2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 20 Hard

As shown in the figure, line segment \overline{AD} trisected by points B and C so that AB=BC=CD=2.Three semicircles of radius 1, \overset{\frown}{AEB}, \overset{\frown}{BFC} and \overset{\frown}{CGD}, have their diameters on \overline {AD}, and are tangent to line EG at  E, F, and G, respectively. A circle of radius 2 has its center on F. The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \frac ab\cdot \pi-\sqrt c+d, where a, b, c, and d are positive integers and a and b are relatively prime. What is a+b+c+d? (2019 AMC 10B Problem, Question#20)

  • A.

    13

  • B.

    14

  • C.

    15

  • D.

    16

  • E.

    17

Answer:E

Divide the circle into four parts: the top semicircle \rm (A); the bottom sector \rm (B), whose arc angle is 120^{\circ} because the large circle's radius is 2 and the short length(the radius of the smaller semicircles) is 1, giving a 30^\circ-60^\circ-90^\circ triangle; the triangle formed by the radii of A and the chord \rm (C), and the four parts which are the corners of a circle inscribed in a square \rm (D), Then the area is A+B-C+D (in B-C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the area of the bottom shaded region).

The area of A is \frac{1}{2}\pi \cdot 2^{2}=2 \pi,

The area of B is \frac{120^{\circ}}{360^{\circ}}\pi \cdot 2^{2}= \frac{4 \pi}{3}.

For the area of C, the radius of 2, and the distance of 1 (the smaller semicicles' radius ) to BC.

creates two 30^\circ-60^\circ-90^\circ triangles, so C's area is 2\cdot \frac 12\cdot 1\cdot\sqrt 3=\sqrt 3.

The area of D is 4\cdot 1-\frac 14\pi\cdot 2^2=4-\pi ,

Hence, finding A+B-C+D. the desired area is \frac {7\pi}{3}-\sqrt 3+4, so the answer is 7+3+3+4=\text {(E)}17.