PUMaC 2009 · 代数（A 组） · 第 8 题
PUMaC 2009 — Algebra (Division A) — Problem 8
题目详情
- The real numbers x , y , z , and t satisfy the following equation: √ 2 2 2 2 x + 4 xy + 3 y − 2 xz − 2 yz + z + 1 = t + y + z − t Find 100 times the maximum possible value for t . 1
解析
- The real numbers x , y , z , and t satisfy the following equation: √ 2 2 2 2 x + 4 xy + 3 y − 2 xz − 2 yz + z + 1 = t + y + z − t Find 100 times the maximum possible value for t . Solution. 125. The equation is equivalent to the following ( ) √ 2 2 2 2 ( z − x − y − 1 / 2) + ( x + y − 1 / 2) + ( y − 1 / 2) + y + z − t − 1 / 2 = 0 Hence y = 1 / 2, x = 0, z = 1 and t = 5 / 4. We would like to give some motivation for this: First, note that you are on the right track once you want to complete squares. The equation is of the form √ a = t + b − t it is very natural to move everything to one side and write it as √ a − b − 1 / 4 + ( b − t ) − b − t + 1 / 4 = 0 3 ( √ ) 2 or also a − b − 1 / 4 + b − t − 1 / 2 = 0. Hence we must examine the expression a − b − 1 / 4 in more detail. In our case this is a non-homogenous polynomial in x, y, z , such that each term is of degree at most 2. 2 2 2 2 x + 4 xy + 3 y − 2 xz − 2 yz + z + 3 / 4 − y − z However, we can actually transform it into a homogenous polynomial using the substitution x = s/v , y = t/v , z = u/v in which case it becomes 2 2 2 2 2 s + 4 st + 3 t − 2 su − 2 tu + u + 3 / 4 v − tv − uv 2 v The reason to do this is that there is a standard theorem about how one can reduce such homogenous quadratic polynomials, called quadratic forms: The Canonical Form Theorem for Symmetric Quadratic Forms. This provides a standard and straightforward method in which you can diagonalize any symmetric quadratic form - basically by completing squares in the appropriate way. Applying that theorem to our little problem may be a bit of an overkill, but at least it shows some connection of this problem to some classical area of mathematics, making it less sketchy. 4