Difference between revisions of "2018 AMC 10B Problems/Problem 24"
(→Solution) |
(→Solution) |
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Line 20: | Line 20: | ||
X=(1/2, sqrt(3)); | X=(1/2, sqrt(3)); | ||
Y=(5/4, sqrt(3)/4); | Y=(5/4, sqrt(3)/4); | ||
− | Z=(-1/ | + | Z=(-1/4, sqrt(3)/4); |
draw(A--B--C--D--E--F--cycle); | draw(A--B--C--D--E--F--cycle); | ||
− | draw(A--C--E); | + | draw(A--C--E--cylce); |
− | draw(X--Y--Z); | + | draw(X--Y--Z--cycle); |
label("$A$",A,NW); | label("$A$",A,NW); |
Revision as of 17:35, 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.