AMC10 2025 B

AMC10 2025 B · Q25

AMC10 2025 B · Q25. It mainly tests Transformations, Geometry misc.

Square $ABCD$ has sides of length $4$. Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD}$, respectively, with $AP=\frac{8}{5}$ and $DQ=\frac{10}{3}$. A path begins along the segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles). If the path hits a vertex of the square, it terminates there; otherwise it continues forever. At which vertex does the path terminate?

正方形 $ABCD$ 边长为 $4$。点 $P$ 和 $Q$ 分别在 $\overline{AD}$ 和 $\overline{CD}$ 上，$AP=\frac{8}{5}$，$DQ=\frac{10}{3}$。一条路径从 $P$ 到 $Q$ 的线段开始，然后在 $ABCD$ 的边上反射（入射角和出射角相等）。如果路径击中正方形的顶点，则在那里终止；否则无限继续。路径在哪个顶点终止？

(A) A A

(B) B B

(D) D D

(E) \text{The path continues forever.} \text{The path continues forever.}

Answer

Correct choice: (B)

正确答案：（B）

Solution

The square can be placed with $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$. Then $P=\left(0,\frac{8}{5}\right)$, $Q=\left(\frac{10}{3},4\right).$ Straight line reflection is equivalent to passing through. The straight line from $P$ to $Q$ is $y=\frac{18}{25}x+\frac{8}{5}$ $25y=18x+40\equiv 4\pmod{18}$ Notice that $25\times 2=7\times 2\equiv -4\pmod{18}$, so $25\times(-2)\equiv 4\pmod{18}$, $y=-2\equiv 16\pmod{18}$, then $x=20$, the line passed through $(20,16)$. From $\left(0,\frac{8}{5}\right)$ to $(20,16)$, $x$ coordinate increases $20$ whole units, $y$ coordinate increases $15$ whole units. Excluding the case where $PQ$ passes through a lattice point at the final grid point, the segment $PQ$ crosses integer grid lines a total of $(20-1)+(15-1)=33$ times. Each time the line crosses an integer grid line, it corresponds to one reflection; every four reflections, the path completes a full rotation within the square. $33\equiv 1\pmod{4}$, after one reflection off side $CD$, the line $PQ$ will pass through point $\boxed{B}$.

该正方形可取为 $A=(0,0)$，$B=(4,0)$，$C=(4,4)$，$D=(0,4)$。则 $P=\left(0,\frac{8}{5}\right)$，$Q=\left(\frac{10}{3},4\right)$。直线反射等价于“穿过”展开后的平面。连接 $P$ 到 $Q$ 的直线为 $y=\frac{18}{25}x+\frac{8}{5}$ $25y=18x+40\equiv 4\pmod{18}$ 注意 $25\times 2=7\times 2\equiv -4\pmod{18}$，因此 $25\times(-2)\equiv 4\pmod{18}$，$y=-2\equiv 16\pmod{18}$，进而 $x=20$，该直线经过 $(20,16)$。从 $\left(0,\frac{8}{5}\right)$ 到 $(20,16)$，$x$ 坐标增加 $20$ 个整单位，$y$ 坐标增加 $15$ 个整单位。排除 $PQ$ 在最终格点处经过某个整点的情况，线段 $PQ$ 穿过整网格线的总次数为 $(20-1)+(15-1)=33$ 次。每穿过一条整网格线对应一次反射；每经历四次反射，路径在正方形内完成一次完整循环。由于 $33\equiv 1\pmod{4}$，在边 $CD$ 处反射一次后，直线 $PQ$ 将经过点 $\boxed{B}$。

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