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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Soviet Union

Problem:

Four lines in the plane intersect in six points. Each line is thus divided into two segments and two rays. Is it possible for the eight segments to have lengths $1, 2, 3, \ldots, 8$ ? Can the lengths of the eight segments be eight distinct integers?

Solution

Answer no, yes

If a triangle has integer sides, one of which is $1$ , then it must be isosceles. So the only candidates for the segment length $1$ are $AB$ and $AE$ . wlog $AB = 1$ , so $BF = AF$ . Hence $\cos DFE = 1 - 1/(2 AF^2)$ . Hence $ED^2 = DF^2 + EF^2 + 2DF \cdot EF (1 - 1/2AF^2) = DF^2 + EF^2 + 2DF \cdot EF/AF^2$ . But the first three terms are integers and the last term is $< 1$ . Contradiction. (Careful, looking at the figure one is tempted to conclude that $ED < AB$ , but a more realistic figure shows that is false.)

Building on the $3,4,5$ triangle we get the figure above.

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