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Difference between revisions of "2025 AMC 8 Problems/Problem 1"

(Undo revision 215690 by Theumbrellaacademy (talk))
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Let m and n be 2 integers such that m > n. Suppose m + n = 20, m² + n² = 328, find m² - n².
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== Problem ==
  
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A) 280
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The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire <math>4\times4</math> grid is covered by the star?
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B) 292
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C) 300
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<asy>
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D) 320
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path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle;
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E) 340
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path y = reflect((0,0),(4,4)) * x;
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path z = (1,0)--(2,1)--(3,0)--(3,1)--(2,2)--(1,1);
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fill(x, gray(0.6));
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fill(rotate(90, (2,2)) * x, gray(0.6));
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fill(rotate(180, (2,2)) * x, gray(0.6));
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fill(rotate(270, (2,2)) * x, gray(0.6));
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fill(y, gray(0.8));
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fill(rotate(90, (2,2)) * y, gray(0.8));
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fill(rotate(180, (2,2)) * y, gray(0.8));
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fill(rotate(270, (2,2)) * y, gray(0.8));
 +
draw(z);
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draw(rotate(90, (2,2)) * z);
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draw(rotate(180, (2,2)) * z);
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draw(rotate(270, (2,2)) * z);
 +
add(grid(4,4));
 +
</asy>
 +
 
 +
<math>\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 80</math>
 +
 
 +
== Solution 1 ==
 +
 
 +
Each of the unshaded triangles has base length <math>2</math> and height <math>1</math>, so they all have area <math>\frac{2 \cdot 1}{2} = 1</math>. Each of the unshaded unit squares has area <math>1</math>. The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or <math>4^2 - 4 \cdot 1 - 4 \cdot 1 = 8</math>. The star is then <math>\frac{8}{16} = \frac{1}{2} = \frac{50}{100}</math>, or <math>\boxed{\textbf{(B)}~50}</math> percent of the entire grid.
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~cxsmi
 +
 
 +
== Solution 2 ==
 +
 
 +
There are <math>16</math> total squares in the diagram and each square has <math>2</math> triangles whose areas are half the area of a unit square. Thus, the total number of triangles in the diagram is <math>4^2 \cdot 2 = 32</math> triangles. There are <math>16</math> shaded triangles in the diagram, so the area of the star is <math>\dfrac{16}{32} = \frac{1}{2} = \frac{50}{100}</math>, or <math>\boxed{\textbf{(B)}~50}</math> percent.
 +
~Pi_in_da_box
 +
 
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== Solution 3 ==
 +
 
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There are <math>4</math> squares that are entirely shaded and <math>4</math> squares that have no shading. The rest of the squares are half-half. Therefore the shaded region is <math>\boxed{\textbf{(B)}~50}</math> percent of the grid.
 +
 
 +
== Solution 4 ==
 +
 
 +
Note that we can move one triangle from each of the four cells in the middle to each of the four corners. This will leave every cell in the grid with one triangle each, and each triangle has an area of half the area of each cell. Thus, our answer must equal to <math>\frac{1}{2}</math>, and so our answer is <math>\boxed{\textbf{(B)}~50}.</math>
 +
~derekwang2048
 +
 
 +
== Solution 5 ==
 +
 
 +
The shaded area is a <math>2 \times 2</math> square in the middle of the figure combined with <math>8</math> small triangles. Since each small triangle is <math>\frac{1}{2}</math> of a unit square, the star's area is equal to the area of <math>4 + 8 \cdot \frac{1}{2} = 8</math> unit squares, which <math>\boxed{\textbf{(B)}~50}</math> percent of the grid.
 +
 
 +
-vockey
 +
 
 +
== Solution 6 ==
 +
Using [[Pick's Theorem]], we see there are 16 boundary points and 1 interior point. <math>1 + \frac{16}{2} - 1 = 8</math>. There are <math>4 \cdot 4 = 16</math> squares, and 8 is <math>\boxed{\textbf{(B)}~50}</math> percent of 16.
 +
 
 +
-leafy
 +
 
 +
== Video Solution 1 ==
 +
 
 +
[//youtu.be/rGEGn7U4uHk ~ ChillGuyDoesMath]
 +
 
 +
== Video Solution 2 by SpreadTheMathLove ==
 +
 
 +
https://www.youtube.com/watch?v=jTTcscvcQmI
 +
 
 +
==Video Solution 3 ==
 +
 
 +
[//youtu.be/VP7g-s8akMY?si=rhpz6FZayHNW--j8&t=7 ~hsnacademy]
 +
 
 +
== Video Solution 4 by Daily Dose of Math ==
 +
 
 +
[//youtu.be/rjd0gigUsd0 ~Thesmartgreekmathdude]
 +
 
 +
== Video Solution 5 by Thinking Feet ==
 +
 
 +
https://youtu.be/PKMpTS6b988
 +
 
 +
==Video Solution 6 by CoolMathProblems==
 +
 
 +
https://youtu.be/SNviHnUR3x4
 +
 
 +
== Video Solution 7 by Pi Academy ==
 +
 
 +
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
 +
 
 +
== Video Solution 8 ==
 +
https://youtu.be/Yc4KSp0aOco
 +
 
 +
==Video Solution(Quick, fast, easy!)==
 +
https://youtu.be/fdG7EDW_7xk
 +
 
 +
~MC
 +
 
 +
== See Also ==
 +
 
 +
{{AMC8 box|year=2025|before=First Problem|num-a=2}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 14:21, 23 May 2025

Problem

The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire $4\times4$ grid is covered by the star?

[asy] path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle; path y = reflect((0,0),(4,4)) * x; path z = (1,0)--(2,1)--(3,0)--(3,1)--(2,2)--(1,1); fill(x, gray(0.6)); fill(rotate(90, (2,2)) * x, gray(0.6)); fill(rotate(180, (2,2)) * x, gray(0.6)); fill(rotate(270, (2,2)) * x, gray(0.6)); fill(y, gray(0.8)); fill(rotate(90, (2,2)) * y, gray(0.8)); fill(rotate(180, (2,2)) * y, gray(0.8)); fill(rotate(270, (2,2)) * y, gray(0.8)); draw(z); draw(rotate(90, (2,2)) * z); draw(rotate(180, (2,2)) * z); draw(rotate(270, (2,2)) * z); add(grid(4,4)); [/asy]

$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 80$

Solution 1

Each of the unshaded triangles has base length $2$ and height $1$, so they all have area $\frac{2 \cdot 1}{2} = 1$. Each of the unshaded unit squares has area $1$. The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or $4^2 - 4 \cdot 1 - 4 \cdot 1 = 8$. The star is then $\frac{8}{16} = \frac{1}{2} = \frac{50}{100}$, or $\boxed{\textbf{(B)}~50}$ percent of the entire grid. ~cxsmi

Solution 2

There are $16$ total squares in the diagram and each square has $2$ triangles whose areas are half the area of a unit square. Thus, the total number of triangles in the diagram is $4^2 \cdot 2 = 32$ triangles. There are $16$ shaded triangles in the diagram, so the area of the star is $\dfrac{16}{32} = \frac{1}{2} = \frac{50}{100}$, or $\boxed{\textbf{(B)}~50}$ percent. ~Pi_in_da_box

Solution 3

There are $4$ squares that are entirely shaded and $4$ squares that have no shading. The rest of the squares are half-half. Therefore the shaded region is $\boxed{\textbf{(B)}~50}$ percent of the grid.

Solution 4

Note that we can move one triangle from each of the four cells in the middle to each of the four corners. This will leave every cell in the grid with one triangle each, and each triangle has an area of half the area of each cell. Thus, our answer must equal to $\frac{1}{2}$, and so our answer is $\boxed{\textbf{(B)}~50}.$ ~derekwang2048

Solution 5

The shaded area is a $2 \times 2$ square in the middle of the figure combined with $8$ small triangles. Since each small triangle is $\frac{1}{2}$ of a unit square, the star's area is equal to the area of $4 + 8 \cdot \frac{1}{2} = 8$ unit squares, which $\boxed{\textbf{(B)}~50}$ percent of the grid.

-vockey

Solution 6

Using Pick's Theorem, we see there are 16 boundary points and 1 interior point. $1 + \frac{16}{2} - 1 = 8$. There are $4 \cdot 4 = 16$ squares, and 8 is $\boxed{\textbf{(B)}~50}$ percent of 16.

-leafy

Video Solution 1

~ ChillGuyDoesMath

Video Solution 2 by SpreadTheMathLove

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

Video Solution 3

~hsnacademy

Video Solution 4 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution 5 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution 6 by CoolMathProblems

https://youtu.be/SNviHnUR3x4

Video Solution 7 by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

Video Solution 8

https://youtu.be/Yc4KSp0aOco

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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