By Ross Honsberger

Booklet by way of Honsberger, Ross

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**Example text**

It is clear that the first number from B, for example, which can appear after the initial value r, is not r + 2 because it would not be preceded by either r + I or r + 3. Similarly, none of r + 3, r + 4, . . , n can be the first member of B to follow r. Repeated application of this argument shows that the members of B must occur in the order r + l , r + 2, . . , n . Similarly for A . 47 DOUBLE SEQUENCES So long as the elements of A and B keep their natural order within their own group, it doesn ' t matter how the groups may be interlaced.

25. reaches an image point B ' such that A I B ' splits the other 2n points of S in half. Let A and B of the given set denote the antecedents of A ' and B'. We observe that A ' B ' cannot pass through the center of inversion 0, for this would mean that A , B, and 0 of the given set would lie on a line (a contradiction). Therefore the antecedent of the line A I B' must be a circle K through 0 (FIG. 26). K, then, passes through the three points A , B, and 0 of the given set. And because, when inverting back the images of S to the points of the given set, all the images on one side of A I B I are taken inside K and all those on the other side of A I B I are carried outside K, we see that K splits the given set in half.

For a thorough treatment of this famous result, see my book Mathematical Gems, Vol. I , Dolciani Mathematical Expositions, Mathematical Association of America, 24-34. It is proved there that the minimum sum PA + PB + PC is given by the segment Ar--- � ____________________ o FIG. 1 5 . 24 MATHEMATICAL MORSELS A r----.. D '"'------------=...... c FIG. 16 8' A FIG. 17. AN OPAQUE SQUARE 25 BB', where B' is the vertex of the equilateral triangle constructed outwardly on the side A C (FIG. 1 7). A B E is isosceles.