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It's obviously difficult to do this via this text box, but if you grab a sheet of paper and a pencil and try to do the following:

- Draw a square
- Inside the square make an X so that it takes up the entire box (touches the corners)
- On each side of the box, make an arc from each corner (the bottom of the box would have something that looks like 'U', the top would look like 'n', etc)

Overall, it should look like a 4 leaf clover with a box in the middle, containing an X.

Now that you've drawn it how it should look, the obejct is to draw that exact design, but without lifting your pencil or retracing an already drawn line (you can cross over them, you just can't draw over the same exact line already made).

A math teacher gave us this puzzle when I was in 8th grade (1985) and I've still not been able to find an answer to it (either why it can or can't be done).

2006-07-09 16:26:09 · 12 answers · asked by gallostravels 1 in Science & Mathematics Mathematics

12 answers

It cannot be done. The reason is that the four corners of the square each have 5 lines intersecting at them. Now, 5 is an odd number. In order to have 5 lines at a point you have to either start or end at that point. If you start at it, you go out from it first. Then you can come in, go out, come in, and go out... that gives you five edges. If you end at it, you come in, go out, come in, go out, and come in. But you can't have it somewhere in the middle. Since there are 4 points that have an odd number of edges at them, it can't be traced without lifting your pencil or retracing any lines.

In general, if a figure has zero or two nodes with an odd number of edges, it is traceable. If it has one or more than two nodes with an odd number of edges, it is not.

2006-07-09 16:30:55 · answer #1 · answered by mathsmart 4 · 0 0

1

2016-12-25 03:37:53 · answer #2 · answered by Anonymous · 0 0

This is a graph theory problem . . .

At each of the corners of the square there are 5 lines that meet. This means that it is not an eulerian circut and is thus not doable (an eulerian circut has only even degree at each vertex).


Do not feel bad, Euler developed this theory after the Königsberg bridge problem. The people in the town wondered about this for a long time, and it even got Euler to have to think about it (meaning it's not an easy thing to figure out on your own).

2006-07-09 16:30:59 · answer #3 · answered by Eulercrosser 4 · 0 0

LOL. It CAN be done. It is a trick question. Draw the picture, when you get stuck, fold the paper so you can write on the folded part without lifting your pen or tracing over any line. Then, keeping your pen on the folded part, move the folded part down until you get to the part of the figure that is left to be drawn. Presto, you have your figure without breaking your teacher's rules.

2006-07-09 17:18:13 · answer #4 · answered by Franky 1 · 0 0

Ya we too were given this problem but it can not be solved without lifting ur pencil.
YEs but there is a tweak........................ yu fold the paper then try to make the figure such that any of 4 u's is on other fold of paper.............then when u reach the point where u cant make line and cant lift ur pencil then only unfold the paper and continue.........srry pal there is only 1 soln and tht is only by tweaking.lol
bye

2006-07-09 18:54:56 · answer #5 · answered by ☼ Magnus ☼ 4 · 0 0

It is not solvable.

You cannot draw the lines without tracing over at least one line twice if there are more than two points where an odd number of lines come together.

This is a fairly well know theorem and I taught it in my high school geometry classes.

2006-07-09 16:49:44 · answer #6 · answered by Anonymous · 0 0

Yep. it is 9! alrighty. to ascertain why, count number the kind of procedures to gather the numbers a million-8 into the 3x3 grid. position the numbers so as, so no association is counted two times. There are 9 squares to settle on for the a million. as quickly because it is placed, there are 8 squares left to position both in. That makes 9*8=seventy 2 procedures to position purely the a million and a couple of. continuing, for each of preparations, there are 7 squares left for the three, then 6 left for the 4 and so on till there are 2 squares left for the 8. the entire count number is 9*8*7*6*5*4*3*2, it truly is 9! (with out the common-yet-ineffective ingredient of one million). even as 8 numbers were placed, there is only a million position left to be sparkling...in spite of the indisputable fact that it is already sparkling. you could evaluate that to be the "lacking ingredient of one million", i guess.

2016-11-30 23:12:35 · answer #7 · answered by ? 4 · 0 0

If you wish to learn to draw an ideal picture all you need is time and Realistic Pencil Portrait Mastery guide from here https://tr.im/BmpCH to stay the proper path.
The lessons from Realistic Pencil Portrait Mastery guide include 208 pages and an overall total of 605 illustrations.  The essential method used is that you begin with a picture, bring a mild outline of the feature, and then shade it in.
Realistic Pencil Portrait Mastery is the right allied to help make the ideal draw.

2016-05-01 06:53:34 · answer #8 · answered by alethea 3 · 0 0

i remember this puzzle, the trick of it is you have to draw this on a sheet of paper folded onto itself, i don't remember when, but at some point you unfold, and finish drawing the puzzle

2006-07-09 16:56:58 · answer #9 · answered by sexy joker 6 · 0 0

Aaaaw! I could prove this if I remembered discrete math better. You can prove it using Hamiltonian Circuits (or Euler Circuits... I forget which).

No, I don't think this is possible.

2006-07-09 16:31:39 · answer #10 · answered by M 4 · 0 0

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