# Maths proof

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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.

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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#2

(Original post by

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.

**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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#3

(Original post by

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.

**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.

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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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.

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.

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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#5

(Original post by

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.

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.

**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.

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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#6

**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.

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#7

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?

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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#8

(Original post by

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.

**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.

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#9

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°

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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#10

(Original post by

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°

**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°

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#11

(Original post by

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°

**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°

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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#12

(Original post by

Isn't it f(x,y) = sin(x) + sin(y) - sin(x+y) ?

**mikesgt2**)Isn't it f(x,y) = sin(x) + sin(y) - sin(x+y) ?

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#13

**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.

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#14

(Original post by

Does this involve partial differentials and Lagrange multipliers?

**Nylex**)Does this involve partial differentials and Lagrange multipliers?

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#15

(Original post by

I hope there's a way to avoid it

**theone**)I hope there's a way to avoid it

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#16

(Original post by

I hope there's a way to avoid it

**theone**)I hope there's a way to avoid it

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#17

(Original post by

theone, what do you think of fermat's proof?

**bono**)theone, what do you think of fermat's proof?

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#18

(Original post by

In all honesty, I think it's fine, but don't particularly like it because it uses calculus.

**theone**)In all honesty, I think it's fine, but don't particularly like it because it uses calculus.

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#19

(Original post by

theone, what do you think of fermat's proof?

**bono**)theone, what do you think of fermat's proof?

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#20

**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.

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