Maths Olympiad Prep

Track / Stage 3 / 191 of 260 #191 of 1964

Problem 191

AMC 10/12, early questions
Geometry Difficulty 3.6 Multiple choice

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 mn+pm \sqrt{n} + p, where mm, nn, and pp are integers and nn is not divisible by the square of any prime.
What is m+n+pm+n+p?

Figure (Asymptote source)
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));

Pick one

Official solution

Refer to the diagram above.
Let the origin be at the center of the square, AA be the intersection of the top and right hexagons, BB be the intersection of the top and left hexagons, and MM and NN be the top points in the diagram.
By symmetry, AA lies on the line y=xy = x. The equation of line ANAN is y=−x3+323−12y = -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 AA by finding the intersection of the two lines:
x=−x3+33−12x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}
x(3+1)=33−12x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}
x=33−12⋅13+1x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}
=33−12(3+1)⋅3−13−1= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}
=10−434= \frac{10 - 4\sqrt{3}}{4}
=52−3= \frac{5}{2} - \sqrt{3}
∴A=(52−3,52−3).\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).
This means that we can find the length ABAB, which is equal to 2(52−3)=(5−232(\frac{5}{2} - \sqrt{3}) = (5 - 2\sqrt{3}. We will next find the area of trapezoid ABMNABMN. The lengths of the bases are 11 and 5−235 - 2\sqrt{3}, and the height is equal to the yy-coordinate of MM minus the yy-coordinate of AA. The height of the hexagon is 3\sqrt{3} and the bottom of the hexagon lies on the line y=12y = \frac{1}{2}. Thus, the yy-coordinate of MM is 3−12\sqrt{3} - \frac{1}{2}, and the height is 23−32\sqrt{3} - 3. We can now find the area of the trapezoid:
[ABMN]=(23−3)(1+5−232)[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)
=(23−3)(3−3)= (2\sqrt{3} - 3)(3 - \sqrt{3})
=63+33−9−6= 6\sqrt{3} + 3\sqrt{3} - 9 - 6
=93−15.= 9\sqrt{3} - 15.
The total area of the figure is the area of a square with side length ABAB plus four times the area of this trapezoid:
Area=(5−23)2+4(93−15)\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)
=37−203+363−60= 37 - 20\sqrt{3} + 36\sqrt{3} - 60
=163−23.= 16\sqrt{3} - 23.
Our answer is 16+3−23=(B) −416 + 3 - 23 = \boxed{\textbf{(B) }-4}.
~mathboy100

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.