2018 AMC 10B Problems/Problem 24
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 30-60-90 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. Unknown error_msg)AXFZXBCY
ZYED
X
Y
Z
\frac{\sqrt{3}}{4}
\frac{5\sqrt{3}}{16}
\frac{15\sqrt{3}}{16}.
ABCDEF
\triangle XYZ
YC = 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 |
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