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

(Solution)
(Solution)
Line 29: Line 29:
 
draw(A--C--E--cycle);
 
draw(A--C--E--cycle);
 
draw(X--Y--Z--cycle);
 
draw(X--Y--Z--cycle);
draw(M--N--O--cycle);
+
draw(M--N--O);
  
 
label("$A$",A,NW);
 
label("$A$",A,NW);
Line 40: Line 40:
 
label("$Y$", Y, ESE);
 
label("$Y$", Y, ESE);
 
label("$Z$", Z, WSW);
 
label("$Z$", Z, WSW);
label("$M$",M, W);
+
label("$M$", M, W);
 
label("$N$", N, NE);
 
label("$N$", N, NE);
 
label("$O$", O, SE);
 
label("$O$", O, SE);

Revision as of 16:56, 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

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); 
M=(0,sqrt(3)/2);
N=(1,3/2);
O=(1,1/2);

draw(A--B--C--D--E--F--cycle);

draw(A--C--E--cycle);
draw(X--Y--Z--cycle);
draw(M--N--O);

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);
label("$M$", M, W);
label("$N$", N, NE);
label("$O$", O, SE);

 (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