Christmas Tree Geometry

Let $[P]$ denote the area of $P$ .

For similar triangles whose dimensions have ratio $r$ , we know that their areas have ratio $r^{2}$ . Therefore we can conclude that $[T_{n}]=\frac{3}{4}[T_{n-1}]$ for $n=2,3,4$ , or that $[T_{n-1}]=\frac{4}{3}[T_{n}]$ . Working our way up the four triangles gets us that $[T_{4}]=27, [T_{3}]=36, [T_{2}]=48, [T_{1}]=64$ .

When dealing with the overlaps, we can see that the region is really composed of $T_{4}$ , plus the frustums that you get when you chop off $T_{n}$ with the base of $T_{n+1}$ for $n=1,2,3$ .

Since the base of $T_{n+1}$ bisects the height of $T_{n}$ , we know that the area of $T_{n}$ that lies inside $T_{n+1}$ is a triangle similar to $T_{n}$ whose dimensions are $\frac{1}{2}$ of the dimensions of $T_{n}$ . It follows that the area of $T_{n}$ outside $T_{n+1}$ is a frustum of area $[F_{n}]=\frac{3}{4}[T_{n}]$ for $n=1,2,3$ . We get that $[F_{1}]=48, [F_{2}]=36, [F_{3}]=27$ .

The total area is $[F_{1}]+[F_{2}]+[F_{3}]+[T_{4}]=48+36+27+27=\boxed{138}$ .