Strange knight's tour on a strange chessboard. kweenie

this problem is really tough. I worked on it the day before yesterday for a whole afternoon. The condition of "not color the $ 12$ centers at corners" makes me prefer to guess that the answer is "YES", and after the hardworking I think I came up with a conformation. BUT, just now I checked it and found it was wrong. Then I suddenly realized that the answer should be "NO"! Assume the broken line exists, denote its symmetric center as $ O$. First step is to prove that $ O$ is the symmetric center of the whole rectangle. Consider the symmetric mapping with center $ O$, then each segment of the broken line is mapped to another segment which is also a part of the broken line, so we deduce that the image of each red center is also a red center. Therefore all $ 96$ red centers are of $ 48$ pairs such that the centers of each pair is symmetric of $ O$. So $ O$ is the centroid of all red centers which means $ O$ is the symmetric center of the whole rectangle. for any red center $ A$, call the red center who is symmetric to $ A$ with center $ O$ as "$ A$'s mate" Now color all red center with black and white such that any two centers whose distance is $ 1$ are of different colors.(since now the word "color" only means black and white, no red anymore) It is easy to check that every red center and its mate are of different colors. And it is also obvious that any two centers whose distance is $ \sqrt 13$ are of different colors. So two endpoints of each segment of the broken line are of different colors. Next if $ XY$ is a segment of the broken line then the segment connecting $ X$'s mate and $ Y$ 's mate are also a part of the broken line. To prove this, note that all segments connecting all red centers have finitely many intersection points, so there exist a point $ Z$ on segment $ XY$ that is not any of those intersection points. Due to the symmetry of the whole rectangle, the symmetric point of $ Z$ with center $ O$ (called $ Z_1$) is also not any of those intersection points.Since $ Z$ is on the broken line and $ O$ is its symmetric center, then $ Z_1$ is also on the broken line that mean $ Z_1$ is on a segment connecting two red centers and that segment is a part of the broken line. Because $ Z$ is on the segment $ XY$ so $ Z_1$ is on the segment connecting $ X$'s mate and $ Y$ 's mate. But $ Z_1$ could be on at most one segment connecting two red centers, so we deduce that the segment connecting $ X$'s mate and $ Y$ 's mate is also a part of the broken line. Now take a look at the endpoints of the broken line. Arrange them on a circle according to the order that they are on the broken line and such that they form a regular $ 96-gon$. Due to the fact we proved just now we deduce that the segment connecting any point of the regular $ 96-gon$ and that point's mate is a diameter of the circle! In other words, for any red center $ A$, there are $ 47$ endpoints between $ A$ and $ A$'s mate on the broken line. That follows the contradiction: since every two adjacent endpoints of the broken line are of different colors, so if two endpoints have odd number of endpoints between them on the broken line, then they are of the same color. However $ A$ and $ A$'s mate are of different colors! a contradiction. Q.E.D.

"I don't want to disturb your circle" as happened to Archimedes but can the broken line not be a kind of figure 8 with crossing in the center O? kweenie

A kind of fig 8, instead of a circle, is NOT possible, observe the following: Number the squares with A to L and 1 to 9. 1) A normal square in the middle can have up to 8 possible segments. If a square has only two possible segments then this two segments are forced. That’s the case for B2 and B8 (also K2 and K8). This gives the following forced path: E4-B2-D5-B8-E6 (also H4-K2-I5-K8-H6) 2) There are only two segments with symmetrical (with respect to O) endpoints to make a kind of fig 8, namely E4-H6 and E6-H4. But then we have a closed circuit with the two forced paths from 1 above, and cannot reach anymore the other squares. Thus I did NOT "disturb your circle". There are also four other forced paths. As D5 (and I5) is full, A3 and A7 (also L3 and L7) have only two possible segments left over. Thus also D1-A3-C6 and its symmetries are forced as well. The remaining squares can have now 3, 4, 5, 6, 7 or 8 possible segments. The number of odd squares (3, 5) is always even. We can delete segments by finding paths between two odd squares, trying to include as much as possible 4-squares, but only once, and not using existing forced paths. This produces other forced paths. Continuing this way gives always (I hope so) a solution. Doiing this in a symmetrical way will also give a symmetrical solution. kweenie