The Surprise Attack in Mathematical Problems

By Louis A. Graham

For decades, professionals in applied mathematics and mathematical puzzle fans from all over the world supplied provocative problems to the puzzle columns of the Graham Dial. Upon the appearance of an interesting problem, readers wrote in to the Dial suggesting new approaches to the solution, some greatly simplifying the problem and others broadening its scope. This collection presents 52 of the finest problems, all featuring the "surprise attack" and abounding in imaginative and interesting solutions.
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.

• Language: English
• Publisher: Dover Publications
• Release date: Mar 17, 2017
• ISBN: 9780486820767

Book preview

Problems

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 f 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.

FIG. 1

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 + bc = ab, and dividing both sides by abc we have 1/b + 1/a = 1/c.

FIG. 2

FIG. 3

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, a/b = 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.

FIG. 4

FIG. 5

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. Come 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 ?

FIG. 6

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.

FIG. 7

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."

FIG. 8

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 points on 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."

FIG. 9

MATHEMATICAL NURSERY RHYME No. 1

Ride a cock horse

To Banbury Cross

And you’ll have plenty of power.

For a horse, so they say,

On any fine day,

Can raise up a ton,

Yes, a full U.S. ton,

A thousand feet high in an hour.

MATHEMATICAL NURSERY RHYME No. 2

Old Boniface he took his cheer,

Then he drilled a hole through a solid sphere—

Clear through the center, straight and strong.

And the hole was just six inches long.

Now tell us, when the end was gained

What volume in the sphere remained ?

Sounds like we haven’t told enough,

But that’s all you need—and it isn’t tough.

MATHEMATICAL NURSERY RHYME No. 3

Simple Simon met a π man

Going to the fair.

Said Simple Simon to the π man

"You have unusual ware.

The π’S I’ve seen before were round

But, gosh, your π’S r²."

MATHEMATICAL NURSERY RHYME No. 4

Hey diddle diddle, the cat and the fiddle,

The cow jumped over the moon ;

Which requires computation of its orbit’s equation

To avoid jumping late or too soon.

If Earth’s mass be m, gravitation be g,

The product must equal indeed

Moon’s distance from Earth (cancel out the moon’s mass)

Multiplied by the square of its speed.

3. Making 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?