plainglazed Posted April 15, 2011 Report Share Posted April 15, 2011 Consider a six and a seven sided simple polygon in/on a unit square. If you maximize the perimeter of each, which has the greatest volume? Quote Link to comment Share on other sites More sharing options...

0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 six sided regular polygon Quote Link to comment Share on other sites More sharing options...

0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 (edited) the simple heptagon, because it has an odd number of sides. To maximize the perimeter of my hexagon I stretched it into a thin capital N. The heptagon had an additional side causing it to open a larger space inside the shape. Oh, I was looking at a unit circle not a unit square. I still stand by my answer though. Edited April 15, 2011 by rlgandy Quote Link to comment Share on other sites More sharing options...

0 HoustonHokie Posted April 15, 2011 Report Share Posted April 15, 2011 "volume" implies a third dimension which is not present in the OP. Being 2-dimensional elements, both have 0 volume. But, assuming you meant "area" instead, I'd have to agree with sks. Maximizing the perimeter of the two shapes is an interesting task. For the 6-gon, I think I'd use the 4 corners of the square, the place a 5th point inside the square very close to one of the corners, and the 6th point on the edge of the square very close to the diagonal opposite corner of the square from the 5th point. In this way, the perimeter would be very nearly 4 + 2√2, which is similar to rigandy's perimeter, but the area is much larger than his thin capital N. The area would be very nearly 1. For the 7-gon, I think I'd use the 4 corners of the square again and 2 points just inside 2 of the square's corners, diagonally opposite from each other. The 7tgh point would be on the perimeter of the square, very close to one of the other 2 corners. In this way, the perimeter maxes out at nearly 6 + √2. In so doing, the 7-gon's area is only about 0.5 Quote Link to comment Share on other sites More sharing options...

0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 "volume" implies a third dimension which is not present in the OP. Being 2-dimensional elements, both have 0 volume. But, assuming you meant "area" instead, I'd have to agree with sks. Maximizing the perimeter of the two shapes is an interesting task. For the 6-gon, I think I'd use the 4 corners of the square, the place a 5th point inside the square very close to one of the corners, and the 6th point on the edge of the square very close to the diagonal opposite corner of the square from the 5th point. In this way, the perimeter would be very nearly 4 + 2√2, which is similar to rigandy's perimeter, but the area is much larger than his thin capital N. The area would be very nearly 1. For the 7-gon, I think I'd use the 4 corners of the square again and 2 points just inside 2 of the square's corners, diagonally opposite from each other. The 7tgh point would be on the perimeter of the square, very close to one of the other 2 corners. In this way, the perimeter maxes out at nearly 6 + √2. In so doing, the 7-gon's area is only about 0.5 hexagon... Start at a corner and go back and forth to the opposite diagonal corner 4 times offseting points just slightly. 6th point at an adjacent corner to the starting point. Perimeter would be nearly 2 + 4√2 which is greater than 4 + 2√2. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 None, since polygons are 2 dimensonal figures... Quote Link to comment Share on other sites More sharing options...

0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 To get the longest perimeter most of the lines should have to run along the diagonal By the 5th point four lines are nearly √2. For the hexagon the 6th point would have to head to an adjacent corner to enclose the polygon. The area would be near 0. For the heptagon the 6th point would head back across the diagonal with the 7th point heading to the adjacent corner. The area would be near 0.5. To counter: Create a hexagon with a perimeter greater than nearly 2 + 4√2 Create a heptagon with a perimeter greater than nearly 2 + 5√2 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 If yo actually meant "which has the greatest AREA" then it would be the heptagon. The highest the n in an n-sided polygon, the best it resembles a circle... Thus, maximizing the possible area occupable Quote Link to comment Share on other sites More sharing options...

0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 If yo actually meant "which has the greatest AREA" then it would be the heptagon. The highest the n in an n-sided polygon, the best it resembles a circle... Thus, maximizing the possible area occupable You missed the part about maximizing the perimeter. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 You missed the part about maximizing the perimeter. Quote Link to comment Share on other sites More sharing options...

0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 Don't understand why you are quoting me without comment... You went for the maximum area first. I believe he was asking for a comparison of the areas of polygons with the maximum perimeters. Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted April 15, 2011 Author Report Share Posted April 15, 2011 To get the longest perimeter most of the lines should have to run along the diagonal By the 5th point four lines are nearly √2. For the hexagon the 6th point would have to head to an adjacent corner to enclose the polygon. The area would be near 0. For the heptagon the 6th point would head back across the diagonal with the 7th point heading to the adjacent corner. The area would be near 0.5. To counter: Create a hexagon with a perimeter greater than nearly 2 + 4√2 Create a heptagon with a perimeter greater than nearly 2 + 5√2 yes sir. not very taxing, I know. just a little distraction. and apologies all re that volume/area botch up . Quote Link to comment Share on other sites More sharing options...

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