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

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\textbf{(D) }24 \qquad
 
\textbf{(D) }24 \qquad
 
\textbf{(E) }32</math>
 
\textbf{(E) }32</math>
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 +
==Solution==
 +
 +
We calculate the area as the area of the red octagon minus the four purple congruent triangles:
 +
<asy>
 +
        import geometry;
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        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);
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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);
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        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 <math>90^\circ</math> with respect to the other, so the angles between any sides is <math>150^\circ</math>. In particular, as the purple triangles are isosceles, they have angles <math>150^\circ,15^\circ</math>, and <math>15^\circ</math>, and the octagon is equiangular (all its angles are <math>135^\circ</math>). 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 <math>\sqrt 3</math>, so the length of a side of the square is <math>2\sqrt 3-1</math>. In particular, the horizontal/vertical sides of the octagon have length <math>1</math>, so the legs of the isosceles triangles are
 +
<cmath>\frac{2\sqrt3-1-1}2=\sqrt3-1</cmath>Thus, the area of the octagon is
 +
<cmath>(2\sqrt3-1)^2-4\cdot\frac 12(\sqrt3-1)^2=5</cmath>Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is <math>(\sqrt 3-1)\sqrt 2</math>, so the area is
 +
<cmath>\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</cmath>Thus, the area of the dodecagon is
 +
<cmath>5-4(7-4\sqrt3)=16\sqrt3-23</cmath>Thus the answer is <math>16+3-23=-4</math>, or <math>\boxed{\textbf{(B)}}</math>.
 +
 +
~cr. naman12
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:21, 19 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 $m+n+p$?

$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$

Solution

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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