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

(Solution)
(Solution)
Line 12: Line 12:
 
<asy>
 
<asy>
  
size(125);
+
import graph;
defaultpen(linewidth(0.8));
+
size(9cm);
path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;
+
pen dps = fontsize(10); defaultpen(dps);
 
+
pair A = (1/2,\sqrt3);
 
+
pair B = (3/2, \sqrt3);
 +
pair C = (2, \sqrt3/2);
 +
pair D = (3/2, 0);
 +
pair E = (1/2, 0);
 +
pair F = (0,\sqrt3/2);
 
<\asy>
 
<\asy>
  

Revision as of 16:22, 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

<asy>

import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair A = (1/2,\sqrt3); pair B = (3/2, \sqrt3); pair C = (2, \sqrt3/2); pair D = (3/2, 0); pair E = (1/2, 0); pair F = (0,\sqrt3/2); <\asy>

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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