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, 17/10);
+
pair D = (0,0);
pair B = (3/2, 17/10);
+
pair F = (1/2,0);
pair C = (2, 17/20);
+
pair C = (1,0);
pair D = (3/2, 0);
+
pair G = (0,1);
pair E = (1/2, 0);
+
pair E = (1,1);
pair F = (0,17/20);
+
pair A = (0,2);
 +
pair B = (1,2);
 +
pair H = (1/2,1);
  
 
</asy>
 
</asy>

Revision as of 16:26, 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 D = (0,0);
pair F = (1/2,0);
pair C = (1,0);
pair G = (0,1);
pair E = (1,1);
pair A = (0,2);
pair B = (1,2);
pair H = (1/2,1);

 (Error making remote request. Unknown error_msg)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png