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Math Forum / Mathematics / General Topics / September 2004I found a great FORMULA !!! pls check if corect !!!
Hello all !! In high school we learn about a concept called 'diagonals' in polygons, which is basically lines that are connecting the dots around the polygon. We have this formula: For triangle (all sort, same length, or different side length) we have no diagonals = the sides dont count. For any square or rectangle (same length or different length) we can have two diagonals, like cross in the middle. For five, we have five, like a star. I dunno how to show without pictures :):):) but i hope you can imaginize. for six, we have nine (again, have to draw *wink* *wink* :)) as my high school teacher used to say, i love to play around with formulas !! so i kind of find a pattern so here is the patern (quite complicated but it take days for me too :)) the number of 'diagonals' is the number of sides multiplied with 0,1/2,1,1 1/2 (one and one half), 2, etc...let me explain how that works: like first 0x3=0 (diagonals in triangle) then 1/2x4=2 (diagonals in square), then 1x5=5 (diagonals in five side) and 1 1/2 x 6=9 !!! see it works perfectly :):) can you check if that formula is true ?? suhaimi. > Hello all !! > > In high school we learn about a concept called 'diagonals' in polygons, which is basically lines that are connecting the dots around the polygon. We have this formula: > For triangle (all sort, same length, or different side length) we have no diagonals = the sides dont count. > [quoted text clipped - 6 lines] > > as my high school teacher used to say, i love to play around with formulas !! so i kind of find a pattern so here is the patern (quite complicated but it take days for me too :)) > the number of 'diagonals' is the number of sides multiplied with 0,1/2,1,1 1/2 (one and one half), 2, etc...let me explain how that works: > > like first 0x3=0 (diagonals in triangle) then 1/2x4=2 (diagonals in square), then 1x5=5 (diagonals in five side) and 1 1/2 x 6=9 !!! see it works perfectly :):) > can you check if that formula is true ?? > > suhaimi. Congratulations! Your formula is correct! So... how many diagonals would a 100-sided polygon have? :)  SignatureClive Tooth http://www.clivetooth.dk M. Suhaimi Ramly wrote: > Hello all !! > [quoted text clipped - 15 lines] > > can you check if that formula is true ?? Congratulations! Is is indeed correct. Now you might want to try taking the next step: Can you prove that it is correct? Regards, Rick On Wed, 22 Sep 2004, M. Suhaimi Ramly wrote: > In high school we learn about a concept called 'diagonals' in polygons, > which is basically lines that are connecting the dots around the > polygon. We have this formula: What's a dot? That's not a geometrical term. Were you learning geometry or art? > For triangle (all sort, same length, or different side length) we have > no diagonals = the sides dont count. [quoted text clipped - 3 lines] > > For five, we have five, like a star. I dunno how to show without Given a regular polygon of n sides, there are n(n-3)/2 diagonals. > On Wed, 22 Sep 2004, M. Suhaimi Ramly wrote: > [quoted text clipped - 14 lines] > > > Given a regular polygon of n sides, there are n(n-3)/2 diagonals. Nice reply, William. I particularly enjoyed the total lack of encouragement that your post contained. Impressive. The way you slapped the poster down for using the incomprehensible term "dot" was masterful. And stopping your quote of the original post in the middle of a sentence... just perfect. However, I was a little disappointed that you did not comment on the absence of the apostrophe in the word "dont". SignatureClive Tooth http://www.clivetooth.dk On 22 Sep 2004, M. Suhaimi Ramly wrote: >Hello all !! > [quoted text clipped - 17 lines] > >suhaimi. Hi Suhaimi, Try to visualise this: Take a (regular!) polygon with n points. Then from each point we can draw N-3 lines to other points. This is because we don't draw a line from the point to itself, and the lines from the point to two neighbours are already drawn by the polygon. This means: from each of the N points there will be N-3 lines: N * (N-3) lines in total. But of course we have counted each of the lines twice now. So the total number of diagonals is (1/2) * N * (N-3). Your rule is: "the number of 'diagonals' is the number of sides multiplied with 0,1/2,1,1 1/2 ...". You will see that for a triangle this means: Number of sides * 0 = N * (1/2)*(3-3) For a square: Number of sides * (1/2) = N * (1/2)*(4-3) ....... Hope you see it. With kind regards, Geert van der Wulp An interesting extension is to find the number of crossings for each polygon, (multiple crossings are not allowed). For a square it is 1, for a pentagon it is 5, for a hexagon it is 15 (I think). The formula for it is not all that easy to find for a beginner. No-one knows the formula for joining the dots together with the least crossings. For 4 dots it is 0, for five it is 1, for 6 it is 3, for 7 it is 9. But you can get a formula which is an upper bound for the fewest crossings from the formula in the first paragraph above. Loosen up, you dorks ! I'm just having fun putting dumb posts to see your replies. I'm a senior math major at MIT. Suhaimi suhaimi@mit.edu |
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