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Area problem

  1. Dec 23, 2009 #1

    Mentallic

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    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.

    1. The problem statement, all variables and given/known data
    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 [Broken]

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

    3. 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: May 4, 2017
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  3. Dec 23, 2009 #2

    Mentallic

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    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:
     
  4. Dec 23, 2009 #3

    Mentallic

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    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).
     
  5. Dec 24, 2009 #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: Dec 24, 2009
  6. Dec 24, 2009 #5

    Mentallic

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    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?
     
  7. Dec 24, 2009 #6

    LCKurtz

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    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.
     
  8. Dec 24, 2009 #7

    Mentallic

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    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: Dec 24, 2009
  9. Dec 24, 2009 #8

    Mentallic

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    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: Dec 24, 2009
  10. Dec 24, 2009 #9

    Mentallic

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    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:
     
  11. Dec 25, 2009 #10

    Mentallic

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    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.
     
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