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Difference between revisions of "2025 AMC 8 Problems/Problem 25"
 
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== Problem ==
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Makayla finds all the possible ways to draw a path in a <math>5 \times 5</math> 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?
 
Makayla finds all the possible ways to draw a path in a <math>5 \times 5</math> 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?
  
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~cxsmi
 
~cxsmi
  
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==Solution 2 (kinda similar to above)==
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==Solution 2==
 
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>
 
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>
  
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~alwaysgonnagiveyouup
 
~alwaysgonnagiveyouup
  
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==Vide Solution 1 by SpreadTheMathLove==
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==Solution 5 Easier to Motivate==
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Most other solutions don't explain how they got all the cases, as well as require an insight that's somewhat hard to think of, so I'll explain another in detail.
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If we rotate the grid <math>45</math> degrees clockwise we can have a 5x5 grid where we can move up and right. There can only be one horizontal line segment in each column at a certain y coordinate. We'll denote the y coordinates as <math>0,1,2,3,4,5</math>. Once we choose the y coordinates for each column, we have a unique path. However, we can't move down which means that columns cannot have a higher y coordinate than the ones to the right. This is the same as distributing 5 balls in six boxes labeled <math>0,1,2,3,4,5</math>. For example, if we get 2 in 0, 2 in 1, and 1 in 4, then the order would be <math>0,0,1,1,4</math>. This is one unique path, and the total number of paths is represented by 6+5-1 choose 6-1 = 252. Also, the sum of the y-coordinates represents the area, which means we want the average sum of the y-coordinates. This is <math>5*(0+6)/2 = 25/2</math>, and <math>25/2 * 252 = \fbox{3150}</math>.
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~Bread10
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==Video Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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== See Also ==
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{{AMC8 box|year=2025|num-b=24|after=Last Problem}}
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{{MAA Notice}}

Latest revision as of 19:02, 30 January 2025

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?

[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

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

Solution 4

If we test this problem on a smaller $2 \times 2$ diamond, we have $6$ ways to go from $A$ to $B$, and the total area is $0 + 1 + 2 + 2 + 3 + 4 = 12$, so the average area is $\frac{12}{6} = 2$, which is also the area of the diamond $2 \times 2 = 4$ divided by 2. If we assume this is true for a $5 \times 5$ diamond, then the average area is $\frac{25}{2}$. The number of paths from $A$ to $B$ is $\binom{10}{5} = 252$, and $252 \cdot \frac{25}{2} = \boxed{\textbf{(B)}~3150}$.

~alwaysgonnagiveyouup

Solution 5 Easier to Motivate

Most other solutions don't explain how they got all the cases, as well as require an insight that's somewhat hard to think of, so I'll explain another in detail.

If we rotate the grid $45$ degrees clockwise we can have a 5x5 grid where we can move up and right. There can only be one horizontal line segment in each column at a certain y coordinate. We'll denote the y coordinates as $0,1,2,3,4,5$. Once we choose the y coordinates for each column, we have a unique path. However, we can't move down which means that columns cannot have a higher y coordinate than the ones to the right. This is the same as distributing 5 balls in six boxes labeled $0,1,2,3,4,5$. For example, if we get 2 in 0, 2 in 1, and 1 in 4, then the order would be $0,0,1,1,4$. This is one unique path, and the total number of paths is represented by 6+5-1 choose 6-1 = 252. Also, the sum of the y-coordinates represents the area, which means we want the average sum of the y-coordinates. This is $5*(0+6)/2 = 25/2$, and $25/2 * 252 = \fbox{3150}$.

~Bread10

Video Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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