2025 AMC 10B Problem 25

Set coordinates as in the figure: $A = (0,0)$, $B = (4,0)$, $C = (4,4)$, $D = (0,4)$. Then $\overline{AD}$ is the left side and $\overline{CD}$ the top side, so
$$P = \left(0, \tfrac85\right), \qquad Q = \left(\tfrac{10}{3}, 4\right).$$

Unfold. Instead of reflecting the ray at a side, reflect the square. Repeatedly reflecting the square in its sides tiles the plane with copies, one in each cell
$$[4i,\,4i+4] \times [4j,\,4j+4],$$
and under this unfolding the bouncing path becomes a single straight ray from $P$. A bounce is just the moment the ray crosses from one cell to the next, and the images of the four vertices of the square are exactly the lattice points $(4m, 4n)$. So:

> the path terminates at a vertex precisely when the straight ray from $P$ first meets a point $(4m, 4n)$.

The ray. Its direction is
$$Q - P = \left(\tfrac{10}{3},\; 4 - \tfrac85\right) = \left(\tfrac{10}{3},\; \tfrac{12}{5}\right), \qquad \text{slope} = \frac{12/5}{10/3} = \frac{18}{25},$$
so the ray is $y = \tfrac85 + \tfrac{18}{25}x$ with $x$ increasing.

When does it hit a lattice point? Put $x = 4m$ and $y = 4n$ with $m, n$ positive integers:
$$4n = \frac85 + \frac{72m}{25} \implies 100n = 40 + 72m \implies 25n = 10 + 18m .$$
Reducing modulo $25$: $18m \equiv -10$. Since $18 \cdot 7 = 126 \equiv 1 \pmod{25}$, multiply by $7$:
$$m \equiv -70 \equiv 5 \pmod{25}.$$
The smallest positive solution is $m = 5$, which gives $n = \dfrac{10 + 90}{25} = 4$. So the ray first reaches a lattice point at $(20, 16)$, and the path does terminate.

Which vertex? Folding back, the copies alternate orientation, so the image point $(4m, 4n)$ corresponds to the corner of the original square whose $x$-coordinate is $0$ when $m$ is even and $4$ when $m$ is odd, and whose $y$-coordinate is $0$ when $n$ is even and $4$ when $n$ is odd. Here $m = 5$ is odd and $n = 4$ is even, so the corner is $(4, 0)$, which is $B$.

The answer is $\boxed{\textbf{(B)}\ B}$.

Reflecting the billiard table rather than the ball is the standard device for any "bounce off the walls" problem: it replaces an unbounded amount of casework by one straight line, and every question about the path becomes a question about that line's intersections with a lattice. Hitting a corner then reads as a single linear congruence, which also settles choice (E) at once — either the congruence has a solution or it does not. The last step, recovering which corner, is pure parity: an even number of reflections in the vertical sides returns the left edge to the left, an odd number swaps it to the right. As a check, $m = 5$ and $n = 4$ say the ray crosses five columns and four rows of cells, so the real path consists of $5 + 4 - 1 = 8$ segments before it dies, which is what a careful hand trace confirms.

• Tracing the bounces one at a time, losing the thread after four or five reflections and concluding from the absence of an early corner that the path runs forever, which is (E).
• Placing $P$ and $Q$ from the wrong ends of their sides: $AP$ is measured from $A$ and $DQ$ from $D$, so $P = \left(0,\tfrac85\right)$ and $Q = \left(\tfrac{10}{3},4\right)$, not $\left(0,\tfrac{12}{5}\right)$ and $\left(\tfrac23, 4\right)$.
• Finding $m = 5$ and $n = 4$ but then reading the corner off as $(0,4) = D$ or $(4,4) = C$; it is the parity of each index separately that decides each coordinate.