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Show all threads Hide all threads Show all messages Hide all messages | WA3 | Zergatul | 2101. Knight's Shield | 12 May 2021 01:10 | 2 | WA3 Zergatul 28 Nov 2020 02:48 Try this test: =========================== -10 -10 10 5 5 10 0 0 --- 42.8571428571 5 | exact solution | ASK | 2101. Knight's Shield | 13 Mar 2018 02:44 | 1 | Possible with doubles, but quite complicated. Easy exact solution is possible with rational numbers (fractions), e.g., in the second case it is 390/43. | Why Wrong answer #5? | Qudrat(TUIT Urgench) | 2101. Knight's Shield | 12 Mar 2018 17:46 | 4 | What is maximum number of the rectangles? According to my idea this is equal 6. I found the surfaces of these rectangles but Wrong answer #5. Edited by author 20.11.2016 19:45 I think that, maximum number of the rectangle is 9. The rectangle has four vertexes, so for each rectangle there is a triangle side that holds two vertexes. It means the opposite side of the rectangle is parallel to that side of the triangle. The hole can be on that (parallel) side of rectangle or on the perpendicular one, thus for each side of the triangle there are at most two rectangles, that is six in total. | wrong ? | Vadim | 2101. Knight's Shield | 20 Nov 2016 11:57 | 2 | Nope. The only inscribed rectangle with area=50 doesn't contain hole. Note, that there are only 4 possible rectangles in the solution. |
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