2020 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 14 Medium

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region - inside the hexagon but outside all of the semicircles?(2020 AMC 10B, Question #14)

  • A.

    6 \sqrt{3}-3 \pi

  • B.

    \frac{9 \sqrt{3}}{2}-2 \pi

  • C.

    \frac{3 \sqrt{3}}{2}-\frac{\pi}{3}

  • D.

    3 \sqrt{3}-\pi

  • E.

    \frac{9 \sqrt{3}}{2}-\pi

Answer:D

Solution 1:

Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, B C=1, since B is the center of the semicircle with radius 1 that C lies on, A B=1, since B is the center of the semicircle with radius 1 that A lies on, and \angle B A C=60^{\circ}, as a regular hexagon has angles of 120 , and \angle B A C is half of any angle in this hexagon. Now, using the sine law, \frac{1}{\sin \angle A C B}=\frac{1}{\sin 60^{\circ}}, so \angle A C B=60^{\circ}. Since the angles in a triangle sum to 180 \circ, \angle A B C is also 60 \circ. Therefore, \triangle A B C is an equilateral triangle with side lengths of 1 .Since the area of a regular hexagon can be found with the formula \frac{3 \sqrt{3} s^{2}}{2}, where S is the side length of the hexagon, the area of this hexagon \frac{3 \sqrt{3}\left(2^{2}\right)}{2}=6 \sqrt{3}. is 2 . Since the area of an equilateral triangle can be found with the formula \frac{\sqrt{3}}{4} s^{2}, where S is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is \frac{\sqrt{3}}{4}\left(1^{2}\right)=\frac{\sqrt{3}}{4}. Since the area of a circle can be found with the formula \pi r^{2}, the area of a sixth of a circle with radius 1 is \frac{\pi\left(1^{2}\right)}{6}=\frac{\pi}{6}. In each sixth of the hexagon, there of a circle with radius 1 is 6 \quad \frac{6}{6}. In each sixth of the hexagon, there sixth of a circle with radius 1 colored white, with an area of \overline{6}. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is 2\left(\frac{\sqrt{3}}{4}\right)+\frac{\pi}{6}, which equals \frac{\sqrt{3}}{2}+\frac{\pi}{6}, and the total area colored white is 6\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right), which equals 3 \sqrt{3}+\pi. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is 6 \sqrt{3}-(3 \sqrt{3}+\pi), which equals (D) 3 \sqrt{3}-\pi

Solution 2:

First, subdivide the hexagon into 24 equilateral triangles with side length 1:

Now note that the entire shaded region is just 6 times this part:

The entire rhombus is just 2 equilatrial triangles with side lengths of 1 , so it has an area of: 2 \cdot \frac{\sqrt{3}}{1}=\frac{\sqrt{3}}{2} \text { area of: } \frac{1}{6} \cdot \pi \cdot 1^{2}=\frac{\pi}{6} Hence, the area of the shaded region in that section is \frac{\sqrt{3}}{2}-\frac{\pi}{6} For a final area 6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3 \sqrt{3}-\pi \Rightarrow(\text{D})

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