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The definition of Ilian (in Chambers under Ilium) is 'From the city of Paris, the king's son', i.e. from Troy, the city that was the home of Paris, the king (of Troy's) son. Wordplay is then NAIL + I rev.

Now, as Meursault quite reasonably asks, can somebody who's willing not to be coy just tell us where to put the rhombus? I worked out - as others have - that it's only the cells at the ends of the lines that we need to pinpoint in their centres, but that hasn't helpmed me.

Sorry - helped me.

Also, what was that post about that said a rhombus was two right triangles plus a rectangle? I thought a rhombus consisted only of four right triangles, but I'm a Bear of Little Brain and mathematicians bother me.

Thanks for the explanation, IonaCarr.

As for the positioning of the Rhombus. 1) I think it's a contrived ending because the setter had difficulty getting either 'rhombus' or 'diamond' into the right configuration. 2) For me, it's another Hermann Hesse moment and I really don't want to spend any more time on it. 3) There are people who come onto this forum to get the assistance they want, then slink away without giving anything back.

Thanks Ionacarr I was just going on my Chambers (on iPad) reference which gave ILIAN =Troy.

Using my playschool method which I know has an area exactly equivalent to 20 squares I have placed my rhombus top line through centres of BUFFOM and the bottom line through ESHENN though I can see no justification in the letters that the lines go through. Is this what others have?

That is what I have gone for too. Letters involved do not matter because there are five shapes spelt out as instructed and this one is identified by correct letters gleaned from clues. Ambiguous preamble and untidy finish I feel - as others have said.

As I've been out I've not been able to comment further until now. I thought my previous post made the maths clear.

The only rhombus with an effective area of 20 cells that will fit the available space has to be five cells wide and a perpendicular height of four cells. A line joining the centres of the first top six cells (BUFFOM) is effectively five cells wide. Ditto ESHENN, which will be perpendicularly four effective cells distant from the top line. Join the B to the E and the M to the N and you have the other two sides.

If you don't belive me look up how to calculate the are of a rhombus on the internet. There are several methods, one of which is base times perpendicular height.

I don't agree with Meursault that the slanting lines are not the same length as the others (so the shape is not a true rhombus). I printed out a large grid and measured the lengths. They are equal.

Once I had refreshed my memory of the area of a rhombus it took only a few minutes to draw it in the space available. People seem to be making it a much harder task than it is.

Hi planks and essira,

I agree with your positioning of the rhombus, but I don't think that the preamble is ambiguous. A straight line is defined by its endpoints (a common informal definition is that a straight line is the shortest distance between two points). So a straight line does not have to go through the centres of all the cells it traverses, it merely has to join the two cell centres at the ends of the line.

For the five thematic items beginning P, Q, R, S and T, the lines that make up these items do indeed pass through all the cell centres, but that's a consequence of the names of these items being spelled out. However, the preamble asks us simply to draw an example of the rhombus. There is no hint that the letters in the cells through which the straight lines constituting the rhombus pass spell out some message (though if the setter had managed this, it would have been great).

Hi Dryden,

We are in agreement regarding the shape and position of the rhombus. You can if you want prove mathematically that the sloping sides have the same length as the horizontal sides (the equivalent of five cells).

If you compare the positions of the top left corner (in the first cell of 1ac, containing the letter B) and the bottom left corner (in the fourth cell of 20ac, containing the letter E) of the rhombus, the bottom left corner is three cells to the right and four cells below the top left corner. Making a triangle of the top left and bottom left corners and the centre of the cell perpendicularly above the bottom left corner (this is the fourth letter of 1ac, containing the letter F), we have a right-angled triangle with sides equivalent to three and four cells, so by Pythagoras' Theorem, the other side must have a length equivalent to five cells, as is the case for the two horizontal sides.

I think that Meursault was trying to draw the sloping sides of the rhombus as passing through the centres of ALL the cells that these sides traversed. It's possible to show mathematically that no rhombus can be constructed that satisfies this condition together with the other conditions of the preamble, but as it's rather late I won't set the proof out.

You don't need to measure the length of the lines to prove they are the same. The top and bottom lines span 5 squares from centre to centre.
The other lines form the longer sides of two right angled triangles. The other sides of these triangles are 3 and 4 respectively using the centre of the cells as a starting point.
Pythagoras says the hypotenuse must be length 5. Square root of sum of the squares of the other two sides.
QED.