Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2025 AMC 8 Problems/Problem 1"
Line 1: Line 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?
 
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?
  
Line 30: Line 31:
  
 
<math>\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 80</math>
 
<math>\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 80</math>
 +
 +
==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

Revision as of 21:46, 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?


[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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_8_Problems/Problem_1&oldid=241058"