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Thread: The signboard problem

  1. #1 The signboard problem 
    Forum Ph.D. Leszek Luchowski's Avatar
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    Hi;

    here is a problem that has been buzzing at the back of my head for years. I think it would be easy to solve it with calculus, but I have been told there is a common-sense solution using just elementary geometry, and that is what I am looking for. But I don't even know where to begin.

    The problem goes like this:

    There is a vertical signboard of height positioned so that its lower edge is at height above the ground. How far from it should I stand so the angular width (heigth, actually) of the sign as seen by my eyes is greatest?

    For simplicity, let's assume my eyes are at ground level.

    TIA, LL.


    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  3. #2  
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    I don't understand the problem as stated, well I understand it, but I believe my understanding is incorrect. The way I look at it now, your distance would need to be infinity to get the way you view the sign to be as close as possible to h - H. I am correct to assume that the problem changes significantly with the vantage point; as an example what would change if your eyes were level with H?


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  4. #3  
    Forum Radioactive Isotope MagiMaster's Avatar
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    The problem makes sense to me. If your eyes were level with the board, the closer you get, the bigger it'd get. But if your eyes were below the board, as you got closer, you'd be looking at it edge on. As you get farther, it'd be shrinking visually. There'd have to be a maximum somewhere in between.
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  5. #4  
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    you have 2 right triangles in this problem, one with legs and one with legs where =height of bottom edge, =height of top edge, d=distance away from the vertical line that the sign lies in. The relevant angles will be the angles opposite the Heights, namely and , respectively. The maximum of the equation is the point. So you will have an equation like in which to play with to find the maximum. Derivatives I believe will come in handy, since is a constant you have an input variable and an output, d and \, respectively.
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  6. #5  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Thanks to MagiMaster for explaining my problem with more clarity.

    Arcane - yes I know this can be solved by looking for a maximum by equating the derivative to zero. And I think I could do it with relatively little effort. But when I first read about this problem it was explicitly stated (don't ask me where or by whom, I don't remember) that it can be done without resorting to calculus. And this is what has been puzzling me so long.

    Cheers, and looking forward,
    Leszek.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  7. #6  
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    You can solve it with trig alone. Identities for angle addition and angle subtraction come in VERY handy, most specifically the one for arctangent.
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    I tried something with the cosine rule it seemed to work for when the sign was on the ground which is a simple case.
    Anyhow I had an equation in d which you make equal a maximum value, there might be an obvious way to get the answer without differentiation


    anyway the equation for the angle was (2d^2 -2h1h2 ) /2(sqrt(h1^2 + d^2)

    where h1 and h2 were the hight of the bottom and top of the sign respectively,
    and d was the distance.


    I think that simplifies to
    = d^2 /(sqrt(h1^2 + d^2)

    you need to make that a maximum, is there an obvious answer to that?

    Anyway I probably screwed up the maths as the answer seems to be independent to the height of the top of the sign which seems wrong. I need too check the maths, but anyhow it did not look to me that there would be a simple answer.

    Well how about you make the distance the square of the height of the bottom of the sign that has the right kind of 'feel' to it.

    I mean I think we are going to be talking in terms of squares here one way or another.
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  9. #8  
    Forum Professor sunshinewarrior's Avatar
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    Fascinating problem. Thanks for that. I shall now spend the whole day/week going quietly insane trying to figure this thing out.

    I can't help feeling that a square (ie an isosceles right triangle) must be involved somehow...

    EDIT (After lots of frantic spreadsheet work)

    My initial intuition was false. Max angle subtended appears to be at a distance just less than the average height of the board (total height+height of base/2).

    Of course, I'm just chucking brute force at it - I presume the geometric solution will be elegant, if and when it presents itself...
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    Definitions; height of the top of the sign, height of the bottom of the sign, d= distance you are from the vertical plane that the sign lies in. is the angle between d and and is the angle between d and . This is a problem with constraints: and the problem of finding the difference of the two angles, making the equation for visual angle of the sign:
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    Forum Radioactive Isotope MagiMaster's Avatar
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    Since arctan is monotonic, you can ignore it and just maximize the equation inside. That simplifies things somewhat, but I still don't see how you'd find the maximum without taking a derivative.
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    assuming

    looks an awful lot like the quotient of 2 quadratics of the same variable. is a constant. I know you can find a max to this equation without derivatives. What we need to find is the minimum of
    A more general equation, for example, would be find the minimum of where a and b are constants.


    I cheated and took the derivative. The answer is the geometric mean of the 2 values, and ,
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  13. #12  
    Forum Professor sunshinewarrior's Avatar
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    Quote Originally Posted by Arcane_Mathematician
    I cheated and took the derivative. The answer is the geometric mean of the 2 values, and ,
    Cool answer - fits in with intuition. Now if only we can demonstrate it without using derivatives...
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    Forum Isotope Bunbury's Avatar
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    Hmm. My intuition said it should be where the angle between eye level and tthe top of the billboard is 45 degrees. No particular reason, it just seemed right, but I guess my intuition ain't worth the paper it ain't written on.
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    Forum Radioactive Isotope MagiMaster's Avatar
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    Well, it would be the point where the geometric mean of the top and bottom would be at 45 degrees.
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  16. #15  
    Forum Professor sunshinewarrior's Avatar
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    Quote Originally Posted by Bunbury
    Hmm. My intuition said it should be where the angle between eye level and tthe top of the billboard is 45 degrees. No particular reason, it just seemed right, but I guess my intuition ain't worth the paper it ain't written on.
    It's where I started, but realised that any solution had to take into account both heights. Second intuition was arithmetic mean, but testing showed that it didn't work.

    Geometric mean just seems right - neither too hot, nor too cold.
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  17. #16  
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    makes you wonder... perhaps this is one of those situations that, in the grand scheme of things, shows you exactly why the geometric mean is the way it is and is as useful as it is, instead of just being an abstract concept in geometry.
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  18. #17  
    Forum Professor sunshinewarrior's Avatar
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    Quote Originally Posted by Arcane_Mathematician
    makes you wonder... perhaps this is one of those situations that, in the grand scheme of things, shows you exactly why the geometric mean is the way it is and is as useful as it is, instead of just being an abstract concept in geometry.
    Not sure about grand schemes of things, but I always love it when connetions like these are pointed out. Perhaps why I infest a science forum.

    Found this fascinating diagram on wiki and wondered if it helps explain (proper mathematicians, anyone?) why the answer to the problem is the geometric mean?
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  19. #18  
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    that's pretty cool! I never thought of the means that way before. I don't know what the Q is though, but the others I understand, and no, it doesn't really help explain this specific situation. Great way to visualize the means.
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  20. #19  
    Forum Professor sunshinewarrior's Avatar
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    Quote Originally Posted by Arcane_Mathematician
    and no, it doesn't really help explain this specific situation. Great way to visualize the means.
    Oh well...

    The reason I was hopeful was, of course, that the way it was laid out was in a sense the signboard problem, but horizontal.

    Back to the drawing board... or signboard, as the case may be.
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  21. #20  
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    Taking the signboard question into the 3rd dimension would also be fun. A billboard that is H' by W' and A' from the road and B' off the ground, perpendicular to the road. In terms of spherical sectors, while traveling down the road when is it the 'biggest'? I'm also curious how we would determine what the 'biggest' is, I'm guessing relative unit-sphere sector area, but I don't know how to determine that either.
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  22. #21  
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    Quote Originally Posted by Arcane_Mathematician
    Taking the signboard question into the 3rd dimension would also be fun. A billboard that is H' by W' and A' from the road and B' off the ground, perpendicular to the road. In terms of spherical sectors, while traveling down the road when is it the 'biggest'? I'm also curious how we would determine what the 'biggest' is, I'm guessing relative unit-sphere sector area, but I don't know how to determine that either.
    Wouldn't that just be when you're directly opposite it, regardless of other considerations?
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  23. #22  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Thanks to all who answered; glad we've had some interesting discussion.

    Now I will spend the rest of my life pondering three questions:

    - Who told me this problem can be solved without calculus?
    - Can it?
    - Did they somehow construe taking the derivative as not being part of calculus?

    Cheers,
    L.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
    History teaches us that we don't learn from history.
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  24. #23  
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    Quote Originally Posted by sunshinewarrior
    Quote Originally Posted by Arcane_Mathematician
    Taking the signboard question into the 3rd dimension would also be fun. A billboard that is H' by W' and A' from the road and B' off the ground, perpendicular to the road. In terms of spherical sectors, while traveling down the road when is it the 'biggest'? I'm also curious how we would determine what the 'biggest' is, I'm guessing relative unit-sphere sector area, but I don't know how to determine that either.
    Wouldn't that just be when you're directly opposite it, regardless of other considerations?
    no, because it would be perpendicular to the road, just like how the signboard is perpendicular to your path.
    Wise men speak because they have something to say; Fools, because they have to say something.
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  25. #24  
    Forum Radioactive Isotope MagiMaster's Avatar
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    BTW, the term you're looking for is solid angle, although I think you'd get the same answer if you used the signs projection onto a specified plane.
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  26. #25  
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    That is exactly what I was talking about, thank you magi. And I had a sudden moment of clarity on the original problem, without the use of calculus (I believe).

    Quote Originally Posted by Arcane_Mathematician
    What we need to find is the minimum of
    A more general equation, for example, would be find the minimum of where a and b are constants.
    the minimum of is where and I can't remember how I know this, but I know it's true. Calculus provides at least one proof (of I hope many that it doesn't). but the equality simplifies to

    and used in this equation we have;
    : : :

    no calculus, afaik
    Wise men speak because they have something to say; Fools, because they have to say something.
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