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

Revision as of 21:41, 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 1

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

Solution 2 (Coord Bash)

$[asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); pair O = (0.5, 0.5); label("$O$", O, N); pair A = dot(O); [/asy]$

Let the origin be the center of the square, $0, 0$ .

Video Solution

https://youtu.be/QYclqXWnxxE

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

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