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

Makayla finds all the possible ways to draw a path in a $5 \times 5$ diamond-shaped grid. Each path starts at the bottom of the grid and ends at the top, always moving one unit northeast or northwest. She computes the area of the region between each path and the right side of the grid. Two examples are shown in the figures below. What is the sum of the areas determined by all possible paths?

Pick an answer.

(A)

2520

(B)

3150

(C)

3840

(D)

4730

(E)

5050

View mode:

Toolkit + CCSS Solution

Understand

Restated: On a $5 \times 5$ 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.

Givens: A $5 \times 5$ diamond grid containing $25$ unit cells; Every step is either one unit NE or one unit NW; Each path runs from the bottom vertex to the top vertex; The two example pictures show paths with right-side areas $11$ and $13$; Answer choices: (A) $2520$, (B) $3150$, (C) $3840$, (D) $4730$, (E) $5050$

Unknowns: The total sum of right-side areas over all valid bottom-to-top paths

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #1 Draw a Diagram

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$\leftrightarrow$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 $1 \times 1$ and $2 \times 2$ diamond before trusting it at size $5$. Tool #2 (Systematic List / Counting) counts the $10$-step sequences as $\binom{10}{5} = 252$. Tool #1 (Diagram) keeps the diamond, the path, and "right side" unambiguous.

Execute — Answer: B

#1 Draw a Diagram 4.G.A.3 Step 1

Sketch the diamond and a single path.
The $5\times5$ square is rotated $45°$, so "NE" and "NW" act like the two grid directions of the un-rotated square.
A path uses $5$ NE moves and $5$ NW moves to climb from the bottom corner to the top corner; the right-side region is the area to the right of the path and to the left of the right edge.

$$5 \text{ NE} + 5 \text{ NW} = 10 \text{ steps total}$$

💡 Drawing the diamond and one example path makes the symmetry axis (the vertical line through the top and bottom vertex) obvious.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count the total number of paths.
Each path is a sequence of $10$ moves with exactly $5$ NE choices, so the count is the number of ways to choose the positions of the NE moves among $10$ slots — a combination.

$$\binom{10}{5} = \dfrac{10!}{5!\,5!} = \dfrac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2\cdot 1} = 252$$

💡 Listing $10$-letter NE/NW words with $5$ of each letter is exactly counting compound outcomes with an organized list.

#9 Solve an Easier Related Problem 4.OA.C.5 Step 3

Verify the symmetry idea on a smaller diamond first.
On a $1 \times 1$ diamond there are only $2$ paths (NE-NW and NW-NE) with right-side areas $1$ and $0$, summing to $1 = 1 \cdot \tfrac{1}{2} \cdot 2$.
On a $2 \times 2$ diamond, the $\binom{4}{2}=6$ paths give right-side areas $\{0,1,1,3,3,4\}$ which sum to $12 = 6 \cdot \tfrac{4}{2}$.
In each case the sum equals (number of paths) $\times$ (half the grid area).
That justifies the same plan at size $5$.

$1 \times 1: \sum A_R = 0+1 = 1 = 2 \times \tfrac{1}{2}.$ $\quad 2 \times 2: \sum A_R = 0+1+1+3+3+4 = 12 = 6 \times \tfrac{4}{2}.$

💡 Trying the same setup on a tiny grid reveals the rule "sum $=$ #paths $\times$ half-area" before we trust it on size $5$.

#16 Change Focus / Count the Complement 8.G.A.1 Step 4

Now apply Tool #16 (change focus).
For any path $P$, define its mirror $P'$ by swapping every NE step with NW.
The mirror $P'$ is again a valid path, and the right-side area of $P'$ equals the left-side area of $P$.
For any single path, the right plus left areas fill the whole diamond, so $A_R(P) + A_L(P) = 25$.

$$A_R(P) + A_L(P) = 25, \qquad A_R(P') = A_L(P)$$

💡 Reflecting a figure (a Grade 8 rigid motion) swaps left and right but keeps every length and area the same.

#16 Change Focus / Count the Complement 6.EE.A.3 Step 5

Add the two equations above to pair up areas.
For each unordered pair $\{P, P'\}$, $A_R(P) + A_R(P') = A_R(P) + A_L(P) = 25$.
Even when $P$ happens to equal $P'$ (a self-symmetric path), the same algebra forces $2A_R(P) = 25$, so that path also contributes $25$ when counted with its (identical) partner.
Either way, every one of the $252$ paths can be paired so its pair-sum is $25$.

$$A_R(P) + A_R(P') = 25 \text{ for every path } P$$

💡 Adding the two area equations cancels the unknown $A_R(P)$ and leaves the clean constant $25$ — the change-of-focus payoff.

#16 Change Focus / Count the Complement 7.NS.A.3 Step 6

Sum over all $252$ paths.
Because every path's right-side area, when added to its mirror's right-side area, gives $25$, the average right-side area is $\tfrac{25}{2}$, and the total sum is the number of paths times that average.

$$\displaystyle\sum_{P} A_R(P) = 252 \cdot \dfrac{25}{2} = 126 \cdot 25 = 3150 \;\Rightarrow\; \textbf{(B)}$$

💡 Multiplying the number of items by the average (which equals half the grid area) gives the total — Grade 7 rational arithmetic.

[1] #1 4.G.A.3 Sketch the diamond and a single path. The $5\times5$ square is rotated $45°$, so

[2] #2 7.SP.C.8 Count the total number of paths. Each path is a sequence of $10$ moves with exac

[3] #9 4.OA.C.5 Verify the symmetry idea on a smaller diamond first. On a $1 \times 1$ diamond t

[4] #16 8.G.A.1 Now apply Tool #16 (change focus). For any path $P$, define its mirror $P'$ by s

[5] #16 6.EE.A.3 Add the two equations above to pair up areas. For each unordered pair ${P, P'}

[6] #16 7.NS.A.3 Sum over all $252$ paths. Because every path's right-side area, when added to it

Review

Reasonableness: 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 $\tfrac{25}{2} = 12.5$, which sits perfectly between $11$ and $13$. Multiplying $252 \times 12.5 = 3150$ matches the small-grid pattern from Step $3$ (sum $=$ #paths $\times$ 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.

Alternative: Tool #5 (Look for a Pattern) using Tool #14 (Finite Differences) on the small cases gives the same answer without explicit symmetry. Computing sums for $n \times n$ diamonds yields $1, 12, 90, 560, 3150, \ldots$ for $n = 1, 2, 3, 4, 5$. Each value equals $\binom{2n}{n} \cdot \tfrac{n^2}{2}$ (number of paths times half the grid area), confirming $\binom{10}{5}\cdot\tfrac{25}{2} = 252 \cdot 12.5 = 3150$.

CCSS standards used (min grade 8)

4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Identifying that the diamond has a vertical line of symmetry through the top and bottom vertices, which is the axis used to mirror each path.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Generating the small-case sums on $1\times1$ and $2\times2$ diamonds to test and trust the rule "total $=$ paths $\times$ half-area" before applying it to $5\times5$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Combining $A_R(P)+A_L(P)=25$ with $A_R(P')=A_L(P)$ to deduce $A_R(P)+A_R(P')=25$ by substitution.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the $\binom{10}{5}=252$ NE/NW step sequences as compound outcomes organized in a systematic list.)
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers (Multiplying $252 \times \tfrac{25}{2} = 3150$ to sum the $252$ areas with rational-number arithmetic.)
8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Reflecting each path across the diamond's vertical axis to produce its area-swap partner, the key move behind the symmetry pairing.)

⭐ 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 $25$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

More like this