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

m (Vide Solution 1 by SpreadTheMathLove)
m (Problem)
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fill(rotate(270, (2,2)) * y, gray(0.8));
 
fill(rotate(270, (2,2)) * y, gray(0.8));
  
draw((1,1)--(3,3));
+
draw((1,1)--(q293579105798023
 +
));
 
draw((3,1)--(1,3));
 
draw((3,1)--(1,3));
  

Revision as of 23:29, 29 January 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?


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)--(q293579105798023
));
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);
 (Error making remote request. Unknown error_msg)

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

Solution 1

Each of the side triangles has base length $2$ and height $1$, so they all have area $\frac{1}{2} \cdot 2 \cdot 1 = 1$. Each of the four corner squares has side length $1$ and hence area $1$. Then the area of the shaded region is $4^2 - 4 \cdot 1 - 4 \cdot 1 = 8$. Since $\frac{8}{16}=\frac{1}{2}$, our answer is $\frac{1}{2} \cdot 100 = \boxed{\textbf{(B)}~50}$. ~cxsmi

Video Solution 1 by SpreadTheMathLove

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