Track / Stage 5 / 45 of 400 #645 of 1964

Problem 645

AIME late

Algebra Difficulty 5.0 Find the answer Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO · Italy

The dimensions of a television screen are $60~\mathrm{cm} \times 45~\mathrm{cm}$ . A camera films the entire television, and sends the image back onto the television itself, so that inside this television another one can be seen, and so on. The largest television seen inside the screen has an area equal to half the area of the screen. Assuming that a person watches the television seated at a distance such that they cannot distinguish images with an area smaller than $1~\mathrm{cm}^2$ , how many televisions does he see inside the screen?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution:

The answer is 11. The area of the screen is $60~\mathrm{cm} \times 45~\mathrm{cm} = 2700~\mathrm{cm}^2$ . The largest television framed has an area equal to half the area of the screen. The second one has an area equal to half of the first, that is $1/4 = 1/2^2$ of the area of the screen; similarly, the third television has an area equal to $1/2^3$ , the fourth an area equal to $1/2^4$ of the area of the screen, and so on. If $n$ is the number of televisions distinguishable by the viewer, then we have $2700/2^n \geq 1$ and $2700/2^{n+1} < 1$ . Analyzing the successive powers of 2, we see that $n=11$ is the number sought, since $2^{11} = 2048$ , $2^{12} = 4096$ , so $\frac{2700}{2048} > 1$ while $\frac{2700}{4096} < 1$ .

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