Russian problems Problem 32 Find the smallest value x such that, given any point inside an equilateral triangle of side 1, we can always choose two points on the sides of the triangle, collinear with the given point and a distance x apart. Solution
: 2/3. Let O be the center of ABC. Let AO meet BC at D, let BO meet CA at E, and let CO meet AB at F. Given any point X inside ABC, it lies in one of the quadrilaterals AEOF, CDOE, BFOD. Without loss of generality, it lies in AEOF. Take the line through X parallel to BC. It meets AB in P and AC in Q. Then PQ is shorter than the parallel line MON with M on AB and N on AC, which has length 2/3. If we twist the segment PXQ so that it continues to pass through X, and P remains on AB and Q on AC, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than 2/3. So at some intermediate position its length will be 2/3. To show that no value smaller than 2/3 is possible, it is sufficient to show that any segment POQ with P and Q on the sides of the triangle has length at least 2/3. Take P on MB and Q on AN with P, O, Q collinear. Then PQ cos POM = MN - QN cospi/3 + PM cos pi/3. But PM>QN (using the sine rule, PM = OM sinPOM/sinOPM and QN = ON sinQON/sinOQN, but OM=ON, angle POM = angle QON, and angle OQN = angle OPM + pi/3 > angle OPM), and hence PQ > MN sec POM > MN. Russian 31-40 (C) John Scholes jscholes@kalva.demon.co.uk 19 Sep 1998