AMC10 2020 A

AMC10 2020 A · Q16

AMC10 2020 A · Q16. It mainly tests Probability (basic), Geometric probability (basic).

A point is chosen at random within the square in the coordinate plane whose vertices are $(0,0)$, $(2020,0)$, $(2020,2020)$, and $(0,2020)$. The probability that the point lies within $d$ units of a lattice point is $\frac{1}{2}$. (A point $(x,y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth?

在坐标平面内的一个正方形中随机选择一点，该正方形的顶点为$(0,0)$、$(2020,0)$、$(2020,2020)$和$(0,2020)$。该点距离格点$d$单位以内的概率为$\frac{1}{2}$。（点$(x,y)$是格点当且仅当$x$和$y$均为整数。）$d$的最接近的十分位数是多少？

(A) 0.3 0.3

(B) 0.4 0.4

(D) 0.6 0.6

(E) 0.7 0.7

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Without loss of generality, it can be assumed that the point falls within the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. The probability that the point is within $d$ units of a lattice point is the total area of the four quarter-circles of radius $d$ centered at the corners of the square, provided $d<\frac12$. This is the shaded region in the figure. Thus $\pi d^2=\frac12$, so $d=\frac{1}{\sqrt{2\pi}}<\frac12$. Estimating gives $$ d=\frac{1}{\sqrt{2\pi}}\approx\frac{1}{\sqrt{6.28}}\approx\frac{10}{\sqrt{625}}=\frac{10}{25}=0.4. $$ For a rigorous justification of this estimate, it must be shown that $\frac{1}{\sqrt{2\pi}}$ satisfies $$ \frac{7}{20}=0.35<\frac{1}{\sqrt{2\pi}}<0.45=\frac{9}{20}, $$ which is equivalent to $$ \frac{49\pi}{400}<\frac12<\frac{81\pi}{400}. $$ Indeed, $$ \frac{49\pi}{400}<\frac{50\pi}{400}=\frac{\pi}{8}<\frac12 $$ and $$ \frac{81\pi}{400}>\frac{80\pi}{400}=\frac{\pi}{5}>\frac12. $$

答案（B）：不失一般性，可假设该点落在顶点为 $(0,0)$、$(1,0)$、$(1,1)$、$(0,1)$ 的单位正方形内。该点距离某个格点不超过 $d$ 的概率，等于以正方形四个角为圆心、半径为 $d$ 的四个四分之一圆的总面积（要求 $d<\frac12$）。这就是图中的阴影区域。因此 $\pi d^2=\frac12$，所以 $d=\frac{1}{\sqrt{2\pi}}<\frac12$。估算得到 $$ d=\frac{1}{\sqrt{2\pi}}\approx\frac{1}{\sqrt{6.28}}\approx\frac{10}{\sqrt{625}}=\frac{10}{25}=0.4。 $$ 为了严格证明该估算，需要证明 $\frac{1}{\sqrt{2\pi}}$ 满足 $$ \frac{7}{20}=0.35<\frac{1}{\sqrt{2\pi}}<0.45=\frac{9}{20}, $$ 这等价于 $$ \frac{49\pi}{400}<\frac12<\frac{81\pi}{400}。 $$ 确实有 $$ \frac{49\pi}{400}<\frac{50\pi}{400}=\frac{\pi}{8}<\frac12 $$ 以及 $$ \frac{81\pi}{400}>\frac{80\pi}{400}=\frac{\pi}{5}>\frac12。 $$

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