Maths proof Watch

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john !!
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#1
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Prove that the triangle of greatest area that fits completely inside a circle is an equilateral triangle.

I just randomly thought up this question. I had a go but got confused. I'll try later as well, but I'll put it up here if anyone else wants to try.
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Ralfskini
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(Original post by mik1a)
Prove that the triangle of greatest area that fits completely inside a circle is an equilateral triangle.

I just randomly thought up this question. I had a go but got confused. I'll try later as well, but I'll put it up here if anyone else wants to try.

If you thought the question up randomly then how do you know that the triangle of greatest area that can fit inside a circle IS an equilateral triangle.
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[email protected]
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(Original post by Ralfskini)
If you thought the question up randomly then how do you know that the triangle of greatest area that can fit inside a circle IS an equilateral triangle.
It is! I don't know why.
Would you have to get into some calculus? Maximising area etc. I've done a q similar to this before. I'll try and dig it out.
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john !!
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Good point. I added "or provide a counterexample", but decided to remove it because 'it obviously is an equilateral triangle'. That was my thought process I just quoted. :cool:

Maximum implies calculus. You need to set a limit of the traingle's dimensions so that it is bound by a circle, substitute something, and then differentiate and put the gradient to zero. But the problem is, unlike with a rectangle/square, there aren't two lengths, but three. So the part before the substitution seems like it'll be quite complicated.
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Ralfskini
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(Original post by mik1a)
Good point. I added "or provide a counterexample", but decided to remove it because 'it obviously is an equilateral triangle'. That was my thought process I just quoted. :cool:

Maximum implies calculus. You need to set a limit of the traingle's dimensions so that it is bound by a circle, substitute something, and then differentiate and put the gradient to zero. But the problem is, unlike with a rectangle/square, there aren't two lengths, but three. So the part before the substitution seems like it'll be quite complicated.
The two limits to use would be perpendicular height and the length of the side from which the perpendicular is dropped.
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theone
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(Original post by mik1a)
Prove that the triangle of greatest area that fits completely inside a circle is an equilateral triangle.

I just randomly thought up this question. I had a go but got confused. I'll try later as well, but I'll put it up here if anyone else wants to try.
Well one way I've thought of amounts to showing the expression f(x,y) = sin(x+y) + sin(x) + sin(y), with x and y < 180, is maximised when x = y = 120. I'll think about proving this later, but I'm busy now.
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4Ed
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i haven't really got time to think of a proof, as I'm at work atm, but I reckon you could break down the proof into two parts.

First, prove that for a given chord on a circle, the maximum area generated by a triangle using that chord as a base occurs when the other two lines are of equal length (an isoceles traingle).

Then, look at an isoceles triangle from a given point, where the base of the triangle varies in length. and prove that the maximum area will occur when the base length is equal to the other two sides.

Do people reckon that's a good way to think about it?
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mikesgt2
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(Original post by theone)
Well one way I've thought of amounts to showing the expression f(x,y) = sin(x+y) + sin(x) + sin(y), with x and y < 180, is maximised when x = y = 120. I'll think about proving this later, but I'm busy now.
Isn't it f(x,y) = sin(x) + sin(y) - sin(x+y) ?
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Fermat
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#9
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A simple geometrical proof will show that for any triangle within a circle, its max area is given when it is isosceles.

Continuing from there.

Looking at Fig 1. and taking similar triangles.

AD/DB = DC/AB

AD²=DB.DC

Also.

DC=D-DB, where D is the diameter of the circle

Area of triangle CEA,

A=½base x height
A=½.AE.DC
A=¼.AD.DC

Substituting for AD,

4A=√DB.√DC.DC
4A=√DB.(D-DB)^(3/2)

Now differentiate and equate to zero to find the maximum.

d(4A)/d(DB)=½.DB^(-½)(D-DB)^(3/2) -(3/2).DB^(½).(D-DB)^(½)

Since d(4A)/d(DB)=0, we get,

1/(2√DB).(D-DB)^(3/2) = (3DB)/(2√DB).(D-DB)^(1/2)
(D-DB) = 3DB
D=4DB
DB=D/4
======

DB=D/4
=> DC=3D/4
==========

AD²=(D/4).(3D/4)=3D²/16
AD=√3D/4
========

tan CAD = DC/AD = (3D/4)/(√3D/4)
tan CAD = (3/4).(4/√3) = √3

=> CAD = 60°
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Nima
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#10
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(Original post by Fermat)
A simple geometrical proof will show that for any triangle within a circle, its max area is given when it is isoceles.

Continuing from there.

Looking at Fig 1. and taking similar triangles.

AD/DB = DC/AB

AD²=DB.DC

Also.

DC=D-DB, where D is the diameter of the circle

Area of triangle CEA,

A=½base x height
A=½.AE.DC
A=¼.AD.DC

Substituting for AD,

4A=√DB.√DC.DC
4A=√DB.(D-DB)^(3/2)

Now differentiate and equate to zero to find the maximum.

d(4A)/d(DB)=½.DB^(-½)(D-DB)^(3/2) -(3/2).DB^(½).(D-DB)^(½)

Since d(4A)/d(DB)=0, we get,

1/(2√DB).(D-DB)^(3/2) = (3DB)/(2√DB).(D-DB)^(1/2)
(D-DB) = 3DB
D=4DB
DB=D/4
======

DB=D/4
=> DC=3D/4
==========

AD²=(D/4).(3D/4)=3D²/16
AD=√3D/4
========

tan CDA = DC/AD = (3D/4)/(√3D/4)
tan CDA = (3/4).(4/√3) = √3

=> CDA = 60°
Wow, superb.
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4Ed
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#11
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(Original post by Fermat)
A simple geometrical proof will show that for any triangle within a circle, its max area is given when it is isosceles.

Continuing from there.

Looking at Fig 1. and taking similar triangles.

AD/DB = DC/AB

AD²=DB.DC

Also.

DC=D-DB, where D is the diameter of the circle

Area of triangle CEA,

A=½base x height
A=½.AE.DC
A=¼.AD.DC

Substituting for AD,

4A=√DB.√DC.DC
4A=√DB.(D-DB)^(3/2)

Now differentiate and equate to zero to find the maximum.

d(4A)/d(DB)=½.DB^(-½)(D-DB)^(3/2) -(3/2).DB^(½).(D-DB)^(½)

Since d(4A)/d(DB)=0, we get,

1/(2√DB).(D-DB)^(3/2) = (3DB)/(2√DB).(D-DB)^(1/2)
(D-DB) = 3DB
D=4DB
DB=D/4
======

DB=D/4
=> DC=3D/4
==========

AD²=(D/4).(3D/4)=3D²/16
AD=√3D/4
========

tan CAD = DC/AD = (3D/4)/(√3D/4)
tan CAD = (3/4).(4/√3) = √3

=> CAD = 60°
so following my idea worked then! woohoo

glad to see my maths brain cells havent died off completely (tho i probably wouldn't know how to do the above )
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theone
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#12
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(Original post by mikesgt2)
Isn't it f(x,y) = sin(x) + sin(y) - sin(x+y) ?
I don't think so, since sin(360-x-y) = sin(x+y)?
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Nylex
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#13
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#13
(Original post by theone)
Well one way I've thought of amounts to showing the expression f(x,y) = sin(x+y) + sin(x) + sin(y), with x and y < 180, is maximised when x = y = 120. I'll think about proving this later, but I'm busy now.
Does this involve partial differentials and Lagrange multipliers?
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theone
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(Original post by Nylex)
Does this involve partial differentials and Lagrange multipliers?
I hope there's a way to avoid it
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Nima
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#15
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(Original post by theone)
I hope there's a way to avoid it
theone, what do you think of fermat's proof?
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Nylex
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#16
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(Original post by theone)
I hope there's a way to avoid it
You seem to know your stuff though.
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theone
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#17
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(Original post by bono)
theone, what do you think of fermat's proof?
In all honesty, I think it's fine, but don't particularly like it because it uses calculus.
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4Ed
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#18
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(Original post by theone)
In all honesty, I think it's fine, but don't particularly like it because it uses calculus.
can you think of a non calculus way to solve it? i probably would have ended up using calculus myself...
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makesomenoise
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#19
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(Original post by bono)
theone, what do you think of fermat's proof?
The margin ain't big enough :rolleyes:
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davey_boy
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#20
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#20
(Original post by mik1a)
Prove that the triangle of greatest area that fits completely inside a circle is an equilateral triangle.

I just randomly thought up this question. I had a go but got confused. I'll try later as well, but I'll put it up here if anyone else wants to try.
If you want to go further - prove that for any polygon with a given number of sides, the polygon of greatest area that fits completely inside a circle is a regular polygon (i.e. equilateral triangle for a polygon with 3 sides, square for a polygon with 4 sides, regular pentagon etc.....).
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