Rhombus coordinates

**prasadcad** · December 31st 2009, 12:42 AM

Can someone find a solution of this problem....

I have 500mmX500mm rhombus ( Rhombus A ). This is divided into 2500 equal size rhombuses of size 10mmX10mm ( R1 - R2500) . Now what I need to find is if I place a point(x,y) in Rhombus A, I have to locate in which of the rhombus of R1-R2500 the point(x,y) is placed.

Thanks in advance

**Grandad** · December 31st 2009, 06:08 AM

Hello prasadcad

Welcome to Math Help Forum!

Originally Posted by prasadcad

Can someone find a solution of this problem....

I have 500mmX500mm rhombus ( Rhombus A ). This is divided into 2500 equal size rhombuses of size 10mmX10mm ( R1 - R2500) . Now what I need to find is if I place a point(x,y) in Rhombus A, I have to locate in which of the rhombus of R1-R2500 the point(x,y) is placed.

Thanks in advance

There are a few details that you haven't included in your description, so in the attached diagram, I'm assuming that:

$\theta$ is the acute angle between two adjacent sides of the rhombus

one side lies along the positive $x$ -axis, with a vertex at the origin

the small rhombuses are numbered from the origin, in rows, so that R1 - R50 lie along the $x$ -axis.

Then let's suppose that the point $(x, y)$ lies in the rhombus which is in the $m^{th}$ column and the $n^{th}$ row. Then, with the assumption I've made above, this is rhombus number $50(n-1)+m$ . It now remains to find the values of $m$ and $n$ .

Next, you will need to make a decision about what to do if $(x,y)$ lies on one or more of the edges of one of the small rhombuses. In my working below, I have in this case taken the rhombus to the right and/or above the point.

In the diagram, then, you will see that the rhombus containing $(x, y)$ is bounded by four lines, whose equations I have shown. The point $(x,y)$ therefore satisfies the following inequalities:

$y \ge 10(n-1)\sin \theta$

$y < 10 n\sin\theta$

$y \le (x+10m+10)\tan\theta$

$y > (x-10m)\tan\theta$

The values of $m$ and $n$ that satisfy these inequalities can be expressed using the floor function, where $\lfloor z \rfloor$ , is the largest integer less than or equal to $z$ . Then (if my working is correct - I have checked it with some simple data, but you need to check this with some data of your own) we get:

$n = \left\lfloor \frac{y}{10\sin\theta}\right\rfloor + 1$

$m = \left\lfloor \frac{1}{10}\left(x - \frac{y}{\tan\theta}\right)\right\rfloor +1$

Grandad

**prasadcad** · January 3rd 2010, 07:25 AM

@Grandad
Thanks for the reply.

The details you have asked for are
1. The acute angle is 52Degrees.
2. X-Axis lies along the diagonal of the Rhombus A.

with these values I am not getting appropiate results.
What changes need to be done in deducing the equations for m and n?

**Grandad** · January 4th 2010, 12:51 AM

Hello prasadcad

Originally Posted by prasadcad

@Grandad
Thanks for the reply.

The details you have asked for are
1. The acute angle is 52Degrees.
2. X-Axis lies along the diagonal of the Rhombus A.

with these values I am not getting appropiate results.
What changes need to be done in deducing the equations for m and n?

I have re-drawn the diagram so that the $x$ -axis lies along the diagonal of the rhombus, and have written down the equations of the lines bounding the small rhombus containing the point $(x, y)$ . You will notice that I have now called the angle between the sides $2\theta$ . So in the case you mention, you will need $\theta = 26^o$ .

Using the same conventions that I assumed before, namely, that:

the small rhombuses are numbered from the bottom left-hand corner, with R1 - R 50 lying along the bottom edge;

if the point $(x, y)$ lies exactly on one or more edges of a small rhombus, then we take the rhombus to the right and/or above it;

then the rhombus containing $(x, y)$ is number $50(n -1)+m$ , where:

$m = \left\lfloor\frac{x\tan\theta-y}{20\sin\theta}\right\rfloor+1$

and

$n = \left\lfloor\frac{x\tan\theta+y}{20\sin\theta}\rig ht\rfloor+1$

Let me know if this works OK.

Grandad

**prasadcad** · January 4th 2010, 08:47 AM

Hi Grandad,

Thanks a lot for the solution. It absolutely works fine.

But I need to understand how can we make it more generic. I had given some constant values like 500 mm X 500mm. If we consider the length of each side of Rhombus A is L1 and it is divided into P and small rhombus length is l1.

In the question I had posted the values I had given were L1 = 500mm, P = 2500 and l1= 10.
I would like to know how these values were used to deduce the equations? And how we can use it make it Generic equations?

Please excuse for my little knowledge about geometry.

**Grandad** · January 4th 2010, 11:47 AM

Hello prasadcad

Originally Posted by prasadcad

Hi Grandad,

Thanks a lot for the solution. It absolutely works fine.

But I need to understand how can we make it more generic. I had given some constant values like 500 mm X 500mm. If we consider the length of each side of Rhombus A is L1 and it is divided into P and small rhombus length is l1.

In the question I had posted the values I had given were L1 = 500mm, P = 2500 and l1= 10.
I would like to know how these values were used to deduce the equations? And how we can use it make it Generic equations?

Please excuse for my little knowledge about geometry.

First, note that $L_1, P$ and $l_1$ are not independent. The number of small rhombuses, $P$ , will be the square of the number that will fit along one edge of the large rhombus. In other words:

$P = \left(\frac{L_1}{l_1}\right)^2$

and, of course, if $P$ is to be an integer, then $L_1$ will have to be a multiple of $l_1$ .

The formula for the number of the small rhombus containing $(x,y)$ that I gave you, $50(n-1) + m$ , will then become:

$\frac{L_1(n-1)}{l_1}+m$

and the formulae for $m$ and $n$ will become:

$m = \left\lfloor\frac{x\tan\theta-y}{2l_1\sin\theta}\right\rfloor+1$

and

$n = \left\lfloor\frac{x\tan\theta+y}{2l_1\sin\theta}\r ight\rfloor+1$

Grandad

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