Solving a Square in the x-y Plane: Finding m with s, d and n

In summary, the problem asks for an expression for m in terms of s,d and n. TheAttempt at a Solution found that m is restricted to 0<m<s/d. Any ideas to possibly help nudge me in a direction are welcome. Anything is helpful, even if you don't feel like you're right
  • #1
Mentallic
Homework Helper
3,802
95
This isn't homework, so there's no rush, or a necessity to find the answer but I'd like input into helping me solve this problem. I constructed this problem from an Australian Mathematics competition that instead used simple numbers in place of the variables I'm replacing them with.

Homework Statement


I'll describe it as best as I can:
In the x-y plane, there is a square of side lengths s that has it's closest vertex located a distance d from the origin, and it is sitting on the x-axis in the first quadrant. There is a line y=mx that cuts the square in the ratio [tex]\frac{A_1}{A_2}=n[/tex] where A1 is the area of the section in the square that is above the line, and A2 the area below the line. n can take any value that is meaningful, in this case, all real positive values (thus A1 and A2 will have a value larger than 0).

http://img198.imageshack.us/img198/7304/pfsquaren.png

Find an expression for m in terms of s,d and n.

The Attempt at a Solution


I've barely scratched the surface of this problem...
All I can really think of is that m is restricted to [itex]0<m<s/d[/itex].

Any ideas to possibly help nudge me in a direction are welcome. Anything is helpful, even if you don't feel like you're right :smile:
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
I have tried something:

[tex]\frac{A_1}{A_2}=n[/tex]

[tex]A_1=nA_2[/tex] (1)

[tex]A_1+A_2=s^2[/tex] (2)

(1) [itex]\rightarrow[/itex] (2)

[tex]A_2(1+n)=s^2[/tex] (3)

[tex]A_2=\frac{s}{2}(md+m(s+d))[/tex] (4) - by area of parallelogram, left vertical side length of [itex]A_1[/itex] is [itex]md[/itex] and the right side is [itex]m(s+d)[/itex]

(3) [itex]\rightarrow[/itex] (4)

[tex]\frac{s^2}{1+n}=\frac{s}{2}(md+m(s+d))[/tex]

solving for m...

[tex]m=\frac{2s}{(2d+s)(1+n)}[/tex]


But I tested this formula for some random values, and found it doesn't work. For e.g. if I use d=2, s=1 and n=1 (so it cuts the square into 2 equal pieces) the formula gives me [itex]y=2x/5[/itex]

oh wait... while writing this I rechecked my substituting and found it's actually [itex]y=x/5[/itex]... and to think I failed at the easiest part in this problem.

Never mind, solved :smile:
 
  • #3
Actually... looking at it again, it seems like as [itex]n\rightarrow 0[/itex] the value of m should be such that it touches the top-left vertex of the square, and this gradient is s/d, but the formula doesn't suggest that...

I'm thinking that my error is in the part where I assumed the area of A2 is a parallelogram, because while it is for the most part, A1 and A2 become a triangle and truncated square respectively after [tex]m\geq\frac{s}{s+d}[/tex] (the top-right vertex of the square).
 
  • #4
I got the same equation for m:

[tex]m=\frac{2s}{(2d+s)(n+1)}[/tex].

What I did is by sectioning the areas into rectangles and and triangles and use the the ratio to solve for m.
 
Last edited:
  • #5
Well that supports it then. And I've tried the formula on a few more examples, and it works unless [tex]\frac{s}{d}>m>\frac{s}{s+d}[/tex]

For this "part" of the formula, I've calculated and checked that the answer is the solution to this equation:

[tex]d^2m^2-\frac{2s}{n+1}\left(sn+dn+d\right)m+s^2=0[/tex]

which seems pretty complicated if I were to try solve it...

What is up with this? I've never had 2 formulas needed to answer the 1 same question. Is it possible to combine both formulas to create 1 that answers my question entirely?
 
  • #6
I agree with everyone on the first part. I don't have time to check the second case right now but I assume you are right. As to getting one simple formula, I wouldn't expect similar or easily combined forumulas because the nature of the problem changes when the line cuts the top side of the square. Until then the areas were two trapezoids and now they are a pentagon and a triangle. A two-piece formula is probably as appropriate as anything. You could always jerry-rig up a single formula using step functions or some other trick, but it wouldn't add any clarity.
 
  • #7
Thanks LCKurts. A step function will be the way I'd go, but the only odd thing I find about that is that if anyone wanted to use the formula, they'd first have to figure out if their gradient is [itex]m>s/(s+d)[/itex] or [itex]m\leq s/(s+d)[/itex]... I've just never been required to do such a thing before. Oh well, I guess it seems kind of neat to learn something new :smile:

By the way, I've tried solving m for the 2nd case just so I can try and see if the formula is correct by testing it with a few random cases. This is what I got:

[tex]m=\frac{s}{d^2(n+1)}\left(sn+dn+d\pm \sqrt{(sn+dn+d)^2-d^2(n+1)^2}\right)[/tex]

What do I do about the [itex]\pm[/itex]? Of course the gradient couldn't be two answers. I'm guessing the positive is the correct answer, and the negative is some void imaginary solution, but I have no way of telling from just looking at it.

I could always try a random question and use both m to see which is more suitable...
 
Last edited:
  • #8
I've encountered another problem.

Since I want to find out what m is, and I have the value of s, d and n, how am I meant to know which formula to use?

If

[tex]m>\frac{s}{s+d}[/tex] then I use
[tex]m=\frac{s}{d^2(n+1)}\left(sn+dn+d\pm \sqrt{(sn+dn+d)^2-d^2(n+1)^2}\right)[/tex]

but if

[tex]m\leq \frac{s}{s+d}[/tex] then I use
[tex]m=\frac{2s}{(2d+s)(n+1)}[/tex]

But I don't know what m is, so I won't know which formula to use...??
 
Last edited:
  • #9
Ok I think I've solved my second problem on figuring out which formula to use.
I needed to find n in terms of s and d. (btw, is this expressed as n=f(s,d)?)

If I take the line y=mx such that it cuts the top-right vertex of the square (this is the seperator between the first and second formula) I have this:

[tex]\frac{A_1}{A_2}=n[/tex]

[tex]A_1=\frac{s}{2}\left(s-md\right)[/tex]

[tex]A_2=smd+A_1[/tex]

Hence, [tex]n=\frac{\frac{s}{2}\left(s-md\right)}{smd+\frac{s}{2}\left(s-md\right)}[/tex]

simplifying...

[tex]n=\frac{s-md}{s+md}[/tex]

but at this point, [tex]m=\frac{s}{s+d}[/tex]

so [tex]n=\frac{s-\frac{s}{s+d}d}{s+\frac{s}{s+d}d}[/tex]

simplifying...

[tex]n=\frac{s}{s+2d}[/tex]

And now going by logic,

if
[tex]n<\frac{s}{s+2d}, m>\frac{s}{s+d}[/tex]

[tex]n\geq \frac{s}{s+2d}, m\leq\frac{s}{s+d}[/tex]

So then I can know which formula to use, I hope :smile:
 
  • #10
I've created a scenario where the line would cut the top part of the square, so that [itex]m>s/(s+d)[/itex] and I found that the formula works when I take the negative of the [itex]\pm[/itex]. I don't understand the logic of this, but I can't complain.

Anyway, I believe I have this question answered.
 

Related to Solving a Square in the x-y Plane: Finding m with s, d and n

1. What is the formula for solving a square in the x-y plane?

The formula for solving a square in the x-y plane is m = (s + d + n)/2, where m represents the length of the side of the square, s represents the area of the square, d represents the distance from the center of the square to any of its vertices, and n represents the number of vertices.

2. How do I find the length of the side of a square using the given parameters?

To find the length of the side of a square, simply plug in the values for s, d, and n into the formula m = (s + d + n)/2 and solve for m. The resulting value of m will be the length of the side of the square.

3. Can I find the area of a square using this formula?

No, this formula is used to find the length of the side of a square. To find the area of a square, you can use the formula A = s^2, where A represents the area and s represents the length of the side.

4. How many vertices does a square have?

A square has 4 vertices, which are the corners of the square where the sides meet.

5. What is the significance of finding the distance from the center of the square to its vertices?

The distance from the center of the square to its vertices is important because it helps determine the size and positioning of the square. It also plays a role in calculating the length of the side of the square using the given formula.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
17
Views
406
  • Precalculus Mathematics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
842
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
2K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
13
Views
2K
  • Precalculus Mathematics Homework Help
2
Replies
39
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
650
Back
Top