The Crease: Torsten Sillke, 1999-05-04
last update 2002-06-02
Problem (A): [B. J. Battersby, J. S. Madachy] [Mathematical Quickies Q 224]
Two opposite vertices of an 'a' by 'b' rectangle are brought into
coincidence and the rectangle is flattened out to form a crease.
Find the length of the crease.
Problem (A1):
Apply the formula for (A) to a sheet of paper with DIN A4 format.
Problem (A2):
Apply the formula for (A) to a german 10DM bank note (series II or III).
The size is (a,b) = (130mm, 65mm) +- 0.6mm. Calculate the exact,
the minimal and maximal length of the crease.
Problem (B): [Longley-Cook]
Two opposite vertices of an rectangle are brought into coincidence
and the rectangle is flattened out to form a crease. The crease
divides the width of the rectangle with ratio 2:1.
Find the ratio of crease:width.
Problem (C): [B. Barwell]
A rectangular sheet of paper measures 84 centimeters wide.
When folded so that one of the corners meets the midpoint
of the opposite short side, the crease is excactly 87.5
centimeters long. What is the lenght of the longer side
of the paper?
Problem (D): [R. R. Rowe] [Mathematical Quickies Q 107]
A rectangular card is folded through one corner so that the adjacent
corner falls on a side, thus forming three right triangles with areas
in arithmetic progression. If the area of the smallest triangle is
3 square inches, what is the area of the largest? (Difficult)
Problem (E): [H. E. Dudeney][Martin Gardner, Chap. 16], a shortest crease
Fold a page, so that the bottem outside corner touches the inside edge
and the crease is the shortest possible. The crease should go through
adjacent edges of the rectangle.
Problem (F): [L. R. Chase]
Why is it that when we fold a sheet of paper the crease is a straight line?
Problem (G):
Fold any rectangular piece of paper into 5 pieces of equal area,
without using any instrument.
Problem (H): [J. Kato]
Given a triangle with side length 2,3,4.
Make a tetrahedron by folding three lines.
Problem (I): [Mathematical Quickies Q 79]
Can any sealed rectangular envelope, after a single straight cut,
be folded into two congruent tetrahedrons?
Further problems of this kind are mentioned in the references.
Hint (A):
Determine the lenght of the diagonal.
Then look for similar triangles and find the relation
diagonal(a,b) : crease(a,b) = a : b
Therefore we have diagonal(1,1/x) = crease(x,1).
Hint (A1):
Definition of the DIN formats:
- halving the sheet of paper gives two similar ones.
- the area of DIN A0 is 1 m^2.
Hint (C):
Double the paper lenght and folded as in problem (A).
Let the dimension of the double paper be (a,b) and the
crease c, show: a:b = b:sqrt(c^2-b^2).
Solution (A): Alan Wayne
D---Q-------C
| \ |
| O |
| \ |
A-------P---B
Let the rectangle be ABCD, with length a=AB=CD, and width
b=AD=BC, where a>=b. Let O be the center of the rectangle.
Folding A up to C means a reflection A<->C on the crease PQ.
Therefore the crease PQ is perpendicular to AC at O.
From the similarity of triangles APO and ABC, it follows that
PO : BC = AO : AB, or PQ : AC = BC : AB. That is
PQ = (b/a) * sqrt(a^2 + b^2) = b * sqrt(1 + (b/a)^2)
Solution (A1):
crease(A4) = diagonal(A5) = sqrt(3) / sqrt(2)^5.5 meter
= 0.25747 meter.
Solution (A2):
crease = 65/2 sqrt(5) mm = 72.67 mm.
Now look at the perturbated rectangle (a',b') with
a' = a*(1-x) and b' = b*(1+y). The crease is monotonic
increasing in x and y. The first order error term is
crease(x,y) = crease * (1 + y + b^2/(a^2 + b/2)*(x+y))
For (x,y) = 0.6*(1/130, 1/65) we get crease(x,y) = crease * 1.012
Solution (B):
Using the notation of the diagram of solution (A) we see
CP = CQ by symmetry and PQ = CP by congruent triangles BCP
and QPR where R is the orthogonal projection of P on CQ.
Therefore the ratio of crease:width is 3:2.
Solution (C):
a = 288 cm. The longer side measures 144 cm.
Solution (D):
Area = 5 + 2 sqrt(13) = 12.2111 [square inch]
R. R. Rowe, Civil Engineering-ASCE 18, Februar 1948, p. 70.
Solution (E): Heinrich Hemme
| E
| / |
| / |
| / | The triangles CAE and CDE are congruent.
D / | The crease CE is perpendicular to AD at F.
| / |
| F |
| / |
B----C--------A
Let AB = b, AF = DF = h, BD = x, CF = c1, EF = c2, and the crease CE = c.
The triangles BAD, FEA, and FAC are similar
(1) b : x = c2 : h = h : c1
and with Pythagorean thm we have
(2) (2h)^2 = x^2 + b^2 right triangle BAD.
Now the crease c = c1 + c2 and eliminating c1 and c2 by (1) gives
(3) c = h*(x/b + b/x) = h*(x^2 + b^2)/(bx)
(4) c^2 = (x^2 + b^2)^3 / (4 x^2 b^2) by (2)
Applying the arithmetic geometric mean inequality (x1+x2+x3)^3 >= 27 x1 x2 x3
to the numerator with (x1, x2, x3) = (x^2, b^2/2, b^2/2) we get
(5) c^2 >= 27/16 b^2 with equality if and only if x^2 = b^2/2.
So the minimum is at
(x, c) = (b/sqrt(2), 3b*sqrt(3)/4).
Solution (E): Bernhard Wiezorke
| E
| / |
| / |
| / | The triangles CAE and CDE are congruent.
D / | The crease CE is perpendicular to AD at F.
| / |
| F |
| / |
B----C--------A
Let x = AD, y = BD, z = CE, a = AC, b = AE, and set AB = 1.
We know that AF = DF = AD/2 = x/2.
The triangles BAD, AEC, and FAC are similar
(1) x : 1 = z : b = a : x/2
and with Pythagorean thm we have
(2) z^2 = a^2 + b^2 right triangle AEC.
In (2) a and b can be eliminated with (1) giving
(3) z^2 = x^4/4 + z^2/x^2.
(4) z^2 = x^6 / ( 4 (x^2 - 1) )
Applying the arithmetic geometric mean inequality (x1+x2+x3)^3 >= 27 x1 x2 x3
to the numerator with (x1, x2, x3) = (x^2-1, 1/2, 1/2) we get
(5) z^2 >= 27/16 with equality if and only if x^2 - 1 = 1/2 <=> x^2 = 3/2.
So the minimum is at
(x, y, z, a, b) = (sqrt(6)/2, sqrt(2)/2, 3sqrt(3)/4, 3/4, 3sqrt(2)/4).
Solution (E): Torsten Sillke
| E
| / |
| / |
| / | The triangles CAE and CDE are congruent
D / |
| / |
| / |
| / |
B----C--------A
We want to minimize the ratio crease:width that is CE/AB.
The crease CE folds A to D.
Let phi = angle(AEC) = angle(DEC) = angle(CAD) = angle(CDA) then
psi = angle(BCD) = 2phi as the outer sum angle(BCD) = angle(CAD) + angle(CDA).
Let the length of the crease CE = 1.
Then AC = sin(phi) = s and BC = CD * cos(psi) = AC * cos(2phi).
AB = AC + BC = AC * ( 1 + cos(2phi) )
= 2 s (1 - s^2) as 1 + cos(2phi) = 2 cos^2(phi) = 2 ( 1 - sin^2(phi) ).
Our aim is to maximize AB.
Solve dAB/ds = 0 for 1/sqrt(2) >= s >= 0.
From dAB/ds = 2 - 6 s^2 we conclude s^2 = 1/3.
So we have AC = s = 1/sqrt(3) and AB = 4*sqrt(3)/9 and the ratio AC:BC = 3:1.
The crease:width ratio CE/AB is 3*sqrt(3)/4 = 1.2990381057.
Furthermore as AE = cos(phi) we get AE/width = 3*sqrt(2)/4 = 1.0606601718.
Boundary cases:
s = 0 AB = 0 (Then AC = 0)
s = 1/sqrt(2) AB = 1/sqrt(2) (Then BC = 0)
Conclusion: For height:width >= 3*sqrt(2)/4 take the local maximum.
Otherwise select s as large as possible, the crease meats the upper corner.
Variation: Maximizing AB^2 = 4 s^2 (1 - s^2)^2
Applying the arithmetic geometric mean inequality x1 x2 x3 <= (x1+x2+x3)^3/27
to (x1, x2, x3) = (2s^2, 1-s^2, 1-s^2) we get for 0 <= s^2 <= 1
AB^2 <= 16/27 with equality if and only if 2s^2 = 1 - s^2 <=> s^2 = 1/3.
And at s=1/sqrt(3) is indead the maximal value.
Note: If we use DIN-paper then BD : AE : height = 2 : 3 : 4.
So the crease is parallel to one diagonal.
Solution (E) Dankwart Vogel:
F E
| / |
| / |
| / | The triangles CAE and CDE are congruent
D / |
| / |
| / |
| / |
B----C--------A
Let x = BC, y = CE, a = BD, b = DF, and AB = EF = 1 then AE = BF = a + b.
The triangles DFE and CBD are similar
(1) b : 1 = x : a
and with Pythagorean thm we have
(2) y^2 = (1-x)^2 + (a+b)^2 right triangle CAE
(3) a^2 = (1-x)^2 - x^2 = 1 - 2x right triangle CBD.
In (2) b can be eliminated by (1) and then a^2 by (3)
(4) y^2 = (1-x)^2 + (a + x/a)^2
= (1-x)^2 + a^2 + 2x + x^2/a^2
= (1-x)^2 ( 1 + 1/(1 - 2x) )
= 2 (1-x)^3 / (1 - 2x).
Applying the arithmetic geometric mean inequality (x1+x2+x3)^3 >= 27 x1 x2 x3
to the numerator of (4) with (x1, x2, x3) = (1/2 - x, 1/4, 1/4) we get for x <= 1/2
(5) y^2 >= 27/16 with equality if and only if 1/2 - x = 1/4 <=> x = 1/4.
So the minimum is at
(x, y, a) = (1/4, 3sqrt(3)/4, 1/sqrt(2)).
Solution (E) [Martin Gardner, chap 16, p144-145]:
F E
| / |
| / |
| / | The triangles CAE and CDE are congruent
D / |
| / |
| / |
| / |
B----C--------A
Let AC = CD = x, BD = y and AB = a then BC = a-x.
with the Pythagorean thm we have
(1) y^2 + (a-x)^2 = x^2 right triangle CBD.
That is y^2 = a (2x - a)
[Gardner skips some steps]
The crease z is
(4) z^2 = x^3 / (x - a/2)
Applying the arithmetic geometric mean inequality (x1+x2+x3)^3 >= 27 x1 x2 x3
to the numerator of (4) with (x1, x2, x3) = (x - a/2, a/4, a/4) we get for x >= a/2
(5) z^2 >= 27/16 with equality if and only if x - a/2 = a/4 <=> x = 3/4 a.
Alternate Solution (E):
F E
| / |
| / |
| / | The triangles CAE and CDE are congruent
D / |
| / |
| / |
| / |
B----C--------A
Let x = CA, y = BD, z = DF, w = AB = EF,
a = AC = CD, b = AE = DE, c = CE then
(1) x^2 + y^2 = a^2 right triangle DBC
(2) w^2 + z^2 = b^2 right triangle EFD
(3) a^2 + b^2 = c^2 right triangle CAE
(4) w - x = a
(5) y + z = b
First eliminate the variables a and b.
(6) x^2 + y^2 + z^2 + w^2 = c^2 <=> (1) + (2) + (3)
(7) w^2 - y^2 = 2wx <=> (4)^2 - (1)
(8) w^2 - y^2 = 2yz <=> (5)^2 - (2)
Now try to find the minimal c^2 with the constraints (7) and (8).
Find the minimum with Lagrange multipiers p and q.
(9) L(x,y,z,p,q) = x^2 + y^2 + z^2 + w^2 + p (w^2 - y^2 - 2wx) + q (w^2 - y^2 - 2yz)
Now the partial differential quotients must be zero.
dL/dx = 2x - 2pw = 0 that is
(10) p = x/w
dL/dz = 2z - 2qy = 0 that is
(11) q = z/y
dL/dy = 2y + 2py + 2q(y+z)
= 2y + 2py + 2qy(1+q) = 0 and dividing by y gives
(12) 1 - p - q - q^2 = 0
(13) w^2 - y^2 = 2p w^2 by eliminating x from (7) and (10)
(14) w^2 - y^2 = 2q y^2 by eliminating y from (8) and (11)
Now (13) - (14) give 2p w^2 = 2q y^2 which is
(15) p/q = y^2/w^2.
(16) q - p = 2pq by eliminating y^2 from (13) and (15) (or (14) and (15)).
(17) 2q^3 + 3q^2 - 1 = (2q-1)(q+1)^2 = 0 Eliminating p from (16) and (12).
The only positive root is q=1/2 which gives the solution
(p, q, y/w) = (1/4, 1/2, 1/sqrt(2)).
The negative root gives (1, -1, sqrt(-1)).
Solution (F): [L. R. Chase], [M. Gardner, chap 16]
Solution (G): Marco Manfredini marco@taris.de
A-+-+-+-+-H-+-+-+-+-D
+-+-+-+-+-+-+-+-+-+-+
+-+-+-+-+-+-L-+-+-+-+
+-+-+-+-+-+-+-+-+-+-+
+-+-I-+-+-+-+-+-+-+-+
E-+-+-+-+-+-+-+-+-+-G
+-+-+-+-+-+-+-+-K-+-+
+-+-+-+-+-+-+-+-+-+-+
+-+-+-+-J-+-+-+-+-+-+
+-+-+-+-+-+-+-+-+-+-+
B-+-+-+-+-F-+-+-+-+-C
Fold AB on DC to get HF and unfold.
Fold AD on BC to get EG and unfold.
Make a crease along AF, BG, CH, and DE.
You get four right triangles AID, BJA, CKB, and DLC
plus a square IJKL in the middle. Each figure has
the same area.
Notice that this is the inverse of the old puzzle
dissecting the cross (5 squares) and build a square.
Further notice that horizontal lines through the points
of the inner square gives you five stripes of equal hight.
Marco Manfredini made a nice picture of this configuration
http://tunix.is-bremen.de/~taris/marco/foldingpaper.gif
The problem to determine the ratio of the square to the
inner square in this configuration is much older.
[W.J.C. Miller, Educational Times 59, 1893, p51]
A related construction of (G):
==> geometry/dissections/square.five.p <==
Can you dissect a square into 5 parts of equal area with just a straight edge?
==> geometry/dissections/square.five.s <==
1. Prove you can reflect points which lie on the sides of the square
about the diagonals.
2. Construct two different rectangles whose vertices lie on the square
and whose sides are parallel to the diagonals.
3. Construct points A, A', B, B' on one (extended) side of the square
such that A/A' and B/B' are mirror image pairs with respect to another
side of the square.
4. Construct the mirror image of the center of the square in one
of the sides.
5. Divide the original square into 4 equal squares whose sides are
parallel to the sides of the original square.
6. Divide one side of the square into 8 equal segments.
7. Construct a trapezoid in which one base is a square side and one
base is 5/8 of the opposite square side.
8. Divide one side of the square into 5 equal segments.
9. Divide the square into 5 equal rectangles.
Solution (H): [J. Kato]
Solutions can be found in
Ed Pegg www.mathpuzzle.com/katotriangle.htm and
Eppstein www.ics.uci.edu/~eppstein/junkyard/toshikato.html
This disproves [Martin Gardner, 1971, Chapter 19 Problem 2]
where it is stated that only acute triangles can be folded
into a tetrahedron.
The triangle ABC with AB=2, AC=3, BC=4.
The three folds are BE, EF, and DF.
Further lengths are BF = CF = 2 and AE = DE = 1/2.
A
/ E
/ D
/ \
/ \
/ \
B--------------F--------------C
Solution (I): [AMM 56 (June 1949) 410][Mathematical Quickies Q 79]
Since the areas of congruent tetrahedrons are equal, the envelope must
be cut so that the two pieces will have equal areas. Hence, the cut
must pass through the center of the rectangle. First strongly crease
the envelope along its diagonals and along the line through the center
which is perpendicular to the longer side. Then make any straight cut
through the center. In each case, when the cut halves are folded out
along the diagonals and in along the edges and the long side
perpendicular so as to join the extremities of the cut, two congruent
isosceles tetrahedrons are formed.
The single exception occurs when the envelope is square. Then instead
of tetrahedrons, two square envelopes result from the cutting and
folding. (A flat tetrahedron)
This construction shows that the lateral surface of a zylinder can
be folded into a tetrahedron if circumference : height < 4 : 1.
The ratio 4:1 gives the flat tetrahedron again.
A dollar bill has almost the ratio sqrt(3):4 and folds therefore
in a regular tetrahedron. See [MM 61:2 (Apr 1988) 101-102].
This zylindrical construction was used by Tetra Pak to sell their
beverage Sunkist (in Germany). See [Martin Gardner, 1971, Chapter 19].
References:
Stephen Barr;
Mathematical Brain Benders 2nd Miscellany Puzzles,
New York, 1982, Dover Publ. ISBN 0-486-24260-9
Problem 12: Paper-Folding (make a 3-4-5 triangle of a 1:2 or 1:3 rectangle)
Problem 25: More Origametry (fold a square with 3/4 of the area)
Problem 31: More \Phi Origametry (Fold a Golden Rectangle)
Problem 32: The Bookmark
Problem 45: To Cover a Circle
Problem 54: The Vanadium Steel Clothesline
Brian Barwell;
Problem 1414: Rectangle Dissection,
Journal of Recreational Puzzles 17:3 (1984) 216 (problem)
Journal of Recreational Puzzles 18 (1985) 301-302 (solution)
(problem ( ) solution by T. M. Campbell)
Brian Barwell;
Problem 1732: An Origami Problem,
Journal of Recreational Puzzles 22:3 (1990) 232-233
(problem (C) solution by the proposer)
Bernard J. Battersby, Joseph S. Madachy;
Recreational Mathematics Journal No. 6 (Dec 1961) p47 (problem (A))
Recreational Mathematics Journal No. 7 (Feb 1962) p53 (Solution (A))
G. Brandreth;
The Puzzle Mountain, 1981
(problem (E) p 29, 207, reprint of Dudeney)
L. R. Chase;
American Mathematical Monthly (June-July 1940)
- the crease of a folded paper is a straight line
Henry Ernest Dudeney;
Modern Puzzles and How to Solve Them,
London 1926,
(problem (E) = problem 139, p55, solution p144, without derivation)
Martin Gardner;
The Second Scientific American Book of Mathematical Puzzles and Diversions,
Simon & Schuster (1961)
(chapter 16: Origami
p 138 problem (F)
p 142 Figure 68 shows problem (E) solution p144-145)
Martin Gardner;
M. Gardner's Sixth Book of Mathematical Games from Scientific American,
Freeman (1971) San Francisco
(german: Mathematisches Labyrinth, Vieweg, 1979, ISBN 3-528-08402-2)
chapter 19: tetrahedrons
Heinrich Hemme;
Heureka,
G\"ottingen, 1988, Vandenhoeck & Ruprecht, ISBN 3-525-40732-7
(problem (B) = problem 72, p37, 85)
Heinrich Hemme, Mathias Schwoerer;
Mathematischer Denkspass,
Augsburg, 1998, Weltbild Verlag, ISBN 3-89604-615-2
Problem 30: Der Knick im Geldschein
(problem (A2) with (a,b) = (145mm, 70mm)
serie III German 10 DM note: (a,b) = (130mm, 65mm) +- 0.6mm ratio 2:1
crease = 72.7mm crease_u = 73,5mm crease_l = 71.8mm
serie III German 20 DM note: (a,b) = (138mm, 68mm) +- 0.6mm
serie III German 50 DM note: (a,b) = (146mm, 71mm) +- 0.6mm)
Donovan A. Johnson;
Paperfolding for the Mathematics Class,
National Council of Teachers of Mathematics, 1957
L. H. Longley-Cook;
New Math Puzzle Book,
New York 1970
(problem (B) p96, 102-103)
R. R. Rowe;
Civil Engineering-ASCE 18, Februar 1948, p. 70.
(problem (D))
Charles W. Trigg;
Mathematical Quickies, 1967
(reprint: Dover Publ. 1985, ISBN 0-486-24949-2)
Problem 79: Envelope into Tetrahedrons
Problem 107: A Folded Card
Problem 224: Creased Rectangle
Alan Wayne;
School Science and Mathematics 64 (March 1964) 241
solution to problem 2916 (Proposer: C. W. Trigg)
(problem (A))
IES;
http://www.ies.co.jp/math/java/geo/origami/origami.html
A java applet to fold a square paper. Shows hints for the proof.
Peter Andree;
http://www.ksk.ch/mathematik/mathonline/problemcorner/puzzle.htm
Problem May-June 2000 No 1
Fold a square paper. Show there is a triangle with constant perimeter.
Appendix:
Solving Minima and Maxima Problems:
Schupp [Schupp 1992, p59] writes that [D\"orrie 1919] introduced
the arithmetic geometric mean method to solve extreamal problems.
Other colections of problems which are solved with this method are
[Natanson 1960] and [Niven 1981]. The packing of 4 rectangles
of size a x b into a square of length a+b which shows (a+b)^2 >= 4ab
is well known. The 3 variable case is known as Hoffman's cube.
This consists of 27 blocks, a x b x c, to make into a cube a+b+c on a side.
It was first proposed by Dean Hoffman at a conference at Miami Univ. in 1978.
See [Winning Ways 1982] and [Hoffman 1981].
The arithmetic mean geometric mean inequality:
The arithmetic mean A and the geometric mean G of the n variables x1, ..., xN
which are not negative is defined as
A = (x1 + x2 + ... + xN) / N,
G = (x1 * x2 * ... * xN)^(1/N)
Theorem: A-G-Inequality
A >= G
with equality if and only if all varaibles are equal.
The Babylonean knew the case n = 2
((x + y)/2)^2 = xy + ((x-y)/2)^2
and this means
(AG2) (x + y)/2 >= sqrt( x y ) with equality if and only if x = y.
Another way of proofing (AG2) is (x + y)^2 >= 4xy <=> (x - y)^2 >= 0.
Cauchy gave an elegant proof for the general case using the
induction n -> 2n and n -> n-1.
William Barnier, Douglas Martin;
Unifying a Family of Extrema Problems,
College Math. Journal 28:5 (1997) 388-391
- proof of the arithmetic-geometric means inequality
Elwyn R. Berlekamp, John H. Conway, Richard K. Guy;
Winning Ways,
for your mathematical play, Vol 2, (Part 4: Diamond)
Academic Press, London, 1982
(german: Gewinnen, Strategien f\"ur mathematische Spiele,
Band 4: Solitairspiele, Vieweg Verlag, Braunschweig, 1985)
pp. 739-740 & 804-806. Shows all 21 inequivalent solutions of Hoffman's cube.
Ralph P. Boas, Jr. and Murray S. Klamkin;
Extrema of Polynomials,
Mathematics Magazine 52:2 (1977) 75-78
- by the arithmetic-geometric mean inequality
P. S. Bullen, D. S. Mitrinovics, P. M. Vasic;
Means and their Inequalities,
Reidel, Dordrecht, 1988
(more than 50 proofs for the arithmetic-geometric mean inequality are listed)
Augustin Louis Cauchy;
Analyse algebrique,
(Note 2; Oeuvres completes, Serie 2, Vol 3, p375-377; Paris, Gauthier-Villars 1897)
- gives an elegant elemanetary proof of the AM-GM-Inequality
Heinrich D\"orrie;
Ein neues elementares Verfahren zur L\"osung von Extremwertaufgaben,
MU 18 (1972) Heft 5, 23-51
(Reprint of: Zeitschrift f\"ur den mathematischen und naturwissenschaftlichen
Unterricht aller Schulgattungen, 50 (1919) 153-177)
S. Dvoryaninov, E. Yasinovyi;
Obtaining symmetric inequalities,
Quantum (Nov. Dec. 1999) 44-48.
- Muirhead's theorem (1903) by examples.
(AG3) via Muirhead:
2(x^3+y^3+z^3) - (x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y)
= (y + z)(y - z)^2 + (x + y)(x - y)^2 + (z + x)(z - x)^2 >= 0
(x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y) - 6 x y z
= x (y - z)^2 + z (x - y)^2 + y (z - x)^2 >= 0
and therefore
2(x^3+y^3+z^3) - 6 x y z = (x+y+z)( (y - z)^2 + (z - x)^2 + (x - y)^2 ) >= 0
Dean G. Hoffman;
Packing problems and inequalities.
In: The Mathematical Gardner, ed. by David Klarner;
Prindle, Weber & Schmidt/Wadsworth, 1981. Pp. 212-225
In: Mathematical Recreations - A Collection in Honor of Martin Gardner,
Dover Publ. 1998, p212-225.
(packing proofs for the inequality of the arithmetic and geometric means.)
Includes photos of Carl Klarner assembling the first set of the blocks.
Asks if there are analogous packings in n dimensions.
Murray S. Klamkin;
On two Classes of Extremum Problems without Calculus,
Mathematics Magazine 65:2 (1992) 113-117
- by the arithmetic-geometric mean inequality
P. P. Korowkin;
Ungleichungen,
Berlin, Deutscher Verlag der Wissenschaften, 19??
Kleine Erg\"anzungsreihe zu den Hochschulb\"uchern der Mathematik, Band 4.
(russion edition: Moskow 1952)
- gives D\"orrie's proof of the AM-GM-Inequality.
D. S. Mitrinovics;
Elementary Inequalities,
P. Noordhoff Ltd, Groningen, The Netherlands, 1964
I. P. Natanson;
Einfachste Maxima- und Minimaaufgaben,
Berlin, Deutscher Verlag der Wissenschaften, 1960
Kleine Erg\"anzungsreihe zu den Hochschulb\"uchern der Mathematik, Band 9.
(russion edition: Moskow 1951)
- gives Cauchy's proof of the AM-GM-Inequality.
Wilhelm Ness;
Anwendung des Satzes vom arithmetischen und geometrischen Mittel
auf Extremwertaufgaben,
Der mathematische und naturwissenschaftliche Unterricht, 20:7 (1967) 266
Ivan Niven;
Maxima and Minima without Calculus,
The Mathematical Association of America, Washington, 1981
Hans Rademacher, Otto Toeplitz;
Von Zahlen und Figuren,
Heidelberger Taschenb\"ucher Band 50,
Springer Verlag, Berlin, New York, 1968 (1st edition 1930) ISBN 3-540-04190-7
- chapter 3: Einige Maximumaufgaben
(Euklid Book VI Theorem 27: ((x+y)/2)^2 >= xy.
The area of the regular n-gon is bigger than
the area of any other cyclic n-gons.)
Karl Schuler;
Maxima-Minima-Aufgaben,
Archimedes, Sonderheft 1966
- gives Schl\"omilch's (1858) proof of the AM-GM-Inequality.
Hans Schupp;
Optimieren,
Extremwertbestimmung im Mathematikunterricht,
BI-Wiss.-Verlag, Mannheim, 1992, ISBN 3-411-15771-2
Series: Lehrb\"ucher und Monographien zur Didaktik der Mathematik Band 20
- gives D\"orrie's proof of the AM-GM-Inequality.
Jerry (= G. K.) Slocum, Jack Botermans;
Puzzles Old & New _ How to Make and Solve Them,
Univ. of Washington Press, Seattle, 1986
(german: Geduldspiele der Welt, Hugendubel, 1987)
- p43: the Hoffman cube
--
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http://www.mathematik.uni-bielefeld.de/~sillke/