2025 AMC 8 Problems/Problem 25

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?

$[asy] unitsize(9); real a = 0.7071; path w = (0,0)--(2a, 2a)--(-a,5a)--(a,7a)--(-a,9a)--(0,10a); fill(w--(5a,5a)--cycle, gray(0.8)); draw(w, linewidth(1.5)); path x = (10,0)--(10-a,a)--(10+2a,4a)--(10-2a,8a)--(10,10a); fill(x--(10+5a,5a)--cycle, gray(0.8)); draw(x, linewidth(1.5)); add(rotate(45, (0,0)) * grid(5,5)); add(rotate(45, (10,0)) * (shift((10,0)) * grid(5,5))); dot((0,0)); dot((0,7.07106)); dot((10,0)); dot((10,7.07106)); label("area = 11", (0,-1), S); label("area = 13", (10,-1), S); [/asy]$

$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 3150 \qquad \textbf{(C)}\ 3840 \qquad \textbf{(D)}\ 4730 \qquad \textbf{(E)}\ 5050$

Solution 1

Consider a given path. The region below it, and hence its area, is accounted for in exactly one way (that is, with the given path; there are no other ways to get the same region below the path). However, the region ABOVE the path is also accounted for by exactly one path (the one found by rotating the entire grid $180^{\circ}$ )! Hence, between these two paths, the entire area of the grid (that being $25$ square units) is accounted for. The total number of paths is $\binom{10}{5}=252$ (we choose $5$ of our steps to be northeast steps, determining the northwest steps). Each pair is accounted for exactly twice (once for each member of the pair). Hence, our answer is $252 \cdot \frac{1}{2} \cdot 25 = \boxed{\textbf{(B)}~3150}.$ ~cxsmi

Solution 2 (kinda similar to above)

Consider a valid path. Observe that we can make another valid path by swapping the directions of the original paths (for example, if a valid path could be NW NW NE NE NE NW NW NW NE NE, then a swapped path would be NE NE NW NW NW NE NE NE NW NW). Observe that the area formed by the valid path and the 'swapped' path is exactly 25. However, we need to divide our answer by 2 as we are counting the swapped path as a valid path, so our answer is $\binom{10}{5} \times 25 * \frac{1}{2}$ , which is $\boxed{\textbf{(B)}~3150}.$

~aleyang

Solution 3

Note that we can consider the area over the axis to get another configuration that works. These two configurations sum to 25, so the average area is $\frac{25}{2}$ . Therefore, our answer is $\frac{25}{2} \cdot \binom{10}{5} = \boxed{\textbf{B) 3150}}$

~MathCosine

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

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2025 AMC 8 Problems/Problem 25

Contents

Solution 1

Solution 2 (kinda similar to above)

Solution 3

Vide Solution 1 by SpreadTheMathLove