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

(Solution)
(Solution)
Line 11: Line 11:
  
 
<asy>
 
<asy>
 +
pair A,B,C,D,E,F,W,X,Y,Z;
 +
A=(0,0);
 +
B=(1,0);
 +
C=(3/2,sqrt(3)/2);
 +
D=(1,sqrt(3));
 +
E=(0,sqrt(3));
 +
F=(-1/2,sqrt(3)/2);
 +
W=(4/3,2sqrt(3)/3);
 +
X=(4/3,sqrt(3)/3);
 +
Y=(-1/3,sqrt(3)/3);
 +
Z=(-1/3,2sqrt(3)/3);
 +
draw(A--B--C--D--E--F--cycle);
 +
draw(W--Z);
 +
draw(X--Y);
 +
draw(F--C--B--E--D--A);
  
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,ESE);
 +
label("$D$",D,NE);
 +
label("$E$",E,NW);
 +
label("$F$",F,WSW);
 +
label("$W$",W,ENE);
 +
label("$X$",X,ESE);
 +
label("$Y$",Y,WSW);
 +
label("$Z$",Z,WNW);
 
</asy>
 
</asy>
  

Revision as of 17:29, 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=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A);  label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); [/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

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