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

(→‎Video Solution 1 by SpreadTheMathLove)
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== Problem ==
 
== Problem ==
  
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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?
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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?
  
 
<asy>
 
<asy>
 
path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle;
 
path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle;
 
path y = reflect((0,0),(4,4)) * x;
 
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(x, gray(0.6));
 
fill(rotate(90, (2,2)) * x, gray(0.6));
 
fill(rotate(90, (2,2)) * x, gray(0.6));
 
fill(rotate(180, (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(rotate(270, (2,2)) * x, gray(0.6));
−
 
 
fill(y, gray(0.8));
 
fill(y, gray(0.8));
 
fill(rotate(90, (2,2)) * y, gray(0.8));
 
fill(rotate(90, (2,2)) * y, gray(0.8));
 
fill(rotate(180, (2,2)) * y, gray(0.8));
 
fill(rotate(180, (2,2)) * y, gray(0.8));
 
fill(rotate(270, (2,2)) * y, gray(0.8));
 
fill(rotate(270, (2,2)) * y, gray(0.8));
−
 
+
draw(z);
−
draw((1,1)--(3,3));
+
draw(rotate(90, (2,2)) * z);
−
draw((3,1)--(1,3));
+
draw(rotate(180, (2,2)) * z);
−
 
+
draw(rotate(270, (2,2)) * z);
 
add(grid(4,4));
 
add(grid(4,4));
−
 
−
path w = (1,0)--(2,1)--(3,0);
 
−
 
−
draw(w);
 
−
draw(rotate(90, (2,2)) * w);
 
−
draw(rotate(180, (2,2)) * w);
 
−
draw(rotate(270, (2,2)) * w);
 
 
</asy>
 
</asy>
  
Line 34: Line 26:
 
== Solution 1 ==
 
== Solution 1 ==
  
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Each of the side triangles has base length <math>2</math> and height <math>1</math>, so they all have area <math>\frac{1}{2} \cdot 2 \cdot 1 = 1</math>. Each of the four corner squares has side length <math>1</math> and hence area <math>1</math>. Then the area of the shaded region is <math>4^2 - 4 \cdot 1 - 4 \cdot 1 = 8</math>. Since <math>\frac{8}{16}=\frac{1}{2}</math>, our answer is <math>\frac{1}{2} \cdot 100 = \boxed{\textbf{(B)}~50}</math>.
+
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.
 
~cxsmi
 
~cxsmi
  
 
== Solution 2 ==
 
== Solution 2 ==
  
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There are <math>16</math> total squares in the diagram and each square can have <math>2</math> triangles. Thus, the total number of triangles in the diagram is <math>16 \cdot 2 = 32</math> triangles. There are <math>16</math> shaded triangles in the diagram (you can just count this up), so the percentage of the shaded triangles is <math>\dfrac{16}{32} \cdot 100 =</math> <math>\boxed{\textbf{(B)}~50}</math>.
+
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
 
~Pi_in_da_box
  
 
== Solution 3 ==
 
== Solution 3 ==
  
−
Quite simply, you can "rearrange" the triangles to make a 2 column section in the middle, which equates to <math>\boxed{\textbf{(B)}~50}</math>. Probably the most accurate in-test solution.
+
Quite simply, we can rearrange the triangles to make a two-column section which is <math>\boxed{\textbf{(B)}~50}</math> percent of the entire grid.
 
~shreyan.chethan
 
~shreyan.chethan
  
 
== Solution 4 ==
 
== 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 half of the area of each cell. Thus, our final answer is <math>\frac12=\boxed{\textbf{(B)}~50}.</math>
+
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
 
~derekwang2048
  
 
== Solution 5 ==  
 
== Solution 5 ==  
  
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The first thing that we can see is that the shaded region is a <math>2</math>x<math>2</math> square(in the middle of the figure) combined with <math>8</math> small triangles. Since each small triangle is <math>1 \cdot \frac{1}{2} = \frac{1}{2}</math> unit squares, we get <math>4 + 8 \cdot \frac{1}{2} = 8</math> unit squares. Now we can easily see the answer is <math>\boxed{\textbf{(B)}~50}</math>.
+
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
 
-vockey
  
−
== Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
+
== Solution 6 ==
−
Youtube Link ⬇️
+
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 ==
  
 
https://youtu.be/rGEGn7U4uHk
 
https://youtu.be/rGEGn7U4uHk
  
−
~ ChillGuyDoesMath :)
+
~ ChillGuyDoesMath
  
−
== Video Solution by SpreadTheMathLove ==
+
== Video Solution 2 by SpreadTheMathLove ==
  
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
  
−
==Video Solution (A Clever Explanation You’ll Get Instantly)==
+
==Video Solution 3 ==
 +
 
 
https://youtu.be/VP7g-s8akMY?si=rhpz6FZayHNW--j8&t=7
 
https://youtu.be/VP7g-s8akMY?si=rhpz6FZayHNW--j8&t=7
 
~hsnacademy
 
~hsnacademy
  
−
== Video Solution by Daily Dose of Math ==
+
== Video Solution 4 by Daily Dose of Math ==
  
 
https://youtu.be/rjd0gigUsd0
 
https://youtu.be/rjd0gigUsd0
Line 79: Line 76:
 
~Thesmartgreekmathdude
 
~Thesmartgreekmathdude
  
−
== Video Solution by Thinking Feet ==
+
== Video Solution 5 by Thinking Feet ==
  
 
https://youtu.be/PKMpTS6b988
 
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
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:15, 3 February 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

Quite simply, we can rearrange the triangles to make a two-column section which is $\boxed{\textbf{(B)}~50}$ percent of the entire grid. ~shreyan.chethan

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

https://youtu.be/rGEGn7U4uHk

~ ChillGuyDoesMath

Video Solution 2 by SpreadTheMathLove

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

Video Solution 3

https://youtu.be/VP7g-s8akMY?si=rhpz6FZayHNW--j8&t=7 ~hsnacademy

Video Solution 4 by Daily Dose of Math

https://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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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