Difference between revisions of "2025 AMC 8 Problems/Problem 1"
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== Solution 3 == | == Solution 3 == | ||
| − | 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. | + | There are <math>4</math> squares that are entirely shaded and <math>4</math> squares that have no shading. This cancels them out. 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 == | == Solution 4 == | ||
| Line 50: | Line 50: | ||
== Solution 6 == | == Solution 6 == | ||
| − | + | You can move the extra half a squares and make 4 squares so there are 8/16. The answer is <math>\boxed{\textbf{(B)}~50}</math> percent of 16. | |
| − | + | William | |
| − | == Video Solution 1 | + | == Video Solution 1 (Detailed Explanation) 🚀⚡📊 == |
| − | https:// | + | https://www.youtube.com/watch?v=rGEGn7U4uHk |
| + | ~ ChillThingz :) | ||
| − | == Video Solution 2 == | + | == Video Solution 2 Super Easy to understand and efficient!== |
| − | + | https://youtu.be/dZkFuq0DZyk | |
== Video Solution 3 by SpreadTheMathLove == | == Video Solution 3 by SpreadTheMathLove == | ||
| Line 79: | Line 80: | ||
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==Video Solution | + | ==Video Solution 7 by CoolMathProblems== |
https://youtu.be/SNviHnUR3x4 | https://youtu.be/SNviHnUR3x4 | ||
| − | == Video Solution | + | == Video Solution 8 by Pi Academy == |
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | ||
| − | == Video Solution | + | == Video Solution 9 == |
https://youtu.be/Yc4KSp0aOco | https://youtu.be/Yc4KSp0aOco | ||
Latest revision as of 19:28, 16 September 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Video Solution 1 (Detailed Explanation) 🚀⚡📊
- 9 Video Solution 2 Super Easy to understand and efficient!
- 10 Video Solution 3 by SpreadTheMathLove
- 11 Video Solution 4
- 12 Video Solution 5 by Daily Dose of Math
- 13 Video Solution 6 by Thinking Feet
- 14 Video Solution 7 by CoolMathProblems
- 15 Video Solution 8 by Pi Academy
- 16 Video Solution 9
- 17 Video Solution(Quick, fast, easy!)
- 18 See Also
Problem
The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire 
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]](http://latex.artofproblemsolving.com/5/6/6/56638188764158bdffa266bc933d11b6acd2fb76.png)
Solution 1
Each of the unshaded triangles has base length 
and height 
, so they all have area 
. Each of the unshaded unit squares has area . The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or 
. The star is then 
, or 
percent of the entire grid.
~cxsmi
Solution 2
There are 
total squares in the diagram and each square has triangles whose areas are half the area of a unit square. Thus, the total number of triangles in the diagram is 
triangles. There are shaded triangles in the diagram, so the area of the star is 
, or percent.
~Pi_in_da_box
Solution 3
There are 
squares that are entirely shaded and squares that have no shading. This cancels them out. The rest of the squares are half-half. Therefore the shaded region is 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 
, and so our answer is 
~derekwang2048
Solution 5
The shaded area is a 
square in the middle of the figure combined with 
small triangles. Since each small triangle is of a unit square, the star's area is equal to the area of 
unit squares, which percent of the grid.
-vockey
Solution 6
You can move the extra half a squares and make 4 squares so there are 8/16. The answer is percent of 16.
William
Video Solution 1 (Detailed Explanation) 🚀⚡📊
https://www.youtube.com/watch?v=rGEGn7U4uHk
~ ChillThingz :)
Video Solution 2 Super Easy to understand and efficient!
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 4
Video Solution 5 by Daily Dose of Math
Video Solution 6 by Thinking Feet
Video Solution 7 by CoolMathProblems
Video Solution 8 by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
Video Solution 9
Video Solution(Quick, fast, easy!)
~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. 
