AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 14 Medium

Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of \overline{AB}. One side of rectangle D is tangent to both circles A and B, and its opposite side is tangent to circle C. The other two sides are tangent to circles A and B, respectively. What is the area of the part that is inside the rectangle but not inside the circle?

  • A.

    12-3\pi

  • B.

    16-3\pi

  • C.

    10-2\pi

  • D.

    14-2\pi

  • E.

    10-3\pi

Answer:C

By the principle of inclusion-exclusion, it can be deduced that the desired area is equal to the area of the rectangle D minus the sum of the areas of the three circles A,B, and C, plus the overlapping area between circle C and circles A and B.

Then, we can compute the shaded area as the area of half of C plus the area of the rectangle minus the area of the two sectors created by A and B. This is \dfrac{\pi(1)^{2}}{2}+(2)(1)-2 \cdot \dfrac{\pi(1)^{2}}{4}=2.

The area of the rectangle is 4\times 3=12, so the area is 12-2\pi-2=10-2\pi.