A puzzle (for those who are bored!)

1: Cos(30) = 1/r therefore r = 2/sqrt3 and r^2 = 4/3

2:Area of equil Triang = (1/2)*(2)*sqrt((2^2)-(1^2))=sqrt3 therefore Area of rectangle = xy = sqrt3

3 Draw lines to the opposite vertices of the rectangle then add the area of the 4 triangles you just made inside the rectangle = total area:

2*area Tri A= 2*[(1/2)(x)sqrt(4/3 - (x^2)/4)

2*area Tri B= 2*[(1/2)(y)(x/2)]

4, Solve simultaneous equation to get x and y:

xy = sqrt3

(x)sqrt(4/3 - (x^2)/4) + (1/2)xy = sqrt3

--> If doing manually you substitute the first equation into the second then solve for x, followed by y, but since I'm lazy I used Wolfram alpha to do this for me and got

x≈2.16662 and y≈0.799424

Oh, cool!  Thanks!

I thought this was a really cool question.  Thanks for providing it Alan.

Isn't any one of you high school students going to show us what you are made of and answer it.

You could at least give it a try :/    

That was nice!!

Correct numerical answer Bro, though it would be nice to see your full expressions for x and y!  The reason for this is that there are several ways to solve this problem, each of which can result in expressions that look different, but are, of course, equivalent.

.

I did like Brodudedoodebrodude in principle (even if I´m a little late.)

x:

$${\frac{{\sqrt{{\mathtt{3}}}}}{{\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}}}$$

y:

$${\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}$$

This has probably been done to death but too bad.

The area of the triangle is

 $$\\3\times \frac{1}{2}\times r^2*sin \frac{2\pi}{3}\\\\
=\frac{3r^2}{2}sin\frac{\pi}{3}\\\\
=\frac{3r^2}{2}\times\frac{\sqrt{3}}{2}\\\\
=\frac{3\sqrt{3}r^2}{4}\\\\$$

 $$\\Area\\
=4\times\frac{1}{2}\times r^2sin\theta\\\\
=2r^2sin\theta\\\\
so\\\\
2r^2sin\theta=\frac{3\sqrt3*r^2}{4}\\\\
sin\theta=\frac{3\sqrt3}{8}\\\\
If\;opp=3\sqrt3\;\;and\;\;hyp=8\;\;then\;\;adj=\sqrt{64-27}=\sqrt{37}\\\\
so \;\;cos\theta=\frac{\sqrt{37}}{8}\\\\
-----\\
x^2=2r^2-2r^2cos\theta\\
x^2=2r^2-2r^2\times\frac{\sqrt{37}}{8}\\
x^2=\frac{8r^2}{4}-\frac{r^2\sqrt{37}}{4}\\
x^2=\frac{r^2}{4}\left(8-\sqrt{37}\right)\\\\
x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies$$

$$\\xy=\frac{3\sqrt3r^2}{4}\\\\
y=\frac{3\sqrt3r^2}{4}\div x\\\\
y=\frac{3\sqrt3r^2}{4}\times \frac{1}{x} \\\\
y=\frac{3\sqrt3r^2}{4}\times \frac{2}{r\sqrt{8-\sqrt{37}}} \\\\
y=\frac{3\sqrt3r}{2}\times \frac{1}{\sqrt{8-\sqrt{37}}} \\\\
y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\$$

Is this answer the same as the others?

This is fine as far as it goes Melody, and if you calculate r you can take it all the way!

.

Yes I forgot about that Alan.   Maybe this is better   

$$\\4=2r^2-2r^2cos120\\
4=2r^2-2r^2*-0.5\\
4=2r^2+r^2\\
4=3r^2\\
\frac{4}{3}=r^2\\
r=\frac{2}{\sqrt3}\\\\
y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\
y= \frac{3*\frac{2}{\sqrt3}\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\
y= \frac{6}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\
y= \frac{3}{\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\\\
x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\
x=\frac{\frac{2}{\sqrt3}*\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\
x=\frac{\sqrt{3\left(8-\sqrt{37}\right)}}{3}\;\;unit\;thingies$$

The area of the equilateral triangle = (1/2) 4 sin120 = 2(√3)/ 2  = √3

The radius of both circles is given by  √[ 1^2 + (1/√3)^2 ]  =   √[1 + 1/3]   =  2/ √3

So....the diagonal of the rectangle = 2 ( 2/ √3 )  =  4 / √3

And, using the Pythagorean Thorem, "y" in the rectangular figure = √ [( 4/ √3) ^2 - x^2 ] = √[16/3 - x^2]

So......since the area of the triangle and the rectangle is equal, we have...

y * x  = √3

√[16/3 - x^2] * x =  √3    square   both sides

(16/3)x^2 - x^4  = 3        multiply through by 3

16x^2 - 3x^4  = 9            rearrange

3x^4 - 16x^2  + 9 = 0         let m= x^2   ....  so we have

3m^2 - 16m + 9 = 0

Using the quadratic formula [and an assist from the on-site solver], we have

$${\mathtt{3}}\left[{{m}}^{{\mathtt{2}}}\right]{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{m}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,-\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\
{\mathtt{m}} = {\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{0.639\: \!079\: \!156\: \!567\: \!260\: \!1}}\\
{\mathtt{m}} = {\mathtt{4.694\: \!254\: \!176\: \!766\: \!073\: \!2}}\\
\end{array} \right\}$$

So  x = about .7994   or x = about 2.1666

If x = .799, y = √[16/3 - .7994^2] = .2166    ...but this is impossible....since y is smaller than x

If x = 2.1666, y = √[16/3 - 2.1666^2]  = about . 7994

So....x =about .2166 and y = about .7994

Nice answer Chris  

Yours looks a bit easier than mine.

I suppose I should include my approx answers so that the answersc an be compared.

This was a really good question Alan, thanks for posting it :)   

Thanks, Melody.....I realized that making that quadratic substitution  would prevent some "messy" radicals...!!!!