AMC10 2012 A

AMC10 2012 A · Q24

AMC10 2012 A · Q24. It mainly tests Systems of equations, Manipulating equations.

Let a, b, and c be positive integers with a ≥ b ≥ c such that $a^2 - b^2 - c^2 + ab = 2011$ and $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$. What is a?

设 a, b, c 为正整数且 a ≥ b ≥ c，使得 $a^2 - b^2 - c^2 + ab = 2011$ 且 $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$。a 是多少？

(A) 249 249

(B) 250 250

(D) 252 252

(E) 253 253

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Adding the two equations gives $2a^2+2b^2+2c^2-2ab-2bc-2ac=14,$ so $(a-b)^2+(b-c)^2+(c-a)^2=14.$ Note that there is a unique way to express 14 as the sum of perfect squares (up to permutations), namely, $14=3^2+2^2+1^2$. Because $a-b$, $b-c$, and $c-a$ are integers with their sum equal to 0 and $a\ge b\ge c$, it follows that $a-c=3$ and either $a-b=2$ and $b-c=1$, or $a-b=1$ and $b-c=2$. Therefore either $(a,b,c)=(c+3,c+1,c)$ or $(a,b,c)=(c+3,c+2,c)$. Substituting the relations in the first case into the first given equation yields $2011=a^2-c^2+ab-b^2=(a-c)(a+c)+(a-b)b=3(2c+3)+2(c+1).$ Solving gives $(a,b,c)=(253,251,250)$. The second case does not yield an integer solution. Therefore $a=253$.

答案（E）：将两个方程相加得到 $2a^2+2b^2+2c^2-2ab-2bc-2ac=14,$ 因此 $(a-b)^2+(b-c)^2+(c-a)^2=14.$ 注意：把 14 表示为完全平方数之和（不计排列）的方法是唯一的，即 $14=3^2+2^2+1^2$。由于 $a-b$、$b-c$、$c-a$ 是整数，且它们的和为 0，并且 $a\ge b\ge c$，可推出 $a-c=3$，并且要么 $a-b=2$ 且 $b-c=1$，要么 $a-b=1$ 且 $b-c=2$。因此要么 $(a,b,c)=(c+3,c+1,c)$，要么 $(a,b,c)=(c+3,c+2,c)$。将第一种情形的关系代入题目给出的第一个方程，得到 $2011=a^2-c^2+ab-b^2=(a-c)(a+c)+(a-b)b=3(2c+3)+2(c+1).$ 解得 $(a,b,c)=(253,251,250)$。第二种情形不能得到整数解。因此 $a=253$。

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