bonanova Posted March 5, 2009 Report Share Posted March 5, 2009 Related to previous puzzle, which seems mired. There are many ways to arrange 12 unit-length matches to form a polygon whose area is [exactly] an integer. What is the smallest non-zero area you can achieve? Quote Link to comment Share on other sites More sharing options...
0 Yoruichi-san Posted March 5, 2009 Report Share Posted March 5, 2009 (edited) Any constraints? Otherwise... ...should be 1, the smallest non-zero integer...there's a number of ways to do this... example: rhombus with sides of length 3, and height of 1/3, which gives the small angle to be arctan(1/9). Edit: corrected my trig Yes - the constraint is that you don't have a protractor. Clearly a unit-area 12-perimeter rhombus exists - but you couldn't make it without just guessing the angles. Edited March 5, 2009 by bonanova Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 ______ |_|_|_| __|_| then the area = 3 unit sq Quote Link to comment Share on other sites More sharing options...
0 Yoruichi-san Posted March 5, 2009 Report Share Posted March 5, 2009 Any constraints? Otherwise... ...should be 1, the smallest non-zero integer...there's a number of ways to do this... example: rhombus with sides of length 3, and height of 1/3, which gives the small angle to be arctan(1/9). Edit: corrected my trig Yes - the constraint is that you don't have a protractor. Clearly a unit-area 12-perimeter rhombus exists - but you couldn't make it without just guessing the angles. Okay. I guess this isn't a lateral-thinking puzzle, but actually I wouldn't necessarily need a protractor...I could do it with a string or a sheet of paper (something I could fold into thirds) or a semi-solid substance (that I could measure out, assuming I know the width of the match) and a clear slide... Anyways, another question: do all the matches have to be used in the perimeter? Can they be stacked on top of each other? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 5, 2009 Author Report Share Posted March 5, 2009 Okay. I guess this isn't a lateral-thinking puzzle, but actually I wouldn't necessarily need a protractor...I could do it with a string or a sheet of paper (something I could fold into thirds) or a semi-solid substance (that I could measure out, assuming I know the width of the match) and a clear slide... Anyways, another question: do all the matches have to be used in the perimeter? Can they be stacked on top of each other? No stacking. End-to-end. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 5, 2009 Author Report Share Posted March 5, 2009 ______ |_|_|_| __|_| then the area = 3 unit sq Yes. Is it possible to make one with Area=4? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 Yes. Is it possible to make one with Area=4? __ _ | __|_ |_| |_| Quote Link to comment Share on other sites More sharing options...
0 Yoruichi-san Posted March 5, 2009 Report Share Posted March 5, 2009 Okay. I guess this isn't a lateral-thinking puzzle, but actually I wouldn't necessarily need a protractor...I could do it with a string or a sheet of paper (something I could fold into thirds) or a semi-solid substance (that I could measure out, assuming I know the width of the match) and a clear slide... Actually...using no other objects than the matches and pure geometry...make my rhombus by putting 3 matches end to end horizontally and 1 match vertically at the end, and 1 match vertically at the end of the first match, then use more matches to form a straight line from the beginning of the first match to the vertical match at the end. This line will intersect the vertical match at the end of the first match at exactly 1/3 unit length, and then I can use more matches to draw a straight line perpendicular to the base to mark this 1/3 unit height for my rhombus, then make place my 3 unit side to intersect with this 1/3 unit height line...then remove my marker matches and use them to form the other two sides... Sorry...but I'm really confused about what the rules are...it seems that we can 'eyeball' 90 degree angles (or else woon's solution wouldn't have worked)...what geometric properties are we allowed to use in conjunction with that? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 I'd also argue that woon's solution is three separate polygons. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 5, 2009 Author Report Share Posted March 5, 2009 I'd also argue that woon's solution is three separate polygons. Yes. I'm having a hard time ruling out rhombus solutions. You need to make angles whose sines have rational values - tough to know when you have done that. But to be fair, you have to eyeball horizontal and vertical to know when a square is exact. OK. Truce. Let's burn the matches, and state the idea differently. You are givena sheet of ruled paper, with horizontal and vertical lines 1" apart.no other notion of distance or angle [don't tear off a piece of the paper to use as a ruler]an unmarked straight edgea sharp pencil - which writes only on paper [i.e. won't write on straight edges]Draw a polygonwhose diameter everywhere is positive definite [ruling out woon's polygons and a 0x6 rectangle]all of whose sides are integral lengthwhose perimeter is 12whose area is integralWhat is the minimum area you can achieve? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 I come up with 1. If you make it a 3D cube you use all 12 matches and when seen from the side it only ocupies 1^2 interger. I think this follows the rules set into place. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 5, 2009 Author Report Share Posted March 5, 2009 I come up with 1. If you make it a 3D cube you use all 12 matches and when seen from the side it only ocupies 1^2 interger. I think this follows the rules set into place. A unit cube has area 6 and volume 1. And it's a polyhedron, not a polygon. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 Five. It will look like a short staircase or "W"-ish. I haven't figured out how to space out my matchsticks to draw the diagram. Reducing any further seems to always cause either an overlap of matchsticks or it divides into multiple polygons connected by a common point. Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted March 5, 2009 Report Share Posted March 5, 2009 (edited) I have a solution with an area of 3 (which I won't give away yet), but I have another question... [*]a sharp pencil - which writes only on paper [i.e. won't write on straight edges]Do I have an eraser? Or does everything I write need to be part of the polygon? (i.e., can I cross lines drawn with the straight edge to find a point, then erase the lines?) Or, equivalently, can I use more than one (infinitely long, infinitesimally thin) straight edge? Edited March 5, 2009 by EventHorizon Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted March 5, 2009 Report Share Posted March 5, 2009 I have a solution with an area of 3 (which I won't give away yet), but I have another question... Do I have an eraser? Or does everything I write need to be part of the polygon? (i.e., can I cross lines drawn with the straight edge to find a point, then erase the lines?) Or, equivalently, can I use more than one (infinitely long, infinitesimally thin) straight edge? Oops....ignored that the area still needs to be integral. If that was removed and I could use many straight edges, I could get a polygon with as small an area as I want. My solution of 3 still works, but it's going to be harder to get lower than I previously thought. Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted March 5, 2009 Report Share Posted March 5, 2009 (edited) Oops....ignored that the area still needs to be integral. If that was removed and I could use many straight edges, I could get a polygon with as small an area as I want. My solution of 3 still works, but it's going to be harder to get lower than I previously thought. Alright. Can I hold the straight edge fixed at one point to create a circle? If so, then the area is 1. Otherwise, I'm stuck at 3 for the moment. Edited March 5, 2009 by EventHorizon Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 5, 2009 Report Share Posted March 5, 2009 Related to previous puzzle, which seems mired. There are many ways to arrange 12 unit-length matches to form a polygon whose area is [exactly] an integer. What is the smallest non-zero area you can achieve? Is this a polygon? Quote Link to comment Share on other sites More sharing options...
0 Yoruichi-san Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) Thanks Bonanova for stating the rules clearly...sorry for making you burn the matches...I honestly just wanted to know what the rules were so I could figure out the best solution Anyways... A 3 sq. unit polygon. It's basically a the 3,4,5 Pythagorean triangle (which has a area of 6 and perimeter of 12), but instead of having a convex right angle, I have 'carved' out of it 3 sq. units. Trying to depict it as best as possible...7 unit lines arranged as follows (a figure with a total length of 4 and height of 3) _| __| _| with a line from the bottom left point to the top right point that is 5 units long. Edit: Oops...sorry EH if you had the same solution...I was actually thinking about this when I was tossing and turning in bed last night... Edited March 6, 2009 by Yoruichi-san Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted March 6, 2009 Report Share Posted March 6, 2009 Thanks Bonanova for stating the rules clearly...sorry for making you burn the matches...I honestly just wanted to know what the rules were so I could figure out the best solution Anyways... A 3 sq. unit polygon. It's basically a the 3,4,5 Pythagorean triangle (which has a area of 6 and perimeter of 12), but instead of having a convex right angle, I have 'carved' out of it 3 sq. units. Trying to depict it as best as possible...7 unit lines arranged as follows (a figure with a total length of 4 and height of 3) _| __| _| with a line from the bottom left point to the top right point that is 5 units long. Edit: Oops...sorry EH if you had the same solution...I was actually thinking about this when I was tossing and turning in bed last night... that is not the exact one I had, but it's equivalent. also uses the 3-4-5 triangle. Start at one vertex, draw right one, then a line of length 5 up 3 and over 4, draw left one, and connect with another line of length 5. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 6, 2009 Author Report Share Posted March 6, 2009 EventHorizon: Eraser: yes - as if you had matches, you could move some. Two straight edges: No. The straight edge is just to align the matches, if we were using matches, so it's just to draw lines in this case. Use it to draw a circle: no. the intent is not to construct a figure. G Threat: Yes it's a polygon but I don't know how you'd guarantee the area is integral Y-san: Apology was sweet, but not necessary. Since I don't smoke or otherwise set fire to things, the matches are no loss. And I agree, the limits should be clear. The best puzzles, I've found, have a bit of genius in the wording. I didn't have the needed genius on this one. Now to the answers. EH: I'd love to see a sketch, cuz I got lost when you mentioned two lines of length 5. Y-san: One word - Bingo. Still a bit of a puzzle to word the question and require this solution. Can't rule out others by saying convex or rectilinear. I still think this is the only solution that you could really do with matches. Fairly easy to eyeball vertical 3, horizontal 4, line up the 5, then dent in the corner. Rhombus seemed qualitatively harder, to me at least. I read the puzzle a couple days ago and liked it, even tho it didnt come with a solution. In fact, it asked to make A=4, which I envisioned as 345 with two bottom squares missing. So You did it one better. I had to draw it to convince myself. Nice. Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted March 6, 2009 Report Share Posted March 6, 2009 EH: I'd love to see a sketch, cuz I got lost when you mentioned two lines of length 5. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 6, 2009 Author Report Share Posted March 6, 2009 That's a 3. And you could do that with matches. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
Related to previous puzzle, which seems mired.
There are many ways to arrange 12 unit-length matches to form a polygon whose area is [exactly] an integer.
What is the smallest non-zero area you can achieve?
Link to comment
Share on other sites
21 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.