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Difference between revisions of "2022 AMC 10A Problems/Problem 21"

(→‎Problem)
(→‎Solution 1)
Line 60: Line 60:
  
 
==Solution 1==
 
==Solution 1==
−
We extend some sides of two adjacent hexagons to form equilateral triangles in the planes of the hexagons, as shown below:
+
We extend line segments <math>\ell,m,</math> and <math>n</math> to their point of concurrency, as shown below:
 +
<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;
 +
 
 +
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);
  
−
<b>DIAGRAM IN PROGRESS. WILL FINISH REALLY SOON.
+
draw(uVsl[1]--uVsr[0],red+2bp);
 +
draw(uVsr[3]--uVsl[4],red+2bp);
 +
draw(lVs[0]--mVs[0],red+2bp);
  
−
LEAVE IT TO ME. THANKS</b>
+
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>
 
Note that the blue line segments in the plane of the octagon are perpendicular, and their lengths are each <math>1.</math> Essentially, we are adding a right tetrahedron to each corner of the bowl.
 
Note that the blue line segments in the plane of the octagon are perpendicular, and their lengths are each <math>1.</math> Essentially, we are adding a right tetrahedron to each corner of the bowl.
  
Line 71: Line 132:
 
<asy>
 
<asy>
 
/* Made by AoPS; edited by MRENTHUSIASM */
 
/* Made by AoPS; edited by MRENTHUSIASM */
−
unitsize(5cm);
+
size(225);
 
real r = 1/3;
 
real r = 1/3;
−
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
+
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);
 
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);
 
draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp);

Revision as of 05:00, 16 November 2022

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] 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] Note that the blue line segments in the plane of the octagon are perpendicular, and their lengths are each $1.$ Essentially, we are adding a right tetrahedron to each corner of the bowl.

The 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 \[3^2-4\cdot\left(\frac12\cdot1\cdot1\right)=\boxed{\textbf{(B) }7}.\]

~MRENTHUSIASM

Solution 2

Note that the octagon is equiangular by symmetry, but it is not equilateral. $4$ of it's 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 degree angle:

Screenshot 20221112 010724.png

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: \[\left(\sqrt{2} + 2 \cdot \frac{\sqrt{2}}{2}\right)^2 \implies 2\sqrt{2}^2 = 8\] 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 $7=\boxed{B}$

~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 $1$, the hypotenuse of the triangles would be $\sqrt2$. 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 $\sqrt2$. Next, we have $5$ $1$ by $1$ squares. The triangles each have an area of $\frac{1}{2}$, and the squares each have an area of $1$.

Then, we add these up, so we get $\boxed{\textbf{(B) }7}.$

~orenbad

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

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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