Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{3}{4}$ is applied to triangle $XYZ,$ to obtain triangle $X'Y'Z'.$ Let $A$ be the area of the region that is contained in both triangles $XYZ$ and $X'Y'Z'.$ Find $\frac{A}{[XYZ]}.$
Since the dilation is centered at G, G is also the center of triangle X'Y'Z'. So, the area of triangle X'Y'Z' is (-3/4)^2 = 9/16 times the area of triangle XYZ. The area of the region that is contained in both triangles is the area of triangle XYZ minus the area of triangle X'Y'Z'. So, A/[XYZ] = 1 - 9/16 = 7/16.
