AMC 10 · 2025 · #25
Grade 8 geometry-2dSquare ABCD has sides of length 4. Points P and Q lie on AD and CD, respectively, with AP=58 and DQ=310. 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?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A ball starts at point $P$ on the left side of a square with side $4$, travels in a straight line to point $Q$ on the top side, then keeps bouncing off the walls like light in a mirror-box (equal in and out angles). It stops only if it lands exactly on a corner. Find which corner it reaches.
Givens: Square $ABCD$ has side length $4$, placed as $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$; $P$ is on $\overline{AD}$ with $AP=\frac{8}{5}$, so $P=(0,\frac{8}{5})$; $Q$ is on $\overline{CD}$ with $DQ=\frac{10}{3}$, so $Q=(\frac{10}{3},4)$; The path starts along $P\to Q$ and reflects off each wall with equal incoming and outgoing angles; The path stops the instant it hits a vertex; otherwise it continues forever
Unknowns: Which vertex ($A$, $B$, $C$, or $D$) the bouncing path finally lands on, or whether it never hits one
Understand
Restated: A ball starts at point $P$ on the left side of a square with side $4$, travels in a straight line to point $Q$ on the top side, then keeps bouncing off the walls like light in a mirror-box (equal in and out angles). It stops only if it lands exactly on a corner. Find which corner it reaches.
Givens: Square $ABCD$ has side length $4$, placed as $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$; $P$ is on $\overline{AD}$ with $AP=\frac{8}{5}$, so $P=(0,\frac{8}{5})$; $Q$ is on $\overline{CD}$ with $DQ=\frac{10}{3}$, so $Q=(\frac{10}{3},4)$; The path starts along $P\to Q$ and reflects off each wall with equal incoming and outgoing angles; The path stops the instant it hits a vertex; otherwise it continues forever
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #9 Solve an Easier Related Problem, #4 Introduce a Variable, #1 Draw a Diagram
Chasing a ray as it zig-zags off four walls is hopeless to track by hand. Tool #17 (Visualize Spatial Relationships) supplies the key mental move: instead of bending the ray at each wall, reflect the whole square across that wall and let the ray fly straight into the mirrored copy. The reflected copies tile the plane into a $4\times4$ grid, and the bounced path becomes one straight line. That is Tool #9 (Solve an Easier Related Problem): a hard bouncing problem turns into an easy straight-line problem. Tool #4 (Introduce a Variable) puts coordinates on the picture so the straight line has an equation, and Tool #1 (Draw a Diagram) anchors which mirrored corner is which.
Execute — Answer: B
6.NS.C.8 Step 1 Put the square on coordinates
- Place the square with $A$ at the origin, so $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$.
- Point $P$ sits on the left side $\overline{AD}$ a height $AP=\frac{8}{5}$ above $A$, giving $P=(0,\frac{8}{5})$.
- Point $Q$ sits on the top side $\overline{CD}$ a distance $DQ=\frac{10}{3}$ from $D$, giving $Q=(\frac{10}{3},4)$.
💡 Naming every point by an $(x,y)$ pair turns the picture into numbers you can compute with.
8.EE.B.6 Step 2 Write the line through P and Q
- Find the slope of $\overline{PQ}$ as rise over run.
- The rise is $4-\frac{8}{5}=\frac{12}{5}$ and the run is $\frac{10}{3}$, so the slope is $\frac{12}{5}\div\frac{10}{3}=\frac{12}{5}\cdot\frac{3}{10}=\frac{18}{25}$.
- Since the line starts at $P=(0,\frac{8}{5})$, its $y$-intercept is $\frac{8}{5}$, giving the equation of travel.
💡 Slope is just how much you climb for each step right, so it fixes the whole direction of the shot.
8.G.A.3 Step 3 Unfold the bounces into one straight line
- Reflecting a ray off a wall looks the same as sending it straight into a mirror-image of the room.
- So instead of bending the path, reflect the square across each wall it meets.
- The mirrored copies tile the plane into a grid of $4\times4$ squares, and the entire bouncing path straightens into the single line $25y=18x+40$.
- A real bounce ends at a corner exactly when this straight line reaches a grid corner, a point $(x,y)$ whose coordinates are both multiples of $4$, because every mirrored copy of a corner still lands on such a grid point.
💡 A mirror bounce and a straight shot into the mirror image trace the very same picture.
8.EE.C.7 Step 4 Find the first grid corner the line reaches
- Substitute $x=4a$ and $y=4b$ into $25y=18x+40$: this gives $100b=72a+40$, and dividing by $4$ leaves $25b=18a+10$.
- Look for the smallest whole number $a>0$ that makes $b$ a whole number.
- Testing multiples: $a=5$ gives $18\cdot5+10=100=25\cdot4$, so $b=4$.
- No smaller positive $a$ works, so the very first grid corner on the ray is $(x,y)=(20,16)$.
💡 Turning the geometry question into one equation lets you hunt for the first exact corner by arithmetic.
8.G.A.3 Step 5 Fold the grid corner back to a real vertex
- Now read which real vertex sits at grid corner $(20,16)$ by undoing the mirroring.
- Along $x$: $20=4\cdot5$, and reflecting an odd number of $4$-wide strips flips the left edge to the right, so $x=20$ folds back to the real $x=4$.
- Along $y$: $16=4\cdot4$, an even number of strips, which folds back to the real $y=0$.
- The real corner is $(4,0)$, which is vertex $B$.
- So the path terminates at $\textbf{(B)}\ B$.
💡 Each strip you cross flips a coordinate, so the parity of how many strips you crossed tells you which corner you landed on.
6.NS.C.8 Place the square with $A$ at the origin, so $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D= 8.EE.B.6 Find the slope of $\overline{PQ}$ as rise over run. The rise is $4-\frac{8}{5}=\ 8.G.A.3 Reflecting a ray off a wall looks the same as sending it straight into a mirror- 8.EE.C.7 Substitute $x=4a$ and $y=4b$ into $25y=18x+40$: this gives $100b=72a+40$, and di 8.G.A.3 Now read which real vertex sits at grid corner $(20,16)$ by undoing the mirrorin Review
Reasonableness: Check the landing point on the original line: at $x=20$, $y=\frac{18}{25}\cdot20+\frac{8}{5}=\frac{360}{25}+\frac{40}{25}=\frac{400}{25}=16$, so $(20,16)$ really is on the ray. Both $20$ and $16$ are multiples of $4$, so it is a genuine grid corner, and it is the first one because $a=5$ is the smallest positive solution of $25b=18a+10$. The parity read-off ($x$: odd strips $\to 4$; $y$: even strips $\to 0$) gives $(4,0)=B$, a real vertex, not choice (E). The path does terminate, ruling out (E), and lands at (B).
Alternative: Count reflections instead of folding coordinates. From $P=(0,\frac{8}{5})$ to $(20,16)$ the straight line crosses the vertical walls $x=4,8,12,16$ (four crossings) and the horizontal walls $y=4,8,12$ (three crossings), for $7$ reflections before it reaches the corner. Starting the labels at $A$ and applying $7$ bounces cycles the corner labels the same number of quarter-turns and again lands on $B$, confirming the answer.
CCSS standards used (min grade 8)
6.NS.C.8Solve problems by graphing points in the coordinate plane (Placing the square, $P$, and $Q$ at coordinates so the geometry becomes computable.)8.EE.B.6Use slope and derive the equation y = mx + b for a line (Computing the slope $\frac{18}{25}$ of $\overline{PQ}$ and writing its equation $y=\frac{18}{25}x+\frac{8}{5}$.)8.G.A.3Describe the effect of reflections on figures using coordinates (Unfolding each bounce into a reflected copy of the square and folding the grid corner $(20,16)$ back to the real vertex.)8.EE.C.7Solve linear equations in one variable (Reducing $25y=18x+40$ to $25b=18a+10$ and finding the first whole-number corner $(20,16)$.)
⭐ To follow a ball bouncing in a mirror-box, don't bend the path — reflect the box and let the ball fly straight, then see which mirrored corner it reaches first.
⭐ To follow a ball bouncing in a mirror-box, don't bend the path — reflect the box and let the ball fly straight, then see which mirrored corner it reaches first.
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