Tunnel Vision

Geometry Level 5

Inside a unit square, there is an infinite series of squares, each with sides $\frac{3}{4}$ of the sides of the square one size larger. Their positions are at the top, on the right, at the bottom, on the left, at the top again, and so on, centered on the side they are on. As the size of the squares goes to zero, their location converges to a point $(x, y)$ , where the origin of the coordinate system is the lower left corner of the original square.

If the ratio $\dfrac xy$ can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, submit your answer as $a+b$ .

The answer is 57.

2 solutions

Brian Moehring
Mar 20, 2017

Let $z_0=x+iy$ denote the point in question on the Argand plane. Then we consider the following composition of maps:

$z \mapsto z-\frac{1}{8}-i$ : Moves the top-left corner of the first green square to the bottom left of the original yellow square.
$z \mapsto \frac{4}{3}z$ : Stretches the green square to a unit square (which now sits below the yellow)
$z \mapsto iz$ : Rotates the green square to overlay the yellow square so that the new orange square sits on top of the old green square, the orange square sits on top of the old grey square, etc.

In particular, this means that the composition of these maps, if we restrict our attention to the unit square and ignore colors, fixes the image. Therefore, the point $z_0$ we want to find is fixed by the composition of these maps, giving the equation: $z_0 = \frac{4}{3}i\left(z_0-\frac{1}{8}-i\right)$

Solving for $z_0$ yields $z_0 = \frac{14}{25} + \frac{29}{50}i$ and $\frac{x}{y} = \frac{14/25}{29/50} = \frac{28}{29}$

I get the maps, the fact that they fix the image, the fact that $z_0$ is fixed, and the solution. It all works. I still have a problem though. It fixes the entire image, not just $z_0$ , so how come the solution gives us just that one point?

Marta Reece - 4 years, 2 months ago

I may be using the wrong term there saying the image is "fixed".

Rather, I just mean to use the symmetry of self-similarity. This is analogous to how a square has 90 degree rotational symmetry, but only the centroid of the square is fixed by this rotation.

More explicitly, note that $\frac{4}{3}i\left(z-\frac{1}{8}-i\right) = \frac{4}{3}i\left(z-z_0\right)+z_0$

This shows that the map I gave can be rewritten as a 90 degree rotation about $z_0$ followed by a stretch about the point $z_0$ , so in this context, it obviously is the unique fixed point.

Brian Moehring - 4 years, 2 months ago

Marta Reece
Mar 20, 2017

Let's track the position of the center as it changes from one square to the next smaller one and so on. Both coordinates start at $\frac{1}{2}$ . $y$ is the first to increase, by the difference between $\frac{1}{2}$ and $\frac{3}{4}$ of $\frac{1}{2}$ , that is by $\frac{1}{8}$ . Next only $x$ changes. By the time it's $y$ 's turn, the starting square is only $(\frac{3}{4})^2=\frac{9}{16}$ of the original. The change is a decrease and it is $\frac{1}{8}$ of this figure. The sequence for $y$ is:

$y=\frac{1}{2}+\frac{1}{8}\times\sum_{n=0}^{\infty}(-\frac{9}{16})^n=\frac{1}{2}+\frac{1}{8}\times\frac{16}{25}=\frac{29}{50}$

The sequence of locations for $x$ starts with the same $\frac{1}{2}$ but the additions and subtractions are all $\frac{3}{4}$ of those for $y$ .

$x=\frac{1}{2}+\frac{1}{8}\times \frac{3}{4}\times\sum_{n=0}^{\infty}(-\frac{9}{16})^n=\frac{1}{2}+\frac{1}{8}\times \frac{3}{4}\times\frac{16}{25}=\frac{14}{25}$ .

So the ratio $\frac{x}{y}$ goes to $\frac{28}{29}$ .

Tunnel Vision

The answer is 57.

2 solutions

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