Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

From the point (x,y)(x, y), a legal move is a move to (x3+u,y3+v)\left(\frac{x}{3}+u, \frac{y}{3}+v\right), where uu and vv are real numbers such that u2+v21u^{2}+v^{2} \leq 1. What is the area of the set of points that can be reached from (0,0)(0,0) in a finite number of legal moves?

A number or a short expression. Spacing and $ signs are ignored.

Solution

We claim that the set of points is the disc with radius 32\frac{3}{2} centered at the origin, which clearly has area 9π4\frac{9 \pi}{4}. First, we show that the set is contained in this disc. This is because if we are currently at a distance of rr from the origin, then we can't end up at a distance of greater than r3+1\frac{r}{3}+1 from the origin after a single move. Since r3+1<32\frac{r}{3}+1<\frac{3}{2} if r<32r<\frac{3}{2}, we will always end up in the disc of radius 32\frac{3}{2} if we start in it. Since the origin is inside this disc, any finite number of moves will leave us inside this disc. Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, after one move we can get all points within a distance of 1. After two moves, we can get all points within a distance of 43\frac{4}{3}. After three moves, we can get all points within a distance of 139\frac{13}{9}. In general, after nn moves we can get all points within a distance of 32123k1\frac{3}{2}-\frac{1}{2 \cdot 3^{k-1}}. This means that for any distance d<32d<\frac{3}{2}, we will eventually get all points within a distance of dd, so all points in the disc of radius 32\frac{3}{2} can be reached after some number of moves.

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