Timeline for Distance of a barycentric coordinate from a triangle vertex
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 5, 2013 at 21:28 | answer | added | Guest | timeline score: 2 | |
| Jun 26, 2012 at 18:37 | answer | added | ragnar | timeline score: 1 | |
| Aug 26, 2010 at 6:09 | answer | added | sleepless in beantown | timeline score: 0 | |
| Aug 26, 2010 at 5:23 | answer | added | Aaron Meyerowitz | timeline score: 6 | |
| Aug 25, 2010 at 17:46 | comment | added | Suresh Venkat | not AFAIK. but it looks like you got an answer, so it's all good :) | |
| Aug 25, 2010 at 16:33 | vote | accept | timday | ||
| Aug 25, 2010 at 14:30 | comment | added | timday | Sorry! Was completely unaware of math.stackexchange.com! Suspected I was a bit out of my depth here. Can it be migrated over ? | |
| Aug 25, 2010 at 14:27 | vote | accept | timday | ||
| Aug 25, 2010 at 14:28 | |||||
| Aug 25, 2010 at 14:12 | comment | added | Suresh Venkat | better yet, try it on math.stackexchange.com | |
| Aug 25, 2010 at 13:40 | comment | added | Benoît Kloeckner | This question seems not suitable for MO (not research level, this only uses Pythagore theorem), please read the FAQ. There is a formula, that you can devise by assuming $A$ is at the origin, $B$ is on the $x$-axis. The coordinates of $C$ are then easy to compute in terms of $a,b,c$, than you get the coordinates of your point. The distance is then given by the usual Euclidean formula. | |
| Aug 25, 2010 at 13:38 | answer | added | Robin Chapman | timeline score: 4 | |
| Aug 25, 2010 at 12:44 | history | asked | timday | CC BY-SA 2.5 |