AMC 8 · 2025 · #25

Grade 8 countinggeometry-2d
combinations-basicreflection-symmetryarea-rectanglessystematic-enumeration complementary-countingeasier-related-problempattern-recognition ↑ Prerequisites: combinations-basicarea-rectanglesreflection-symmetry
📏 Long solution 💡 4 insights 📊 Diagram
Problem
On a 5 × 5 diamond-shaped grid (a square tilted 45°), every lattice path starts at the bottom vertex and ends at the top vertex, moving one unit northeast (NE) or northwest (NW) at each step. For each such path, measure the area of the region trapped between the path and the right edge of the diamond. Sum that area over every possible path.

Pick an answer.

(A)
2520
(B)
3150
(C)
3840
(D)
4730
(E)
5050

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Adding the 252 right-side areas one by one is hopeless. Tool #16 (Change Focus) saves us: instead of computing each A_R, pair every path P with its mirror image P' (NE⇔NW). Since A_R(P') = A_L(P) = 25 - A_R(P), each pair contributes exactly 25, no matter what the individual areas are. Tool #9 (Easier Problem) is used first to test the idea on a tiny 1 × 1 and 2 × 2 diamond before trusting it at size 5. Tool #2 (Systematic List / Counting) counts the 10-step sequences as C(10, 5) = 252. Tool #1 (Diagram) keeps the diamond, the path, and "right side" unambiguous.

1STEP 1

Sketch the diamond and one path: tilting the 5×5 square 45° makes NE/NW the two grid directions, and each path climbs up in 10 steps.

5 NE + 5 NW = 10 steps total
2STEP 2

Each path is 10 moves with 5 NE chosen among them, so the number of paths is C(10,5) = 252.

C(10, 5) = 10!5!5!\frac{10!}{5! 5!} = (109876)(54321)\frac{(10· 9· 8· 7· 6)}{(5· 4· 3· 2· 1)} = 252
3STEP 3

Test the idea small: the 1×1 diamond sums to 1 and the 2×2 sums to 12, both matching sum = (#paths) × (half the grid area).

1 × 1: Σ A_R = 0+1 = 1 = 2 × 12\frac{1}{2}. 2 × 2: Σ A_R = 0+1+1+3+3+4 = 12 = 6 × 42\frac{4}{2}.
4STEP 4

Change focus: mirror any path (NE↔NW) to swap its right and left regions; since one path splits the whole diamond, A_R + A_L = 25.

A_R(P) + A_L(P) = 25, A_R(P') = A_L(P)
5STEP 5

Add the two facts so the unknown cancels: for each mirror pair the right areas total A_R(P) + A_R(P') = 25.

A_R(P) + A_R(P') = 25 for every path P
6STEP 6

Sum over all paths: the average right area is 252\frac{25}{2}, so the total is 252 × 252\frac{25}{2} = 3150 → (B).

Σ_P A_R(P) = 252 · 252\frac{25}{2} = 126 · 25 = 3150 → (B)
Answer
3150
The two example paths in the figure already show right-side areas of 11 and 13, both close to half of 25. The average area for all 252 paths must also be exactly half of 25 by symmetry, giving 252\frac{25}{2} = 12.5, which sits perfectly between 11 and 13. Multiplying 252 × 12.5 = 3150 matches the small-grid pattern from Step 3 (sum = #paths × half-area) and is choice (B). The other choices fail this sanity test: (A) 2520 would force an average area of 10, (C) 3840 would force 15.24, neither of which can hold under the reflection symmetry.
💡Key takeaway

This AMC 8 problem only needs Grade 8 reflection symmetry you already know — flip each path across the middle to pair up areas that always add to 25!