Hard Geometry

Yes, I would like to see you using your method in solving this problem. Thanks!

civonamzuk 

ok but I am not sure where you want me to start.

I will assume for starters that you accept my diagram.

Since I said to let the sides of all the little euqilatoreral triangles be 2x, The height of the little tianges will be

\(h=\sqrt{(2x)^2-x^2}\\ h=\sqrt{3x^2}\\ h=\sqrt{3}\;x\\ Area\; of \;the \;little \;triangles\; = \sqrt3\; x^2 \)

Area of the sector ADD'

 \(=\frac{60}{360}*\pi * (2x)^2\\ =\frac{\pi}{6}*4x^2\\ =\frac{2\pi x^2}{3}\)

Area of the little segment

\(=\frac{2\pi x^2}{3}-\sqrt3\;x^2\\ =\frac{2\pi x^2-3\sqrt3\;x^2}{3}\\ =\frac{x^2}{3}(2\pi -3\sqrt3)\\\)

The area of the triangle outside the circle 

\(=\frac{3\sqrt3 x^2 }{3}-\frac{2\pi x^2-3\sqrt3 x^2}{3}\\ =\frac{3\sqrt3 x^2 -2\pi x^2+3\sqrt3 x^2 }{3}\\ =\frac{ x^2}{3}*(6\sqrt3 -2\pi )\\\)

But we are told that this area is 1

\(\frac{ x^2}{3}*(6\sqrt3  -2\pi )=1\\ x^2=\frac{-3}{2\pi-6\sqrt3}\\ x=\sqrt{\frac{-3}{2\pi-6\sqrt3}}\\ AB=4x\\ AB=4\sqrt{\frac{-3}{2\pi-6\sqrt3}}\approx 3.4178 \quad (edited: \;due \;to\; a \;copy \; error \;41 \;not \;14) \)

This answer looks really awfull so it is quite likely that I made a careless error somewhere.

You need to check my working and logic.

LaTex:

h=\sqrt{(2x)^2-x^2}\\
h=\sqrt{3x^2}\\
h=\sqrt{3}\;x\\
Area\; of \;the \;little \;triangles\;  = \sqrt3\; x^2

=\frac{60}{360}*\pi * (2x)^2\\
=\frac{\pi}{6}*4x^2\\
=\frac{2\pi x^2}{3}

=\frac{2\pi x^2}{3}-\sqrt3\;x^2\\
=\frac{2\pi x^2-3\sqrt3\;x^2}{3}\\
=\frac{x^2}{3}(2\pi -3\sqrt3)\\

=\frac{3\sqrt3 x^2 }{3}-\frac{2\pi x^2-3\sqrt3 x^2}{3}\\
=\frac{3\sqrt3 x^2 -2\pi x^2+3\sqrt3 x^2 }{3}\\
=\frac{ x^2}{3}*(6\sqrt3  -2\pi )\\

\frac{ x^2}{3}*(6\sqrt3  -2\pi )=1\\

x^2=\frac{-3}{2\pi-6\sqrt3}\\

x=\sqrt{\frac{-3}{2\pi-6\sqrt3}}\\
AB=4x\\
AB=4\sqrt{\frac{-3}{2\pi-6\sqrt3}}\approx 3.4178 (edited \;due \;to\; a \;copy\; error) 

I call you out on that.

This actually may work:

1/  We may find the ratio of the area of the double-segment to the yellow area. (That ratio doesn't change with the circle size)

2/   We can use that ratio and calculate the double-segment area.

3/   If CD = 2 then Yellow area = 1.369706513

4/   If CD = 2 then  Double-segment area = 0.724688589

PROPORTION:

1.369706513 : 0.724688589 = 1 : x

x = 0.529083115

Area of a sector = 1 + x = 1.529083115

Area of a circle = 6(1 + x) = 9.17449869

AB = 2[sqrt(9.17449869 / pi)] = 3.417796737

This is very hard to explain. I don't blame you if you do not understand it...

I checked these numbers, and it works. 

Our answers agree Jugoslav.  Good going.    :))

In geometry, there's always more than one way to solve a problem. Although our answers are the same, your method 

is more mathematical than mine. Good job. Melody! 

Hi Jugoslav,

I'm not really convinced that our answers are that much different.

You have written

If CD = 2 then Yellow area = 1.369706513

Where did this come from? 

I think you have mostly just estimated right from the beginning and left out the working that I included. 

No, no... I didn't use your numbers.

Post #5 is mine too; 

In my post #6, I wrote that "it may work" because I knew that that would work!

CD can be any number; it doesn't have to be 2.

I had to give the dimension to CD in order to find out the ratio between the yellow area and the area of the rest of the sector (I named it a double-segment area)

CD is a radius of a circle.

After that, it was easy, using the proportion, to calculate the area of the double-segment when the yellow area is 1 square unit.

I'm very bad at explaining things...

I used proportions too, in a more algebraic way.

I was not saying that you were using my numbers or that either of us was copying the other.

But I think our methods are more similar than you realize. 

It doesn't matter anyway,  We both got the same answer.  That is always a bonus :)