Timeline for Equal volume and projections
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2019 at 23:44 | comment | added | STrick | Any inscribed quadrilateral (inscribed in the parallelogram) has orthogonal projections equal to 1. We can easily choose one of them to have area 1. | |
| Feb 28, 2019 at 3:35 | comment | added | erz | but these are orthogonal projection, not parallel to $v_1$ and $v_2$. Orthogonal projections of your parallelogram will be greater than $1$. | |
| Feb 27, 2019 at 19:21 | comment | added | STrick | Thanks for the reply, but I think what you claimed in the end is not true. You can find $K'=P_{span(v_1,v_2)}K$ with $|K'|=1$ and both projections on to $v_1$ and $v_2$ having length equal to 1. Just draw two strips of width one. Their intersection is a parallelogram and has volume bigger than 1. So inside you can find a set with volume 1 and having both projections equal to 1. | |
| Feb 22, 2019 at 3:00 | comment | added | erz | By considering $v_1,v_2,v_3$ biorthogonal to $v_1,v_2,v_3$ you can get rid of the orthogonal complements. Then $P_{v_1}=P_{v_1}P_{span (v_1,v_2)}$, and so $1=|P_{v_1}P_{span (v_1,v_2)}K|$. Analogously, $1=|P_{v_2}P_{span (v_1,v_2)}K|$. Since $|P_{span (v_1,v_2)}K|=1$ this means that the projections of a set of area $1$ on two lines have length $1$, and so these lines have to be orthogonal. Thus, $v_1,v_2,v_3$ is an orthonormal basis, and so the same is true for $u_1,u_2,u_3$. Or am I missing something? | |
| Feb 21, 2019 at 22:06 | history | edited | YCor |
edited tags
|
|
| Feb 21, 2019 at 21:00 | history | asked | STrick | CC BY-SA 4.0 |