# trigonometry conundrum

• Dec 11th 2009, 07:48 AM
rainer
trigonometry conundrum
Please see attached image.

1) For what value(s) of b does

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

(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 $\displaystyle \frac{A_1}{A_2}$ 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.
• Dec 11th 2009, 02:54 PM
Defunkt
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:

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

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

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

So:

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

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

$\displaystyle \frac{1-b}{b} = 2b-1 \Rightarrow 2b^2-b = 1-b \Rightarrow b^2 = \frac{1}{2} \Rightarrow \boxed{b = \frac{1}{\sqrt{2}}}$
• Dec 11th 2009, 11:34 PM
simplependulum
Quote:

Originally Posted by Defunkt

$\displaystyle \frac{1-b}{b} = 2b-1 \Rightarrow 2b^2-b = 1-b \Rightarrow b^2 = \frac{1}{2} \Rightarrow \boxed{b = \frac{1}{4}}$

but $\displaystyle ( \frac{1}{4} )^2 = \frac{1}{16 }$ ...
• Dec 12th 2009, 02:30 AM
Defunkt
Quote:

Originally Posted by simplependulum
but $\displaystyle ( \frac{1}{4} )^2 = \frac{1}{16 }$ ...

$\displaystyle b=\frac{1}{\sqrt{2}}$ (Giggle)
• Dec 12th 2009, 08:51 AM
rainer
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

$\displaystyle y=x\tan{\theta}$

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 $\displaystyle \frac{1}{\sqrt{2}}$ as a trig function valid for all theta? (This is the same problem I posted in the puzzles forum)
• Dec 12th 2009, 01:04 PM
Defunkt
I don't see why $\displaystyle \frac{1}{\sqrt{2}}(cos^2(\theta)+sin^2(\theta))$ is not a valid answer. I also don't see the point in trying to find a trigonometric function that is constant!
• Dec 12th 2009, 02:57 PM
rainer
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?