Question

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 $\frac{A}{[XYZ]}.$

Answer 1

The area divided by XYZ is 16/9

i.e. $\frac{A}{[XYZ]} = \frac{16}{9}$

The given parameters can be represented as:

$r = -\frac{4}{3}$ --- the ratio of dilation

The area of the second triangle is as follows:

$A =r^2 * \triangle XYZ$

Make $r^2$ as subject

$r^2 = \frac{A}{\triangle XYZ}$

Substitute value for $r^2$

$(\frac{4}{3})^2 = \frac{A}{\triangle XYZ}$

$\frac{16}{9} = \frac{A}{\triangle XYZ}$

Rewrite as:

$\frac{A}{\triangle XYZ} = \frac{16}{9}$

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