2018 AMC 10B Problems/Problem 24
Contents
Problem
Let 
be a regular hexagon with side length 
. Denote 
, 
, and 
the midpoints of sides 
, 
, and 
, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of 
and 
?
Solution 1
![[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); Q=(9/8, 3sqrt(3)/8); R=(0,sqrt(3)/4); fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray); 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, NNW); label("$Q$", Q, ESE); label("$R$", R, SW); [/asy]](http://latex.artofproblemsolving.com/9/8/a/98a0f283425961b0831f348ed7e5985563e4c9bc.png)
The desired area (hexagon 
) consists of an equilateral triangle (
) and three right triangles (
, 
, and 
).
Notice that 
(not shown) and 
are parallel. 
divides transversals and 
into a 
ratio. Thus, it must also divide transversal 
and transversal 
into a ratio. By symmetry, the same applies for 
and 
as well as 
and 
In , we see that ![$\frac{[MNO]}{[ACE]} = \frac{1}{4}$](http://latex.artofproblemsolving.com/6/8/4/68490453244fed9770c615d1a795c111339d7c98.png)
and ![$\frac{[MPN]}{[ACE]} = \frac{1}{8}$](http://latex.artofproblemsolving.com/c/d/5/cd54752f3f3f6ba754f03b56a485fb3f32f77f48.png)
. Our desired area becomes
![\[(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}\]](http://latex.artofproblemsolving.com/4/9/e/49ea2b5904e94896ced03049fe0c056bc4fa688e.png)
~Llamabonobo
Solution 2
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). 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 
(by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of 
for a total area of 
(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 giving us an area of 
per triangle, and a total area of 
. Adding the two areas together, we get 
. Finding the total area, we get 
. Taking the complement, we get 
--- 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 | |
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