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        riddles >> easy >> Equilateral triangle
        
(Message started by: NickH on Jun 20th, 2004, 1:13pm)
Title: Equilateral triangle
Post by NickH on Jun 20th, 2004, 1:13pm
Prove that, of all triangles with a given perimeter, the equilateral has the greatest area.
Title: Re: Equilateral triangle
Post by Leonid Broukhis on Jun 20th, 2004, 10:46pm
Doesn't it follow from the formula S = [sqrt]p(p-a)(p-b)(p-c), where p = (a+b+c)/2 ?
Or you want us to prove the formula?
Title: Re: Equilateral triangle
Post by Sir Col on Jun 21st, 2004, 12:42am
Is it really that obvious how it follows from the formula?  ???
Title: Re: Equilateral triangle
Post by THUDandBLUNDER on Jun 21st, 2004, 3:11am

on 06/21/04 at 00:42:50, Sir Col wrote:
Is it really that obvious how it follows from the formula?  ???

By symmetry, the maximum must be when a = b = c

Title: Re: Equilateral triangle
Post by Eigenray on Jun 21st, 2004, 6:12am
Perhaps more elementarily, consider [hide]the ellipse[/hide].
Title: Re: Equilateral triangle
Post by Barukh on Jun 21st, 2004, 9:03am
Like it always happens with Nick's problems, this one is very elegant.


on 06/20/04 at 22:46:41, Leonid Broukhis wrote:
Doesn't it follow from the formula S = [sqrt]p(p-a)(p-b)(p-c), where p = (a+b+c)/2 ?
Or you want us to prove the formula?

If Nick doesn't want, I do! Would you?  ;)
Title: Re: Equilateral triangle
Post by Leonid Broukhis on Jun 22nd, 2004, 1:52pm
Let a be the base of the triangle, and x be the fraction of the base that lies on the 'b' side of the height. Then h[sup2] = b[sup2] - x[sup2]a[sup2] = c[sup2] - (1-x)[sup2]a[sup2]. Solve for x (it is linear), then insert into S[sup2] = h[sup2]a[sup2]/4
Title: Re: Equilateral triangle
Post by Barukh on Jun 25th, 2004, 3:49am

on 06/22/04 at 13:52:01, Leonid Broukhis wrote:
Let a be the base of the triangle, and x be the fraction of the base that lies on the 'b' side of the height. Then h[sup2] = b[sup2] - x[sup2]a[sup2] = c[sup2] - (1-x)[sup2]a[sup2]. Solve for x (it is linear), then insert into S[sup2] = h[sup2]a[sup2]/4

If I am not mistaken, this approach requires rather unpleasant simbolic manipulations. Here's a purely geometrical approach.

All that's needed is drawn in the attached picture. The small circle is the incircle, and the big one is the escribed circle, or excircle (it touches one side of the triangle, and the continuations of two others).

Let IY = r, IaYa = ra. It is not difficult to see that AYa = p, AY = p-a (why?). Also, the area S = rp.

Then, there are two pairs of similar triangles: AIY and AIaYa; ICY and CIaYa. From the first pair, we have the relation r/(p-a) = ra/p. The second pair gives r/CY = CYa/ra, or r/(p-c) = (p-b)/ra. Combining the two:
S2 = rprp = rpra(p-a) = p(p-a)(p-b)(p-c).

I saw this proof a few years ago on triangle geometry forum. It really impressed me.
Title: Re: Equilateral triangle
Post by Grimbal on Jun 25th, 2004, 4:58am
I just thought of a more visual proof.

Imagine you fix one side and you can vary the 2 others.  It is like having a loop over 2 pegs on a horizontal line.  You complete the triangle with your finger.  To get the maximum surface, the finger must be as high as possible, which is obviously in the middle.  So, whatever the length of one side, the 2 other sides must be equal to reach the maximum.  So, the maximum is when all 3 sides are equal.
Title: Re: Equilateral triangle
Post by Sir Col on Jun 29th, 2004, 5:02am

on 06/25/04 at 03:49:07, Barukh wrote:
AY = p-a (why?)

Good question!  ???
(and could you also explain why the area S=rp?)

And whilst I'm being incredibly stupid...

Grimbal, if I am understanding your method (and perhaps I am not), doesn't it only demonstrate that an isosceles triangle maximises the area? If the sum of the other two lengths is not double the fixed length, won't your proof fail?
Title: Re: Equilateral triangle
Post by Eigenray on Jun 29th, 2004, 5:30am

on 06/29/04 at 05:02:43, Sir Col wrote:
Good question!  ???

I'll tell you why AY = p-a, if you tell me why AYa = p. ;)
Let X be the point of intersection of the circle I with side AB, and X' the intersection of circle I with side BC.
Then  AX = AY, CY=CX', BX=BX', so
AY + CY + CYa = AYa = p = AY + CY + BX'
iff CYa = BX'.
So that's true iff AY = AYa - CY - CYa = p - (CX' + BX') = p - a.


Quote:
(and could you also explain why the area S=rp?)

Divide it into smaller triangles you can find the areas of.


Quote:
Grimbal, if I am understanding your method (and perhaps I am not), doesn't it only demonstrate that an isosceles triangle maximises the area? If the sum of the other two lengths is not double the fixed length, won't your proof fail?

It demonstrates that no matter which two sides we pick, they must be equal.
Title: Re: Equilateral triangle
Post by Sir Col on Jun 29th, 2004, 10:32am
Aha! Thanks, Eigenray.


on 06/29/04 at 05:30:17, Eigenray wrote:
I'll tell you why AY = p-a, if you tell me why AYa = p. ;)

I labelled the point where the large circle meets the tangent through AB, as Z. Using the labels on the diagram...

As CX = CYa, AYa = b + CX
As BX = BZ, AZ = c + BX
As AYa = AZ, 2AYa = AYa + AZ = b + c + CX + BX = a + b + c
Hence AYa = (a + b + c)/2 = p
Title: Re: Equilateral triangle
Post by rmsgrey on Jun 30th, 2004, 12:35pm

on 06/29/04 at 05:30:17, Eigenray wrote:
It demonstrates that no matter which two sides we pick, they must be equal.

It certainly demostrates that the equilateral triangle is a local optimum, but I'm not convinced that it shows directly that the equilateral triangle is globally optimal. On the other hand, the extra work required is fairly small: take an arbitrary triangle, pick a longest side, and make the other two equal (and shorter than the longest - with the shortest side getting longer unless the two were already equal) then fix one of those two sides and equalise the longest and the other (shortening the longest and lengthening the other) You now have a triangle with longest side is shorter than the original (except when the original was equilateral), likewise, the shortest side is longer, and the area has increased (since each equalising operation increases the length) so by doing this repeatedly a sequence of ever larger triangles is produced, whose side lengths are bounded above and below by quantities that converge inwards. They converge to equality (if they don't, then taking the limits to which they converge as side lengths for longest and shortest sides, L and S then they should be unchanged by a single iteration - since each iteration after the first starts with two joint longest sides, the side lengths must start as L,L,S->L,(L+S)/2,(L+S)/2->(3L+S)/4,(3L+S)/4,(L+S)/2 which is a change when L[ne]S - contradiction) so the limit triangle is the equilateral triangle. Since whatever triangle you start at you end with the equilateral triangle as the limit of a sequence of area-increasing transformations, the equilateral triangle must be the largest.
Title: Re: Equilateral triangle
Post by Eigenray on Jun 30th, 2004, 1:25pm
Oh, I see.  I was assuming that a largest triangle exists.  Then if any two sides aren't equal, you have a contradiction.
Existance should be clear though: we can parameterize triangles of a given perimeter by two angles, and the set of angle pairs which define a (possibly degenerate) triangle {[alpha],[beta] | [alpha]+[beta][le][pi], [alpha],[beta][ge]0} is a compact subset of R2.  Since the area becomes 0 anywhere on the boundary (when one of the three angles vanishes), the global maximum must occur somewhere in the interior.


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