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AMC10 2024 B

AMC10 2024 B · Q19

AMC10 2024 B · Q19. It mainly tests Coordinate geometry, Counting in geometry (lattice points).

In the following table, each question mark is to be replaced by "Possible" or "Not Possible" to indicate whether a nonvertical line with the given slope can contain the given number of lattice points (points both of whose coordinates are integers). How many of the 12 entries will be "Possible"? \[ \begin{array}{|c|c|c|c|c|} \hline & \text{zero} & \text{exactly one} & \text{exactly two} & \text{more than two} \\ \hline \text{zero slope} & ? & ? & ? & ? \\ \hline \text{nonzero rational slope} & ? & ? & ? & ? \\ \hline \text{irrational slope} & ? & ? & ? & ? \\ \hline \end{array} \]
在下表中,每个问号都要用 “Possible” 或 “Not Possible” 替换,用来表示:具有给定斜率的一条非竖直直线,是否可能包含给定数量的格点(横纵坐标均为整数的点)。12 个表格项中有多少个会是 “Possible”? \[ \begin{array}{|c|c|c|c|c|} \hline & \text{零个} & \text{恰好一个} & \text{恰好两个} & \text{多于两个} \\ \hline \text{零斜率} & ? & ? & ? & ? \\ \hline \text{非零有理斜率} & ? & ? & ? & ? \\ \hline \text{无理斜率} & ? & ? & ? & ? \\ \hline \end{array} \]
(A) 4 4
(B) 5 5
(C) 6 6
(D) 7 7
(E) 9 9
Answer
Correct choice: (C)
正确答案:(C)
Solution
Let's look at zero slope first. All lines of such form will be expressed in the form $y=k$, where $k$ is some real number. If $k$ is an integer, the line passes through infinitely many lattice points. One such example is $y=1$. If $k$ is not an integer, the line passes through $0$ lattice points. One such example is $y=1.1$. So we have $2$ cases. Let's now look at the second case. The line has slope $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers. The line has the form $y = \frac{p}{q}x + m$. If the line passes through lattice point $(a,b)$, then the line must also pass through the lattice point $(a+kq, b+kp)$, where $a,b,k$ are all integers. Therefore, the line can pass through infinitely many lattice points but it cannot pass through exactly $1$ or $2$. The line can pass through $0$ lattice points, such as $y=x+0.5$. This contributes $2$ more cases. If the line has an irrational slope, it can never pass through more than $2$ lattice points. We prove this using contradiction. Let's say the line passes through lattice points $(a,b)$ and $(c,d)$. Then the line has slope $\frac{d-b}{c-a}$, which is rational. However, the slope of the line is irrational. Therefore, the line can pass through at most $1$ lattice point. One example of this is $y=\sqrt{2}x$. This line contributes $2$ final cases. Our answer is therefore $\boxed{6}$.
先看斜率为零的情况。此类直线形式为$y=k$,$k$为实数。如果$k$是整数,该直线通过无穷多个晶格点,如$y=1$。如果$k$非整数,则通过$0$个晶格点,如$y=1.1$。所以有$2$种情况。 现在看有理斜率的情况。斜率为$p/q$,$p,q$互质。直线形式$y = \frac{p}{q}x + m$。如果通过晶格点$(a,b)$,则也通过$(a+kq, b+kp)$,$a,b,k$为整数。因此,直线通过无穷多个晶格点,但不能恰好$1$或$2$个。可以通过$0$个,如$y=x+0.5$。贡献$2$种情况。 如果斜率无理,则最多通过$2$个晶格点。用反证法:假设通过$(a,b)$和$(c,d)$,则斜率$(d-b)/(c-a)$为有理数,与无理斜率矛盾。因此最多通过$1$个,如$y=\sqrt{2}x$。贡献$2$种情况。 总计$\boxed{6}$。
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