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So suppose you draw a line through the centres of cell 'BUFFO', another through the centres of cells 'SHENN' and join up the other sides, from the centres of the cells, making a 5X5 object. The first thing is that the slanted lines are longer than the horizontals, so the shape is not a rhombus. The second is that, without any calculation, the area, row by row, must be 2,4,4,4 and 2. Since, in the case of BUFFO, the middle 3 cells only count half a cell each, and the corner cells combined only count half a cell. In the case of SEASA, the area is clearly 4 cells, etc.

Hi, Meursault,

If the area, row by row, is 2, 4, 4, 4 and 2, you've accounted for only 16 cells. Try 2.5, 5, 5, 5 and 2.5.

Thanks all for the thoughts.. I have been working along those lines but find the centre of cell requirement imposes severe constraints and haven't got there yet. Maybe I will sleep on it, that often works!

You could do that if you stretched the top left corner of the rhombus outside the grid, but the preamble requires straight lines joining cell centres...

There's an ambiguity that I'd like to mention in case it's thwarting others - it held me up for some time. "...formed of straight lines joining cell centres." I interpreted this to mean that the lines went through the centres of ALL cells they entered, ie that included angles in the shapes had to be a multiple of 45 degrees. Not so - in fact I think not soluble that way.

Meursault,

If you define the top of the rhombus as a line from the centre of the B in BUFFO to the centre of the O, and the bottom as a line from the centre of the S in the fifth cell of 20ac to the N which is the fourth letter of 22ac, then those sides of the rhombus are only four cells long, and the sloping sides are roughly 5.66 (square root of 32) cells long, so this is not, as you correctly say, a rhombus. However, this parallelogram has a height of 4 cells and a base of 4 cells so its area is equivalent to only 16 cells.

A line 5 cells long beginning in the centre of the cell containing the B in BUFFO doesn't end in the cell containing the O, but rather the first cell of 5dn.

I think you are right, unless one takes it that the requirement is that the lines start and end in the centre of a cell but do not necessarily pass through the centre of all the cells they traverse, then it is impossible. Since all the other shapes do use centre of all cells I think that the ambiguity in preamble is unfair............or maybe there is another solution!

I've been enjoying the good weather today in the south so haven't finished the clues yet!
I don't understand the definition in 42ac. Presumably won changes to son but I'm still none the wiser! Any explanation much appreciated.

Hi Wintonian, If you take the same slanted side from the start of 5, it touches parallelogram, so is invalidated. Perhaps a little more information might be useful for the solvers who come on here. This site isn't answerbank, and really should be doing its best to compensate for answerbank not doing its job...

Unclued, I have ILIAN, though it wasn't in my Chambers. Like you, I assumed that won become son, but never did work out the wordplay.