Length of an equilateral traingle within a circle

The Law of Cosines is this

c^2  = a^2 + b^2 - 2(a)(b)cos(theta)

Where c is the opposite side to the angle "theta" ....and "theta" is the included angle between sides a and b.

So...... in the circle, a and b are radii = 9  ....and since there are three equal sides of the triangle, each side spans an arc of 120 degrees.

We can use the Law of Cosines to solve this..letting x be the side length, we have...

x^2 =  (4.5)^2 + (4.5)^2 - 2(4.5)(4.5)cos120

x= √60.75  = 7.7942286340599478  = 4.5√3

Thanks, Mathematician...for seeing my previous error...!!!

Can you give me the formula, and then the explanation? And also how to get an appriximation of the answer? I'm confused by your explanation (Probably because I'm unfamiliar with Cosines.)

Just one final question. Can you explain it without Trigonmetry (cosines)? I haven't learned that yet.

I do not know Mathematician.

Maybe another mathematician can come up with something.   

Sorry, Mathematician....I'm kinda' 'stuck" with using Trig.....if I think of some geometric way to do it...I'll let you know...!!!

I just realized that the answer doesn't make sense; shouldn't the side be less than the diameter to fit within the circle?

Eh...you're right..I read "diameter" as "radius'.....let me go back and correct that....sorry!!!

Note in the pic below that FH = (1/2)s    where s is the side length

And also that DA =  AF = r = 4.5

So we have, by the Pythagorean Theorem

AF^2 =  (1/2)s^2 + AH^2

4.5^2 = 20.25 = (1/2)s^2 + AH^2

AH^2 = 20.25 - (s/2)^2

AH = √(20.25 - (s/2)^2 )

Therefore, again using the Pythagorean Theorem, we have

s^2 = (s/2)^2 + [DA + AH]^2

s^2 = (s/2)^2 + ((4.5 + √(20.25 - (s/2)^2 ))^2

(3/4)s^2 = 20.25 + 9 √(20.25 - (s/2)^2 ) + 20.25 - (s/4)^2   (add (s/4)^2 to each side)

s^2 = 40.5 + 9 √(20.25 - (s/2)^2 )  subtract 40.5 from each side

s^2 - 40.5  = 9 √(20.25 - (s/2)^2 )   square both sides

(s^2 - 40.5)^2 = 81 (20.25 - (s^2)/4)

s^4 - 81s^2 + 1640.25 = 1640.25 - (81/4)s^2

s^4 - 81s^2 + (81/4)s^2 = 0

s^4 - (243/4)s^2 = 0

Factor

(s^2) (s^2 - (243/4)) =0   one solution is s = 0 .... reject this

So we have

(s^2 - 243/4) = 0

s = (1/2)√243 = (1/2)(9)√3 = 4.5√3 = 7.7942286340599478

Here's a pic....(the angles are not exact..it's just to give you a feel for the geometry of the problem)

I had a thought Chris,  we (well Alan or Heureka) could probably do this with calculus as well.   

I've done it with Calculus, before......but...since "Mathematician" hasn't had trig......I doubt that Calculus would do him much good, either......LOL!!!

Yes I was't thinking that it would be much use for Mathematician.  LOL

My solution below is basically the same as Chris's, but laid out slightly differently:

.

What is the side length of an equilateral triangle that fits exactly inside a circle with a 9 inch diameter ?

I.

$$\cos{(60\ensurement{^{\circ}})}=\frac{r-h}{r}=\frac{1}{2} \quad | \quad \small{\text{h = the height of the chord}} \\
\Rightarrow \boxed{h=\dfrac{r}{2}}$$

II.

Intersecting chord theorem:

$$\small{\text{
$
h*(d-h)= (\frac{x}{2})*(\frac{x}{2}) = (\frac{x}{2})^2
$
}}\\
\Rightarrow \boxed{x = d* \frac{ \sqrt{3} } {2}= 4.5* \sqrt{3}}$$

okay I have looked at CPhill's answer and I decided to reproduce it - but it is still CPhill's answer.

I left labelling the same as CPhill's except that I called the side length 2x inches.  

First I want to find AH in terms of x

$$\\AH^2=4.5^2-x^2\\
AH^2=20.25-x^2\\
AH=\sqrt{20.25-x^2}\\$$

Now consider the big right angled triangle DEH

$$\begin{array}{rll}
DH^2&=&DE^2-EH^2\\
(DA+AH)^2&=&(2x)^2-x^2\\
(4.5+\sqrt{20.25-x^2})^2&=&4x^2-x^2\\
(4.5+\sqrt{20.25-x^2})^2&=&3x^2\\
\sqrt{(4.5+\sqrt{20.25-x^2})^2}&=&\sqrt{3x^2}\\
4.5+\sqrt{20.25-x^2}&=&\sqrt{3}\;x\\
\sqrt{20.25-x^2}&=&\sqrt{3}\;x-4.5\\
(\sqrt{20.25-x^2})^2&=&(\sqrt{3}\;x-4.5)^2\\
20.25-x^2&=&3x^2-2*4.5\sqrt{3}\;x+4.5^2\\
20.25-x^2&=&3x^2-9\sqrt{3}\;x+20.25\\
-4x^2&=&-9\sqrt{3}\;x\\
x^2&=&\frac{9\sqrt{3}\;x}{4}\\
x&=&\frac{9\sqrt{3}}{4}\\
2x&=&\frac{9\sqrt{3}}{2}\\
\end{array}$$

So each side is exactly   $$\frac{9\sqrt{3}}{2}$$    inches long

One can turn the equilateral triangle, thus one side is vertical, then the height (h) of the curve (chord) is half the radius.

I do not understand Heureka.   

                                          +

                                          |   .      +

                                          |      .            +

                                 x/2    |          .                +

                                          |            r                      +

                                          |                .                         +

                                          |           60      .                              +

            arc     (<----r/2---->+<----r/2----> C <------------r--------->) arc

                       <-      h   -> | <-   r - h    ->                            +

                                          |                                          +

                                          |                                   +

                              x/2       |                            +

                                          |                     +

                                          |            +

                                          |      +

                                          +

I was just looking at what Heureka was saying.  It took me a while to work it out.

                                                   B

This is the triangle that Heureka was describing

The radius that bisects the chord is perpendicular to the chord so <AHE=90 degrees

which means <AEH=30 degrees (Angle sum of a triangle)

Using trig I know that AH=0.5*AE=2.25  but I am not allowed to use trig.  

Now Heureka says that AH=HB  

Edited: that is because AB is one radius and AH is 1/2 radius so  HB must also be a 1.2 radius 

Join EB forming a new triangle HEB

Now triangle HEB is congruent to triangle HEA because all the angles are congruent and they have a common corresponding side.

therefore triangle AEB is equilateral  and HA=HB

therefore all the sides are  4.5 inches

therefore HA=HB=2.25 inches

 $$\\x^2=4.5^2-2.25^2\\
x^2=20.25-5.0625\\
x^2=15.1875\\
x^2=151875/10000\\
x=\sqrt{151875/10000}\\
x=\sqrt{25*25*81*3/10000}\\
x=(25*9/100)\sqrt{3}\\
x=(2.25)\sqrt{3}\\
side = 4.5\sqrt{3}\;\;inches$$

Thanks Heureka 

Now Heureka says that AH=HB  

This certainly looks right but at present I can't see how to prove it.

Given that AH = r/2, AD = r and DB = 2r  where r is the radius, then HB = DB - AH - AD = 2r - r/2 - r  = r/2 = AH

.

Hi Melody,

i thank you very much.  Sorry i  can't  Insert pictures. I don't know why.

bye.

Heureka, have you tried the process explained here:  http://web2.0calc.com/questions/how-to-upload-an-image-nbsp_1 ?

.

Hi Alan,

Tiny Pic Upload Service says:

"This IP address has been banned for violating our Terms of Use" ????????????????????

Thank you Alan.  Yes is usual for two half radii to equal 1 radii.  I guess I had a brain freeze  

Why doesn't some smart person show their skill and do this with calculus.   

Hi Melody,

my calculation with your picture of the triangle without trigonometry:

I.  The triangle BED is an right angle triangle ( Thales Theorem ). See : http://www.mathopenref.com/thalestheorem.html

II.  The triangle AEB is equilateral so the side EB = r ! ( You have worked out this )

III. We have the Altitude on Hypotenuse Theorem in a right triangle : EB * EB = BH * BD . See: http://www.dummies.com/how-to/content/how-to-solve-problems-with-the-altitude0nhypotenus.html

with EB = r and BD = d = 2r  we have: r*r = BH * 2r   or   r = BH * 2   or  $$\small{\text{
$
\boxed{BH = \frac{r}{2}}
$
}}$$

IV. Now we have the Geometric mean Theorem in a right triangle:  BH * HD = x * x

with BH = r/2 and HD = d - r/2 = 2r - r/2. See: https://en.wikipedia.org/wiki/Geometric_mean_theorem

$$\small{\text{
$
\dfrac{r}{2}\left(
2r-\dfrac{r}{2}
\right)=x^2,\ \quad
r^2-\dfrac{r^2}{4} = x^2 ,\ \quad
\dfrac{3}{4} r^2 = x^2 ,\ \quad
\dfrac{\sqrt{3}}{2}r = x
$
}}\\\\
\small{\text{
The side is
$
2x = \sqrt{3}*r
$
}}
\small{\text{
and $ r = \frac{9}{2} = 4.5 $ we have the side
= 4.5*\sqrt{3}
$
}}$$

Thank you Heureka   

This boils down to finding the side length of an equiateral triangle inscribed in a circle. Of course, the area is maximized when the side lengths are maximized....!!!

Applying the Calculus to our given problem where r = 4.5 and x = s

We have

A = .5x(4.5 + √(20.25 - .25x^2))

A = 2.25x + .5x√(20.25 - .25x^2)

dA/dx = 2.25 +.5(20.25 - .25x^2)^(.5) - (.125x^2)(20.25 - .25x^2)^(-.5) = 0

18 + 4(20.25 - .25x^2)^(.5) - x^2(20.25 - .25x^2)^(-.5) = 0  rearrange

18 + 4(20.25 - .25x^2)^(.5) = x^2(20.25 - .25x^2)^(-.5)

18 (20.25 - .25x^2)^(.5) + 4 (20.25 - .25x^2) = x^2

18(20.25- .25x^2)^(.5) = x^2 - 4 (20.25 - .25x^2)     square both sides

324(20.25- .25x^2) = x^4 - 8 (20.25 - .25x^2)x^2 + 16(20.25-.25x^2)^2

6561 - 81x^2 = x^4 - 162x^2 + 2x^4 + 16(6561/16 - 61/8x^2 + 1/16x^4)

6561 - 81x^2 = 4x^4 - 324x^2  +6561

4x^4 - 243x^2 = 0   factor

x^2 (4x^2 - 243) = 0      reject x =0

4x^2 - 243 - 0

4x^2 = 243

x^2 = 243/4

x = √(81 * 3) / 2 =   (9/2)√3  = 4.5√3 = s

And this result can be generalized to r√3 for any given radius

Thank you Chris  :D

More for me to think about :)))

Mathematician, I bet you didn't think that your question would attract THIS much attention  LOL  

I don't think that "Fermat's Last Theorem" got as much attention as this one did...!!!

{But..it took a lot longer to prove....over 350 years....}

Regardless...it was a good one to play around with, Mathematician.....many different approaches by our "professionals"

And even some bumbling and stumbling by me, too!!!

Yes, a very interesting question. Someone commented I was a part of the Illuminati when they saw it on my project... it seems like that craze has taken over my school.

The "Illuminati".......LOL!!!

BTW....do you recognize this symbol????.....if so....you may REALLY be part of  "THE ILLUMINATI"