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

Revision as of 16:19, 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>

size(125); defaultpen(linewidth(0.8)); path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;

<\asy>

2018 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

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

Revision as of 16:19, 16 February 2018

Problem

Solution

See Also

@@ Line 11: / Line 11: @@
 <asy>
 size(125);
 defaultpen(linewidth(0.8));
 path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;
-fill(hexagon,lightgrey);
-for(int i=0;i<=5;i=i+1)
 <\asy>