Question

设

$\tiny S_1 = \{(x, y) \mid \log_{10}(1 + x^2 + y^2) \le 1 + \log_{10}(x + y)\}$

并且

$\tiny S_2 = \{(x, y) \mid \log_{10}(2 + x^2 + y^2) \le 2 + \log_{10}(x + y)\}.$

$S_2$ 的面积与 $S_1$ 的面积之比是多少？

Let

$\tiny S_1 = \{(x, y) \mid \log_{10}(1 + x^2 + y^2) \le 1 + \log_{10}(x + y)\}$

and

$\tiny S_2 = \{(x, y) \mid \log_{10}(2 + x^2 + y^2) \le 2 + \log_{10}(x + y)\}.$

What is the ratio of the area of $S_2$ to the area of $S_1?$

$98$

$99$

$100$

$101$

$102$

答案：E

知识点：对数配方法圆面积

难度评级：2180

解答：

对 $j = 1, 2,$ 条件变为 $j + x^2 + y^2 \le 10^j(x + y),$ 即 $\begin{gathered} \left(x - \frac{10^j}{2}\right)^2 \\ {}+ \left(y - \frac{10^j}{2}\right)^2 \\ \le \frac{10^{2j}}{2} - j. \end{gathered}$

这是两个圆盘， $S_1$ 的半径平方为 $\tfrac{100}{2} - 1 = 49$ ， $S_2.$ 的半径平方为 $\tfrac{10000}{2} - 2 = 4998$ 。

对数要求 $x+y>0.$ 对每个圆盘，其圆心到直线 $x+y=0$ 的距离平方为 $\tfrac{10^{2j}}{2},$ 比半径平方恰好多 $j$ 。所以两个圆盘都完全位于 $x+y>0,$ 内，改写不等式时没有加入额外点。面积之比为 $\dfrac{4998}{49} = 102.$

所以正确答案是 E。

For $j = 1, 2,$ the condition becomes $j + x^2 + y^2 \le 10^j(x + y),$ i.e. $\begin{gathered} \left(x - \frac{10^j}{2}\right)^2 \\ {}+ \left(y - \frac{10^j}{2}\right)^2 \\ \le \frac{10^{2j}}{2} - j. \end{gathered}$

These are disks with squared radii $\tfrac{100}{2} - 1 = 49$ for $S_1$ and $\tfrac{10000}{2} - 2 = 4998$ for $S_2.$

The logarithms require $x+y>0.$ For each disk, the squared distance from its center to the line $x+y=0$ is $\tfrac{10^{2j}}{2},$ which is $j$ more than the squared radius. Hence both disks lie entirely in $x+y>0,$ so no points were added when the inequalities were rewritten. The area ratio is $\dfrac{4998}{49} = 102.$

Thus, the correct answer is E.

2006 AMC 12A 第 21 题

答案：E

难度评级：2180

解答：

其他年份的第 21 题