2025 AMC 10B Problem 25

Problem 25 of 25HarderGeometry Number Theory

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

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?

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Solution

Place

A = (0,0),

B = (4,0),

C = (4,4),

D = (0,4),

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,

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.

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