Difference between revisions of "2018 AMC 10B Problems/Problem 24"
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | Let <math>ABCDEFG</math> be a regular hexagon with side length 1. Denote X, Y, and Z the midpoints of sides <math>AB</math>, <math>CD</math>, and <math>EF</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>? | + | Let <math>ABCDEFG</math> be a regular hexagon with side length <math>1</math>. Denote <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>AB</math>, <math>CD</math>, and <math>EF</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>? |
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math> | <math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math> | ||
Revision as of 16:30, 16 February 2018
Problem
Let 
be a regular hexagon with side length 
. Denote 
, 
, and 
the midpoints of sides 
, 
, and 
, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of 
and 
?
Answer: 
Solution
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
