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

Mathematician
Jan 28, 2015

#3**+10 **

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.

CPhill
Jan 28, 2015

#1**+5 **

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...!!!

CPhill
Jan 28, 2015

#2**+5 **

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.)

Mathematician
Jan 28, 2015

#3**+10 **

Best Answer

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.

CPhill
Jan 28, 2015

#4**+5 **

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

Mathematician
Jan 28, 2015

#5**+5 **

I do not know Mathematician.

Maybe another mathematician can come up with something.

Melody
Jan 28, 2015

#6**0 **

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

CPhill
Jan 28, 2015

#7**+5 **

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

Mathematician
Jan 28, 2015

#8**0 **

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

CPhill
Jan 28, 2015

#9**+5 **

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)

7.7942286340599478#sthash.QLAWaiN0.dpuf

7.7942286340599478#sthash.QLAWaiN0.dpuf

CPhill
Jan 28, 2015

#10**+5 **

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

Melody
Jan 28, 2015

#11**+5 **

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!!!

CPhill
Jan 28, 2015

#13**+5 **

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

.

Alan
Jan 28, 2015

#14**+5 **

**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}}$$

heureka
Jan 28, 2015

#15**+5 **

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**

Melody
Jan 28, 2015

#16**+5 **

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

heureka
Jan 28, 2015

#18**+5 **

+

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

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

| +

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x/2 | +

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+

heureka
Jan 28, 2015

#19**+8 **

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**

Melody
Jan 28, 2015

#20**+5 **

**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

.

Alan
Jan 28, 2015

#21**+5 **

Hi Melody,

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

bye.

heureka
Jan 28, 2015

#22**+5 **

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

.

Alan
Jan 28, 2015

#23**+5 **

Hi Alan,

Tiny Pic Upload Service says:

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

heureka
Jan 28, 2015

#24**0 **

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

Melody
Jan 29, 2015

#26**+5 **

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}

$

}}$$

heureka
Jan 29, 2015

#28**+5 **

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

CPhill
Jan 29, 2015

#30**+5 **

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

Melody
Jan 29, 2015

#31**+5 **

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!!!

CPhill
Jan 29, 2015

#32**+5 **

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.

Mathematician
Feb 3, 2015