An equilateral triangle ABC is partly inscribed in a circle with diameter AB (See diagram).

The area of a triangle outside of a circle is equal to 1 square unit. Find the length of AB.

civonamzuk Feb 7, 2022

#1**+1 **

Hi civonamzuk,

You can break that triangle up into 4 conguent smaller triangles, all similar to the big one.

Let the side length of the equilatieral little triangles be 2x

find all the necessary areas in terms of x.

then put the segment = 1 and solve for x.

I can give more help if you need it..

Melody Feb 7, 2022

#3**+1 **

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

civonamzuk

Guest Feb 8, 2022

#4**+2 **

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)

Melody
Feb 8, 2022

#2**0 **

You can solve this using the proportion. This is how I've done it.

But I won't give you the answer.

This problem is really good! Kudos, civonamzuk!

Guest Feb 8, 2022

#6**+2 **

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.

jugoslav
Feb 8, 2022

#8**0 **

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!

jugoslav
Feb 8, 2022

#9**0 **

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.

Melody
Feb 8, 2022

#10**0 **

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...

jugoslav
Feb 8, 2022

#11**0 **

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 :)

Melody
Feb 9, 2022