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MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3

C. B. F. A. E. D. MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3. 4. 5. 7. 6. 1. 3. 2. J. K. E. D. 7. 10. F. B. A. H. C. 5. G. Polar 1 = [EFG] 2 = [DFH] 3 = [DEK] 4 = [BCG] 5 = [ACH] 6 = [ABK] 7 = [ADJ] 8 = [BEJ] 9 = [CFJ]

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MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3

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  1. C B F A E D MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3

  2. 4 5 7 6 1 3 2

  3. J K E D 7 10 F B A H C 5 G Polar 1 = [EFG] 2 = [DFH] 3 = [DEK] 4 = [BCG] 5 = [ACH] 6 = [ABK] 7 = [ADJ] 8 = [BEJ] 9 = [CFJ] 10 = [GHK] Pole A B C D E F G H K J

  4. Here is the dual of the Solomon’s Seal geometry depicted on the preceding page. Note there are only six points in this geometry. As in graph theory, the other places where it looks like two lines intersect are not actually points in the space.

  5. Listed below are the 13 lines with initial point assignments as suggested by the discussion on the preceding page. Here are the 78 possible point pairs, and I have crossed out the ones which have already been assigned. EF EG EH EI EJ EK EL EM HI HJ HK HL HM GH GI GJ GK GL GM CD CE CF CG CH CI CJ CK CL CM FG FH FI FJ FK FL FM DE DF DG DH DI DJ DK DL DM AB AC AD AE AF AG AH AI AJ AK AL AM BC BD BE BF BG BH BI BJ BK BL BM 1 = [ABCD] 2 = [AEFG] 3 = [AH I J] 4 = [AKLM] 5 = [BE x x] 6 = [BF x x] 7 = [BG x x] 8 = [CE x x] 9 = [CF x x] 10 = [CG x x] 11 = [DE x x] 12 = [DF x x] 13 = [DG x x] IJ IK IL IM JK JL JM KL KM LM Once I obtain my final solution I can double-check, verifying that each point is on exactly 4 lines: Notice in the “Sudoku” puzzle above I used the row EFG from line 2 and ran it down three times in the column for lines 5 through 13. I could try that again using HIJ and KLM from lines 3 and 4, but once I but those in lines 5 through 7 I won’t be able to put them in the same order in the remaining six lines. However, I could permute them, say IJH for lines 8, 9, 10 and JHI for lines 11, 12, 13. Then all that remains is to determine the permutations of KLM in the last six lines. Here is my final solution: 1 = [ABCD] 2 = [AEFG] 3 = [AH I J] 4 = [AKLM] 5 = [BEHK] 6 = [BF I L] 7 = [BGJM] 8 = [CE I M] 9 = [CFJK] 10 = [CGHL] 11 = [DEJL] 12 = [DFHM] 13 = [DG I K] You can also see in the above list that each of the numbers 1 through 13 is used exactly 4 times, and the pattern of numbers given above is the same as the pattern of letters listed at left. Of course, this makes sense because this geometry is self-dual. Now what does this geometry look like? Proceed to the next page and get preparedto be blown away . . .

  6. A B B A C C D D I E G H L J K M F PG(2,3) It took me about an hour of searching the internet, but I finally found a graph of PG(2,3) which I have reproduced here. The graph I found looks nicer than mine, but it is in black and white, while I have color-coded my graph to make it easier to see all the lines. If I can find that website again I will post the link to it on our website. It this graph, the red circles are supposed to both represent the same point A, the green circles both represent the point B, and so on. The black line around the outside is the line 1 = [ABCD]. The other three lines through A (2, 3, 4) have been colored in red, the lines through B have been colored in green, and the other lines are similarly color coded. I used different shades of blue and yellow/orange for the lines through C and D so you can tell which line is which. Note some of these are actually loops in the graph, for example the line 10 = [GCHL].

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