### Point common to two similar rectangles: What is this about? A Mathematical Droodle

(In the applet below, the eight blue vertices and the blue rectangle are draggable.)

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Explanation  Given a rectangle and its smaller replica (meaning the two rectangles at hand are similar, one smaller, the other bigger.) There's an association between the points of two rectangles, which becomes quite natural if you think of the two as two maps of the same region: a big map and a small map.

Now toss the small rectangle onto the bigger one, so that it lies completely in the interior of the big rectangle. The curious fact is that in the small rectangle there is always a point that happens to be exactly on top of its associated point in the big rectangle.

In general terms, due to Brouwer's Fixed Point theorem, at least one such point should exist. In our case of two similar rectangles, the point is unique and, as such, it displays some properties peculiar to the configuration.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

First, let's assume the rectangles have their associated sides parallel. Then the point in question is the center of similitude -- homothety -- that maps one rectangle onto the other. Indeed, denote the rectangles ABCD and A'B'C'D'. Since we assumed the rectangles to have different sizes, AB ≠ A'B'. Since AB||A'B', AA' and BB' can't be parallel and, therefore, intersect at a finite point. Call it P. Triangles ABP and A'B'P are similar, so that, say, AP/A'P = AB/A'B'. Similarly, let Q be the intersection of AA' and DD'. Then the triangles ADQ and A'D'Q are similar and, in particular, AQ/A'Q = AD/A'D'. From the similarity of the given rectangles we also have AB/A'B' = AD/A'D', which leads to AP/A'P = AQ/A'Q and finally to P = Q.

In this manner we see that P is the point common to all four lines: AA', BB', CC', and DD'. We conclude that the similarity mapping between the two rectangles is exactly the homothety with center P. Note that the center of (a non-trivial) homothety is its only fixed point so that P is exactly the sought point.

Next, we shall consider the case where the associated sides of the rectangles are not parallel. For a pair of sides, say, AB and A'B' we can find a spiral similarity that maps one on the other. As a matter of fact, this transformation rotates the whole of the plane around a point and then dilates the plane radially with the point as the center of homothety.

Since all pairs of associated sides of the two rectangles are inclined towards each other under the same angle, the transformation we just found is exactly the similarity that maps one of the rectangles onto the other. Obviously, we may have started with any other pair of the associated sides. The constructions - the circles involved - would have been different, but the result - the spiral similarity - would have been the same. We thus got four circles associated with two similar triangles that pass through the same point - the center of the spiral similarity that maps one of the rectangles onto the other.

We may also consider two analogous cases of equal rectangles. If their associated sides are parallel, they could be obtained from each other by either translation or a half-turn, i.e. central symmetry. Otherwise, there exists a rotation that maps one onto the other.

### References

1. C. Pritchard, General Introduction, in The Changing Shape of Geometry, edited by C. Pritchard, Cambridge University Press, 2003
2. I. M. Yaglom, Geometric Transformations I, MAA, 1962
3. I. M. Yaglom, Geometric Transformations II, MAA, 1968 