Difference between revisions of "2018 AMC 10B Problems/Problem 24"
Blueberry2 (talk | contribs) m (→Problem) |
(→Solution) |
||
Line 9: | Line 9: | ||
==Solution== | ==Solution== | ||
+ | |||
+ | <asy> | ||
+ | size(125); | ||
+ | defaultpen(linewidth(0.8)); | ||
+ | path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle; | ||
+ | fill(hexagon,lightgrey); | ||
+ | for(int i=0;i<=5;i=i+1) | ||
+ | |||
+ | <\asy> | ||
==See Also== | ==See Also== |
Revision as of 17:18, 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
<asy> size(125); defaultpen(linewidth(0.8)); path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle; fill(hexagon,lightgrey); for(int i=0;i<=5;i=i+1)
<\asy>
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.