AMC12 2011 B

AMC12 2011 B · Q21

AMC12 2011 B · Q21. It mainly tests Quadratic equations, Fractions.

The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x - y|$?

两个不同的正整数 $x$ 和 $y$ 的算术平均数是一个两位整数。$x$ 和 $y$ 的几何平均数是通过反转算术平均数的数字得到的。求 $|x - y|$？

(A) 24 24

(B) 48 48

(D) 66 66

(E) 70 70

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer: (D) $\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$, $0\le b\le 9$. $\sqrt{xy} = 10 b+a$ Squaring the first and second equations, $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$ $xy = 100b^2 + 20ab + a^2$ Subtracting the previous two equations, $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$ $|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$ Note that for x-y to be an integer, $(a^2 - b^2)$ has to be $11n$ for some perfect square $n$. Since $a$ is at most $9$, $n = 1$ or $4$ If $n = 1$, $|x-y| = 66$, if $n = 4$, $|x-y| = 132$. In AMC, we are done. Otherwise, we need to show that $n=4$, or $a^2 -b^2 = 44$ is impossible. We can arrive at $|x-y| = 6\sqrt{11(a^2 - b^2)}$ using the method above. Because we know that $|x-y|$ is an integer, it must be a multiple of 6 and 11. Hence the answer is $66.$

答案：(D) $\frac{x + y}{2} = 10 a+b$，其中 $1\le a\le 9$, $0\le b\le 9$。 $\sqrt{xy} = 10 b+a$ 将第一式与第二式分别平方，得 $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$ $xy = 100b^2 + 20ab + a^2$ 用前一式减去后一式，$\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$ $|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$ 注意要使 $x-y$ 为整数，$(a^2 - b^2)$ 必须为 $11n$，其中 $n$ 为某个完全平方数。由于 $a$ 最大为 $9$，所以 $n = 1$ 或 $4$。若 $n = 1$，则 $|x-y| = 66$；若 $n = 4$，则 $|x-y| = 132$。在 AMC 中到此即可。否则需要说明 $n=4$（即 $a^2 -b^2 = 44$）不可能。由上述方法可得 $|x-y| = 6\sqrt{11(a^2 - b^2)}$。因为已知 $|x-y|$ 是整数，它必须同时是 6 和 11 的倍数。因此答案是 $66$。

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