Difference between revisions of "2025 AMC 8 Problems/Problem 25"
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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 <math>180^{\circ}</math>)! Hence, between these two paths, the entire area of the grid (that being <math>25</math> square units) is accounted for. The total number of paths is <math>\binom{10}{5}=252</math> (we choose <math>5</math> of our steps to be northeaast steps, determining the northwest steps). Each pair is accounted for exactly twice (one for each member of the pair). Hence, our answer is <cmath>252 \cdot \frac{1}{2} \cdot 25 = \boxed{\textbf{(B)}~3150}.</cmath> | 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 <math>180^{\circ}</math>)! Hence, between these two paths, the entire area of the grid (that being <math>25</math> square units) is accounted for. The total number of paths is <math>\binom{10}{5}=252</math> (we choose <math>5</math> of our steps to be northeaast steps, determining the northwest steps). Each pair is accounted for exactly twice (one for each member of the pair). Hence, our answer is <cmath>252 \cdot \frac{1}{2} \cdot 25 = \boxed{\textbf{(B)}~3150}.</cmath> | ||
~cxsmi | ~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 <math>\binom{10}{5} \times 25 * \frac{1}{2}</math>, which is <math>\boxed{\textbf{(B)}~3150}.</math> | ||
| + | |||
| + | ~aleyang | ||
Revision as of 20:58, 29 January 2025
Problem
Makayla finds all the possible ways to draw a path in a 
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?
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 
)! Hence, between these two paths, the entire area of the grid (that being 
square units) is accounted for. The total number of paths is 
(we choose 
of our steps to be northeaast steps, determining the northwest steps). Each pair is accounted for exactly twice (one for each member of the pair). Hence, our answer is ![\[252 \cdot \frac{1}{2} \cdot 25 = \boxed{\textbf{(B)}~3150}.\]](http://latex.artofproblemsolving.com/e/1/8/e185bf016f25681d38532cbbe070b44a6c6ddbe5.png)
~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 
, which is 
~aleyang
