HMMT 二月 2000 · ORAL 赛 · 第 5 题

HMMT February 2000 — ORAL Round — Problem 5

[45] Show that it is impossible to find a triangle in the plane with all integer coordinates such that the lengths of the sides are all odd.

解析

Suppose there is such a triangle with vertices at ( m , n ) , ( m , n ) , ( m , n ). Then we 1 1 2 2 3 3 √ 2 2 2 2 are given ( m − m ) + ( n − n ) is an odd integer, so that ( m − m ) + ( n − n ) 1 2 1 2 1 2 1 2 is 1 (mod 4). Hence exactly one of m − m and n − n must be odd and the other 1 2 1 2 must be even. Suppose w.lo.g m − m is odd and n − n is even. Similarly, exactly 1 2 1 2 one of m − m and n − n must be odd and the other must be even. If m − m is odd 1 3 1 3 1 3 and n − n even, then m − m and n − n are both even, which is a contradiction 1 3 2 3 2 3 2 2 sice we want ( m − m ) + ( n − n ) to be 1 (mod 4). Similarly if m − m is even and 2 3 2 3 1 3 2 2 n − n is odd then m − m and n − n are both odd, and ( m − m ) + ( n − n ) is 1 3 2 3 2 3 2 3 2 3 2 (mod 4), a contradiction. 1 Alternate solution : By Pick’s theorem ( A = I + B − 1), the area is an integer or a 2 √ 1 half-integer. But by Hero’s formula, the area is product of 4 odd integers. this is a 4 contradiction. 3 3