Difference between revisions of "2018 AMC 10B Problems/Problem 24"
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Contents
Problem
Let be a regular hexagon with side length . Denote by , , 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
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 and . Our desired area becomes
Solution 2 (Alternate Geometrical Approach to 1)
Instead of directly finding the desired hexagonal area, can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. Since and are equilateral (easily proven), , so . As is a transversal running through (use your imagination) and , .
Then, is a 306090 triangle. By HL congruence, . . Then, the area of is \frac{\sqrt{3}}{32}\triangle XYZ\triangle XYZ\frac{3}{2}\frac{9sqrt{3}}{16}\frac{9sqrt{3}}{16}\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32} \implies \boxed{C}$~BJHHar
whoof
==Solution 3 == Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids ($ (Error compiling LaTeX. ! Missing $ inserted.)AXFZXBCYZYEDXYZ\frac{\sqrt{3}}{4}\frac{5\sqrt{3}}{16}\frac{15\sqrt{3}}{16}.ABCDEF\triangle XYZYC = 1/2.\frac{\sqrt{3}}{4}\frac{\sqrt{3}}{32}3\frac{\sqrt{3}}{32}\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}\frac{3\sqrt{3}}{2}  \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}120^{\textrm{o}}60^{\textrm{o}}1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}ACE\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}\frac{9\sqrt{3}}{16}\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}\triangle MNO\triangle PMN[MPNQOR]=\dfrac{5}{8}[ACE][ACE] = \dfrac{1}{2}[ABCDEF][MPNQOR]=\dfrac{5}{16}[ABCDEF]5\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}5[ABCDEF]6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}$.
Video Solution
https://www.youtube.com/watch?v=yDbn9Mx2myw
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.