 z9 A9 B9 C are real and A + B + C is an integral multiple of IT. Prove
that if Fj*.F2 = o, then F =o for all positive integral r.
L\t The inscribed sphere of a given tetrahedron touches each of the four
faces of the tetrahedron at their respective centroids. Prove that the
tetrahedron is regular.
5, If 1 >.a9b9e> o, prove that
T - ^ T + — ~ + -4-r + (1 -a)(l -b){l -o) < 1. £+<?+l c?+a+l a+b+1
THE SIXTEENTH BRITISH MATHEMATICAL OLYMPIAD
March 13, 1980 - 3| hours
1, Prove that the equation xn + yn^zn9 where n is an integer greater than 1,
has no solution in integers x9y9z with o<x<n9 o<y<n.
2, Find a set #={rc} of 7 consecutive positive integers for which a polynomial
P(x) of the 5th degree exists with the following properties: (a) all the coefficients in P(x) are integers; (b) P(n)=n for 5 members of S9 including the least and greatest;
(c) P(n) = o for one member of s.
3, On the diameter AB bounding a semicircular region there are two points P
and Q, and on the semicircular arc there are two points R and S such that
- 178 -
PQRS is a square. C is a point on the semicircular arc such that the areas of
the triangle ABC and the square PQRS are equal.
Prove that a straight line passing through one of the points R and S and
through one of the points A and B cuts a side of the square at the incentre of
the triangle.
J|, Find the set of real numbers a0 for which the infinite sequence {a } of real numbers defined by
an+l = ^ ~Zan* n = °» l s 2 >"-
is strictly increasing, that is, a <a . for n > 0 . J 3 n n+1 5, In a party of ten persons, among any three persons there are at least
two who do not know each other. Prove that at the party there are four
persons none of whom knows another of the four. *
SOLUTIONS TO PRACTICE SET 13
13-11 In ^-dimensional Euclidean space £^, determine the least and greatest
distances between the point A= (a1 9a2*...»a ) and the w-dimensional
rectangular parallelepiped whose vertices are (±vx 9±vZ9... ,±v ) with v.>o. (Some may find it helpful first to do the problem in E3 or even in E2.)
Solution.
By symmetry, the desired distances will not change if we replace a. by |a.|. Consequently, we may assume that a.^o without loss of generality.
Case 1. The point A lies on the parallelepiped. Here one or more of the a.'s
are equal to the corresponding T K ' S , while a.<v. for the remaining a.'s. The least distance, D . , is obviously zero. The furthest point of the figure from point A is the vertex (-i>,,-v ,-v ). Thus the greatest distance, D , is
1 2 n max given by
DLV = (ax+vx)
2 + (a2+v2)2 + ... + (a^ + v ) 2 . (1)
niu x n n Case 2. The point A lies in the interior of the figure. Here a. <v. for all
i. The least distance is then given by Dm1n = l5&.(W'
this being the distance from A to the point
- 179 -
where r is any one of the values of i such that v. -a. is a minimum. The furthest t t
point from A is again (-v19-v29...9-v ) and the corresponding greatest distance is
again given by (l).
Case 3. The point A lies outside the figure* Here there is at least one i such
that a^>vj.. If the point X = (x19x2,... 9x ) on the figure is nearest to A, then we must have
(a.9 if 0 < a. < v.9
This can conveniently be written as a single equation with the Heaviside unit
function H defined by H(t) = o if t < o and H(t) = 1 if n o :
x. = a. + (t>. - a.)E{a. - v .)»
from which
n
mm A % v* i ^ t=l
The greatest distance is again given by (l).
13-2. If A,B,C are positive angles whose sum does not exceed ir, and such
that the sum of any two of the angles is greater than the third,
show that
(a) there exists a triangle with sides sin A, sinB, sinC;
(b) 4 + £sin 2 A esc2 Bcsc2 C < 2 £c s c 2 A (cyclic sums).
Solution,
(a) We have
sinA + sinB = 2 s i n M C o s M > 2 s i n M C o s ^ - = sin (A+B)
and
sin (A+B)-sinC = 2 c o s ^ ^ s i n ^ ^ > 0 .
Thus sin A +sin B > sin C, and the other two triangle inequalities are proved in the
same way.
Alternatively, consider a tetrahedron 0-PQR with face angles 2A,2B,2C at
vertex 0 and 0P = 0Q = 0R = i. Such a tetrahedron exists since O < A + B + C5TT and
B + C>A, etc. Then it is easy to show that triangle PQR has sides sin A, sinB, sinC.
- 180 -
(b) The inequality to be proved is equivalent to
4Jlsin2 A ^ . ,.. 2 E s i V & s i n 5 C - £ s 1 V A ~ ls (1)
where the sums and product are cyclic over A,B,C. We recall the familiar relations
aba = 4Afl, 16A2 = 2lb2o2 -lah
for a triangle with sides a9b9o9 area A, and circumradius R. If we apply these
relations to the triangle PQR of part (a), then (l) is equivalent to
^ l l l f = ̂ 2 ̂ i or R z i, (2)
and this is clearly true. Equality occurs in (2) in the degenerate case when 0 is
the circumcenter of triangle PQR, that is, when A + B + C = TT.
13-3. (a) If o<a;.<a, £ = l,2,...,n, determine the maximum value of
n
Z i-l u ±<i<j<n
(b) If 0 < # . < 1 , i = l,2,...,n and x = #,, determine the maximum value of % n+1 l
n n
Solution,
We assume throughout that rc>2.
(a) Since A is linear in each x.9 the maximum value is taken on for the
extreme values of each x.. By letting k of the #.!s = aand the remaining n-k of the a?.fs = o we obtain the following set of possible extreme values:
t
A = Y x. - Y x.x..
Mx = a for k = 1,
M2 = 2 a - Q ) a2 for fc = 2f
M3 = 3 a - ( ^ 2 for fc = 3,
It now follows easily by comparing M. with M. that
181
m̂i
2a
3a tax
-©• -©•
(b) We show by induction that
may £
i f a z i ,
i f 1 > a > j 9
i f | > a 4 ,
i f n - 1 '
n = 2 , 3 , 4 , (1)
where the square brackets denote the greatest integer func t ion .
For n = 2, we have
B 2 = Xt +X2 - (X1X2 +3^2^! ) = 1 - Xj^X2 - (1 - X1)(l ~ X2) < 1 ,
and max B2 = l is attained when one of xx 9x2 is o and the other is l; so (l) holds
for rc = 2. And it follows from part (a) with a = l that maxB3 = l, a value attained
at least once (when one of the x.'s is l and the other .r.'s are o); so (1) holds
also for n = 3.
Suppose now that (l) holds for some n = fc>2; then, for any choice of k + 2
numbers #., we have i
Bk+2 = Bk*xk+l +xk+2+xkxi ~xkxk+l ~xk+lxk+2 ~xk+2xi
< \~\+a + b+ed-ca-ab-bd9
where fo r convenience we wr i te a% = a , #x,+2 = Z : ,> ^ = c ' ^i = c?- T h e Equality
B k . ! S M " SI + l
wi 11 then fo l low from a + jb + cd-<?a-a£>-fc<£< l or
(1 ~a)(l -b) + bd + oa > ad.
(2)
(3)
This follows immediately if b>c or azd, so assume b<c and a<d. Inequality (3)
is then still valid since it is equivalent to
(l-a)(l-b)+db > (d-a)io-b).
Finally, we must show that equality can occur in (2). For this, just let #. = l
for all odd i and x.~o for all even i. D
As an extension of this problem, consider the constraints o<x.<a instead of
- 182 -
o<x. <;i.
Editor's note. All communications about this column should be sent to Professor M.S. Klamkin, Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada T6G 2G1.
ft ft ft
SOLUTIONS TO "TWO POLYHEDRAL PROBLEMS" (see page 176)
Solution to Problem 1. It is not true. The regular dodecahedron (a solid
having 12 faces) has a greater volume than the regular icosahedron (a solid having
20 faces), and the regular hexahedron or cube (a solid having 6 faces) has a greater
volume than the regular octahedron (a solid having 8 faces). These unanticipated
results are easily shown from tabular information about the regular polyhedra, as
given, for example, in the CEC Standard Mathematical Tables, 24th edition, pp. 14-15.
Solution to Problem 2. It is true, as can be shown from the tabular information cited in the solution to Problem 1. It is interesting that a regular dodecahedron
and a regular icosahedron inscribed in the same sphere have a common inscribed sphere.
The same is true of a cube and a regular octahedron inscribed in the same sphere.
Also, the circumcircles of the faces of a regular dodecahedron and a regular icosa
hedron inscribed in the same sphere are equal, and the same is true of a cube and a
regular octahedron inscribed in the same sphere.
HOWARD EVES ft ft ft
A MATHEMATICS CALENDAR
The Department of Mathematics and Statistics of Ottawa's Carleton University
has just published a mathematics calendar for the academic year 1980*1981. It
runs from July 1980 to June 1981. It will be of interest to any individual who is
intrigued by the complexity and rare functional beauty of the mathematical world.
The calendar measures 17x11 inches when opened. It opens to present a topic
in mathematics on each upper leaf, with the calendar month and a "problem of the
month" below. In addition, items of mathematical interest are highlighted at
various days of the year.
Copies at $3.00 each (cheques payable to Carleton University) can be ordered
from: The Mathematics Calendar, Mathematics and Statistics, Carleton University,
Ottawa, Ontario, Canada K1S 5B6. ft ft *
- 183 -
P R O B L E M S - - P R O B L E M E S
Problem proposals and solutions should be sent to the editor, whose address appears on the front page of this issue. Proposals should, whenever possible, be accompanied by a solution, references, and other insights which are likely to be of help to the editor. An asterisk (it) after a number indicates a problem submitted without a solution.
Original problems are particularly sought. But other interesting problems may also be acceptable provided they are not too well known and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted by somebody else without his permission.
To facilitate their consideration, your solutions, typewritten or neatly handwritten on signed, separate sheets, should preferably be mailed to the editor before November 1, 1980, although solutions received after that date will also be considered until the time when a solution is published.
5511 Proposed by Sidney Kravitz3 Dover, New Jersey.
Here is another alphametical Cook's tour for which no visas are needed:
FRANCE GREECE
FINLAND '
552 • Proposed by V.N. Murtys Pennsylvania State University, Capitol Campus3
Middletown, Pennsylvania.
Given positive constants a9b9a and nonnegative real variables x9y9z subject to
the constraint x + y + z = -n9 find the maximum value of
f(x9y 9z) = a cos x + b cos y + c cos z.
553 • Proposed by Leroy F. Meyers3 The Ohio State University.
Let X be an angle strictly between o° and 45°. Show that 2X is an acute
angle of a Pythagorean triangle (a right triangle in which all sides have integer
lengths) if and only if X is the smallest angle of a right triangle in which the
legs (but not necessarily the hypotenuse) have integer lengths.
55^, Proposed by G.C. Giri3 Midnapore College, West Bengal, India.
A sequence of triangles {A0, tsl9 A2, ...} is defined as follows: A0 is a
qiven triangle and, for each trianqle A in the sequence, the vertices of A are - n n\\
the points of contact of the incircle of A with its sides. Prove that A "tends to" v n n
an equilateral triangle as rc + °°.
- 184 -
5551 Proposed by Michael W. Ecker, Pennsylvania State Universitys
Worthington Soranton Campus.
An n-persistent number is an integer such that, when multiplied by l,2,...,n,
each product contains each of the ten digits at least once. (A persistent number
is one that is n-persistent for each n = 1,2,3 It is known ["1979: 163] that
there are no persistent numbers. The number
N = 526315789473684210
given there is n-persistent for I<n<l8, but is not 19-persistent.)
(a) Find a 19-persistent number.
(b) Prove or disprove: for each n, there exist n-persistent numbers.
5561 Proposed by Paul Erdos, Mathematical Institute, Hungarian Academy of
Sciences.
Every baby knows that
(n+l)(n+2)...(2n) n(n-l)...2.1
is an integer. Prove that for every k there is an integer n for which
n(n-±)...(n~k+l) u ;
is an integer. Furthermore, show that if (1) is an integer, then k = o(n)> that is,
k/n -* 0.
5571 Proposed by Hayo Ahlburg, Benidorm3 Alicante3 Spain.
For the geometric progression 2, 14, 98, 686, 4802, we have
(2 + 14 + 98 + 686 + 4802)(2 - 14 + 98 - 686 + 4802) = 22 + 142 + 982 + 6862 + 48022 .
Prove that infinitely many geometric progressions have this property.
5581 Proposed by Andy Liu* University of Regina.
(a) Find all n such that an n xn square can be tiled with L-tetrom1noes,
(b) What if X-pentominoes are also available, in addition to L-tetrominoes?
(The term PENTOMINOES has been, since 15 April 1975, a registered trademark of
Solomon W, Golomb.)
5591 Proposed by Charles W. Trigg, San Diego, California. k
Are there any positive integers k such that the expansion of 2 in the
decimal system terminates with kl
- 185 -
5601 Proposed by Basil C. Eennies James Cook University of North Queensland*
Australia.
Take a complete quadrilateral. On each of the three diagonals as diameter,
draw a circle. Prove that these three circles are coaxal. * * *
S O L U T I O N S
No problem is ever permanently closed. The editor will always be pleased to consider for publication new solutions or new insights on past problems,
!̂64i C1979: 2001 Proposed by J. Chris Fisher and E.L. Koh5 University of
Eegina.
(a) If the two squares ABCD and AB'C'D' have vertex A in common and are
taken with the same orientation, then the centers of the squares together with the
midpoints of BD' and BfD are the vertices of a square.
(b) What is the analogous theorem for regular n-gons?
I. Solution by Leroy F. Meyerss The Ohio State University.
The problem can be considerably generalized. In the following theorem, the
polygons need not be regular, nor convex, nor even simple. In fact, they may be
degenerate.
THEOREM. Let AjA2...A and AJA^.-.A' be similar and similarly oriented poly
gons in the plane. Let Bj, B2, ..., B be points in the plane such that the (pos
sibly degenerate) triangles A. A IB. for £ = l,2,.,.,n are similar and similarly
oriented. TJten B,B,...B and A,A,,.,A are similar and similarly oriented. 1 z n l z n
Proof. We imbed the configuration in a complex plane and we identify the
points A., A*., and B. with their affixes. The similarity of AiA?...A to AJA^...A' % % % n n
is equivalent to the existence of a complex constant a such that
A'. -A{ = ce(A. - A x ) , i = 1,2,... ,n.
The similarity of the triangles A ^ A ^ is equivalent to the existence Of a complex
constant 8 such that B. - A. = B(A'. - A.) or
B. = (1 -B>A. +BA1, i = 1,2,....n. t i t -
Hence B^-Rj = ( l - G K A ^ - A j + B f A l - A J ) = (i-g + a B)(A - A , ) , t = l,2,,..,n.
- 186 -
Since l-3 + a$ is constant, B^.-.B is similar and similarly oriented to
A1A2...AM. D
One answer to part (b) of our problem is the special case in which
AjA2...An is a regular polygon and 8 = 1. Part (a) is a special case of t h i s special case for n = 4. Observe here that then B1B2B3B«t is a square whether or
not the two given squares have a common vertex. But if l\t = Ag, then B! and B3
are the centers of the given squares. We then obtain the very special case of
part (a) by the identification
AiAzAaA,, = ABCD and A{A2A|Ai = C'D'AB*.
II. Solution and comment by M.S. Klamkin, University of Alberta.
More general results are already known and these are the subject of a joint
incomplete paper of myself and Leon Bankoff. For earlier references to related
results, see El] and [2], One general result is the following:
Let A.,B .,C .,.. .3 i = 1929.., 9m, be complex numbers denoting the vertices of % is is
a set of n directly similar m-gons, all lying in the same "plane. Also, let there
be weights ap., 3p-j YP.j etc., at vertices A.,B.,C5 etc., respectively, for V is is Is Is Js
i =1,2,...,m. Finally; let 0. denote the centroid of the weights at vertex i
of A.9B.,C.j etc. Then, for i = l929...9m, the $. are also vertices of an m-gon
directly similar to the m-gons A.9B.,C.j etc. Proof. The centroids ^ are given by
aA^ + BB^ + yZi + ... *i= Ct + tf+Y-f ... ' ^ = 1,2,...,777. ( 1 )
By similarity, for r = 2,3,...,
A , -A = z (A -A J , B -B = z (B -B J , C ,-C = z (C -C J , etc. r+1 r rv r r-1 r+1 r rK r r-1 r+1 r r r r-1'
Then by (1), $ - $ = 3 ($ - $ .)
r+1 r r r r-1
and thus the polygon of vertices $. is directly similar to the polygons of vertices
A^B..,C., etc. D
Note that part (a) of the proposed problem just corresponds to the special
case 777 = 4, n = 2, a = B = Y = • • • • Also solved by LEON BANKOFF, Los Angeles, California; JORDI DOU, Escola Tecnica
Superior Arquitectura de Barcelona, Spain; MILTON P. EISNER, J. Sargeant Reynolds Community College, Fichmond, Virginia; HOWARD EVES, University of Maine; J. CHRIS FISHER, University of Regina (in addition to the joint solution with his coproposer);
- 187 -
JACK GARFUNKEL, Flushing, N.Y.; G.P, HENDERSON, Campbellcroft, Ontario? J.A. McCALLUM, Medicine Hat, Alberta; LEROY F. MEYERS, The Ohio State University (second solution); KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; DAN SOKOLOWSKY, Antioch College, Yellow Springs, Ohio; and the proposers.
Editor's comment.
Part (a) of the problem, as given with the red herrinq stipulation of a common
vertex, can easily be proved synthetically. I outline a proof given by several
solvers (make a figure!):
Let P and Q be the centres of ABCD and AB'C'D', and R and S the midpoints of
BD' and B'D. We see that PS and QR are both parallel to and equal to one-half of
BB', and that PR and QS are both parallel to and equal to one-half of DD'. Since
a rotation of 90° with centre A takes BB* into DD', it follows that BB' is equal and
perpendicular to DD', and hence that PSQR is a square. D
The proposers mentioned that proofs of part (a) (different from the above) are
given in Finsler and Hadwiger T31 and in Neumann T41. As their answer to part (b),
they gave the following theorem (which will also appear in the forthcoming joint
paper [5] by one of the proposers):
For any n > 3 the set of regular, convex, counterclockwise-oriented n~gons in
the Euclidean plane is a two-dimensional vector space over the complex numbers.
Part (a) easily follows from this theorem for if the vertices are properly
associated, with ABCD-*•—>-C'D'AB', then the new figure is half the sum of the two
squares and is therefore a square itself.
Finally, as a generalization of part (b), Eves gave without reference the
following theorem of H. Van Aubel: If the lines joining corresponding points of
two directly similar figures are divided in the same ratio, the points of section
form a figure directly similar to the given ones,
REFERENCES
1. American Mathematical Monthly, 39 (1932) 46, 291, 535, 559; 40 (1933) 36,
157; 41 (1934) 330, 370; 44 (1937) 525; 50 (1943) 64; 76 (1969) 698.
2. J.G. Mauldon, "Similar Trianqles", Mathematics Magazine, 39 (May 1966)
165-174.
3. P. Finsler and H. Hadwiger, "Einige Relationen im Dreieck," Comment.
Math. Helvetici,10 (1937) 316-326.
4. B.H. Neumann, "Some Remarks on Polygons," Journal of The London Mathematical
Society, 16 (1941 "> 230-245 (Theorem 3,5).
5. J.C. Fisher, D. Ruoff, and J. Shilleto, "Polygons and Polynomials," to
appear in the Proceedings of the Coxeter Symposium (May 1979).
- 188 -
4651 C1979: 200] Proposed by Peter A. Lindstrom, Genesee Community College,
Batavia, N.Y.
For positive integer n, let a(n) = the sum of the divisors of n and T(n)«the number of divisors of n. Show that if oin) is a prime then -r(n) is a prime.
Solution by Andy Liu, University of Regina.
By hypothesis a(n) is prime, which implies n > \ since cr(l) = l. The arithmetic
function a is multiplicative; so we may assume that the canonical factorization of
n is n = pa for some prime p and positive integer a, for otherwise o(n) would be composite. Thus we have
a+l
0(n) = £ Z— and t i n ) = a + l. p -1
Suppose x(n) is composite, say a + l=rs, with r > l and s>l. Then
and ain) is composite, for each of the two factors on the right in (l) is an integer greater than l. This contradicts the hypothesis, so x(n) is prime.
Also solved by MICHAEL ABRAMSON, 14, student, Benjamin N. Cardozo H.S., Bayside, N.Y.; RICHARD BURNS, East Longmeadow H.S., East Longmeadow, Massachusetts; CLAYTON W. DODGE, University of Maine at Orono; HOWARD EVES, University of Maine? ALLAN WM. JOHNSON JR., Washinton, D.C.; FRIEND H. KIERSTEAD, JR., Cuyahoga Falls, Ohio; ROBERT and JUDITH KNAPP, Herkimer, N.Y.; VIKTORS LINIS, University of Ottawa; LEROY F. MEYERS, The Ohio State University; GREGG PATRUNO, student, Stuyvesant H.S., New York; FERRELL WHEELER, student, Forest Park H.S., Beaumont, Texas; KENNETH M. WILKE, Topeka, Kansas; KENNETH S. WILLIAMS, Carleton University, Ottawa; and the proposer. In addition, two badly flawed solutions were received.
Editor's comment.
One of the flawed solutions purported to show that if oipa) is prime for some prime p * 2 , then necessarily a = 2 and x(pa) = 3. This is contradicted by
a(36) = the prime 1093 and T ( 3 6 ) = 7.
The other contained a flaw of heroic proportions: the solver convinced himself that if ain) is prime, then n is also prime and x(n) = 2, which deserves to be followed by a non-factorial I. We all have our bad moments.
A * *
i[66i T1979: 200] Propose" par Roger Fisehler, Universite Carleton, Ottawa.
Soient AB et BC deux arcs d'un cercle tels que arc AB >arc BC, et soit D le point de milieu de 1'arc AC (voir la figure l). Si DExAB, montrer que AE = EB + BC. (Ce thgor£me est attribue 5 Archim£de.)
- 189 -
Fiqure 1
Fiqure 2
I. Solution by Gregg Patrunoj students Stuyvesant E.S.3 New York.
Since DA = DC and /DAB = ZDCB (see Figure 2), a rotation of ADBC around D takes
C into A and B into a point B' on AB. Now ADBB' is isosceles, so EB = EB' and
AE = EB' +AB' = EB + BC.
II. Comment by Howard Eves3 University of Maine.
This theorem, known as the "Theorem on the Broken Chord," is of historical
interest, for it has been credited by Arabic scholars to Archimedes, and the attri
bution could well be correct since many of the works of Archimedes have become lost
to us. Tropfke T7] has suggested that this theorem may have served Archimedes as
a formula analogous to our
sin (x- y) - sin x cos y - cos x sin y.
To show the correspondence, one has merely to set arc AD = 2x and arc DB = 2y9 and
then successively show that, in a unit circle,
AD = 2 sinx, DB = 2 sin*/,
BC = 2$in (x -y), AE = 2sin#cost/, EB = 2cos#sinz/.
One can also use the theorem to obtain the identity
sinOr + z/) = sin x cos y + cos x sin y.
In short, Archimedes may have found the Theorem on the Broken Chord a useful tool
in his work in astronomy.
- 190 -
Also solved by MICHAEL ABRAMSON, 14, student, Benjamin N. Cardozo H.S., Bayside, N.Y.; HAYO AHLBURG, Benidorm, Alicante, Spain; LEON BANKOFF, Los Angeles, California and CHARLES W. TRIGG, San Diego, California (jointly - 7 solutions); W.J. BLUNDON, Memorial University of Newfoundland; JORDI DOU, Escola Tecnlca Superior Arquitectura de Barcelona, Spain; MILTON and GAIL EISNER, Richmond, Virginia (jointly); HOWARD EVES, University of Maine; JACK GARFUNKEL, Flushing, N.Y.; G.C. GIRI, Midnapore College, West Bengal, India; ALLAN WM. JOHNSON JR., Washington, D.C.; FRIEND H. KIERSTEAD, JR., Cuyahoga Falls, Ohio; JOE KONHAUSER, Macalester College, Saint Paul, Minnesota; VIKTORS LINIS, University of Ottawa; ANDY LIU, University of Regina; L.F. MEYERS, The Ohio State University; NGO TAN, student, J.F. Kennedy H.S., Bronx, N.Y.; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; DAN SOKOLOWSKY, Antioch College, Yellow Springs, Ohio; KENNETH M. WILKE, Topeka, Kansas; KENNETH S. WILLIAMS, Carleton University, Ottawa; JOHN A. WINTERINK, Albuquerque Technical-Vocational Institute, New Mexico; and the proposer.
Editor's comment.
According to Boyer T3] and Eves O ] , this theorem was attributed to Archimedes
by (unidentified) Arabic scholars. According to the proposer, Suter [6] attributes
the attribution to Abufl Raihan al-Biruni (973-1048) and goes on to give 13 proofs
of the theorem. Avishalom [l] stated the theorem without reference or proof, and
this caused Bankoff to offer it as a challenge problem [2], for which solutions by
Bankoff, Konhauser, and Trigg were later published [5].
Of all the solutions the editor has seen, both in and out of the literature,
none, in his opinion, rivaled the classic simplicity of our solution I.
REFERENCES
1. Dov Avishalom, "The Perimetric Bisection of Triangles," Mathematics Magazine,
36 (January-February 1963) 60-62.
2. Leon Bankoff, Proposal 195, Pi Mu Epsilon Journal, 4 (Fall 1967) 296.
3. Carl B. Boyer, A History of Mathematics, Wiley, New York, 1968, pp. 149-150.
4. Howard Eves, An Introduction to the History of Mathematics, Fourth Edition,
Holt, Rinehart and Winston, New York, 1976, p. 157.
5. Solution of Proposal 195, Pi Mu Epsilon Journal, 4 (Fall 1968) 389.
6. Heinrich Suter, "Das Buch der Auffindung der Sehnen im Kreise von Abufl
Raihan Muh al-Biruni," Bibliotheca Mathematica, 11(1910) 11-78, esp. p. 13.
7. Johannes Tropfke, "Archimedes und die Trigonometrie," Archiv fur die
Geschichte der Mathematik, 10 (1927-1928) 432-463. a -:t *
/|67. T1979: 200] Proposed by Harold N. Shapiro„ Courant Institute of
Mathematical Sciencess New York University.
- 191 -
Let n p.,., n, be given positive integers and form the vectors W 1 9..., <£,)
where, for each t = l,..., k9 d. is a divisor of n,. Letting x{d) = the number of divisors of d9 the number of these vectors is t(nl)x(n2)...x(n.). How many of
these have the property that their components are relatively prime in pairs?
Solution by L.F. Meyers3 The Ohio State University.
Let e in) be the exponent of the prime p in the canonical (or prime-power)
factorization of n. Then
Tin) = n(e (n)+l). P P
(Here the product is extended over all primes p, and makes sense since all but
finitely many of the factors are l.)
The components of the vector (dx 9dz ,...9d.) are relatively prime in pairs if and only if any prime p occurs as a factor of at most one of the components.
If d. iis such a component, then e id.) can be any integer from 1 through e (n.), U Tp Ts jp is
inclusively. Since p need not divide any component, the total number of distinct
ways in which p can occur in any acceptable vector is
e (n, ) + e (n0) + ...+ e (rc7 ) + 1 = e (N) + 1, V V V * V
where N-nln2...n,» Hence the total number of acceptable vectors is
P P
Also solved by GREGG PATRUNO, student, Stuyvesant H.S., New York; and the proposer.
'468. C1979: 200] Proposed by Viktors Linis> University of Ottawa.
(a) Prove that the equation
k\ k.2 "K, axx + a2x + ... + a x ™ - l = o,
where a,,...,a are real and k,9...9k are natural numbers, has at most n positive 1 n l n
roots. (b) Prove that the equation
k, . jp , Z, A sq m. . .r . /A. ax (x + lr +bx (x + lr + ex (x + 1) -1 = 0, (1)
where a, b9 e are real and k9 l9 m9 p, q9 r are natural numbers, has at most 14 positive roots.
- 192 -
Editor's conment.
Our proposer found this problem in the Russian journal Kvant (No. 4, 1977,
p. 30, Problem MM-39), where it was proposed by A. Kusnarenko, and a solution by
L. Limanov was published in a later issue (No. 1, 1978, pp. 31-33). Part (b) of
Limanov's solution, which is given below, was edited from a translation supplied
by V. Linis.
I. Solution of part (a) by Friend H. Rierstead* Cuyahoga Falls9 Ohio.
The equation has rc+l terms, so it can have at most n changes of sign; hence,
by Descartes' Rule of Signs, it has at most n positive roots.
II. Solution of part (b) by L. Limanov.
We will need the following fact, which is easily checked directly: if p (x) 777
is a polynomial of degree m and u,v are arbitrary real numbers, then
x (x + 1) P (x) 777
has a derivative of the form
x (x + 1) Fm+1^-
Suppose the given equation (1) has more than 14 positive roots. Then the
equation obtained by differentiating (l) has more than 13 positive roots and is
of the form
x (x + lr At(x)+x (x + ly B1(x)+x (x + 1) Cx(x) = 0, (2)
where A.(x)9 B.(x)9 Cr(x) are polynomials of degree i. If we divide (2) by
x (x + l) ~ , the resulting equation
xk'm(x + l)P~rAl(x)+xl~m(x + l)q'rBl(x)+Cl(x) = 0 (3)
has the same positive roots as (2). After two differentiations, (3) becomes
x (x + lr A3(x)+x (x + lr B3(x) - 0, (4)
which has more than 11 positive roots. We repeat the process. Dividing (4) by
xl~m~2(x + l)q''r~2 yields the equation
xk~l(x + l)p'qA3(x) +B3(x) = 0, (5)
which has the same positive roots as (*0. Finally, differentiating (5) four times
yields the equation
x (aM lr H A7(x) = 0,
which has more than 7 positive roots. This is clearly a contradiction.
- 193 -
Also partially solved (part (a) only) by MILTON P. EISNER, J. Sargeant Reynolds Community College, Richmond, Virginia; and KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India.
Editor's comment,
Limanov's solution was presumably the best of those submitted to the editors of
Kvant* His proof of part (b) is admirable, but his proof of part (a) was a proper
horror,, nearly as long as that of part (b). The Russians have apparently never
heard of Descartes' Rule of Signs; otherwise they would surely have put Descartes
before the horrors. * * *
4691 C1979: 200] Proposed by Gali Salvatores Perkins, Que'bec.
Of the conies represented by the equations
±x2 ± 2xy ± y2 ± 2x ± 2y ± 1 = 0,
how many are proper (nondegenerate)?
Solution by the proposer.
Since two quadratic forms with opposite signs give rise to the same conic,
there are only 32 distinct conies to investigate. These are given by
ax2 + 2hxy + by2 + 2gx + 2fy + 1 = 0, (1)
where each of a9b9f9g9h = ±l. The proper conies are those for which A * o , where
\a h g\
A = \h b f\
\g f il
or, as f2 =g2 =h2 = 1,
A = (a-l)(b-l)+2(fgh~l).
We have fgh-l for 16 of the 32 conies and fgh= -l for the remaining 16.
Case 1: fgh =1. Here
A ^ o «=» a = b = -1.
This occurs exactly four times, so we have 4 proper (and 12 degenerate) conies.
Case 2: fgh~ -1. Here
A = 0 «=> a = b = -1.
This occurs exactly four times, so we have 4 degenerate (and 12 proper) conies.
- 194 -
To sum up, there are 16 proper and 16 degenerate conies.
A conic of the form (l) is a hyperbola, a parabola, or an ellipse according
as ab~h2 is negative, zero, or positive. Since here
ab - h2 = ab - 1 = 0 or -2
according as a,b are equal or unequal, the 4 proper conies of Case l are all parabolas; and of the 12 proper conies of Case 2, 4 are parabolas and 8 are hyperbolas. So our
16 proper conies are yery equitably divided into 8 parabolas and 8 hyperbolas. There
is order in the universe.
Also solved by W.J. BLUNDON, Memorial University of Newfoundland; HOWARD EVES, University of Maine; G.C. GIRI, Midnapore College, West Bengal, India; F.G.B. MASKELL, Algonquin College, Ottawa; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; and FERRELL WHEELER, student, Forest Park H.S., Beaumont, Texas. One incorrect solution was received.
ft & ft
^70, T1979: 201] Proposed by Allan ton. Johnson Jr., Washington, B.C.
Construct an integral square of eleven decimal digits such that, if
each digit is increased by unity, the resulting integer is a square. (A four-digit
example is 452 = 2025 •*• 3136 = 562.)
Solution by Ferr ell Wheelers students Forest Park H.S.S Beaumonts Texas.
Let n2 and m2 be the required and the augmented squares, respectively. Then
m2 -n2 = (m-n)(m + n) = /?n,
where i?n is the repunit 11,111,111,111. It is known [1] that the prime factorization
of Rn is 21649 • 513239. So a solution (m,n)9 if there is one, must come from one
of the systems
m\n - Rlx m+n - 513239 or
m-n = 1 m-n = 21649.
The first system does not give values of m and n which have ll-digit squares, and the second gives 0,n)= (267444, 245795), from which we get the unique solution to
our problem, because here m2 and n2 both have 11 digits and n2 does not include the
digit 9:
n2 = 2457952 = 60415182025,
m2 = 2674442 = 71526293136.
Also solved by HAYO AHLBURG, Benidorm, Alicante, Spain; RICHARD BURNS, East Longmeadow H.S., East Longmeadow, Massachusetts; CLAYTON W. DODGE, University of
- 195 -
Maine at Orono? FRIEND H, KIERSTEAD, JR., Cuyahoga Falls, Ohio; J,A. McCALLUM, Medicine Hat, Alberta; L.F. MEYERS, The Ohio State University; GREGG PATRUNO, student, Stuyvesant H.S., New York; BOB PRIELIPP, The University of Wisconsin-Oshkosh; CHARLES W. TRIGG, San Diego, California; KENNETH M. WILKE, Topeka, Kansas; and the proposer. One incorrect solution was received.
Editor's comment.
Known factors of repunits can also be found in Beiler [21.
The proposer noted that the example given in the proposal is a special case
of the following rule, which Dickson [31 credits to R. Goormaghtigh: if 4= 55...56
and B- 44...45 (each to n digits), then A2 -B2 -R . Goormaghtigh also gave the
following examples for 5- and 7-digit squares:
n2 = 1152 = 13225 n2 = 22052 = 4862025 and
m2 = 1562 = 24336 m2 = 24442 = 5973136
The repunit # n has a curious property which it is almost certain has never
been noticed before. Gali Salvatore, who discovered it and communicated it to the
editor, wishes to dedicate it to that prince of digit delvers, Charles W. Trigg. Let p be the nth prime. The factors of /? M are
21649 = p 2 k 3 1 and 513239 =p I f 2 5 3 6.
The ranks of the factors of Rlx in the sequence of primes3 2431 and 42536., are
permutations of the consecutive digits 1 to 4 and 2 to 6* respectively.
Knowledge of this fact will not bring peace to the Middle East, but it should
make Trigg very happy.
REFERENCES
1. Samuel Yates, "Prime Divisors of Repunits," Journal of Recreational Mathematics, 8 (1975) 33-38.
2. Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New York,
1964, p. 84.
3. Leonard Eugene Dickson, History of the Theory of Numbers, Chelsea, New
York, 1952, Vol. I, p. 464. s'« s'« sV
^711 [1979: 228] Proposed by Alan Waynes Pasco-Hernando Community
College, New Fort Richey, Florida,
Restore the digits in this multiplication, based on a sign in a
milliner's shop window:
WE DO
TAM HAT TRIM
- 196 -
Solution by the proposer*
Clearly, none of W, D, T, H, E, 0, A, R can be zero; so either M = o or 1 = 0.
Suppose M = o. Then 0 = 5 and E is even. It follows that T is even and must
equal 2 or 4. If T = 2, then H = l and there is no value for R. If T = 4, then
A = 7, 1 = 1, and R = 2, so that TRIN = 42lo. But this integer does not factor into
two two-digit integers. Thus M * o .
Therefore 1 = 0, whence R = l and only W or H can be 5. If H = 5, then T = 6
and TRIM is one or more of 6102, 6103, 6107, 6108, or 6109. But none of these is
factorable into two two-digit integers. So W= 5 .
Then (T, H, A) = (3, 2, 7) or (4, 3, 6), The first of these leads to 0 = 6,
M = 4 or 8, with E = 9 and D = 3 = T. Hence (T, H, A) = (4, 3, 6). Finally, 0 = 9,
E = 2, M = 8, and D = 7.
The solution is, uniquely:
52 79
468 364 4108.
Also solved by MILTON P. EISNER, J. Sargeant Reynolds Community College, Richmond, Virginia; J.A.H. HUNTER, Toronto, Ontario; ALLAN WM. JOHNSON JR., Washington, D.C.; J.A. McCALLUM, Medicine Hat, Alberta; NGO TAN, J.F. Kennedy H.S., Bronx, N.Y.; CHARLES W. TRIGG, San Diego, California; FERRELL WHEELER, student, Forest Park H.S., Beaumont, Texas; and KENNETH M. WILKE, Topeka, Kansas.
ft & *
^72 i C1979: 2281 Propose* par Jordi Dou3 Esoola Tecnica Superior Arquiteotura
de Barcelona* Espagne.
Construire un triangle connaissant le c6te* b, le rayon R du cercle circonscrit, et tel que la droite qui joint les centres des cercles inscrit et circonscrit soit
parall£le au c6te* a.
Solution by Howard Evess University of Maine.
We will use the following theorem of Carnot CI], for which a proof can be
found in Altshiller Court C2]:
The sum of the distances of the circumcenter of a triangle from the three
sides of the triangle is equal to the circumradius increased by the inradius.
For a triangle ABC with circum- and in-radii R and **, Carnot's Theorem says that
i?cos A +i?cos B +i?cos C = R + r. (1)
- 197 -
Suppose a solution triangle exists. Then
i?cosA = r and (l) yields cosB + cosC = l. Thus
B is acute and b < 2/?.
Conversely, we show there is a unique
solution triangle if b<2R. Draw a circle
OCR) and mark off in it a chord AC of length
b (see figure). Draw the radius OMN perpen
dicular to AC to cut AC in M and circle 0(i?)
in N. Draw circle 0(MN), and then draw the
chord AB of circle 0(i?) tangent to circle
0(MN) and such that 0 lies within angle BAC.
Triangle ABC is the sought triangle.
The proof follows easily from Carnot's
Theorem.
Also solved by LEON BANKOFF, Los Angeles, California; J.T. GROENMAN, Arnhem, The Netherlands; LEROY F. MEYERS, The Ohio State University; FRANCISCO HERRERO RUIZ, Madrid, Spain; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India? and the proposer.
REFERENCES
1. L.N.M. Carnot, Geometrie de position* Paris, 1803, p. 168.
2. Nathan Altshiller Court, College Geometry, Barnes & Noble, New York,
1952, p. 83. it * it
4731 £1979: 229] Proposed by A. Lius University of Regina.
The set of all positive integers is partitioned into the (disjoint)
subsets Tx , Tz, T3, ... as follows: for each positive integer m9 we have m e T^
if and only if k is the largest integer such that m can be written as the sum of k
distinct elements from one of the subsets. Prove that each T^ is finite.
(This is a variant of Crux 342"[1978: 133, 2971.)
Editor fs comment.
No solution was received for this problem, which therefore remains wide open.
As the proposer noted, the problem is closely related to Crux 3tf2*Tl978: 133> 297]»
which is also unsolved. We have received for No. 342 two partial solutions, each
of which shows that the proposed theorem is true for all odd n but draws no
conclusion for even n. We make a final appeal to readers to find a complete
solution to either or both of these problems. Failing that, a partial solution to
No. 342 will be published in a few months. J'J iV :'c
- 398 -
4741 C1979: 229] Proposed by James Propp3 Harvard Colleges Cambridge,
Massachusetts.
Suppose (s ) is a monotone increasing sequence of natural numbers satisfying s =3rc for all n. Determine all possible values of s 1 9 7 g.
n
Solution by Gregg Patruno, student, Stuyvesant H.S.S New York.
We use functional notation for typographical convenience. Suppose such a
sequence s exists. Since
s(m) = s(n) => 3m = s(s(m)) = s(s(n)) = 3n *=*> m-n9
s is injective and hence strictly increasing. If s(n)^n for some n, then
3n- s(s(n)) <,s(n) <n, a contradiction; hence for all positive integers n we have
77 < s(n) < s(s(n)) = 3n.
In particular, l< e(l) < s(s(l)) = 3, from which s(l) = 2 and s(2)-3. With the
help of s(3n) -s(s{s(n))) =3s(n ) 9 easy inductions show that
s(3k)*2'3k and e(2 • 3*) = 3k+19 fc = 0,1,2
k k k k+1 Since each of the sets Av = {rc|3 <n < 2 • 3 } and B. = {n\2 • 3 <rc < 3 } contains
k exactly 3 distinct elements, we have
s(Ak) = Bk and e(Bk) = s(s{Ak)) = 3Ak = {3j|j€ilfc}.
More precisely,
(n + 3k9 if 3^ < n < 2 • 3k
9
s(n) = { , . t, , fc = 0,l,2,... . (1) (3(n-3*), if 2 -3* £ n < 3 ,
Conversely, it is easy to verify that (1) satisfies the conditions of the problem,
so it is the one and only sequence defined by the problem.
Now 2 • 36 <1979 < 37, so we have uniquely s(1979) = 3(1979 -3 6) = 3750.
Also solved by JORDI DOU, Escola Tecnica Superior Arquitectura de Barcelona, Spain; MICHAEL W. ECKER, Pennsylvania State University, Worthington Scranton Campus; MILTON P. EISNER, J. Sargeant Reynolds Community College, Richmond, Virginia; FRIEND H. KIERSTEAD, JR., Cuyahoga Falls, Ohio? LEROY F. MEYERS, The Ohio State University; FRANCISCO HERRERO RUIZ, Madrid, Spain; JAN VAN DE CRAATS, Leiden University, The Netherlands; FERRELL WHEELER, student, Forest Park H.S., Beaumont, Texas; and the proposer. Incomplete solutions were submitted by W.J. BLUNDON, Memorial University of Newfoundland; and CLAYTON W. DODGE, University of Maine at Orono.
Editor's comment. As a lagniappe, Dou calculated that s(l)+s(2)+...+s(1979) = 3342077. This sum is a prime, the 239701th prime, in fact.
