No. 14 is only on one line. Well, I am not sure whether I can arrange to make 8, 12, and 4 be on one line. It might need some boring computation, though it looks possible. I guess higher than 13 or 14 is not so interesting. And PostDoc is not really higher than PHD. |
Then my contruction is surely wrong. Maybe you can specify the problem strictly and mathematically. Thanks. Anyway, it is kind of comlpicated. |
Assistant Professor: 15 trees, 12 rows, each with 4 trees Associate Professor: 16 trees, 15 rows, each with 4 trees Full Professor: 18 trees, 18 rows, each with 4 trees |
there should not be five trees in a row in any case, correct? |
16 trees problem, it's 2 five stars overlapped, and put one tree at the center. 15 trees, remove one outer vertex. |
remove any outer vertex to get answer for 15 trees |
I think you mean one must plant a tree at each intersection and any line should contain exactly four trees. The picture I drew satisfies the second condition but not the first. |
1. on any straight line, there can be no more than 4 trees. 2. it's not necessary to plant a tree on the intersection of two straight lines, where each of which contains eactly four trees. your pic for 14 tree solution is not correct due to the point 1 above, not because of point 2 above. |
There are no five trees on one row. Maybe my picture is bad, but no. 14 is on the same line as 4, 12, and 8. Trees 1, 4, and 14 are not on the same row. As I said, I am not sure whether I can make 4, 12, and 6 be on the same row unless I do a careful computation, though it looks possible. |
based on my solution to the 13 -tree-on-9-row problem. Assuming that 8, 12, and 4 are on the same row, my solution should be correct. There are no five trees on any row. |
I think you put point 14 on a wrong place, thus causing the confusion. 14 is supposed to becolinear with 5, 9 13, and 8,12, correct? |
This is for 13 trees. Points ABCDE are random, all the rest are determined by the intersections on AD and DE. you have to be very precise in drawing the picture, since it used a property of similar triangle. For 14 trees problem, any tree on the line of BDC will generate a new row. |
Obviously, 5, 9 13, and 8,12 are not colinear. If 14 were with 5, 9, 13, 4, there is no point to put a new line. Noet that except the line for 14, 4, 8, and 12, the picture is a solution for 13-9. No. 4 is the intersetion of three lines now. |
tree on a single line. |
then your answer is wrong. based on your answer, here are three lines 1-2-3-4, 5-9-13-4, and 8-12-4-14 i don't think you can adjust your picture to make that happen. line 1-2-3 and 5-9-13 don't have to meet on at the same point (4) on the line 8-12-14, actually it will most likely not happen |
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