2022 AMC 10A Problems/Problem 21

Problem

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . 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? $[asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy]$ $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution 1

We extend line segments $\ell,m,$ and $n$ to their point of concurrency, as shown below: $[asy] /* Made by AoPS; edited by MRENTHUSIASM */ import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; draw(surface(uVsl[0]--uVsr[0]--uVsl[1]--uVsr[1]--uVsl[2]--uVsr[2]--uVsl[3]--uVsr[3]--cycle),yellow); for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); draw(uVsl[1]--uVsr[0],red+2bp); draw(uVsr[3]--uVsl[4],red+2bp); draw(lVs[0]--mVs[0],red+2bp); draw(uVsr[0]--uVsr[0]+uVsr[0]-uVsl[1],red+dashed+2bp); draw(uVsl[4]--uVsl[4]+uVsl[4]-uVsr[3],red+dashed+2bp); draw(mVs[0]--mVs[0]+mVs[0]-lVs[0],red+dashed+2bp); label("$\ell$",midpoint(lVs[0]--mVs[0]),(1,2,0),red); label("$m$",midpoint(uVsr[3]--uVsl[4]),(1,0,-2),red); label("$n$",midpoint(uVsl[1]--uVsr[0]),(-1,0,-2),red); [/asy]$ We claim that lines $\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\ell$ and $m$ must intersect, and lines $\ell$ and $n$ must intersect. In addition, line $\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\ell,m,$ and $n$ 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 $1.$ In the top plane of the bowl, we know that $\overleftrightarrow{m}\perp\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$

Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below: $[asy] /* Made by AoPS; edited by MRENTHUSIASM */ size(225); real r = 1/3; draw((0,r)--(0,0)--(r,0)^^(1-r,0)--(1,0)--(1,r)^^(1,1-r)--(1,1)--(1-r,1)^^(r,1)--(0,1)--(0,1-r),red+2bp+dashed); fill((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,yellow); draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp); label("$1$",(0.5,0),S); label("$1$",(1,0.5),E); label("$1$",(0.5,1),N); label("$1$",(0,0.5),W); label("$\sqrt2$",(1-r/2,r/2),NW); label("$\sqrt2$",(1-r/2,1-r/2),SW); label("$\sqrt2$",(r/2,1-r/2),SE); label("$\sqrt2$",(r/2,r/2),NE); label("$1$",(1-r/2,0),S); label("$1$",(1,r/2),E); label("$1$",(1,1-r/2),E); label("$1$",(1-r/2,1),N); label("$1$",(r/2,1),N); label("$1$",(0,1-r/2),W); label("$1$",(0,r/2),W); label("$1$",(r/2,0),S); [/asy]$ The area of the octagon is $3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{\textbf{(B) }7}.$

~MRENTHUSIASM

Solution 2 (Symmetry and Similarity)

(This is an alternate way of analyzing the red extension line segments drawn in Solution 1.)

The perimeter of the square bottom of the bowl is $4$ . Halfway up the bowl, the boundary is still a square, with perimeter $4$ times the hexagon circumradius, aka $4 \times 2 = 8$ times the hexagon (also square) side length (1), an increase of $4$ .

Extending the bottom half of the bowl to twice its height (full height of the bowl) would increase the perimeter by the same amount again, forming a square with perimeter $4 + (8-4)\times 2= 12$ . Thus the top octagon is cut out of a square of side length $12\div4=3$ and thus area $9$ .

The difference between the above-constructed square and the octagon is four right triangles, and (by rotational and reflection symmetry), each is isosceles with equal-length perpendicular bases of length $(3-1)/2 = 1$ , and thus having area $\frac12$ . Therefore the area of the octagon is $9-(4\times 1/2) = \boxed{\textbf{(B) }7}$ .

-oinava

Solution 3

Note that the octagon is equiangular by symmetry, but it is not equilateral. $4$ of its sides are shared with the hexagon's sides, so each of those sides have side length $1$ . However, the other $4$ 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 $45-45-90$ 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, a truncated octahedron, and therefore this triangle has a $90^\circ$ -angle:

Now that we have come to this conclusion, by simple Pythagorean theorem, we have that the other $4$ sides of the octagon are $\sqrt{2}$ .

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 $1$ and they are $45-45-90$ triangles because the octagon is equiangular, so each has dimensions $\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},1$ . The side length of the square is $\sqrt{2}$ for the larger sides of the octagon, and adding $2$ of $\frac{\sqrt{2}}{2}$ for each width of the triangle. Therefore, the area of the square is: $\left(\sqrt{2} + 2 \cdot \frac{\sqrt{2}}{2}\right)^2 \implies \left(2\sqrt{2}\right)^2 = 8$ The area of each triangle is $\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{4}$ and there are $4$ of them, so we subtract $1$ from the area of the square. The area of the octagon is thus $\boxed{\textbf{(B) }7}$ .

~KingRavi

Solution 5 (Rigorous? Simple!)

Denote a square $ABCD, AB = 1,$ regular hexagons $ABFKSE, BCHMLF, CDGPNH, ADGQRE,$ triangles $FKL, ESR, GPQ, HMN.$

The main diagonal of each regular hexagon $EF = 2 \implies EFGH$ is square with side $2$ parallel to $ABCD.$ The area of this square $[EFGH] = 4 \implies [EFGH] = [ABCD] + 3.$

This difference is the area of the projection of 4 half of hexagons on the plane of $ABCD.$

So the area of the projections of another $4$ half of hexagons is $3.$

It is evident (may be not only for me) that projections of the coincide sides of hexagons are along diagonals of $ABCD$ (for example $A,F,C,$ and $H$ are collinear.)

So the projections on this plane of the coincide sides of hexagons and triangles are perpendicular to this lines $(SE \perp AE).$

Therefore in this plane projections of points $S,E,$ and $R$ are collinear and plane of $\triangle ESR$ is perpendicular to the plane $ABCD.$

Then, we get $1 +3 +3 = \boxed{\textbf{(B) }7}.$

vladimir.shelomovskii@gmail.com, vvsss

Animated Video Solution

https://youtu.be/yk4euPCYhd0

~Star League (https://starleague.us)

Video Solution By ThePuzzlr

https://youtu.be/br11LJJD4-c

~ MathIsChess

Video Solution by OmegaLearn (Using Equiangular Hexagon Properties)

https://youtu.be/-QHhR2r9HgQ

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

https://youtu.be/nxssWgtWl9E

~MRENTHUSIASM

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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