2025 AMC 8 Problems/Problem 1

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?

$[asy] path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle; path y = reflect((0,0),(4,4)) * x; fill(x, gray(0.6)); fill(rotate(90, (2,2)) * x, gray(0.6)); fill(rotate(180, (2,2)) * x, gray(0.6)); fill(rotate(270, (2,2)) * x, gray(0.6)); fill(y, gray(0.8)); fill(rotate(90, (2,2)) * y, gray(0.8)); fill(rotate(180, (2,2)) * y, gray(0.8)); fill(rotate(270, (2,2)) * y, gray(0.8)); draw((1,1)--(3,3)); draw((3,1)--(1,3)); add(grid(4,4)); path w = (1,0)--(2,1)--(3,0); draw(w); draw(rotate(90, (2,2)) * w); draw(rotate(180, (2,2)) * w); draw(rotate(270, (2,2)) * w); [/asy]$

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

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

Solution 2

There are $16$ total squares in the diagram and each square can have $2$ triangles. Thus, the total number of triangles in the diagram is $16 \cdot 2 = 32$ triangles. There are $16$ shaded triangles in the diagram (you can just count this up), so the percentage of the shaded triangles is $\dfrac{16}{32} \cdot 100 =$ $\boxed{\textbf{(B)}~50}$ . ~Pi_in_da_box

Video Solution 1 by SpreadTheMathLove

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

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2025 AMC 8 Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Video Solution 1 by SpreadTheMathLove