i have tried so many times on this problem:
Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{4}{3}$ 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 $\frac{A}{[XYZ]}.$
i have so many wrong answers:
4/9
11/27
20/27
73/200
73/100
5/9
17/27
please help T^T
tysm!!
HOLY C**P--- THANK YOU SO MUCH!! I was able to understand the explanation that the website gave me ^^
tysm!! you're a lifesaver :))
+118687
I let GF=1 (just for ease)
CX=2 (30,60,90 triangle)
This means that the sides of the original triangle are \(2\sqrt3\;units\)
So the area of the original triangle is
\(Area\;XYZ = \frac{1}{2}*2\sqrt3*2\sqrt3*sin60\\ Area \;XYZ= 6*\frac{\sqrt3}{2}\\Area \;XYZ= 3\sqrt3\;\;u^2\)
CG=4/3
So GX=2 - 4/3 = 2/3
\(tan60=\frac{XG}{GT}\\ \sqrt3=\frac{2}{3*GT}\\ GT=\frac{2}{3\sqrt3}\\ \)
\(\text{Area of }\triangle STX = GT*GX = \frac{2}{3\sqrt3}*\frac{2}{3}\\~\\ \text{Area of }\triangle STX = \frac{4}{9\sqrt3}\\~\\ \text{Area of intersection}=\triangle XYZ-3*\triangle XST\)
\(Intersection\; area = 3\sqrt3-3*\frac{4\sqrt3}{27}=\frac{27\sqrt3-4\sqrt3}{9}=\frac{23\sqrt3}{9}\\~\\\)
\(\displaystyle \frac{\text{intersection area}}{\text{area of original triangle XYZ}}=\frac{23\sqrt3}{9}\times \frac{1}{3\sqrt3}=\frac{23}{27} \)
