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Difference between revisions of "2016 AMC 10B Problems/Problem 23"

(Created page with "==Problem== In regular hexagon <math>ABCDEF</math>, points <math>W</math>, <math>X</math>, <math>Y</math>, and <math>Z</math> are chosen on sides <math>\overline{BC}</math>, ...")
 
(→‎Problem)
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<math>\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}</math>
 
<math>\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}</math>
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 +
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==Solution==
 +
<math>\textbf{(C)}\ \frac{11}{27}</math>
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 +
solution by ngeorge

Revision as of 09:27, 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

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

solution by ngeorge

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