+180 Let G be the center of equilateral triangle ABC A dilation centered at G with scale factor -3/5 is applied to triangle ABC to obtain triangle A'B'C' Let K be the area of the region that is contained in both triangles ABC and A'B'C' Find K/[ABC]
+743
Because the scale factor is negative, triangle A'B'C' is obtained by rotating 180 degrees triangle ABC about point G and then shrinking it. Hence, the overlap between triangles ABC and A'B'C' is the smaller triangle A'BC'. (The region consisting of the intersection of triangle ABC with dilated triangle A'B'C' is actually a kite.) We know that the side lengths of triangle A'B'C' are 3/5 the side lengths of triangle ABC. Since the area of a triangle is proportional to the square of its side lengths, the area of triangle A'B'C' is (3/5)2 times the area of triangle ABC. Therefore, the area of the overlap K is (1−(3/5)2) times the area of triangle ABC. Hence, \begin{align*} \frac{K}{ABC} &= 1-(3/5)^2 \ &= 1 - 9/25 \ &= \boxed{\frac{16}{25}}. \end{align*}
The answer is 16/25.
