. A thin uniform equilateral triangle of side length is then added and it has a vertex at the center of the circle and it has the same density as the circular disk. The triangle and the circle overlap partially and exist in the same plane. This area of overlap between the triangle and the circle is then cut out leaving part of the original disk and the additional part of the triangle. The center of mass of this resulting shape is a distance from the center of mass of the original uniform disk. This distance can be expressed in the form where and are integers and is the radius of the uniform circular disk. What is the value of ?
Consider a thin uniform circular disk of radius
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Dissect this geometrical figure into 1) unit circle 2) small equilateral triangle gap 3) 60° circular segment gap, 4) large equilateral triangle of side 2. Let the masses be m1, m2, m3, m4, and their respective centroid distances be x1, x2, x3, x4. (For convenience, we can let x1 = 0). Then the centroid of this geometrical figure is:
(m1x1 - 2m2x2 - 2m3x3 + m4x4) / (m1 - 2m2 - 2m3 + m4) = 4 / (2π + 3√3)
The reason why masses m2 and m3 are subtracted twice is because of the overlap of the circle and the large equilateral triangle; such masses have to be subtracted "not once, but twice".
Finding the centroids of the circle and the triangles are obvious. Meanwhile, the centroid of a circular segment can be found in a list of centroids of common shapes.