On Parametrizing Polygons, 3
Dec. 10th, 2004 06:06 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowNow let's do something maybe a little more interesting.
It's been known for millennia that the medians of a triangle - the lines connecting each vertex to the midpoint of the opposite side - intersect in a single point, the centroid of the triangle, and that the centroid divides the median in a 2:1 ratio. For a little more than a century, the following has also been known: if ABC is a triangle, it's possible to draw another triangle DEF whose sides are parallel to, and the same length as, the medians of ABC, and if you repeat this construction on DEF the new triangle will be similar to ABC.
Now, if you have a polygon with an odd number of sides, you can still define medians, but usually they won't intersect in a single point. It's also true that you can create a new polygon from the medians (as DEF is created from ABC above), but the second half doesn't work - if you repeat the procedure, there's no obvious relation between the shape of the third polygon and that of the first. However, if your polygon happens to be class 1, the medians will intersect in a single point, again called the centroid. The centroid divides the medians in a fixed ratio - not 2:1, but a ratio depending only on the number of sides. And, finally, if you apply the median construction to a class 1 polygon, the result is again class 1, and doing it twice, again, yields a polygon similar to the original.
How does this relate to my parametrization? If G is a class 1 polygon with an odd number of sides and G' is the polygon constructed from its medians, then G and G' have the same eccentricity, and the symmetry of G' is the negative of the symmetry of G. In other words, the median construction has the effect of flipping the parameter space over the e-axis. In particular, this opens the door to figuring out when G and G' are similar; this happens precisely when the symmetry of G is 0. Computation reveals that this happens when, and only when, the square of one side of G is the average of the squares of the other sides. (I find this condition aesthetically pleasing, for some reason.)
Let's turn to another construction, which can be applied to any polygon. If you start with any polygon and mark the midpoints of the sides, those midpoints are the vertices of another polygon, which I call the midgon. Once again, in general there is no predictable relationship between a polygon and its midgon - unless, that is, the polygon is class 1. There are (as you might expect) three cases.
First, suppose the number of sides is odd. This isn't very interesting; the fact that the centroid divides the medians in a constant ratio implies that the midgon is automatically similar to the original polygon.
If the number of sides is even, then it turns out that the polygon and its midgon have the same eccentricity and (surprise!) their symmetries are the negatives of each other - just as happened with the median construction. Again, we can conclude that a class 1 polygon with an even number of sides is similar to its midgon if and only if its symmetry is 0. There are two possibilities, depending on whether n is a multiple of 4, or, equivalently, whether m=n/2 is even.
If m is odd, something very nice happens. Number the vertices, moving (say) clockwise. The even-numbered vertices form a polygon with m sides, and it turns out to be class 1. The odd-numbered vertices also form a class 1 polygon. In fact, these two polygons are simply the reflection of each other in the centroid. (When the number of sides is even, the centroid is the intersection of the diagonals connecting pairs of opposite vertices. Usually these diagonals don't intersect in a point, but if the polygon is class 1 they do.) Furthermore, the centroid of the original polygon is also the centroid of both smaller polygons. (So, if you want to construct, say, a class 1 hexagon, there's an easy way. Take any triangle, find its centroid, and reflect the triangle in the centroid. The two triangles form something like a distorted Star of David, and if you connect the six vertices you get a class 1 hexagon.) On top of all this, the parameters e and s of the original polygon are equal to those of the smaller polygons!
On the other hand, if, instead of taking every other vertex you take every other side, extending them until they intersect, you'll get another pair of m-sided polygons, with the original polygon as their intersection. Again, each is the reflection of the other in the centroid (which is the centroid of all three). This time, the outer polygons have the same eccentricity as the original, but their symmetry is the negative of the symmetry of the original. (For hexagons, make that distorted Star of David again, but this time look at the hexagon in the center.)
Now here's what happens. The outer polygons are similar to the median polygons of the inner ones, and vice versa. The midgon of the original polygon swaps the two - its inner polygons are similar to the outer polygons of the original, and vice versa. What this all works out to is this: if n is even, but not a multiple of 4, then a class 1 polygon with n sides is similar to its midgon if and only if (ta-da!) the square of one side is the average of the squares of the others.
The remaining case is when n is a multiple of 4. Sadly, there doesn't seem to be a nice interpretation of symmetry 0. The best I've been able to come up with has to do with the relative lengths of the sides and diagonals emanating from some one vertex. In the particular case n=4, though, it's fairly simple. A parallelogram is similar to its midgon if and only if one diagonal is the square root of two times one of the sides (and if that happens, the other diagonal is the square root of two times the other side).
I'll post on this one more time, to talk about triangles.
It's been known for millennia that the medians of a triangle - the lines connecting each vertex to the midpoint of the opposite side - intersect in a single point, the centroid of the triangle, and that the centroid divides the median in a 2:1 ratio. For a little more than a century, the following has also been known: if ABC is a triangle, it's possible to draw another triangle DEF whose sides are parallel to, and the same length as, the medians of ABC, and if you repeat this construction on DEF the new triangle will be similar to ABC.
Now, if you have a polygon with an odd number of sides, you can still define medians, but usually they won't intersect in a single point. It's also true that you can create a new polygon from the medians (as DEF is created from ABC above), but the second half doesn't work - if you repeat the procedure, there's no obvious relation between the shape of the third polygon and that of the first. However, if your polygon happens to be class 1, the medians will intersect in a single point, again called the centroid. The centroid divides the medians in a fixed ratio - not 2:1, but a ratio depending only on the number of sides. And, finally, if you apply the median construction to a class 1 polygon, the result is again class 1, and doing it twice, again, yields a polygon similar to the original.
How does this relate to my parametrization? If G is a class 1 polygon with an odd number of sides and G' is the polygon constructed from its medians, then G and G' have the same eccentricity, and the symmetry of G' is the negative of the symmetry of G. In other words, the median construction has the effect of flipping the parameter space over the e-axis. In particular, this opens the door to figuring out when G and G' are similar; this happens precisely when the symmetry of G is 0. Computation reveals that this happens when, and only when, the square of one side of G is the average of the squares of the other sides. (I find this condition aesthetically pleasing, for some reason.)
Let's turn to another construction, which can be applied to any polygon. If you start with any polygon and mark the midpoints of the sides, those midpoints are the vertices of another polygon, which I call the midgon. Once again, in general there is no predictable relationship between a polygon and its midgon - unless, that is, the polygon is class 1. There are (as you might expect) three cases.
First, suppose the number of sides is odd. This isn't very interesting; the fact that the centroid divides the medians in a constant ratio implies that the midgon is automatically similar to the original polygon.
If the number of sides is even, then it turns out that the polygon and its midgon have the same eccentricity and (surprise!) their symmetries are the negatives of each other - just as happened with the median construction. Again, we can conclude that a class 1 polygon with an even number of sides is similar to its midgon if and only if its symmetry is 0. There are two possibilities, depending on whether n is a multiple of 4, or, equivalently, whether m=n/2 is even.
If m is odd, something very nice happens. Number the vertices, moving (say) clockwise. The even-numbered vertices form a polygon with m sides, and it turns out to be class 1. The odd-numbered vertices also form a class 1 polygon. In fact, these two polygons are simply the reflection of each other in the centroid. (When the number of sides is even, the centroid is the intersection of the diagonals connecting pairs of opposite vertices. Usually these diagonals don't intersect in a point, but if the polygon is class 1 they do.) Furthermore, the centroid of the original polygon is also the centroid of both smaller polygons. (So, if you want to construct, say, a class 1 hexagon, there's an easy way. Take any triangle, find its centroid, and reflect the triangle in the centroid. The two triangles form something like a distorted Star of David, and if you connect the six vertices you get a class 1 hexagon.) On top of all this, the parameters e and s of the original polygon are equal to those of the smaller polygons!
On the other hand, if, instead of taking every other vertex you take every other side, extending them until they intersect, you'll get another pair of m-sided polygons, with the original polygon as their intersection. Again, each is the reflection of the other in the centroid (which is the centroid of all three). This time, the outer polygons have the same eccentricity as the original, but their symmetry is the negative of the symmetry of the original. (For hexagons, make that distorted Star of David again, but this time look at the hexagon in the center.)
Now here's what happens. The outer polygons are similar to the median polygons of the inner ones, and vice versa. The midgon of the original polygon swaps the two - its inner polygons are similar to the outer polygons of the original, and vice versa. What this all works out to is this: if n is even, but not a multiple of 4, then a class 1 polygon with n sides is similar to its midgon if and only if (ta-da!) the square of one side is the average of the squares of the others.
The remaining case is when n is a multiple of 4. Sadly, there doesn't seem to be a nice interpretation of symmetry 0. The best I've been able to come up with has to do with the relative lengths of the sides and diagonals emanating from some one vertex. In the particular case n=4, though, it's fairly simple. A parallelogram is similar to its midgon if and only if one diagonal is the square root of two times one of the sides (and if that happens, the other diagonal is the square root of two times the other side).
I'll post on this one more time, to talk about triangles.
