Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

atriangle

if atriangle has sides of length a,band c,  and the area is S
prove that
4S √3  ≥ a²+b²+c²

with equality holding if and only if the triangle is equilateral

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: atriangle

Proof it holds for equilateral:

Might play around with equality breaking later

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: atriangle

Yay I neary got it! But I'm having trouble with the end!

Last edited by Identity (2007-12-04 01:20:43)

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: atriangle

Hmm... something else:

Last edited by NullRoot (2007-12-04 01:46:26)

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: atriangle

Last edited by JaneFairfax (2007-12-04 02:57:37)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: atriangle

4S √3  ≥ a²+b²+c²

Are you are you’ve got the inequality sign the right way round?

Last edited by JaneFairfax (2007-12-04 07:32:03)

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: atriangle

Yeah he's crazy, isn't he?

I thought it must be a bit of confusion in his wording and took it to mean "Prove the inequality holds if and only if ABC (having sides a,b,c) is equilateral."

We managed to prove algebraically that the inequality holds for an equilateral triangle, and that it fails for a right-angled triangle, but we haven't proved it holds if and only if the triangle is equilateral.

Help, Jane!

In light of the mistake pointed out by Jane, here's another reason why it doesn't hold for right-angled triangles:

However, if he actually meant prove that 4S√3 ≤ a²+b²+c² then we've proven it's true for right angle triangles and 4S√3 = a²+b²+c² if the triangle is equilateral. We can work on generalizing that next

Last edited by NullRoot (2007-12-04 04:59:48)

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: atriangle

The correct inequality is

Now, what we do … we let

The Cauchy–Schwarz inequality says

with equality holding if and only x and y are linearly dependent – which in this case comes down to a[sup]2[/sup] = b[sup]2[/sup] = c[sup]2[/sup], or (since they’re sides of a triangle) a = b = c.

So, plug in the values

Now we use Heron’s formula for the area of a triangle:

From the C–S inequality

Hence

QED.

Last edited by JaneFairfax (2007-12-18 14:50:08)

tony123

Member

Registered: 2007-08-03

Posts: 229

Re: atriangle

I am sorry for all  This large error

you are right jane 4S √3  ≥ a²+b²+c² is  not right
4S √3  ≤ a²+b²+c² this is is right

A pologise

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: atriangle

In my proof I said

were linearly dependent if and only if a = b = c. Perhaps I’d better justify this. Now one way round is easy: if a = b = c. then x and y are clearly linearly dependent. Now assume conversely that they are linearly dependent, so we have h,k not both 0 such that hx+ky = 0. (Indeed, since neither x npr y is 0, if one of h and k is not 0, the other won’t be 0 either.)

The only real solution to this is x = −1, i.e. h = −k. It follows that a[sup]2[/sup] = b[sup]2[/sup] = c[sup]2[/sup], and so (since they’re all positive) a = b = c.
­

Last edited by JaneFairfax (2007-12-04 08:36:35)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: atriangle

*clapclapclap* nice proof, i can never remember the cauchy swarzy thing

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: atriangle

* bows *

Thanks. Maybe there’s a simpler proof, but Cauchy–Schwartz and Heron’s formula were what came into my mind.

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: atriangle

Nice one. Far more elegant than my number juggling.

Jane's got sabers and Null's got chainsaws

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: atriangle

Wow, that was an IMO problem! But then, all your problems are of very high difficulty. Here is another solution I found.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: atriangle

In fact i have a pdf full of inequality techniques which I can barely understand at all, but it may be of interest to you:

http://www.artofproblemsolving.com/Foru … h&id=5880]

100'

Member

Registered: 2007-12-06

Posts: 8

Re: atriangle

*clapclapclap* nice proof, i can never remember the cauchy swarzy thing

remember it from the basic dot product formula:
a.b = |a||b|cos t
where a and b are vectors and t is the angle between them.