+100 Let G be the center of equilateral triangle XYZ. A dilation centered at G with scale factor \(-\frac34\) 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 \(\frac{A}{[XYZ]}\)
+1632 There were no constraints on where G could be located, so I just placed G on the origin, thus the center of triangle XYZ is (0,0).
First, a dilation with a negative is kind of reflecting it across the origin, but literally nobody cares because it is on the origin already (it will make two triangles on top of each other with a 60 degree angle). It won't change the area. However, 3/4 will be scaling it down that much, so if the original side length was 4s, now the side length is 3s.
If a is the area of the region contained, in both triangles, then it would be a hexagon of length s (assuming the original length of the equilateral triangle was 3s), before scaling.
The area of the equiangular hexagon after scaling would be 2/3 x 81/16 x s^2 x sqrt(3)/4
\(2/3 * 81/16 * s^2 * \sqrt{3}/4\) = \(27s^2\sqrt{3}\over 32\)
AXY has area of \(9s^2\sqrt{3}\over4\)... Remember that the original length of the equilateral triangle was 3s. A/XYZ would be:
\({27s^2\over8}\over9s^2\) (i did some simpifying), continuing gets to 3/8
\({A\over[XYZ]} = {3\over8}\)
