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Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:

Rectangle $R_{0}$ has sides of lengths $3$ and $4$ . Rectangles $R_{1}$ , $R_{2}$ , and $R_{3}$ are formed such that:
- all four rectangles share a common vertex $P$ ,
- for each $n=1,2,3$ , one side of $R_{n}$ is a diagonal of $R_{n-1}$ ,
- for each $n=1,2,3$ , the opposite side of $R_{n}$ passes through a vertex of $R_{n-1}$ such that the center of $R_{n}$ is located counterclockwise of the center of $R_{n-1}$ with respect to $P$ .

Compute the total area covered by the union of the four rectangles.

Solution

Solution:

Let $ABCD$ be $R_{0}$ such that $\overline{AB}=3$ and $\overline{BC}=4$ . Then, let $\overline{AC}$ be a side length of $R_{1}$ and let the other two vertices be $E$ and $F$ such that $B$ lies on segment $EF$ . Notice that the area of $\triangle ABC$ is both half of the area of $R_{0}$ and half of the area of $R_{1}$ . This means forming $R_{1}$ adds half of the area of $R_{0}$ to the union of rectangles. Similarly, forming $R_{2}$ adds half of the area of $R_{1}$ to the union of all rectangles, and the same for $R_{3}$ . This means the total area of the union of rectangles is given by
$[R_{0}] + \frac{1}{2}[R_{1}] + \frac{1}{2}[R_{2}] + \frac{1}{2}[R_{3}] = [R_{0}] + \frac{1}{2}[R_{0}] + \frac{1}{2}[R_{0}] + \frac{1}{2}[R_{0}] = \frac{5}{2}[R_{0}] = \frac{5}{2}(3 \cdot 4) = 30$
Note that in the above equation, $[X]$ denotes the area of shape $X$ .

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