2022 AMC 10A Problems/Problem 21
Contents
Problem
A bowl is formed by attaching four regular hexagons of side to a square of side
. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
Solution 1
We extend line segments and
to their point of concurrency, as shown below:
We claim that lines
and
are concurrent: In the lateral faces of the bowl, we know that lines
and
must intersect, and lines
and
must intersect. In addition, line
intersects the top plane of the bowl at exactly one point. Since lines
and
are both in the top plane of the bowl, we conclude that lines
and
are concurrent.
In the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length In the top plane of the bowl, we know that
So, the dashed red line segments create an isosceles triangle with leg-length
Note that octagon has four pairs of parallel sides, and the successive side-lengths are as shown below:
The area of the octagon is
~MRENTHUSIASM
Solution 2
Note that the octagon is equiangular by symmetry, but it is not equilateral. of its sides are shared with the hexagon's sides, so each of those sides have side length
. However, the other
sides are touching the triangles, so we wish to find the length of these sides.
Notice that when two adjacent hexagons meet at a side, their planes make the same dihedral angle at the bottom-most point of intersection and at the top-most point of intersection by symmetry. Therefore, the triangle that is wedged between the two hexagons has the same angle as the square at the bottom wedged between the hexagons. Thus, the triangle is a isosceles triangle.
This conclusion can also be reached by cutting the bottom square across a diagonal and noticing that each resulting triangle is congruent to each triangle wedged between the hexagons by symmetry.
Furthermore, notice that if you take a copy of this bowl and invert it and place it on top of this bowl, you will get a polyhedron with faces of hexagons and squares, and therefore this triangle has a
-angle:
Now that we have come to this conclusion, by simple Pythagorean theorem, we have that the other sides of the octagon are
.
We can draw a square around the octagon so that the area of the octagon is the area of the square minus each corner triangle. The hypotenuse of these corner triangles are and they are
triangles because the octagon is equiangular, so each has dimensions
.
The side length of the square is
for the larger sides of the octagon, and adding
of
for each width of the triangle. Therefore, the area of the square is:
The area of each triangle is
and there are
of them, so we subtract
from the area of the square. The area of the octagon is thus
.
~KingRavi
Solution 3 (Not Rigorous)
Through observation, we can reasonably assume that each of the triangles on this shape is a right triangle. Since each side length of the hexagons is , the hypotenuse of the triangles would be
. Now we know the side lengths of the octagon whose area we are solving for. The octagon can be broken into nine pieces. We have four triangles whose side lengths are 1, and their hypotenuse is a side whose length is
. Next, we have
by
squares. The triangles each have an area of
, and the squares each have an area of
.
Then, we add these up, so we get
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution by OmegaLearn (Using Equiangular Hexagon Properties)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
~MRENTHUSIASM
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
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