Difference between revisions of "2025 AMC 8 Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
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>. ~Pi_in_da_box | 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>. ~Pi_in_da_box | ||
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+ | ==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. | ||
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== |
Revision as of 12:35, 30 January 2025
Contents
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?
Solution 1
Each of the side triangles has base length and height , so they all have area . Each of the four corner squares has side length and hence area . Then the area of the shaded region is . Since , our answer is . ~cxsmi
Solution 2
There are total squares in the diagram and each square can have triangles. Thus, the total number of triangles in the diagram is triangles. There are shaded triangles in the diagram (you can just count this up), so the percentage of the shaded triangles is . ~Pi_in_da_box
Solution 3
Quite simply, you can "rearrange" the triangles to make a 2 column section in the middle, which equates to . Probably the most accurate in-test solution.
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
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.