Question

在边长为 $1$ 的正方形内部或边界上任取五个点。设 $a$ 是满足下述性质的最小数：从这五点中总能选出一对，使其距离小于或等于 $a$ 。则 $a$ 为：

Any five points are taken inside or on a square with side length $1.$ Let $a$ be the smallest possible number with the property that it is always possible to select one pair of points from these five such that the distance between them is equal to or less than $a.$ Then $a$ is:

$\dfrac{\sqrt3}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{2\sqrt2}{3}$

$1$

$\sqrt2$

答案：B

知识点：抽屉原理极端原理正方形（几何）

难度评级：1670

小提示：

将单位正方形分成四个全等小正方形

Partition the unit square into four congruent smaller squares

大提示：

为证明界可达到，寻找使最近点对距离达到该界的五个点

To prove sharpness, look for five points whose closest-pair distance reaches the bound

解答：

将正方形分成四个边长为 $\frac{1}{2}$ 的小正方形。五点中有两点落在同一小正方形内，所以它们的距离至多为其对角线 $\frac{\sqrt2}{2}$ 。在四个顶点和中心各放一点可以达到此界：此时最短距离为 $\frac{\sqrt2}{2}$ 。因此所求最小保证值是 $\frac{\sqrt2}{2}$ 。

因此，正确答案是 B。

Divide the square into four squares of side $\frac{1}{2}.$ Two of the five points lie in the same small square, so their distance is at most its diagonal, $\frac{\sqrt2}{2}.$ This bound is attainable by placing points at the four corners and the center: the shortest distance is then $\frac{\sqrt2}{2}.$ Hence the least guaranteed value is $\frac{\sqrt2}{2}.$

Therefore, the correct answer is B.

1961 AMC 12 第 39 题

答案：B

难度评级：1670

小提示：

大提示：

解答：

其他年份的第 39 题