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 "2016 AMC 10B Problems/Problem 23"

m (an answer is not a solution; the user does not deserve credit for this.)
(→‎Solution)
Line 8: Line 8:
 
==Solution==
 
==Solution==
 
<math>\textbf{(C)}\ \frac{11}{27}</math>
 
<math>\textbf{(C)}\ \frac{11}{27}</math>
 +
 +
 +
==See Also==
 +
{{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Revision as of 13:24, 21 February 2016

Problem

In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$


Solution

$\textbf{(C)}\ \frac{11}{27}$


See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&oldid=76704"