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Difference between revisions of "2022 AMC 12B Problems/Problem 25"

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         draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2));
 
         draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2));
 
         draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2));
 
         draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2));
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        draw((-2+sqrt(3),3-sqrt(3))--(3-sqrt(3),-2+sqrt(3)),linewidth(2));
 
 
         draw((1, sqrt(3))--(1,1),linewidth(2));
 
         draw((1, sqrt(3))--(1,1),linewidth(2));
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         label("$O$",(0.5,0.5),SSE);
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         label("$O$",(0.5,0.5),SE);
 
         dot((0.5,0.5));
 
         dot((0.5,0.5));
 
         label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);
 
         label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);

Revision as of 00:55, 22 November 2022

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 $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime. What is $m+n+p$? [asy] 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,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$

Solution 1

[asy] 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,linewidth(2)); draw((0.5,sqrt(3))--(0.5,1-sqrt(3)),linewidth(2)); draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2)); draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2)); draw((1, sqrt(3))--(1,1),linewidth(2)); label("$O$",(0.5,0.5),SE); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (1, sqrt(3)), NE); label("$C$", (1,1), E); label("$D$", (1/2, sqrt(3)), N); [/asy]

Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of $ODBA$ and multiply it by 8. We split $[ODBA]$ into $[ODBC]+[ABC]$.

Knowing the side length of the hexagon is $1$, we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is $2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.$ Thus, $OD=\sqrt{3}-\frac{1}{2}$ and $BC=\sqrt{3}-1.$ Recognizing $DB=\frac{1}{2}$ and $ODBC$ is a trapezoid, \[8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.\]

Next, we aim to find $[ABC]$. By angle chasing, we find $\angle A=105^\circ,$ $\angle B=30^\circ,$ and $\angle C = 45^\circ.$ We can use the law of sines to find $AB$:

\[\frac{\sin(105^\circ)}{\sqrt{3}-1}=\frac{\sin(45^\circ)}{AB}\implies AB=\frac{\sqrt{6}-\sqrt{2}}{2\sin(105^\circ)}.\]

We may not know what $\sin(105^\circ)$ is by memory, but we can cleverly calculate it using a common trig identity:

\begin{align*} \sin(105^\circ)&=\sin(60^\circ+45^\circ),\\ &=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ),\\ &=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2},\\ &=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{align*}

With some simplification, we'll find $AB=4-2\sqrt{3}$. Now, we can easily calculate $8\cdot [ABC]$ as \[8\cdot\frac{1}{2}\cdot(\sqrt{3}-1)(4-2\sqrt{3})\sin(30^\circ)=12\sqrt{3}-20.\]

Thus, the area of the dodecagon is $8\cdot [ODBA] = 8\cdot [ODBC] + 8 \cdot [ABC] =4\sqrt{3}-3+12\sqrt{3}-20=16\sqrt{3}-23.$

Finally, we find \[16+3-23=\boxed{\textbf{(B)}\ -4}.\]

~Indiiiigo

Solution 2 (Coord bash)

[asy] 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,linewidth(1.5)); draw((3-sqrt(3),3-sqrt(3)) -- (3-sqrt(3),sqrt(3)-2) -- (sqrt(3)-2,sqrt(3)-2) -- (sqrt(3)-2,3-sqrt(3)) -- cycle,linewidth(1.5)); label("$O (0, 0)$",(0.5,0.5),S); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (sqrt(3) - 2, 3-sqrt(3)), NW); label("$M$", (0, sqrt(3)), NW); label("$N$", (1, sqrt(3)), NE); [/asy]

Refer to the diagram above.

Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram.

By symmetry, $A$ lies on the line $y = x$. The equation of line $AN$ is $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines: \[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\] \[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\] \[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\] \[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\] \[= \frac{10 - 4\sqrt{3}}{4}\] \[= \frac{5}{2} - \sqrt{3}\] \[\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\]

This means that we can find the length $AB$, which is equal to $2(\frac{5}{2} - \sqrt{3}) = (5 - 2\sqrt{3}$. We will next find the area of trapezoid $ABMN$. The lengths of the bases are $1$ and $5 - 2\sqrt{3}$, and the height is equal to the $y$-coordinate of $M$ minus the $y$-coordinate of $A$. The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y = \frac{1}{2}$. Thus, the $y$-coordinate of $M$ is $\sqrt{3} - \frac{1}{2}$, and the height is $2\sqrt{3} - 3$. We can now find the area of the trapezoid: \[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\] \[= (2\sqrt{3} - 3)(3 - \sqrt{3})\] \[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\] \[= 9\sqrt{3} - 15.\]

The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid: \[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\] \[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\] \[= 16\sqrt{3} - 23.\]

Our answer is $16 + 3 - 23 = \boxed{\textbf{(B) }-4}$.

~mathboy100

Solution 3

We calculate the area as the area of the red octagon minus the four purple congruent triangles: [asy] import geometry; unitsize(3cm); draw((1-sqrt(3),1-sqrt(3))--(1-sqrt(3),sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3),1-sqrt(3))--cycle,dashed); filldraw((0,1-sqrt(3))--(1,1-sqrt(3))--(sqrt(3),0)--(sqrt(3),1)--(1,sqrt(3))--(0,sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--cycle,red*0.2+white,red); filldraw((1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--cycle,purple*0.2+white,blue); filldraw((sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--cycle,purple*0.2+white,blue); filldraw((0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--cycle,purple*0.2+white,blue); filldraw((0,1-sqrt(3))--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,purple*0.2+white,blue); 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)); [/asy] We first find the important angles in the figure. We note that 2 adjacent hexagons are rotated $90^\circ$ with respect to the other, so the angles between any sides is $150^\circ$. In particular, as the purple triangles are isosceles, they have angles $150^\circ,15^\circ$, and $15^\circ$, and the octagon is equiangular (all its angles are $135^\circ$). 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 $\sqrt 3$, so the length of a side of the square is $2\sqrt 3-1$. In particular, the horizontal/vertical sides of the octagon have length $1$, so the legs of the isosceles triangles are \[\frac{2\sqrt3-1-1}2=\sqrt3-1\]Thus, the area of the octagon is \[(2\sqrt3-1)^2-4\cdot\frac 12(\sqrt3-1)^2=5\]Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is $(\sqrt 3-1)\sqrt 2$, so the area is \[\frac 14\mathrm{base length}^2\cdot\tan(\mathrm{base angle})=\frac 14((\sqrt3-1)\sqrt2)^2\cdot\tan15^\circ=\frac 14(8-4\sqrt3)(2-\sqrt3)=7-4\sqrt 3\]Thus, the area of the dodecagon is \[5-4(7-4\sqrt3)=16\sqrt3-23\]Thus the answer is $16+3-23=-4$, or $\boxed{\textbf{(B)}}$.

~cr. naman12

Video Solution

https://youtu.be/QYclqXWnxxE

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2022 AMC 12B (Problems • Answer Key • Resources)
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