okay, so obviously you already know what an arithmagon is, otherwise you wouldn't have asked this question, but because it really needs to be shown to you, a good definition of an arithmagon is on this website; http://www.nzmaths.co.nz/algebra/units/arithmagons.aspx hope this helps!
yesThe way to solve an arithmagon is to take the 3 box numbers, add them up and divide by 2. That gives your "center" number. For each corner circle, take the opposite box and subtract from the center number.As long as you are allowed to have fractions and/or negative numbers, I think this method always works. Do you have an example of an impossible arithmagon?For example:a -- 7 -- b.\ ........ /. 8 ..... 9... \ .. /..... cAdd up 7+8+9 = 24Divide by 2 = 12a = 12 - 9 = 3b = 12 - 8 = 4c = 12 - 7 = 5
okay, so obviously you already know what an arithmagon is, otherwise you wouldn't have asked this question, but because it really needs to be shown to you, a good definition of an arithmagon is on this website; http://www.nzmaths.co.nz/algebra/units/arithmagons.aspx hope this helps!
It is a triangle with 3 rectangles and 3 circles!
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yesThe way to solve an arithmagon is to take the 3 box numbers, add them up and divide by 2. That gives your "center" number. For each corner circle, take the opposite box and subtract from the center number.As long as you are allowed to have fractions and/or negative numbers, I think this method always works. Do you have an example of an impossible arithmagon?For example:a -- 7 -- b.\ ........ /. 8 ..... 9... \ .. /..... cAdd up 7+8+9 = 24Divide by 2 = 12a = 12 - 9 = 3b = 12 - 8 = 4c = 12 - 7 = 5