03-06-2009, 09:46 PM | #23 |
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03-08-2009, 04:05 PM | #25 |
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What I did was I labeled all of the intersections of the lines. From there I envisioned triangles from any three corners where the sides from all points met. For instance, if the outside corners were labeled A-B-C that's one triangle. If the three lines intersecting the line A-B are labeled B1, B2, and B3 then you can have a triangle of A-B1-C, A-B2-C, A-B3-C, B1-B2-C, B1-B3-C, B1-B-C, B2-B3-C, B2-B-C, and B3-B-C. That's nine just on that one side of A-B-C. You get the same on with the line A-C. Then there are those made up of the nine points in the middle. Hope that illuminates the problem.
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03-08-2009, 04:25 PM | #27 |
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03-08-2009, 06:12 PM | #29 |
you know he kills little girls like you
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03-08-2009, 07:10 PM | #30 |
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Maybe you're right. I was just trying to eye it. JH Valley, is there a answer to this?
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03-08-2009, 07:35 PM | #31 |
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03-08-2009, 09:19 PM | #33 |
you know he kills little girls like you
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03-08-2009, 09:49 PM | #34 |
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69 without a doubt.
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03-09-2009, 01:21 AM | #36 |
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i got 24...will count again
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03-09-2009, 01:47 AM | #37 |
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03-09-2009, 10:17 AM | #38 |
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Thanks for the mathmatical formula, I knew there was one math genius on here. Hey I did not get it right the first time but I did not know the formula.
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03-11-2009, 03:45 AM | #40 |
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thats what i got
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03-11-2009, 04:05 AM | #44 |
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40.
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