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

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<asy>
 
<asy>
−
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O;
+
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;
 
A=(0,sqrt(3));
 
A=(0,sqrt(3));
 
B=(1,sqrt(3));
 
B=(1,sqrt(3));
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N=(3/4,3sqrt(3)/4);
 
N=(3/4,3sqrt(3)/4);
 
O=(3/4,sqrt(3)/4);
 
O=(3/4,sqrt(3)/4);
 +
P=(3/8,7sqrt(3)/8);
  
 
draw(A--B--C--D--E--F--cycle);
 
draw(A--B--C--D--E--F--cycle);
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label("$N$", N, NE);
 
label("$N$", N, NE);
 
label("$O$", O, SE);
 
label("$O$", O, SE);
−
 
+
label("$P$", P, NW);
 
</asy>
 
</asy>
  

Revision as of 17:45, 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 $\triangle XYZ$?

$\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad$

Solution

[asy] pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R; 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=(3/4,3sqrt(3)/4); O=(3/4,sqrt(3)/4); P=(3/8,7sqrt(3)/8); draw(A--B--C--D--E--F--cycle); draw(A--C--E--cycle); draw(X--Y--Z--cycle); draw(M--N--O--cycle); 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, NW); label("$N$", N, NE); label("$O$", O, SE); label("$P$", P, NW); [/asy]

Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). Now we know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of $sqrt(3)/4$ (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of $5sqrt(3)/16$ for a total area of $15sqrt(3)/16$ (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be $sqrt(3)/4$ giving us an area of $sqrt(3)/32$ per triangle, and a total area of $3sqrt(3)/32$. Adding the two areas together, we get $15sqrt(3)/16+3sqrt(3)/32=33sqrt(3)/32$. Finding the total area, we get $6*1^2*sqrt(3)/4=3sqrt(3)/2$. Taking the complement, we get $3sqrt(3)/2-33sqrt(3)/32=15sqrt(3)/32$, so the answer is C --- Arpitr20

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

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

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

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

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