First Fruits
Jul. 13th, 2019 05:38 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowIt may well be that this is a known fact, and it would probably be easy to prove with a little linear algebra. [ETA: Yes, it is easy.] Still, I didn't know it, until my investigation of G-gons began.
Let ABCD and PQRS be two quadrilaterals in the same plane. Let A', B', C', D' be the midpoints of AP, BQ, CR, DS, and let A", B", C", D" be the midpoints of AR, BS, CP, DQ. Then A'B'C'D' is a parallelogram if and only if A"B"C"D" is.
(A, B, C, D, P, Q, R, S are the vertices of a (4,2)-gon. Looking for classes of such objects, it turns out to be natural to look at the properties of A'B'C'D' and of A"B"C"D". Doing some calculations, I was surprised to find that "being a parallelogram" for each of them gave rise to the same class of (4,2)-gons. This is what I found among the simplest classes; there are a bunch of slightly more complicated classes that await investigation.)
Let ABCD and PQRS be two quadrilaterals in the same plane. Let A', B', C', D' be the midpoints of AP, BQ, CR, DS, and let A", B", C", D" be the midpoints of AR, BS, CP, DQ. Then A'B'C'D' is a parallelogram if and only if A"B"C"D" is.
(A, B, C, D, P, Q, R, S are the vertices of a (4,2)-gon. Looking for classes of such objects, it turns out to be natural to look at the properties of A'B'C'D' and of A"B"C"D". Doing some calculations, I was surprised to find that "being a parallelogram" for each of them gave rise to the same class of (4,2)-gons. This is what I found among the simplest classes; there are a bunch of slightly more complicated classes that await investigation.)
