| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3773 |
| 3 | Radewoosh | 3646 |
| 4 | ecnerwala | 3624 |
| 5 | jqdai0815 | 3620 |
| 5 | Benq | 3620 |
| 7 | orzdevinwang | 3612 |
| 8 | Geothermal | 3569 |
| 8 | cnnfls_csy | 3569 |
| 10 | Um_nik | 3396 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Um_nik | 163 |
| 2 | cry | 161 |
| 3 | maomao90 | 160 |
| 4 | -is-this-fft- | 159 |
| 5 | awoo | 158 |
| 6 | atcoder_official | 157 |
| 7 | adamant | 155 |
| 8 | nor | 154 |
| 9 | maroonrk | 152 |
| 10 | Dominater069 | 148 |
given the end points of the largest parallel side and the lengths of other three sides of a trapezium how to find other two coordinate points of that trapezium? 
if point A, point B, b,c,d are given how to find point C and D?
Edit: Thnx everyone solved it finding circle-circle intersection points
Auto comment: topic has been updated by travelers (previous revision, new revision, compare).
How about using ternary-search technique? Try to search x-coordinate of point D, then you can find the y-coordinate of point D. Afterwards, we can find the point C (its just C.x = D.x + c, C.y = D.y). Finally, you can find the length of segment CB. Depending on wether CB is greater than b or less, you can calibrate your ternary-search bounds. Hope it helps.
UPD: missed that quadrilateral should have two parallel sides, this comment is irrelevant.
There is no unique solution. One can "skew" trapezium to either left or right while preserving lengths of all sides. You need some extra information.Perhaps he is talking about trapezoid as he mentioned parallel side? (From what I found on Wikipedia, it is just about different English styles)
Makes perfect sense. I'm sorry for missing the point.
Say x is the length of AD projection onto AB, y — the length of CB projection onto AB, h — the height of trapezoid. Then there are three equations:
Auto comment: topic has been updated by travelers (previous revision, new revision, compare).