Difference between revisions of "2025 AMC 8 Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
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>. ~cxsmi | 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>. ~cxsmi | ||
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| + | ==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 | ||
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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. ~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 half of the area of each cell. Thus, our final answer is <math>\frac12=\boxed{\textbf{(B)}~50}.</math> ~derekwang2048 | ||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/rjd0gigUsd0 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:52, 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?
![[asy] path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle; path y = reflect((0,0),(4,4)) * x; 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((1,1)--(3,3)); draw((3,1)--(1,3)); 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]](http://latex.artofproblemsolving.com/6/f/1/6f1cc7540fcf94f876c74993194b3df4fc4a420c.png)
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. ~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 half of the area of each cell. Thus, our final answer is 
~derekwang2048
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution by Thinking Feet
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. 
