Sources page biographical material

6.BT. PLACING OBJECTS IN CONTACT
New section. The objects involved are usually common objects such as coins or cigarettes, etc. The standard recreation is to have them all touching one another. However, the more basic question of how many spheres can touch a sphere goes back to Kepler and perhaps the Greeks. Similar questions have been asked about cubes, etc.
Endless Amusement II. 1826? Problem II, p. 189. "Five shillings or sixpences may be so placed over each other, as to be all visible and all be in contact." Two solutions. The first has two coins on the table, then two coins on top moved far enough onto one of the lower coins that a vertical coin can touch both of them and the two lower coins at once. The second solution has one coin with two coins on top and two slanted coins sitting on the bottom coin and touching both coins in the second layer and then touching each other up in the air. [I have recently read an article analysing this last solution and showing that it doesn't work if the coin is too thick and that the US nickel is too thick.] = New Sphinx, c1840, p. 131.

Will Baffel. Easy Conjuring without apparatus. Routledge & Dutton, nd, 4th ptg [c1910], pp. 103-104. Six matches, each touching all others. Make a V with two matches and place a third match in the notch to make a short arrow. Lie one of these on top of another.

Will Blyth. Money Magic. C. Arthur Pearson, London, 1926. Five in contact, pp. 98-101. Same as Endless Amusement II.

Rohrbough. Puzzle Craft. 1932. Six Nails, p. 22 (= p. 22 of 1940??). As in Baffel.

Meyer. Big Fun Book. 1940. Five coins, p. 543. Same as the first version in Endless Amusement II.

Philip Kaplan. More Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 29, pp. 35 & 92. Second of the forms given in Endless Amusement II.

Ripley's Puzzles and Games. 1966. P. 36. Six cigarettes, as in Baffel.

I recall this is in Gardner and that a solution with 6 cigarettes was improved to 7.

6.BU. CONSTRUCTION OF N-GONS
New Section. This is really a proper geometric topic, but there is some recreational interest in two aspects.

A. Attempts to construct regular n-gons for impossible values of n, e.g. n = 7, either by ruler and compass or by origami or by introducing new instruments -- see 6.BV.

B. Attempts to construct possible cases, e.g. n = 5, by approximate methods.

Aspect A is closely related to the classic impossible problems of trisecting an angle and duplicating a cube and hence some of the material occurs in books on mathematical cranks -- see Dudley. Further, there were serious attempts on both aspects from classic times onward.

Abu’l-Jūd. 11C. He "devised a geometrical method to divide the circle into nine equal parts." [Seyyed Hossein Nasr; Islamic Science -- an Illustrated Study; World of Islam Festival Publishing Co., London?, 1976, p. 82. Q. Mushtaq & A. L. Tan; Mathematics: The Islamic Legacy; Noor Publishing House, Farashkhan, Delhi, 1993, p. 70.]

Pacioli. De Viribus. c1500. These problems are discussed by Mackinnon, op. cit. in 6.AT.3, pp. 167, 169, citing Agostini, p. 5. Let l_n be the side of a regular n-gon inscribed in a unit circle.

Ff. 146r-147r, XXIII afare la 7^a fi^a dicta nonangolo. cioe de .9. lati difficile (XXIII to make the 7th figure called nonagon, that is of 9 sides, difficult) = Peirani 198 199. Asserts l₉ = (l₃ + l₆)/4. Mackinnon computes this gives .6830 instead of the correct .6840.

Ff. 148r-148v, XXV. Documento della 9 fi^a rectⁱ detta undecagono (XXV. on the 9th rectilinear figure called undecagon) = Peirani 200. Asserts l₁₁ = φ (l₃ + l₆)/3, where φ is the golden mean: (1 - 5)/2. Mackinnon computes this gives .5628 instead of the correct .5635.

F. 148v, XXVI. Do. de' .13. (XXVI. on the 13th) = Peirani 200. Asserts l₁₃ = (1 φ)·5/4. Mackinnon computes this gives .4775 instead of the correct .4786.

Ff. 149r-149v, XXVIII. Documento del .17. angolo cioe fi^a de .17. lati (XXVIII on the 17-angle, that is the figure of 17 sides) = (Peirani 201-202). Peirani says some words are missing in the second sentence of the problem and Agostini says the text is too corrupt to be reconstructed. MacKinnon suggests l₁₇ = (l₃ - l₆)/2 which gives .3660 instead of the correct .3675.

Barbaro, Daniele. La Practica della Perspectiva. Camillo & Rutilio Borgominieri, Venice, (1569); facsimile by Arnaldo Forni, 1980, HB. [The facsimile's TP doesn't have the publication details, but they are given in the colophon. Various catalogues say there are several versions with dates on the TP and colophon varying independently between 1568 and 1569. A version has both dates being 1568, so this is presumed to be the first appearance. Another version has an undated title in an elaborate border and this facsimile must be from that version.] Pp. 26-27 includes discussion of constructing a regular heptagon, but it just seems to say to divide the circumference of a circle into seven equal parts -- ??

Christian Huygens. Oeuvres Complètes. Vol. 14, 1920, pp. 498-500: problem dated 1662, "To inscribe a regular heptagon in a circle." ??NYS -- discussed by Archibald.

R. C. Archibald. Notes (to Problems and Solutions section) 24: Problems discussed by Huygens. AMM 28 (1921) 468 479 (+??). The third of the problems discussed is the construction of the heptagon quoted above. Archibald gives an extensive survey of the topic on pp. 470-479. A relevant cubic equation was already found by an unknown Arab writer, c980, and occurs in Vieta and in Kepler's Harmonices Mundi, book I, Prop. 45, where Kepler doubts that the heptagon can be constructed with ruler and compass.

An approximate construction was already given by Heron of Alexandria and may be due to Archimedes -- this says the side of the regular heptagon is approximately half the side of the equilateral triangle inscribed in the same circle. Jordanus Nemorarius (c1230) called this the Indian method. Leonardo da Vinci claimed it was exact. For the central angle, this approximation gives a result that is about 6.5' too small.

Archibald then goes on to consider constructions which claim to work or be good approximations for all n-gons. The earliest seems to be due to Antoine de Ville (1628), revised by A. Bosse (1665). In 1891, A. A. Robb noted that a linkage could be made to construct the heptagon and J. D. Everett (1894) gave a linkage for n-gons.

Italo Ghersi; Matematica Dilettevoli e Curiosa; 2nd ed., Hoepli, 1921; pp. 425-430: Costruzioni approssimate.

He says the following construction is given by Housel; Nouvelles Annales de Mathématiques 12 (1853) 77-?? with no indication of its source. Ghersi says it also occurs in Catalan's Trattato di Geometria, p. 277, where it is attributed to Bion. However, Ghersi says it is due to Rinaldini (probably Carlo Renaldini (1615-1698)). Let AOB be a horizontal diameter of a circle of radius 1 and form the equilateral triangle ABC with C below the diameter. Divide AB into n equal parts and draw the line through C and the point 4/n in from B. Where this line hits the circle, say P, is claimed to be 1/n of the way around the circumference from B. Ghersi obtains the coordinates of P and the angle BOP and computes a table of these values compared to the real values. The method works for n = 2, 3, 4, 6. For n = 17, the error is 36'37".

On pp. 428-430, he discusses a method due to Bardin. Take AOB as above and draw the perpendicular diameter COD. Divide the diameter into n equal parts and extend both diameters at one end by this amount to points M, N. Draw the line MN and let it meet the circle near B at a point P. Now the line joining P to the third division point in from B is claimed to be an edge of the regular n-gon inscribed in the circle. Ghersi computes this length, finding the method only works for n  5, and gives a table of values compared to the real values. This is exact for n = 6 and is substantially more accurate than Renaldini's method. For n = 17, the error is 1'10.32".

The "New" School of Art Geometry, Thoroughly Remodelled so as to Satisfy all the Requirements of the Science and Art Department for Science Subject I. Sections I. and II, Practical Plane and Solid Geometry, (Cover says: Gill's New School of Art Geometry Science Subject I.) George Gill and Sons, London, 1890.

Pp. 26-27, prob. 66 -- To describe any regular Polygon on a given straight line, AB. He constructs the centre of a regular n-gon with AB as one edge. Taking the side AB as 1, the height h_n of the centre is given by h_n  =  (n 4) 3/4     (n-6)/4, while the correct answer is ½ cot π/n. For large n, the relative error approaches 14.99%. He gives no indication that the method is only approximate and doesn't even work for n = 5.

Pp. 74-75, prob. 188 -- To inscribe any Regular Polygon in a given circle. He gives three methods. The first is to do it by trial! The second requires being able to construct the regular 2n-gon! The third construction is Renaldini's, which he does indicate is approximate.

R. C. Archibald, proposer; H. S. Uhler, solver. Problem 2932. AMM 28 (1921) 467 (??NX) & 30 (1923) 146-147. Archibald gives De Ville's construction and asks for the error. Uhler gives values of the error for n = 5, 6, ..., 20, and the central angles are about 1^o too large, even for n = 6, though the error seems to be slowly decreasing.

T. R. Running. An approximate construction of the side of a regular inscribed heptagon. AMM 30 (1923) 195-197. His central angle is .000061" too small.

W. R. Ransom, proposer; E. P. Starke, solver. Problem E6. AMM 39 (1932) 547 (??NX) & 40 (1933) 175-176. Gives Dürer's method for the pentagon and asks if it is correct. Starke shows the central angle is about 22' too large.

C. A. Murray, proposer; J. H. Cross, E. D. Schell, Elmer Latshaw, solvers. Problem E697 -- Approximate construction of regular pentagon. AMM 52 (1945) 578 (??NX) & 53 (1946) 336-337. Describes a method similar to that of de Ville - Bosse and asks if it works for a pentagon. Latshaw considers the general case. The formula is exact for n = 3, 4, 6. For n = 5, the central angle is 2.82' too small. For n > 6, the central angle is too large and the error is increasing with n.

J. C. Oldroyd. Approximate constructions for 7, 9, 11, 13-sided polygons. Eureka 18 (Oct 1955) 20. Gives fairly simple constructions which are accurate to a few seconds.

Marius Cleyet-Michaud. Le Nombre d'Or. Presses Universitaires de France, Paris, 1973. Méthode dite d'Albert Dürer, pp. 45-47. Describes Dürer's approximate method for the pentagon and says it fails by 22'.

Underwood Dudley. Mathematical Cranks. MAA Spectrum, 1992. This book discusses many related problems, e.g. duplication of the cube, trisection of the angle. The chapter: Nonagons, Regular, pp. 231 234 notes that there seem to be few crank constructors of the heptagon but that a nonagoner exists -- Dudley does not identify him. Actually he constructs 10^o with an error of about .0001', so he is an excellent approximater, but he claims his construction is exact.

Robert Geretschläger. Euclidean constructions and the geometry of origami. MM 68:5 (Dec 1995) 357-371. ??NYS -- cited in next article, where he states that this shows that all cubic equations can be solved by origami methods.

Robert Geretschläger. Folding the regular heptagon. CM 23:2 (Mar 1997) 81-88. Shows how to do it exactly, using the result of his previous paper.

Dirk Bouwens, proposer; Alan Slomson & Mick Bromilow, independent solvers. An early protractor. M500 171 (Dec 1999) 18 & 173 (Apr 2000) 16-17. Draw a semicircle on diameter AOB. Draw the perpendicular through O and extend it to C so that BC = BA (the diameter) [this makes c = OC = 3] . If P divides AO in the ratio λ : 1-λ, then draw CP to meet the semicircle at D and OD divides the arc ADE in approximately the same ratio. He finds the exact value of the angle AOD and finds that the maximum error in the process is only .637^o (when λ  .18). Second author provides a graph of the error and says the maximum error is at about 18^o. [I get .637375^o at .181625^o.]

Ken Greatrix. A better protractor. M500 175 (Aug 2000) 14-15. Taking c = 1.67721 in the previous construction gives a more accurate construction, with maximum error about .32^o at about 13^o. [I get .324020^o at .130164^o.]

6.BV. GEOMETRIC CONSTRUCTIONS
New Section. This is really a proper geometric topic, but there is some recreational interest in it, so I will cite some general references.
Robert C. Yates. Geometrical Tools. (As: Tools; Baton Rouge, 1941); revised ed., Educational Publishers, St. Louis, 1949. Pp. 82-101 & 168-191. Excellent survey. After considering use of straightedge and compasses, he considers: compasses only; folds and creases; straightedge only; straight line linkages; straightedge with fixed figure (circle, square or parallelogram); straightedge with restricted compasses (collapsible compass, rigid (or rusty) compass or rigid dividers); parallel and angle rulers; higher order devices (marked ruler, carpenter's square, tomahawk, compasses of Hermes, two right angle rulers, straightedge with compasses and fixed conic); plane linkages in general. Each section has numerous references.
6.BW. DISTANCES TO CORNERS OF A SQUARE
New section. If ABCD is a rectangle, P is a point and a, b, c, d are the distances of P from the corners of the rectangle, then the basic relation a² + c² = b² + d² is easily shown. This leads to a number of problems. A little research has found references back to 1896, but the idea might be considerably older.
AMM 3 (1896) 155. ??NYS -- this is the earliest reference given by Trigg, cf below.

SSM 15 (1915) 632. ??NYS -- cited by Trigg, below.

AMM 35 (1928) 94. ??NYS -- cited by Trigg, below.

SSM 32 (1932) 788. ??NYS -- cited by Trigg, below.

NMM 12 (1937) 141. ??NYS -- cited by Trigg, below.

AMM 47 (1940) 396. ??NYS -- cited by Trigg, below.

NMM 16 (1941) 106. ??NYS -- cited by Trigg, below.

NMM 17 (1942) 39. ??NYS -- cited by Trigg, below.

AMM 50 (1943) 392. ??NYS -- cited by Trigg, below.

SSM 46 (1946) 89, 783. ??NYS -- cited by Trigg, below.

SSM 50 (1950) 324. ??NYS -- cited by Trigg, below.

SSM 59 (1959) 500. ??NYS -- cited by Trigg, below.

"A. Polter Geist", proposer; Joseph V. Michalowicz, Mannis Charosh, solvers, with historical note by Charles W. Trigg. Problem 865 -- Locating the barn. MM 46:2 (Mar 1973) 104 & 47:1 (Jan 1974) 56-59. For a square with an interior point, a, b, c = 13, 8, 5. How far is P from the nearest side? First solver determines the side, s, by applying the law of cosines to triangles BPA and BPC and using that angles ABP and BPC are complementary. This gives a fourth order equation which is a quadratic in s². Second solver uses a more geometric approach to determine s. The distances to the sides are then easily determined. Trigg gives 13 references to earlier versions of the problem -- see above.

Anonymous. Puzzle number 35 -- Eccentric lighting. Bull. Inst. Math. Appl. 14:4 (Apr 1978) 110 & 13:5/6 (May/Jun 1978) 155. Light bulb in a room, with distances measured from the corners. a, b, c = 9, 6, 2. Find d. Solution uses the theorem of Apollonius to obtain the basic equation.

David Singmaster, proposer and solver. Puzzle number 40 -- In the beginning was the light. Bull. Inst. Math. Appl. 14:11/12 (Nov/Dec 1978) 281 & 15:1 (Jan 1979) 28. Assuming P is inside the rectangle, what are the conditions on a, b, c for there to be a rectangle with these distances? When is the rectangle unique? When can P be on a diagonal? Solution first obtains the basic relation, which does not depend on P being in the rectangle. Reordering the vertices if necessary, assume b is the greatest of the distances. Then a² + c²  b² is necessary and sufficient for a rectangle to exist with these distances. This is unique if and only if equality holds, when P = D. If the distances are all equal, then P is at the centre of the rectangle, which can have a range of sizes. If the distances are not all equal, there is a unique rectangle having P on a diagonal and it is on the diagonal containing the largest and smallest distances.

Marion Walter. Exploring a rectangle problem. MM 54:3 (1981) 131-134. Takes P inside the rectangle with a, b, c = 3, 4, 5. Finds the basic relation, noting P can be anywhere, and determines d. Then observes that the basic relation holds even if P is not in the plane of ABCD. Ivan Niven pointed out that the problem extends to a rectangular box. Mentions the possibility of using other metrics.

James S. Robertson. Problem 1147 -- Re-exploring a rectangle problem. MM 55:3 (May 1982) 177 & 56:3 (May 1983) 180-181. With P inside the rectangle and a, b, c given, what is the largest rectangle that can occur? Observes that a² + c² > b² is necessary and sufficient for a rectangle to exist with P interior to it. He then gives a geometric argument which seems to have a gap in it and finds the maximal area is ac + bd.

I. D. Berg, R. L. Bishop & H. G. Diamond, proposers. Problem E 3208. AMM 94:5 (May 1987) 456-457. Given a, b, c, d satisfying the basic relation, show that a rectangle containing P can have any area from zero up to some maximum value and determine this maximum.

Problem 168.2. M500 168 ??? Seven solvers, M500 170 (Oct 1999) 15-20. Given P inside a square and a, b, c = 5, 3, 4, find the side of the square.