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Difference between revisions of "2025 AMC 8 Problems/Problem 25"

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'''Step 1:''' To find the total number of paths, observe that all paths will have <math>10</math> total steps. We have to choose which <math>5</math> of these steps will be NE (the rest will be NW). So the total number of paths is <math>\binom{10}{5}</math>.  
 
'''Step 1:''' To find the total number of paths, observe that all paths will have <math>10</math> total steps. We have to choose which <math>5</math> of these steps will be NE (the rest will be NW). So the total number of paths is <math>\binom{10}{5}</math>.  
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The formula for combinations is: <math>\binom{n}{r} = \frac{n!}{r!(n-r)!}</math>[https://artofproblemsolving.com/wiki/index.php/Combination?srsltid=AfmBOoqEj147I2nN7FICjcpebQ_ER-gXBhrFAfhziauZu7IlxiYk405y] and <math>\binom{10}{5} = \frac{10!}{5!\times5!}=252</math>.
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The formula for [https://artofproblemsolving.com/wiki/index.php/Combination combinations] is: <math>\binom{n}{r} = \frac{n!}{r!(n-r)!}</math> and <math>\binom{10}{5} = \frac{10!}{5!\times5!}=252</math>.
  
 
'''Step 2:''' Each path splits the total area of <math>25</math> in two parts. So, for any path that gives <math>area = A</math>, you can find a unique ‘sister’ path that has an <math>area = 25-A</math> (in other words, the pair of paths have a combined area of 25).  Possible ways to define the ‘sister’ path are:  
 
'''Step 2:''' Each path splits the total area of <math>25</math> in two parts. So, for any path that gives <math>area = A</math>, you can find a unique ‘sister’ path that has an <math>area = 25-A</math> (in other words, the pair of paths have a combined area of 25).  Possible ways to define the ‘sister’ path are:  
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*Each of the <math>252</math> paths gives an area of <math>25</math> if you also count the ‘sister’ paths. Since each ‘sister’ path is also one of the <math>252</math>, you have to divide by <math>2</math> to avoid double counting.  So the total area given by all paths is <math>\frac{252 \times 25}{2}</math>.
 
*Each of the <math>252</math> paths gives an area of <math>25</math> if you also count the ‘sister’ paths. Since each ‘sister’ path is also one of the <math>252</math>, you have to divide by <math>2</math> to avoid double counting.  So the total area given by all paths is <math>\frac{252 \times 25}{2}</math>.
 
*Note that the average area of two ‘sister’ paths is <math>\frac{25}{2}</math>, so you can think about every path having this area ''on average''.  So the total area given by all paths is <math>252 \times \frac{25}{2}</math>.  
 
*Note that the average area of two ‘sister’ paths is <math>\frac{25}{2}</math>, so you can think about every path having this area ''on average''.  So the total area given by all paths is <math>252 \times \frac{25}{2}</math>.  
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Note : This problem has a  bijection (or 1-1 correspondence , Checkout Intermediate Counting and Probabilty Chapter 4 , and Intro to Counting chapter 5.  
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Note : This problem has a  bijection (or 1-1 correspondence , check out [https://artofproblemsolving.com/store/book/intermediate-counting Intermediate Counting & Probabilty, Chapter 4], and [https://artofproblemsolving.com/store/book/intro-counting Introduction to Counting & Probability, Chapter 5].  
 
The final answer is <math>\boxed{\textbf{(B)}~3150}.</math>
 
The final answer is <math>\boxed{\textbf{(B)}~3150}.</math>
  
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~cxsmi<BR>
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~ cxsmi<BR>
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~aleyang<BR>
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~ aleyang<BR>
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~MathCosine
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~ MathCosine<BR>
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~ [[User:Aoum|Aoum]]
  
 
==Solution 2==
 
==Solution 2==

Latest revision as of 21:28, 21 February 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]

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

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Official Video Solution (Simple counting and multiplication!)

https://www.youtube.com/watch?v=Z54kAp8ez5M ~TheMathGeek (Plz sub)

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Solution 1

Step 1: To find the total number of paths, observe that all paths will have $10$ total steps. We have to choose which $5$ of these steps will be NE (the rest will be NW). So the total number of paths is $\binom{10}{5}$. The formula for combinations is: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and $\binom{10}{5} = \frac{10!}{5!\times5!}=252$.

Step 2: Each path splits the total area of $25$ in two parts. So, for any path that gives $area = A$, you can find a unique ‘sister’ path that has an $area = 25-A$ (in other words, the pair of paths have a combined area of 25). Possible ways to define the ‘sister’ path are:

  • Rotate the entire grid $180^{\circ}$
  • Swap each step of the original paths (for example, each NW becomes NE) (this is a reflection over the diagonal)

Step 3: There are a few ways to get from this observation to the total area:

  • There are $252/2 = 126$ pairs of such paths, and the total area of each pair is $25$. So the total area given by all paths is $126 \times 25$.
  • Each of the $252$ paths gives an area of $25$ if you also count the ‘sister’ paths. Since each ‘sister’ path is also one of the $252$, you have to divide by $2$ to avoid double counting. So the total area given by all paths is $\frac{252 \times 25}{2}$.
  • Note that the average area of two ‘sister’ paths is $\frac{25}{2}$, so you can think about every path having this area on average. So the total area given by all paths is $252 \times \frac{25}{2}$.

Note : This problem has a bijection (or 1-1 correspondence , check out Intermediate Counting & Probabilty, Chapter 4, and Introduction to Counting & Probability, Chapter 5. The final answer is $\boxed{\textbf{(B)}~3150}.$

~ cxsmi
~ aleyang
~ MathCosine
~ Aoum

Solution 2

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 3 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 $5*(0+5)/2 = 25/2$, and $25/2 * 252 = \fbox{3150}$.

~Bread10

Solution 4

As found, there are $252$ total paths. However, we can take advantage of symmetry here. We consider the total area of the left side of the grid in each of the $252$ paths. Since the total area of the left side of the grid in each of the $252$ paths is equal to the total area of the right side of the grid in each of the $252$ paths, and the two total areas sum to $(252)(25) = 6300$, then the total area of the right side of the grid is $6300/2 = \fbox{3150}$.

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=VWOqMx55d5ctta1P&t=4042 ~hsnacademy

Video Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution by Dr. David

https://youtu.be/sdZ4x5CBhIc

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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