Scaling a Recursive Exhibition Layout
A gallery builds a fractal display on a square floor of side 1. Stage 1 places a square of side 1/2 at the corner (0,0). Stage 2 adds two squares: one of side 1/4 adjacent along the top edge of the first, and one of side 1/4 along the right edge. Each later stage adds, along each exposed outer edge of the current figure, a square whose side is half the edge it attaches to. The curator claims the total area of all squares added converges to a finite value, and wants to know whether it stays below the floor area of 1. A second constraint: the figure must never overlap itself, so each new square must attach to a strictly outer edge.
1. Model the total area added across all stages. Does the series converge, and to what value?
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Let a_n be the number of new squares at stage n and s_n their side length. Stage 1: 1 square of side 1/2, area 1/4. Stage 2: 2 squares of side 1/4, area 2·(1/16)=1/8. Each stage doubles the count and halves the side, so area per stage is (2^{n-1})·(1/2^n)^2 = 2^{n-1}/4^n = 1/(2^{n+1}). Total = sum_{n>=1} 1/2^{n+1} = (1/4)/(1-1/2)=1/2. So the added area converges to 1/2, comfortably below 1.
2. The curator worries the perimeter grows without bound. Track the total exposed perimeter and decide whether the floor is at risk of overflow.
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At stage n, each new square of side 1/2^n replaces one exposed edge of length 1/2^n with three exposed edges (its outer three sides), a net perimeter gain of 2·(1/2^n) per square. With 2^{n-1} squares added, the perimeter gain at stage n is 2^{n-1}·2/2^n = 1. So the perimeter increases by 1 each stage and diverges linearly, even though area converges. Conclusion: area is bounded (1/2) but perimeter is unbounded - the two measures behave differently, so 'the floor overflows' is not implied by area alone.
3. Suppose the rule changes so each new square has side equal to one-third of its parent edge, with 3 new squares per exposed edge. Recompute convergence of area.
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Now count multiplies by 3 and side by 1/3 each stage. Area per stage: 3^{n-1}·(1/3^n)^2 = 3^{n-1}/9^n = 1/(3^{n+1}). Sum = (1/9)/(1-1/3)=1/6. Converges faster; still finite. The key test is whether (growth factor)·(shrink factor)^2 < 1, i.e. 3·(1/9)=1/3<1.