# Maths proof

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#1
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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16 years ago
#2
(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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16 years ago
#3
(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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#4
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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16 years ago
#5
(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. 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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16 years ago
#6
(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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16 years ago
#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?
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16 years ago
#8
(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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16 years ago
#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.

Also.

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

Area of triangle CEA,

A=½base x height
A=½.AE.DC

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
==========

========

tan CAD = (3/4).(4/√3) = √3

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16 years ago
#10
(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.

Also.

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

Area of triangle CEA,

A=½base x height
A=½.AE.DC

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
==========

========

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

=> CDA = 60°
Wow, superb. 0
16 years ago
#11
(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.

Also.

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

Area of triangle CEA,

A=½base x height
A=½.AE.DC

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
==========

========

tan CAD = (3/4).(4/√3) = √3

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 )
0
16 years ago
#12
(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)?
0
16 years ago
#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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16 years ago
#14
(Original post by Nylex)
Does this involve partial differentials and Lagrange multipliers?
I hope there's a way to avoid it 0
16 years ago
#15
(Original post by theone)
I hope there's a way to avoid it theone, what do you think of fermat's proof? 0
16 years ago
#16
(Original post by theone)
I hope there's a way to avoid it You seem to know your stuff though.
0
16 years ago
#17
(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.
0
16 years ago
#18
(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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16 years ago
#19
(Original post by bono)
theone, what do you think of fermat's proof? The margin ain't big enough 0
16 years ago
#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.....).
0
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