42nd Putnam 1981 Problem A6 A, B, C are lattice points in the plane. The triangle ABC contains exactly one lattice point, X, in its interior. The line AX meets BC at E. What is the largest possible value of AX/XE? Solution
: 5. Take A = (2, 4), B = (0, 0), C = (2, 1). This contains a single lattice point X (1, 1). B is a distance 2 from the line AC, and AC = 3, so ABC has area 3. B is a distance 1 from the line CX, and CX = 1, so XBC has area 1/2. So A is 6x the distance of X from BC. Hence AX = 5 XE. So the value 5 can certainly be achieved. Let L, M, N be the midpoints of BC, CA, AB. Let B have the coordinates (b 1 , b 2 ) and similarly for the other points. Take B' to be such that X is the midpoint of BB'. Then it is also a lattice point, since its coordinates are (2x 1 - b 1 , 2x 2 - b 2 ). So it cannot lie in the interior of the triangle ABC (since X is the only such lattice point). So X cannot lie in the interior of the triangle BNL (which has dimensions half that of BAC and is similar to it). Similarly, X cannot lie in the interior of CLM. Now consider the point L' on the ray LX such that LL' = 3LX. It is also a lattice point, since its coordinates are (3x 1 - (b 1 + c 1 ), 3x 2 - (b 2 + c 2 ) ). So it cannot lie in the interior of ABC. A fortiori, it cannot lie in the interior of LMN. Take M' on LM such that LM' = 1/3 LM, and N' on LN such that LN' = 1/3 LN. Then LM'N' is similar to LMN and 1/3 the dimensions. So X cannot lie inside LM'N' (otherwise L' would lie within LMN). It follows that X must lie above the extended line M'N' (which is parallel to BC). But the distance of this line above BC is 1/6 the distance of A above BC. So if the line AX cuts it at F (and BC at E), then AF/FE = 5. Hence AX/XE ≤ 5. 42nd Putnam 1981 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001