The problems involve measurement of geometrical spaces, probabilities, distances, number systems other than the decimal, interesting number relations, relative motion, and the discovery of the simplest solution through purely logical means. Solutions involve such techniques as arithmetic, algebra and geometry, number theory, statistics, networks, and inversion. Several approaches to the solution accompany each problem, and even the best of these represent not the ideal solution but only the most satisfying — so far.
The problems involve measurement of geometrical spaces, probabilities, distances, number systems other than the decimal, interesting number relations, relative motion, and the discovery of the simplest solution through purely logical means. Solutions involve such techniques as arithmetic, algebra and geometry, number theory, statistics, networks, and inversion. Several approaches to the solution accompany each problem, and even the best of these represent not the ideal solution but only the most satisfying — so far.
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Overview
The problems involve measurement of geometrical spaces, probabilities, distances, number systems other than the decimal, interesting number relations, relative motion, and the discovery of the simplest solution through purely logical means. Solutions involve such techniques as arithmetic, algebra and geometry, number theory, statistics, networks, and inversion. Several approaches to the solution accompany each problem, and even the best of these represent not the ideal solution but only the most satisfying — so far.
Product Details
| ISBN-13: | 9780486218465 |
|---|---|
| Publisher: | Dover Publications |
| Publication date: | 03/17/2017 |
| Edition description: | Revised ed. |
| Pages: | 160 |
| Product dimensions: | 5.50(w) x 8.50(h) x (d) |
About the Author
Read an Excerpt
The Surprise Attack in Mathematical Problems
By Louis A. Graham
Dover Publications, Inc.
Copyright © 1968 Louis A. GrahamAll rights reserved.
ISBN: 978-0-486-21846-5
CHAPTER 1
Graphic Reciprocals
This interesting problem lends itself to at least four different modes of attack, as described below; it turns out, however, that the optimum, or "surprise" approach, can be extended to other related calculations in physics or engineering.
As stated in the Dial, a young student (we have dubbed him Little Euclid in many of our problems) was doing an experiment in the optical laboratory (Fig. 1) in which he was required to find the focal length/ of a convex lens by measuring the distance u from lens to object and the distance v from lens to image, and then applying the well-known relation 1/f = 1/u + 1 v.
To avoid numerical figuring, Euclid decided to make the addition of the reciprocals graphically, laying off his two lengths in each case with a triangle and graduated rule to effect the addition, and then finding the desired answer by simple measurement. How did he do it? We hinted that "there are at least two different ways — one using a right triangle and the other a 60° triangle." Can you find one or both? (Incidentally, finding by graphical means a number whose reciprocal is the sum of the reciprocals of two given numbers applies not only to the above focal length problem, but also to the finding of the equivalent resistance of an electrical circuit if the resistances of two parallel paths are known, or the equivalent capacity of two capacities in series, and so forth.)
Our readers not only uncovered the two methods hinted at, using 60° and right triangles, but also devised a third method, using a 45° triangle, and a fourth method using no triangle at all, but a compass.
Solution. In the solution using a draftsman's 60° triangle (Fig. 2), a and b are laid off with an included angle of 120°, and c is picked off on the bisector of the angle, as shown. For proof: the area of the triangle at the right is b/2 · c sin 60°, the area of the triangle at the left is a (2 · c sin 60° and the area of the large triangle is a/2 · b sin 60°. Equating the large area to the sum of the two small areas, ac + be = ab, and dividing both sides by abc we have 1/b + 1/a = 1/c.
In the second method, using a 45° draftsman's triangle, the triangle is used to lay off a and b (Fig. 3) perpendicular to each other; the angle between them is bisected, and c is found as the distance from the end of the bisector to b. Proof: from similar triangles, ajb = c/(b — c) from which ab — ac = bc, giving 1/c = 1/a + 1/b as required.
The third method avoids use of a draftsman's triangle but involves the construction of a line through a given point parallel to another line, requiring a compass. Here AB is laid off equal to distance a (Fig. 4) and on any intersecting line AC is laid off equal to b and extended to D so that CD = a. DB is joined. CE is drawn parallel to DB. AE is the required length because by similar triangles AE/b = a/(a + b), from which 1/AE = ([a + b)/ab = 1/b + 1/a.
Strangely enough, the fourth method, which uses a right triangle, is not only simplest but uses a construction which is a counterpart of the familiar crossed ladders, which forms the subject of problem No. 6. The two distances a and b are laid off on the " walls" (Fig. 5) and the "ladders" will cross at the required height c above the "ground," regardless of the width of the "lane." This is because, if we call the lane width unity, from similar triangles 1/b = m/c and 1/ a = n/c from which 1/b + 1/a = (n + m)/c = 1/c, as required.
This "crossed ladder" method has the additional above-mentioned advantage, important to the engineer or physicist, that further reciprocals (parallel resistances, and so forth) can be successively added to the previous sum without the error that might be caused by transfer of measurement.
2. Real Estate by Intersections
Thanks to Henry Eckhardt, project engineer, Aerojet General Corporation, Sacramento, California, we present in this problem the surprising incident of Little Euclid and the real estate agent. Euclid and his dad were considering buying a neighbor's farm which was irregular in outline but had straight line boundaries, and went to the agent's office to close the deal. The large map in the agent's office was ruled off into 100-foot squares (that is, the "intersection points" of the vertical and horizontal rulings were 100 feet apart in both directions). Euclid noticed that the boundary points of the farm were all located at "intersection points" on the map (Fig. 6). The price quoted was $200.00 an acre. After a minute or two of figuring, Euclid turned to his dad and said, "That plot will cost us just $5,417.81." "Gome now," replied Dad, "that's just a guess. You couldn't get the accurate figure that fast and without a planimeter." "Yes, I could," replied Euclid, " that's the exact cost. I just counted the number of intersection points that were on the boundary and the number that were within the area and did some simple figuring here on this paper." How did Euclid make this lightning calculation, and why is it infallible?
Solution. Although it is specified that Euclid's rapid calculation was based merely on the number of intersection points (1) on the boundary and (2) within the area, many who tackled the problem did not count these points, which are 16 and 111 respectively, but divided the area into triangles and rectangles. One such solution is given below (Fig. 7), in which from the area 156 × 104 of the 1200' × 1300' inclusive rectangle are successively subtracted the respective areas 10 × 104, 8 × 104, 5 × 104, 7 × 104, and 4 × 104 of the four triangles A, B, C, E, and F and of the square D. Although this, of course, gives the correct area of the plot, it is hardly to be believed — even if it had not been specifically stated otherwise — that Euclid could have arrived at his prompt solution by this means.
Actually, by employing the "surprise attack," whose accuracy is explained below, it is necessary only to add one-half of 16 to 111 minus one, giving the required area of 118 hundred-foot squares, which is readily converted, by dividing by 4.356, into 27.089 acres; this, when multiplied by 200, gives the $5,417.81.
Two derivations of this simple formula — with which, of course, Euclid was evidently familiar — are given below, the first geometrical and the second algebraic. Wrote S. E. Szasz, senior project engineer, Sinclair Research Laboratories, Tulsa, Oklahoma: Let us associate each intersection point' or grid point' on the map with a 100-foot square of which it is the center. There are four kinds of grid points: (1) those such as A (Fig. 8), well in the interior, for which the entire associated square is inside the area of the farm; (2) those grid points such as B inside the plot so close to the boundary that part of the associated square is clipped by the boundary (for these, note that because the boundaries are straight lines joining grid points, there will always be, for reasons of symmetry, an identical area inside the plot which is clipped from another square the grid point C of which is outside the plot, so that each of these grid points is also associated with an area equal to one square inside the plot) ; (3) grid points D on a straight boundary line : exactly one half of their associated square is within the farm, regardless of the orientation of the boundary line; (4) the grid points E at the corners of the polygon. If, at each corner, we extend one of the sides of the polygon beyond the corner and do this all in the same direction, that is, clockwise, we notice that at each corner, the area within the farm is one-half of one square less that portion between one side of the polygon and the extension of the other. These small pieces clipped from the half-squares can be assembled into one complete square, just as in the well-known computation of the sum of the angles in any polygon the exterior angles add up to 360°. Hence Euclid's lightning calculation: the area of the farm is equal to the area of one square (10,000 sq. ft.) times the number of intersection points within the farm added to one-half the number of intersection points on the boundary (including the corner points) minus one. You need a pencil only to convert this into acres.
The algebraic explanation by Robert E. Corby, General Manager, Trio Laboratories, Plainview, New York reads as follows: "Evidently any rectangle with sides a and b (Fig. 9), and corners on intersection points of the chart, has M pointson its perimeter, where M = 2 (a + b). Also, the number of points within the rectangle is N = (a - 1)(b - 1). Since the area A is equal to ab, if we expand the latter equation, a and b can be eliminated, resulting in A = N - 1 + M/2. Connecting two opposite corners, we form two right triangles with the hypotenuse intersecting c points, not including the two end points. Clearly the number of points S inside the triangle and the number of points T on its perimeter are related to N and M by S = (N - c)/2 and T = M/2 - 1 + c. Solving these last two equations for N and M and substituting in the first equation shows that the area of the triangle A/2 is related to N and M by A/2 = S - 1 + T/2, which is the same as the first equation. Now, a group of rectangles and right triangles using this equation can be combined, as in Fig. 9, to build up any polygon whose boundaries terminate in 'intersection points.' This will always work because the result of each combination of two figures will increase N by unity for each pair of points lost in the common side joining the two figures."
CHAPTER 2Making Change
This original problem tells of Little Mary who was given a bright new 50-cent piece to spend. On her way home from shopping she noted the following curious facts: (1) She had spent all of the 50 cents. (2) For no purchase, except of course the last one, did she have the exact change required. (3) The vendors had all made change for her in such a way that at all times she had held the least possible number of coins for the amount of money she had at the time. (4) She had made the greatest possible number of purchases under the above circumstances. How many purchases did she make?
The problem illustrates, better perhaps than any other we have used in the Dial in years, how the mathematically-minded enjoy going beyond the bare demands of the problem as worded, and beyond, too, "the things dreamed of in its author's philosophy." In this case our readers not only extracted much mathematical stimulus from the problem itself, but two of them conjured up interesting variations on it.
Solution. To help compare the various types of possible solutions, we give the names of some readers employing different modes of attack, with their contributions. P. E. Vesilind, mechanical engineer, Koppers Company, Monoca, Pennsylvania, offered a single correct solution: Little Mary made 11 purchases: (1) 50 — 5 = 25 + 10 + 10; (2) 45 - 5 = 25 + 10 + 5; (3) 40 - 4 = 25 + 10 + 1; (4) 36 - 5 = 25 + 5 + 1; (5) 31 - 4 = 25 + 1 + 1; (6) 27 - 5 = 10 + 10 + 1 + 1; (7) 22 - 5 = 10 + 5 + 1 + 1; (8) 17 - 4 = 10 + 1 + 1 + 1; (9) 13 - 5 = 5 + 1 + 1 + 1; (10) 8 - 4=1 + 1 + 1 + 1; (11) 4 - 4 = 0. "The idea," writes Mr. Vesilind, "is to prevent the change from turning into pennies."
Gino J. Mangialardi, Jr., agricultural engineer, U.S. Department of Agriculture, Stoneville, Mississippi, tackled the thing more broadly, to arrive at a variety of solutions. He wrote: "By the time Little Mary had spent her 50 cents she had made 11 purchases. The smallest denomination which can be broken into smaller denominations is a nickel. There are, of course, 10 such nickels. No sale can be made without breaking a nickel. After 10 such sales, there will be less than a nickel left for the eleventh purchase. The size of every purchase is not fixed, although some depend upon others. The important consideration is that one must break 1 nickel for each purchase. Some samples of the manner in which Little Mary could have spent her money follow. Solution No. 1: 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4. Solution No. 2: 4, 5, 2, 8, 4, 4, 7, 4, 4, 4. 4. Solution No. 3: 3, 7, 4, 3, 4, 6, 8, 4, 5, 4, 2."
Hallock C. Campbell, Director of Research, Arcos Corporation, Philadelphia, Pennsylvania, put his solution in verse:
Little Mary quite contrary,
How does your money go?
It's nickels here and 4 pence there,
ELEVEN sales in a row.
Two nickels, 4 cents, 5 cents more,
And then another 4;
Repeat the same, and suddenly
There's only 4 to go.
Contrariwise the start could be
A 4- or 3-cent prize,
But naught is gained in total sales
'Cause Mary shot "snake eyes."
John Stallkamp, Jet Propulsion Laboratory, Pasadena, California, greatly extended the scope of the problem by affirming not only that the maximum number of purchases could be determined as being 11 without first finding any actual sequence of purchases, but that his method would likewise be applicable to other initial sums. He wrote: "Except for that last one, every purchase must be at least 1 cent greater than the number of 1-cent coins in hand before the purchase. Therefore at least one 5-cent coin or its equivalent must be cashed with each such purchase, for example, a quarter replaced by 2 dimes. Thus, for an initial amount m a maximum number of purchases n may be hoped for that is given by nm/5, rounded off to the next lowest whole number. The extra 1 is expected from end effects: the first purchase will be between 1 and 5 cents except for several initial amounts containing 4 pennies and 1 nickel, and the last purchase is expected to be less than 4 cents. When m equals 50 cents, n equals 11; this is verified by the Table (Fig. 10), which gives all initial values up to 99 cents and the appropriate count down."
It is significant that in arriving at the answer of 11 purchases, G. E. Norton, National Castings, Cicero, Illinois, developed the Table (Fig. 11) which unintentionally duplicates the results in the right-hand half of Stallkamp's Table, although Norton was not actually trying to determine — as was Stallkamp — the maximum number of satisfying purchases for initial amounts other than 50 cents. Wrote Mr. Norton: "The 50-cent problem is most easily solved by working backward. I derive the answer of 11 purchases as follows: For any amount remaining, after purchases have been made, and starting with 1 cent, determine the maximum number of purchases which can be made, to exhaust the fund. The attached tabulation (Fig. 11) shows this clearly. 1, 2, 3, and 4 cents are in 'Group 1,' meaning that only 1 more purchase can be made. In 'Group 2 we have 5, 6, 7, and 8 cents, meaning that not more than 1 purchase can be made to reduce these to 'Group 1,' and so on. Certain remainders — 9, 10, 19, 25, 34, 35, and 44 cents — must be avoided, as indicated, either because too few purchases can be made to arrive at these remainders, or too few purchases can be made to reduce these remainders to zero."
(Continues...)
Excerpted from The Surprise Attack in Mathematical Problems by Louis A. Graham. Copyright © 1968 Louis A. Graham. Excerpted by permission of Dover Publications, Inc..
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site.
Table of Contents
1 Graphic Reciprocals 1
2 Real Estate by Intersections 3
3 Making Change 7
4 Disk Distraction 12
5 The Stamp Dispenser 13
6 Searchlight on Crossed Ladders 16
7 Careless Composition 18
8 Tic-Tac-Toe for Gamblers 19
9 Mathman with a Sense of Humor 22
10 Four Four Finale 27
11 Codes and Computers 28
12 Three Given Altitudes 32
13 Simplification by Projection 34
14 Unequal Integers 36
15 Sport Scores 38
16 Measure of Ribbons 39
17 Clock Watching 41
18 Catch and Carries 44
19 Silver Dollar Saver 46
20 Pell by Succession 47
21 Smallest Triangle 49
22 The Todd County Ranch 52
23 New Route to Roots 53
24 Lost Paddle and Passing Trains 56
25 Quickies 58
26 Merry Xmas to All 67
27 Little Euclid's Age 69
28 Right Angle Check 70
29 Tale of a Tile Setter 73
30 Nearest Trisection 74
31 The Will and the Way 76
32 One-Step Double 78
33 Watch Wound Wrong 81
34 The Unknown Center 82
35 The Isosceles Triangle 85
36 Weather in Azuciana 88
37 Bids and Scores in Duplicate Bridge 91
38 The Ranchers' Split 92
39 Philosophy and Probability 93
40 The Blushing Beauties 104
41 The Car Behind 105
42 Right Triangle Construction 106
43 To Win at Las Vegas 109
44 Sums and Factors 110
45 Base and Bisector 113
46 Who Killed Whom? 115
47 Again-The Golden Ratio 116
48 The Fourth Generation 118
49 Product with Reverse 119
50 The Powerless Binary 120
51 Doubles on Reverse 121
52 Three Parallels 124