Question

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

At which vertex does the path terminate?

$A$

$B$

$C$

$D$

The path continues forever.

Answer: B

Concepts:reflection (geometry)parity greatest common divisor coordinate geometry

Difficulty rating: 2520

Solution:

Place $A = (0,0),$ $B = (4,0),$ $C = (4,4),$ $D = (0,4),$ so $P = \left(0, \tfrac85\right)$ and $Q = \left(\tfrac{10}{3}, 4\right).$ The initial direction is $\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18).$ Unfold the billiard into a grid of reflected copies and follow the straight line from $P.$ At a grid corner $(4a,4b),$ the horizontal coordinate gives $t = \tfrac{4a}{25}.$ The vertical coordinate then requires $10 + 18a = 25b,$ so $a \equiv 5 \pmod{25}.$ The first positive possibility is $a = 5,$ giving $b = 4$ and the unfolded corner $(20,16).$ Crossing $5$ cells across (odd) puts it on the side $x = 4,$ and $4$ cells up (even) puts it on $y = 0.$ That's vertex $(4,0) = B.$ Thus, B is the correct answer.

2025 AMC 10B Problem 25

Answer: B

Difficulty rating: 2520

Solution:

Problem 25 in Other Years