2022 AMC 12B Problems/Problem 25
Problem
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as , where
,
, and
are integers and
is not divisible by the square of any prime.
What is
?
Solution 1
We calculate the area as the area of the red octagon minus the four purple congruent triangles:
We first find the important angles in the figure. We note that 2 adjacent hexagons are rotated
with respect to the other, so the angles between any sides is
. In particular, as the purple triangles are isosceles, they have angles
, and
, and the octagon is equiangular (all its angles are
). Thus, we can draw a square around the octagon, and we note that the ``cut out" triangles are all isosceles right triangles.
Now, we calculate the side length of the square. Note that the hexagon has a height of , so the length of a side of the square is
. In particular, the horizontal/vertical sides of the octagon have length
, so the legs of the isosceles triangles are
Thus, the area of the octagon is
Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is
, so the area is
Thus, the area of the dodecagon is
Thus the answer is
, or
.
~cr. naman12
Solution 2 (Coord bash)
import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle); pair O,A,B,C,D; real r = sqrt3 O=(1/2,1/2);A = (3-r,3-r);B = (-2+r,3-r);M = (0,r);N = (1,r); dot((1/2,1/2)); label("$O$",O,NE); dot((3-r,3-r)); label("$A$", A, NE); dot((-2+r,3-r)); label("$B$", B, NW); dot((0,r)); label("$M$", M, NW); dot((1,r)); label("$N$", N, NE); (Error making remote request. Unknown error_msg)
Let the center of the square be , the place on the border where the top and the right hexagons intersect
, and the place on the border where the top and the left hexagons intersect
. Let the origin be
. We would like to find the coordinates of
.
By symmetry, lies on the line
. The equation of the side of the top hexagon that
is on is
. Thus, we can solve for the coordinates of
:
This means that we can find the length , which is equal to
. Next, let the top points of the top hexagon be
and
. We will find the area of trapezoid
. The lengths of the bases are
and
, and the height is equal to the
-coordinate of
minus the
-coordinate of
, which is
. Thus, the area of the trapezoid is
The total area of the figure is the area of a square with side length plus four times the area of this trapezoid:
Our answer is .
~mathboy100
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last problem |
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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