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Extra resources for Challenging mathematical problems with elementary solutions [Vol. II]
If A, B, C are any three points, let d(A; BC) denote the distance from A to the line I containing Band C. Now suppose that not all points of S are collinear; then there exist three points A, B, C in S such that d(A; BC) > O. There are only a finite number of such triplets A, B, C, for S is finite. Hence there is at least one such triplet such that d(A; BC) is minimal. By hypothesis the line I through Band C contains another point D of S (see fig. 38). The foot P of the perpendicular from A to I divides I A B p D Fig.
39b or c, for if we assume that they are, we necessarily arrive at one of the configurations 39b or c for the five points A, B, C, D, E, and in each of these configurations the length of DE is different from a or b. If A, B, C, D are arranged as in fig. 3ge, then so are A, B, C, E, and this is clearly impossible if D and E are to be distinct. Similarly, if A, B, C, D are as in 39d, E must coincide with D (for the four points A, B, C, E must also be at the vertices of a square). B ~--+--';#E B A A D C D A E D a.
An-I> it is distinct from all the lines joining AI' A 2 • As, ... ,An in pairs. We have thus added at least one new line and therefore have a total of at least n + 1 distinct lines. Consequently, we have shown that if our theorem is true for n points, then it is also true for n + 1 points. By mathematical induction it follows that the theorem holds for any number of points. 108•. Let A, B, C, D be the four points. In all, there are six possible distances between the points, namely, AB, AC, AD, BC, BD, CD.