Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.1 AMC 10/12 Find the answer

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$

Figure (Asymptote source)

unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white);

Pick one

Solution

In the diagram above, notice that triangle $OAB$ and triangle $ABC$ are congruent and equilateral with side length $2$ . We can see the radius of the larger circle is $2\overline{OH} + 1$ . Using $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, we know $\overline{OH} = \sqrt{3}$ . Therefore, the radius of the larger circle is $2\sqrt{3}+1$ .
The area of the larger circle is thus $\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\pi$ , and the sum of the areas of the smaller circles is $13\pi$ , so the area of the dark region is $\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$ .

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