Difference between revisions of "2018 AMC 10B Problems/Problem 24"

(Solution)
(Solution)
Line 15: Line 15:
 
size(9cm);
 
size(9cm);
 
pen dps = fontsize(10); defaultpen(dps);
 
pen dps = fontsize(10); defaultpen(dps);
pair A = (1/2,\sqrt{3});
+
pair A = (1/2,1.7);
pair B = (3/2, \sqrt{3});
+
pair B = (3/2, 1.7);
pair C = (2, \sqrt{3}/2);
+
pair C = (2, 1.7/2);
 
pair D = (3/2, 0);
 
pair D = (3/2, 0);
 
pair E = (1/2, 0);
 
pair E = (1/2, 0);
pair F = (0,\sqrt{3}/2);
+
pair F = (0,1.7/2);
  
 
</asy>
 
</asy>

Revision as of 17:25, 16 February 2018

Problem

Let $ABCDEF$ be a regular hexagon with side length $1$. Denote $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?

$\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad  \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad$


Answer: $\frac {15}{32}\sqrt{3}$

Solution


import graph;
size(9cm);
pen dps = fontsize(10); defaultpen(dps);
pair A = (1/2,1.7);
pair B = (3/2, 1.7);
pair C = (2, 1.7/2);
pair D = (3/2, 0);
pair E = (1/2, 0);
pair F = (0,1.7/2);

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See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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