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

(Solution)
(Solution)
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=(-1/2, sqrt(3)/4);  
+
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 $ABCDEF$ 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] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,sqrt(3)); B=(1,sqrt(3)); C=(3/2,sqrt(3)/2); D=(1,0); E=(0,0); F=(-1/2,sqrt(3)/2); X=(1/2, sqrt(3)); Y=(5/4, sqrt(3)/4); Z=(-1/4, sqrt(3)/4);   draw(A--B--C--D--E--F--cycle);  draw(A--C--E--cylce); draw(X--Y--Z--cycle);  label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,ESE); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,WSW); label("$X$", X, N); label("$Y$", Y, ESE); label("$Z$", Z, WSW);  [/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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