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Topic: Match polygons (Read 589 times) |
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BNC
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Match polygons
« on: Sep 20th, 2003, 2:53pm » |
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The square above is built from 12 matches, and its area is 9 square units.
Using 12 whole matches, construct a polygon whose circumference is 12 units, and area is 3 square units.
Repeat for area of 4, 5, and 6 square units.
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| « Last Edit: Sep 20th, 2003, 2:54pm by BNC » |
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Sir Col
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Re: Match polygons
« Reply #1 on: Sep 20th, 2003, 4:06pm » |
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Are we meant to show that it can't be done?
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It can be shown that area of n-gon, with side length, x, is given by, A=(nx2)/(4tan(180/n))
If perimeter, P=nx=12, x=12/n, x2=144/n2.
Substituting this into area formula (and tidying up), A=36/(ntan(180/n)).
Minimum area achieved when n=3, A=36/(3tan60)~=6.93
Maximum area achieved when n=12 (I'm assuming we can't break the matches), A=36/(12tan15)~=11.2.
In fact the only integral area is when n=4, A=36/(4tan45)=9.
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| « Last Edit: Sep 20th, 2003, 4:07pm by Sir Col » |
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Icarus
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Re: Match polygons
« Reply #2 on: Sep 20th, 2003, 5:08pm » |
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Sir Col - If BNC were to add one word to his puzzle, you would be correct. But as it is, you've made an assumption that need not be true. (And obviously is not for the solutions to these problems.)
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Sir Col
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Re: Match polygons
« Reply #3 on: Sep 21st, 2003, 12:38am » |
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Dang! It doesn't have to be regular. Me thinks more thought is required...
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towr
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Re: Match polygons
« Reply #4 on: Sep 21st, 2003, 9:13am » |
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it's hard to draw, but it's pretty easy.. think parallellogram,
with a base\height of 3\1, 4\1, 5\1 (rectangle) and 3\2
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| « Last Edit: Sep 21st, 2003, 9:14am by towr » |
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Icarus
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Re: Match polygons
« Reply #5 on: Sep 21st, 2003, 11:31am » |
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Yep - once you get that idea it's trivial. You can (theoretically) come up with any area below the maximum simply by squeezing your polygon in a little at the corners. From a practical stand point, its more difficult because you are limited in which angles you can create.
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| « Last Edit: Sep 21st, 2003, 7:56pm by Icarus » |
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"Pi goes on and on and on ...
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I wonder: Which is larger
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Sir Col
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Re: Match polygons
« Reply #6 on: Sep 21st, 2003, 11:55am » |
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Tee-hee, I think Icarus needs a lesson in using tags. 
Inspired idea, towr; it seems so obvious now.
In fact, starting with the maximum area (the square with base 3), you can slant the sides to begin collapsing the parallelogram to form any area from 9 to 0.
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BNC
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Re: Match polygons
« Reply #7 on: Sep 21st, 2003, 1:42pm » |
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Care to try solving using n-gons, n being odd?
(I'm aiming at a nice solution, different than Towr's).
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Icarus
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Re: Match polygons
« Reply #8 on: Sep 21st, 2003, 8:24pm » |
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on Sep 21st, 2003, 11:55am, Sir Col wrote:| Tee-hee, I think Icarus needs a lesson in using tags. |
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Actually - a lesson in always previewing before posting would be more useful!
The square is not the absolute max area - that would be the regular dodecagon (Area [approx] 11.2 for unit sidelength).
For n-gons with n odd, you can get every area from [sqrt]3/4 to n/4tan(180/n) (with [sqrt]3/4 possible only for n = 3).
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I wonder: Which is larger
When their digits are reversed? " - Anonymous
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BNC
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Re: Match polygons
« Reply #9 on: Sep 21st, 2003, 10:56pm » |
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Would you mind showing me how you get 3/4?
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Icarus
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The size x of the gap across the "squeeze point" satisfies (if I haven't blown the algebra):
10x4 + 8x3 - 26x2 - 24x + 9 = 0 (where the matchstick length is 1).
x is the unique solution between 0 & 1. I get x [approx] 0.29330
Note that this requires measuring utensils other than the matchsticks themselves.
You can do the same thing with any odd number of sides. The only difference is that the "needle" is longer and thinner.
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I wonder: Which is larger
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Lightboxes
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Re: Match polygons
« Reply #11 on: Sep 22nd, 2003, 9:46pm » |
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I read the previous posts and I believe I didn't see the an answer where you can make a circle and move every other vertex in if there are an even number of sticks. I can't do the math. I'm sure it's a function where there is a max and min but I don't know how to make my own function.
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BNC
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Re: Match polygons
« Reply #12 on: Sep 22nd, 2003, 10:32pm » |
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Icarus,
Have you used the constraint of using "whole" (i.e., not broken) matches only?
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Icarus
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Re: Match polygons
« Reply #13 on: Sep 23rd, 2003, 3:14pm » |
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Yes - 5 matches - all straight - all the same length (any variance in the picture is unintentional). Area = 3/4.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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towr
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Re: Match polygons
« Reply #14 on: Sep 23rd, 2003, 11:47pm » |
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I thought you were supposed to use all twelve matches.. (some of which form one line, so you can get an odd N-gon)
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BNC
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Re: Match polygons
« Reply #15 on: Sep 24th, 2003, 1:58am » |
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on Sep 20th, 2003, 2:53pm, BNC wrote:
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Using 12 whole matches, construct a polygon whose circumference is 12 units....
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I thought it should be clear... yes, use all 12 matches.
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Icarus
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Re: Match polygons
« Reply #16 on: Sep 24th, 2003, 3:33pm » |
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Okay - I misunderstood. I mistook your "n odd" to mean an odd number of matches.
However, the solution still works (though with a different width for the gap). You make the base out of two matches, the sides of the lower triangular section out of one match each. The remaining 8 matchs are used 4 each in the straight sides of the needle. Any area A with 0 < A [le] [sqrt]35 can be obtained in this fashion. So 3/4 certainly can. To get larger areas, break each of the sides at 2 or more locations, and have them flair out before turning back in.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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SWF
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Re: Match polygons
« Reply #17 on: Sep 24th, 2003, 6:54pm » |
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The original phrasing of the question leaves out the restrictions that turn this into a more interesting problem. Sure, any area from 0 to the dodecagon area can be made, but I think you are should be able to construct the shape for area 3, 4, 5, or 6 using some reasonable rules of matchstick geometry.
Basically these rules would mean you can easily make shapes of the required area using just the matchsticks to figure out positions. These rules should permit the assumption that you are able to line up matches perfectly in a row to make a line segment of length 1 to 12 long. Also, you can lock sticks together in position and manipulate the line segments to do things like form a perfect 3, 4, 5 right triangle (which, by the way, is the answer for forming an area of 6 with an odd number sided polygon). Since you can make the 3, 4, 5 triangle, you can construct right angles too.
I think all constructions with straightedge and compass can be done with just matchsticks. However, the solutions hidden below do not use elaborate constructions, and you can easily assemble matchsticks in these shapes. This is a nice problem and I recommend people try before peeking.
For areas of 3 and 5 with odd number sided polygon:Make a right angle between segments of length 1 and 5. Make a right angle between segments of length 1 and 2. I will call them right triangles even though the hypotenuse is imaginary (not made of toothpicks). Join the two triangles at acute vertices to form a hinge point. Rotate about the hinge points until the other acute vertices are 3 toothpicks apart. That makes a pentagon with sides of 1, 1, 2, 3, and 5. If the pentagon was assembled with right angle of the 1-2-sqrt(5) triangle convex on the pentagon, the area is 5. If that right angle is oriented concave then the area is 3. (Right angle on 1-5-sqrt(26) triangle should be convex in both cases).
For area of 4 with odd number sided polygon:As in the other solution, make right angle between segments of length 1 and 2; as well as 2 and 2. Join acute vertices and hinge until the other acute vertices are 5 matchsticks apart. The penagon with sides 1,2,2,2,5 has area 4.
The solutions given above are not unique.
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| « Last Edit: Sep 24th, 2003, 6:55pm by SWF » |
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BNC
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Re: Match polygons
« Reply #18 on: Sep 24th, 2003, 11:04pm » |
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SWF,
1. Thanks! That's just what I tryed to convey, but lacked the words for.
2. My idea for 3,4,5 is, well, even easier to construct.
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