142 142
142
NAW 5/13 nr. 2 juni 2012 ProblemenPr oblemen
ProblemSectionRedactie:
Johan Bosman Gabriele Dalla Torre Jinbi Jin Ronald van Luijk Lenny Taelman Wouter Zomervrucht Problemenrubriek NAW Mathematisch Instituut Universiteit Leiden Postbus 9512 2300 RA Leiden
problems@nieuwarchief.nl www.nieuwarchief.nl/problems
This Problem Section is open to everyone; everybody is encouraged to send in solutions and propose problems. Group contributions are welcome.
For each problem, the most elegant correct solution will be rewarded with a book token worth 20 Euro. At times there will be a Star Problem, to which the proposer does not know any solution.
For the first correct solution sent in within one year there is a prize of 100 Euro.
When proposing a problem, please either include a complete solution or indicate that it is intended as a Star Problem. Electronic submissions of problems and solutions are preferred (problems@nieuwarchief.nl).
The deadline for solutions to the problems in this edition is September 1st, 2012.
Problem A(proposed by Simone Di Marino)
LetP andQbe distinct points in the plane. Letn ≥ 2. Assumendistinct lines throughP but not throughQare given, as well asndistinct lines throughQbut not throughP. LetT be a collection of2nintersection points of these lines. Suppose that the (unoriented) angle between the linesRPandRQis the same for allRinT. Show thatTcan be partitioned into subsets of at least three elements each, such that every subset consists of the vertices of a regular polygon.
Problem B(proposed by Apoloniusz Tyszka)
Show that there exist ann ≥ 1, a polynomialP ∈Z[X, Y1, . . . , Yn]and an infinite setS of positive integers such that the set
n(y1, . . . , yn) ∈Zn:P (k, y1, . . . , yn) = 0o
is empty for allk < 0and has preciselykelements for allk ∈ S.
Problem C(proposed by Gabriele Dalla Torre)
Is it possible to tile a30by30square grid using the following blocks?
Edition 2011-4 We received solutions from Rik Bos, Pieter de Groen, Alex Heinis, Nicky Hekster and Merlijn Staps.
Problem 2011-4/A LetΓbe a finite undirected graph (without loops or multiple edges). Denote the set of vertices byV. Assume that there are a functionf : V →Zand a positive integern such that
X
v
X
w
(f (v) − f (w)) = 2n,
wherevruns over all the vertices ofΓandwover all the neighbours ofv. Show that there are anm ≤ nand a collection ofmedges such that the graph obtained fromΓby removing those edges is not connected.
143 143
Problemen NAW 5/13 nr. 2 juni 2012
143
Oplossingen
SolutionsSolution We received solutions from Pieter de Groen, Alex Heinis and Merlijn Staps. The book token goes to Pieter de Groen. The following is based on the solutions of Alex Heinis and Pieter de Groen.
For a vertexv ∈ Vwe write
δ(v) =X
w
f (v) − f (w),
wherewruns over all the neighbours ofv. Denote byV+the set of verticesvwithδ(v) ≥ 0. SincePv∈Vδ(v) = 0andPv∈V|δ(v)| = 2n, we must havePv∈V+δ(v) = n.
LetMbe the maximal value off. Consider the partitionV = A ∪ BwithA = f−1(M)and B = V \ A. Sincefis not constant, we know thatAandBare non-empty. ClearlyAis a subset ofV+.
LetEbe the set of edges connectingAandB. Clearly the group obtained by removingEfromΓ is not connect. Moreover, we have
|E| = X
e∈E
1 ≤ X
v∈A
δ(v) ≤ X
v∈V+
δ(v) = n,
where the first inequality follows because for every neighbourwof av ∈ Awe havef (v) − f (w) ≥ 1.
Problem 2011-4/B Letǫbe a positive real number. Show that there is a finite groupGthat is not a2-group, but in which the proportion of elements that have2-power order is at least1 −ǫ.
Solution We received solutions from Nicky Hekster, Alex Heinis and Merlijn Staps. The book token goes to Merlijn Staps. The following is based on the solution of Nicky Hekster.
Letn > 0be such that2−n< ǫ. Letpbe a prime number congruent to1modulo2n. LetCbe the unique subgroup of Fp×of order2n. Consider the subgroup
( a b
0 1
!
:a ∈ C, b ∈Fp
)
of the group of invertible 2by2matrices over Fp. Then the order ofG isp2n. For every a ∈ C \ {1}andb ∈Fpthe matrix
a b 0 1
!
is conjugate to the matrix
a 0 0 1
!
and thus has2-power order. Therefore the proportion of elements of2-power order is at least (2n− 1)/2n> 1 − ǫ.
Problem 2011-4/C LetBbe a commutative ring andAa subring ofB. Assume that the additive group ofAhas finite index inB. Show that the unit group ofAhas finite index in the unit group ofB.
Solution We received solutions from Rik Bos, Nicky Hekster and Alex Heinis. The book token goes to Nicky Hekster. All solutions were similar to the following.
144 144
144
NAW 5/13 nr. 2 juni 2012 ProblemenOplossingen
SolutionsThe setI = {a ∈ A : aB ⊂ A}is an ideal in bothAandB. The kernelUof the homomorphism B×→ (B/I)×is contained in1 +I, which is contained inA. BecauseUis a multiplicative group, we findU ⊂ A×, so it suffices to show thatB×/Uis finite. As this injects into(B/I)×, we are reduced to showing that the latter is finite.
Note thatB/Ais a finite group, so the endomorphism ring End(B/A)is finite. Note also that the kernel of the ring homomorphismB →End(B/A)that sendsbto the map(x 7→ bx)is exactlyI, so we get an injectionB/I →End(B/A). It follows thatB/Iis finite as well, and therefore so is (B/I)×.
Correction to Problem 2011-3/A
Problem 2011-3/AFix a pointPin the interior of a face of a regular tetrahedron∆. Show that∆can be partitioned in four congruent convex polyhedra such thatPis a vertex of one of them.
CorrectionWe thank Jan C. Smit (Nieuwegein) for kindly notifying us that our solution to this problem was incomplete. Indeed, referring to the published solution, the polyhedra we constructed are convex only in the case thatPdlies inside the triangleQabQbcQac. Jan Smit also gave the following fix, which is identical to some of the solutions we had received originally. Assume without loss of generality thatPdlies inside the triangleAQabQac. Then the polyhedronAPbPcPdQabQacQadZis not convex. The plane that containsQad,Qbc, Z,PaandPd cuts this polyhedron into two parts. One checks that the union of one of these parts and the image of the other part under the rotation(AD)(BC)is also a convex polyhedron and the images of this union underV4give the desired partition.
