HMMT 二月 2000 · ORAL 赛 · 第 5 题
HMMT February 2000 — ORAL Round — Problem 5
题目详情
英文原题
- [45] Show that it is impossible to find a triangle in the plane with all integer coordinatessuch that the lengths of the sides are all odd.
解析
英文解析
- 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.
Alternate solution : By Pick’s theorem ( A = I + B − 1), the area is an integer or a 1
√2
half-integer. But by Hero’s formula, the area is product of 4 odd integers. this is a 1
contradiction.4
3 3