2018 AMC 10B Problems/Problem 24: Difference between revisions

Revision as of 16:18, 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; fill(hexagon,lightgrey); for(int i=0;i<=5;i=i+1)

<\asy>

2018 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12B (Problems • Answer Key • Resources)
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2018 AMC 10B Problems/Problem 24: Difference between revisions

Revision as of 16:18, 16 February 2018

Problem

Solution

See Also

@@ Line 9: / Line 9: @@
 ==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;
+fill(hexagon,lightgrey);
+for(int i=0;i<=5;i=i+1)
+<\asy>
 ==See Also==