Maths Olympiad Prep

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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,…,81, 2, 3, \ldots, 8? Can the lengths of the eight segments be eight distinct integers?

Solution

Answer no, yes

Figure 1
If a triangle has integer sides, one of which is 11, then it must be isosceles. So the only candidates for the segment length 11 are ABAB and AEAE. wlog AB=1AB = 1, so BF=AFBF = AF. Hence cos⁡DFE=1−1/(2AF2)\cos DFE = 1 - 1/(2 AF^2). Hence ED2=DF2+EF2+2DF⋅EF(1−1/2AF2)=DF2+EF2+2DF⋅EF/AF2ED^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< 1. Contradiction. (Careful, looking at the figure one is tempted to conclude that ED<ABED < AB, but a more realistic figure shows that is false.)

Figure 2
Building on the 3,4,53,4,5 triangle we get the figure above.

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