๐Ÿ”ฅ New! 2025 AMC 10 B

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Problem 25 Easy

Square ABCD has sides of length 4. Points P and Q lie on \overline { AD} and \overline { C D }, respectively, with AP=\frac { 8 } { 5 } and  D Q = \frac { 1 0 } { 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.

    A

  • B.

    B

  • C.

    C

  • D.

    D

  • E.

    The path continues forever.

Answer:B
Solution

We can let the line keep going, making the square reflected.

It suffices to check if line PQ passes (4k, 4\ell) for some (k, \ell) \in \mathbb{N}_+.

Let line PQ: y = mx + \frac{8}{5}, then m = \frac{4 - \frac{8}{5}}{\frac{10}{3}} = \frac{18}{25}.

So PQ: y = \frac{18}{25}x + \frac{8}{5}.

For x = 4k and y = 4\ell: 25\ell = 18k + 10.

The solution (k, \ell) = (5, 4) is a positive integer solution.

It can be proved that (8k, 8\ell) are images of A, (8k+4, 8\ell) are images of B, (8k, 8\ell+4) are images of D, and (8k+4, 8\ell+4) are images of C.

Therefore (20, 16) is an image of B.

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