HMMT 十一月 2020 · GEN 赛 · 第 6 题

HMMT November 2020 — GEN Round — Problem 6

A sphere is centered at a point with integer coordinates and passes through the three points (2 , 0 , 0), (0 , 4 , 0), (0 , 0 , 6), but not the origin (0 , 0 , 0). If r is the smallest possible radius of the sphere, compute 2 r .

解析

A sphere is centered at a point with integer coordinates and passes through the three points (2 , 0 , 0), (0 , 4 , 0), (0 , 0 , 6), but not the origin (0 , 0 , 0). If r is the smallest possible radius of the sphere, compute 2 r . Proposed by: James Lin Answer: 51 Solution: Let ( x, y, z ) be the center of the sphere. By the given condition, we have 2 2 2 2 2 2 2 2 2 ( x − 2) + y + z = x + ( y − 4) + z = x + y + ( z − 6) . 2 2 2 Subtracting x + y + z yields 2 2 2 2 2 2 x − ( x − 2) = y − ( y − 4) = z − ( z − 6) , or 4( x − 1) = 8( y − 2) = 12( z − 3) . Therefore ( x − 1 , y − 2 , z − 3) must be (6 t, 3 t, 2 t ) for some integer t . Checking small values of t yields that the possibilities for ( x, y, z ) closest to (2 , 0 , 0) are ( − 5 , − 1 , 1), (1 , 2 , 3), and (7 , 5 , 5). The second 2 yields a sphere that passes through the origin and is thus forbidden. The other two yield r = 51 and 2 r = 75, so 51 is the answer.