Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it Brazil

Given a sheet of paper and the use of a rule, compass and pencil, show how to draw a straight line that passes through two given points, if the length of the ruler and the maximum opening of the compass are both less than half the distance between the two points. You may not fold the paper.

Solution

Note that we can draw an arbitrarily long line through a given point by repeatedly extending a short line. We can also find the midpoint of an arbitrary line segment. For suppose the segment is PQPQ. Take a distance kk which is less than the maximum opening of the compass and less than the length of the ruler. Starting at PP and using the compasses, mark off qq distances of kk leaving a final distance of r<kr < k to QQ. Now bisect the final segment of rr as usual and a segment length kk. Then mark off qq distances of k/2k/2 and one of r/2r/2 from PP to get the midpoint.

So suppose the points given are AA and BB. Take any lines through AA and BB meeting at CC. Let M1,N1M_1, N_1 be the midpoints of AC,BCAC, BC respectively. Then take M2,N2M_2, N_2 as the midpoints of M1C,N1CM_1C, N_1C respectively, and so on until we get MnNn<kM_n N_n < k. We can now join MnM_n and NnN_n to get a line parallel to the desired line ABAB. That allows us to draw a short line through AA in the right direction. We mark off a point XX on ACAC with AX=MnCAX = M_n C, then draw circles center AA radius MnNnM_n N_n and center XX radius CNnCN_n to intersect at a point YY with AXYAXY congruent to MnCNnM_n C N_n and hence YY on ABAB. Now extend AYAY to get ABAB.

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