AMC 8 2024 Problem 23
AMC8 2024 年第 23 题
坐标几何★★★★★
Rodrigo has a very large piece of graph paper. First he draws a line segment connecting point (0, 4) to point (2, 0) and colors the 4 cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point (2000, 3000) to point (5000, 8000). Again he colors the cells whose interiors intersect the segment. How many cells will he color this time?
Rodrigo有一张非常大的方格纸。他先画了一条线段连接点（O,4)和点（2,0），然后为内部与该线段相交的4个方格上色，如图所示。紧接着Rodrigo又画了一条线段连接点(2000,3000）和点（5000,8000)，并为内部与该线段相交的方格上色。 请问这次Rodrigo要给多少个方格上色？
- A.6000
- B.6500
- C.7000
- D.7500
- E.8000
正确答案 · Correct Answer
C
解析 · Solution
解法1
常规解法·约分后画小图数格
题里先给的小线段 染了 4 格，只是示范"被线段内部穿过的格子要染色"。下面求大线段。
大线段从 (2000,3000) 到 (5000,8000)：横走 5000-2000=3000、竖走 8000-3000=5000。把比 像约分一样约到最简 =3:5（同时除以 1000）。这说明大线段正好是 1000 段相同的小线段首尾接成的——每段"横3竖5"，从一个格点斜走到下一个格点（线段依次经过 这些格子角，把它切成 1000 段）。
一段小的 穿几格？画出来沿线数 =7 格（见下图）。
所以 1000×7=7000。
易错提醒：约的是"位移"（横走、竖走的差），不是端点坐标。若顺手把端点写成 就错了——那是"横2竖3"的另一条线（只穿 2+3-1=4 格）。正确的小版本是把位移 (3000,5000) 或两个端点同时除以 1000，得 或 ，都保住"横3竖5"。
公式·会的人秒杀
格点到格点的线段，内部穿过的格子数 。本题 。
为什么成立：从头走到尾，每穿过一条网格线就进入一个新格子，所以格子数 =1+穿线次数。横向穿 条竖线、纵向穿 条横线；但在线段经过的 个格点处，一条竖线和一条横线在同一点相遇、只进了一个新格却被算两次，要各减一次。合起来 。
看个小例子 （）：
图里 4 个格子已按顺序编号；线段在圈出的格点 (2,1) 处，正好同时穿过一条竖线和一条横线——这一下只进了 1 个新格，却在「穿竖线」和「穿横线」里各被算了一次，所以减掉 1。最后 4+2-2=4 格，与图中数到的一致。
这里的 正是解法1里"复制 1000 份"的那个数。
前置（不会 gcd 的先补）： 即最大公约数。可用约分/短除（3000,5000 都带三个 0 → 都除以 1000，剩 3,5 互质，故 ），或辗转相除（大数除小数取余数，反复到余 0，最后一个非零余数即 ）。
Solution 1
*(Elementary — reduce the ratio, then draw the small case.)* The small segment in the figure colors 4 cells; that just shows the rule — color the cells whose interior the segment passes through. Now the big segment.
From (2000,3000) to (5000,8000) it moves right 5000-2000=3000 and up 8000-3000=5000. Reduce to lowest terms =3:5 (divide both by 1000). So the big segment is exactly 1000 identical small segments laid end to end — each "3 right, 5 up", from one lattice point to the next (it passes through , splitting it into 1000 pieces).
How many cells does one small cross? Draw it and count along the segment: 7 (see below).
So 1000×7=7000.
Common mistake: reduce the displacement (how far right / up), not the endpoint coordinates. Writing is wrong — that is "2 right, 3 up", a different segment (2+3-1=4 cells). The correct small version divides the displacement (3000,5000), or both endpoints, by 1000: or , both keeping "3 right, 5 up".
*(Formula.)* A lattice-to-lattice segment crosses cells. Here .
Why it works: walking from end to end, each grid line you cross puts you in a new cell, so cells =1+(crossings). You cross vertical lines and horizontal lines; but at each of the lattice points the segment passes through, a vertical and a horizontal line meet at the same point — you enter only one new cell but counted it twice, so subtract one each. Total .
A small example ():
The 4 cells are numbered; at the circled lattice point (2,1) the segment crosses a vertical and a horizontal line at the same instant — it enters only one new cell, yet it was counted once among the vertical crossings and once among the horizontal ones, so subtract one. Hence 4+2-2=4, matching the 4 cells in the figure.
This is exactly the "1000 copies" from Solution 1.
Prerequisite (if gcd is new): is the greatest common divisor. Find it by reduction / short division (3000,5000 both end in three zeros → divide by 1000, leaving coprime 3,5, so ), or by the Euclidean algorithm (divide the larger by the smaller, take the remainder, repeat until 0).
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