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Math Help - trigonometry conundrum

  1. #1
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    trigonometry conundrum

    Please see attached image.

    1) For what value(s) of b does

    \sqrt{\frac{A_1}{A_2}}=2b-1 for all \theta ?

    (A1=area of the small triangle on top, A2=area of the bigger triangle on the bottom with base b)


    2) bonus problem:

    Given the foregoing, express \frac{A_1}{A_2} as a hyperbolic function with b in the argument. Express b as trig function with theta in the argument.

    I would never be able to solve this if I hadn't been thinking about it for the past few months.
    Attached Thumbnails Attached Thumbnails trigonometry conundrum-math-problem-2.jpg  
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  2. #2
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    Sort of in a hurry, so this may have mistakes in it, but here is my approach to the first: (Note that I assumed, as per the image, that 1>b>0)

    Let P denote the rightmost side of the triangle A2 and Q denote the rightmost side of the triangle A1. Then:

    tan \theta = \frac{P}{b} = \frac{Q}{1-b} \Rightarrow Q = \frac{1-b}{b}P

    but: A_1 = \frac{(1-b)Q}{2}

    and: A_2 = \frac{Pb}{2}

    So:

    \frac{A_1}{A_2} = \frac{Q(1-b)}{Pb} = \frac{(1-b)^2}{b^2} = (\frac{1-b}{b})^2

    So we want to solve: \pm \frac{1-b}{b} = 2b-1 w.r.t b. However, \frac{1-b}{b} is strictly positive since 1>b, b>0 therefore -\frac{1-b}{b} is obviously not an option. Now we get:

    \frac{1-b}{b} = 2b-1 \Rightarrow 2b^2-b = 1-b \Rightarrow b^2 = \frac{1}{2} \Rightarrow \boxed{b = \frac{1}{\sqrt{2}}}
    Last edited by Defunkt; December 12th 2009 at 02:31 AM. Reason: typo
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  3. #3
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    Quote Originally Posted by Defunkt View Post

    \frac{1-b}{b} = 2b-1 \Rightarrow 2b^2-b = 1-b \Rightarrow b^2 = \frac{1}{2} \Rightarrow \boxed{b = \frac{1}{4}}
    but  ( \frac{1}{4} )^2 = \frac{1}{16 } ...
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  4. #4
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    Quote Originally Posted by simplependulum View Post
    but  ( \frac{1}{4} )^2 = \frac{1}{16 } ...


    b=\frac{1}{\sqrt{2}}
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    That's right Defunkt. Another way to derive the heights of the two triangles is to consider that the equation for the hypotenuse line is

    y=x\tan{\theta}

    then substituting in b and 1 for x.

    So, in five seconds you made mince meat of a problem that has occupied me for a few months!

    But what about the second part? Can you develop an expression for A1/A2 in terms of a hyperbolic function with b in the argument? This is actually the part that has most occupied me.

    And can you express the answer \frac{1}{\sqrt{2}} as a trig function valid for all theta? (This is the same problem I posted in the puzzles forum)
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  6. #6
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    I don't see why \frac{1}{\sqrt{2}}(cos^2(\theta)+sin^2(\theta)) is not a valid answer. I also don't see the point in trying to find a trigonometric function that is constant!
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  7. #7
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    Sorry if I've annoyed you. Of course your expression is valid too. Just wanted to draw a little bit more attention since the finding I posted in the puzzles forum does amount to "original research," however insignificant it may be.

    What about the hyperbolic expression? Any progress there?
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