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Problemen NAW 5/16 nr. 1 maart 2015
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Pr oblemen
ProblemSectionRedactie:
Gabriele Dalla Torre Christophe Debry Jinbi Jin Marco Streng 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 D20. 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 ofD100.
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 1 June 2015.
Problem A(proposed by Raymond van Bommel and Julian Lyczak)
A commutative ringRis charming if every ideal ofRis an intersection of maximal ideals. Prove that a Noetherian charming ring is a finite product of fields. Does there exist a charming ring that is not a product of fields?
Problem B(folklore)
Let S be a set of prime numbers with the following property: for alln ≥ 0 and distinct p1, . . . , pn ∈ S the prime divisors ofp1· · · pn+ 1are also inS. Show thatS contains all primes.
Problem C(proposed by Roberto Stockli)
Determine all pairs(p, q)of odd primes withq ≡ 3mod 8 such that1p(qp−1− 1)is a perfect square.
Edition 2014-3 We received solutions from Pieter de Groen (Brussels), Thijmen Krebs (Noot- dorp), Tejaswi Navilarekallu and Hendrik Reuvers (Maastricht).
All three problems of 2014/3 asked for a construction with origami in a limited number of moves.
More precisely, given a collection of points and lines (or folds) in the plane, a move (cf. the Huzita–Justin–Hatori axioms) consists of adding to the collection one of the following:
− a fold aligning two distinct points;
− a fold aligning two distinct lines;
− if it exists, a fold having two properties of the following types (except for type 3, one may have two distinct alignments of the same type):
1. the fold aligns a point with a line;
2. the fold passes through a point;
3. the fold is perpendicular to a line;
− a sufficiently general fold having at most one property of types 1, 2 and 3.
Problem 2014/3-A. Given three pointsA,BandC, and a linelpassing throughC, construct in at most six moves a pointDon the linelsuch that|CD| = |AB|.
SolutionWe received solutions from Pieter de Groen, Thijmen Krebs, Tejaswi Navilarekallu and Hendrik Reuvers. The book token goes to Thijmen Krebs. The following is based on his solution.
We give a solution in five moves.
We first assume thatACis not perpendicular tol, and thatAdoes not lie onl.
− Make the foldl1aligningAwithC.
LetE = l ∩ l1. (This uses the assumption thatACis not perpendicular tol.)
− Make the foldl2throughAandE.
− Make a foldl3throughAaligningBwithl2.
− Make the foldl4throughBperpendicular tol3.
LetB′ = l2∩ l4. (If l2 = l4, takeB′ = Binstead.) Asl3 is an angular bisector of∠BAB′, andl4= BB′is perpendicular tol3, it follows that△ABB′is isosceles with apexA. Therefore
|AB′| = |AB|.
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NAW 5/16 nr. 1 maart 2015 ProblemenOplossingen
Solutions− Make the foldl5throughB′perpendicular tol1.
LetD = l ∩ l5. (This uses the assumption thatAdoes not lie onl.) Asl1 andl5 both are perpendicular toAC, by the previous argument, it follows that|CD| = |AB′| = |AB|, as desired.
If in the above case,Alies onl, butBdoes not, then we can simply switch the roles ofAandB in the above.
Now assume that eitherACis perpendicular tol, or that bothA, Blie onl.
− Make the foldl1aligningAwithC.
− Make a foldl2aligning bothAandBwithl1.
LetE = l1∩ l2. (Note thatl1is parallel tol2if and only ifABis parallel tol, but in that case we could have constructedDin one move in the first place.)
− Make the foldl3throughBperpendicular tol2. LetF = l1∩ l3.
− Make the foldl4throughEaligningCwithl1, so thatl4is the reflection ofl2inl1.
− Make the foldl5throughFperpendicular tol4.
LetD = l ∩ l5. Moreover, letA′be the auxiliary point that is the reflection ofAinl2. Then, arguing in a similar way as in the previous case, we see that|A′F | = |AB|in both cases. IfA, B lie onl, then we also have|CD| = |A′F | = |AB|by the same argument. IfACis perpendicular tol, thenlandl1 are parallel, and so areA′C and l5; soCDF A′is a parallelogram, and
|CD| = |A′F | = |AB|.
Problem 2014/3-B. Construct a golden rectangle (including its sides) in at most eight moves.
SolutionWe received solutions from Pieter de Groen, Thijmen Krebs, Tejaswi Navilarekallu and Hendrik Reuvers. The book token goes to Tejaswi Navilarekallu. The following is based on his solution, which has similar ideas to those of Pieter de Groen and Thijmen Krebs.
− Make a foldl1.
− Make a foldl2perpendicular tol1.
− Make a foldl3(distinct froml2) perpendicular tol1.
− Make the foldl4aligningl1andl2.
LetA = l1∩ l2,B = l1∩ l3, and letX = l3∩ l4. Then△ABXis an isosceles triangle with apex B, so|AB| = |BX|.
− Make the foldl5aligningBandX.
LetY = l3∩ l5. Then|BY | =12|BX| =12|AB|, so|AY | =12√5.
− Make the foldl6throughYaligningAwithl3such thatAandBlie on the same side ofl6.
− Make the foldl7throughAperpendicular tol6.
LetC = l3∩ l7. Asl6is an angular bisector of∠AY C, andl7is perpendicular tol6, the triangle
△ACY is isosceles with apexY. Therefore|CY | = |AY | = 12
√5, and|BC| = |BY | + |CY | =
1 2+12√
5, i.e.A, B, Cform three vertices of a golden rectangle.
− Make the foldl8throughCperpendicular tol3. NowABCDis a golden rectangle (with sidesl1, l2, l3, l8).
Problem 2014/3-C. Given two pointsAandB, construct in at most four moves the pointCon the segmentABsuch that|AC| = 13|AB|.
SolutionWe received solutions from Pieter de Groen, Thijmen Krebs, Tejaswi Navilarekallu and Hendrik Reuvers. The book token goes to Hendrik Reuvers. The following solution is based on that of Thijmen Krebs.
− Make the foldl1aligningAwithB.
− Make a foldl2aligningAwithl.
LetP = l1∩ l2. LetA′be the reflection ofAinl. (Note thatA′is not a point that we have constructed; we only use it as an auxiliary point for the following arguments.) Then|A′B| = |AB|, and aslis the perpendicular bisector ofAB, it follows thatAA′Bis equilateral, and thatPis its centroid. Therefore, ifB′is the midpoint ofAA′, we have|PB′| =13|BB′|.
− Make the foldl3throughPperpendicular tol2.
− Make the foldl4throughAandB.
LetC = l3∩ l4. Asl3is parallel toAA′, it follows that|AC| =13|AB|, as desired.
