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

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Kyriegon (talk | contribs)
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Kyriegon (talk | contribs)
117 edits
Line 18: Line 18:
E=(0,0);
E=(0,0);
F=(-1/2,sqrt(3)/2);
F=(-1/2,sqrt(3)/2);
X=(1/2, sqrt(3));
Y=(5/4, sqrt(3)/4);
Z=(-1/2, sqrt(3)/4);


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


label("$A$",A,NW);
label("$A$",A,NW);
Line 27: Line 33:
label("$E$",E,SW);
label("$E$",E,SW);
label("$F$",F,WSW);
label("$F$",F,WSW);
label("$X$", X, N);
label("$Y$", Y, ESE);
label("$Z$", Z, WSW);


</asy>
</asy>

Revision as of 16:34, 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=(-1/2, sqrt(3)/4); draw(A--B--C--D--E--F--cycle); draw(A--C--E); draw(X--Y--Z); 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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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