2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 12 Easy

Line segment \overline{AB} is a diameter of a circle with AB=24. Point C, not equal to A or B , lies on the circle. As point C moves around the circle, the centroid (center of mass) of \triangle ABC traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve? (2018 AMC 10B Problem, Question#12)

  • A.

    25

  • B.

    38

  • C.

    50

  • D.

    63

  • E.

    75

Answer:C

Let A=(-12,0), B=(12,0). Therefore, C lies on the circle with equation x^2+y^2=144. Let it have coordinates (x, y). Since we know the centroid of a triangle with vertices with coordinates of (x_1,y_1), (x_2,y_2), (x_3,y_3) is \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right), the centroid of \triangle ABC is \left(\frac{x}{3},\frac{y}{3}\right). Because x^2+y^2=144, we know that \left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16, so the curve is a circle centered at the origin. Therefore, its area is 16\pi\approx\rm (C)~50.

We know the centroid of a triangle splits the medians into segments of ratio 2:1, and the median of the triangle that goes to the center of the circle is the radius (length 12), so the length from the centroid of the triangle to the center of the circle is always \frac{1}{3}\cdot 12=4 . The area of a circle with radius 4 is 16\pi, or around \rm(C)~50.