Nineteen Problems on Elementary Geometry

Armando Machado

Several years ago s o m e b o d y posed a p rob lem on ele- men t a ry geomet ry that seemed quite innocent but that resisted the more obvious approaches. It mus t be rather well known as it appears repeatedly in mathemat ica l cir- cles. In Fig, 1, we have an isosceles triangle: ,~ = 20 ~ a = 60 ~ and fl = 50 ~ are given, and we m u s t look for the val- ues of "y and 6.

After the more obvious calculations, we obtain most of the angles miss ing in the figure but not the ones we want ; all we conclude is that "~ + 6 = a + fl = 110 ~ At this point I felt that, a l though the p rob lem was clearly well posed, there was pe rhaps no solution by the me thods of e lementary geometry. I then used some t r igonometry and a pocket calculator to de termine "), and 6: To m y as-

Figure I

tonishment, the values were quite nice, namely, ~ = 80 ~ and 6 = 30 ~ With such values there should exist an ele- men ta ry solution! I r emember that I used the same )~ and a second pair of values for a and fl (I don ' t k n o w which any more) and that I again obtained integer values for "y and 6. I was beginning to believe that this was a gen- eral phenomenon , but a third trial told me that it was not so. For example , with & = 20 ~ a = 20 ~ and fl = 70 ~ we obtain -~ = 2~ and 6 = 87~ . . . . I tried all the mult iples of 10 ~ for ~, a , and r , ignoring trivial cases, like those where a = r , as well as those that were symmetr ic to previous ones; the only data that gave nice values for "r and 6 were those in Table 1. (From n o w on, I will omit the degree symbol.)

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Table I Table 2

)~ a fl 7 6

20 50 20 60 10 20 50 40 60 30 20 60 30 80 10 20 60 50 80 30 20 70 50 110 10 20 70 60 110 20

The natural conjecture was that in all these cases there should exist an e lementary solution to the problem, that is to say, one that does not involve tr igonometric formu- las. Indeed, this was verified by one of my colleagues, Margarita Ramalho. The interesting phenomen o n was that for each case one had to present a different proof, a situation not very usual in mathematics. For example, the first two cases are quite trivial, a l though different f rom one another, and the fourth, the original one, can be solved by superposing its figure with that of the first case, mirrored in the vertical axis, and remarking that one equilateral and several isosceles triangles show up.

Recently, I was playing around with the mathematical software M a t h e m a t i c a when somebody raised the same problem. I decided to use this software to look for other initial data that could lead to e lementary solutions. I asked M a t h e m a t i c a to try every integer value of & and every integer or half-integer value of a and fl, choos- ing the cases where the corresponding values of "y and 6 were integers or half-integers. This exper iment led to many more candidates for an elementary solution. Fol- lowing an idea of David Gale, other candidates showed up, involving 7 as a divisor of the right angle. This led eventual ly to the Table 2.

To be strict, not all these cases are different: Cases 3A, 8A, and 12 have all the same solution and are spe- cial cases of a series where 0 < ,k < 60 is arbitrary, a = 45 + ),/4, fl = ~, "~ = 45 + 3)~/4, and 6 = )~/2. In the same way, cases 3B, 8B, and 12 admit the same so- lution and are special cases of a series where 0 < & < 60 is arbitrary, a = 45 + ,V4, fl = 45 - )~/4, 3, = 45 + 3)~/4, and 6 = 45 - 3&/4. Table 2 thus lists 36 problems with possibly different solutions. The number of different problems can be fur ther reduced. From Table 2 a kind of duality is very easily discovered: To each problem ()~, a, fl, % 6), we can associate a dual ()~', a ' , fl', "~', 6'), wi th )~' = ,k, a ' = a , f l ' = a - 6, " / = % and 6' = a - f l

(problem 12 is self-dual). We, thus, have 18 problems once we prove this duality, this proof being the nine- teenth problem referred to in the title; it can be solved by superposing the figures corresponding to (;~, a, fl, % 6) and ()~', a ~, fl', "y~, 6~), after mirroring one of them in the vertical axis, and then applying a special case of Pappus 's theorem from projective geometry. I must confess that I did not solve each one of these new cases by elementary

1 8 THE MATHEMATICAL INTELLIGENCER VOL. 17, NO. 1, 1995

Problem )~ a fl 7 6

1A 60/7 390/7 150/7 480/7 60 /7 1B 60/7 390/7 330/7 480/7 240/7 2A 12 42 18 48 12 2B 12 42 30 48 24 3A 12 48 12 54 6 3B 12 48 42 54 36 4A 12 57 33 75 15 4B 12 57 42 75 24 5A 12 66 42 96 12 5B 12 66 54 96 24 6A 12 69 21 87 3 6B 12 69 66 87 48 7A 12 72 42 108 6 7B 12 72 66 108 30 8A 20 50 20 60 10 8B 20 50 40 60 30 9A 20 60 30 80 10 9B 20 60 50 80 30

10A 20 65 25 85 5 10B 20 65 60 85 40 11A 20 70 50 110 10 11B 20 70 60 110 20 12 36 54 36 72 18 13A 45 45 15 52.5 7.5 13B 45 45 37.5 52.5 30 14A 360/7 240/7 120/7 270/7 90 /7 14B 360/7 240/7 150/7 270/7 120/7 15A 360/7 345/7 150/7 435/7 60/7 15B 360/7 345/7 285/7 435/7 195/7 16A 72 39 21 48 12 16B 72 39 27 48 18 17A 72 42 24 54 12 17B 72 42 30 54 18 18A 72 48 24 66 6 18B 72 48 42 66 24 19A 72 51 39 81 9 19B 72 51 42 81 12 20A 120 24 12 30 6 20B 120 24 18 30 12

methods (in fact, the answer given by the computer is not a true solution, as will be evident below). This is perhaps a good activity for dead periods in academic life, such as boring meetings of the academic staff.

This is also a useful illustration of the care that one must take wi th computer applications in mathematics. I was using the software M a t h e m a t i c a with its default precision, which gives about six correct digits, and my program tested whether a number was an integer or half- integer or not by a method that amounts to looking at the first five decimal digits of its double. Beside the cases in Table 2, the computer also proposed the ones in Table 3.

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Program in Mathemat ic s Table 3

A a fl 7

5 74.5 49 117 6.5 5 74.5 68 117 25.5 8 33.5 5.5 35 4 8 33.5 29.5 3 5 28

23 46 15.5 53 8.5 23 46 37.5 53 30.5 35 33.5 18 37 14.5 39 33.5 11.5 36.5 8.5 39 33.5 25 36.5 22 41 67 13 79.5 0.500003 59 33.5 4 35 2.5 59 33.5 31 35 29.5 61 32.5 7.5 35 5 61 32.5 27.5 35 25 67 32 4 33.5 2.5 67 32 29.5 33.5 28 68 46 28.5 62 12.5 68 46 33.5 62 17.5 68 55.5 41.5 95.5 1.5 77 49.5 14 62.5 0.999998 78 44 25.5 60.5 9 97 23 1.5 23.5 0.999998 97 23 22 23.5 21.5

129 21.5 7 26 2.5

There are several strange things in Table 3: First, there are three values for 6 that are almost integers or half- integers but are not exactly so; second, six dual problems are missing. This could be caused by some roundoff er- rors, so I used the ability of Mathematica to work with an arbitrary number of digits and tested each of the val- ues in Tables 2 and 3 with 50-digit approximation. Every value in Table 2 remained correct, but all the values in Table 3, as well as their missing duals, appeared only as approximate integers or half-integers, although with a very good approximation. For example, the first value for ~ in Table 3 becomes

6.5000016063958834535205203604298488368773 . . . .

At that moment I was astonished by what seemed an incredible coincidence: Several results that were half- integers within five decimals but were not real half- integers. In fact there was no occasion for astonishment: The computer had tried about 950,000 triples, many more than the number of groups of five digits.

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