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Topic: Equilateral triangle (Read 747 times) |
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NickH
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Equilateral triangle
« on: Jun 20th, 2004, 1:13pm » |
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Prove that, of all triangles with a given perimeter, the equilateral has the greatest area.
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Leo Broukhis
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Re: Equilateral triangle
« Reply #1 on: Jun 20th, 2004, 10:46pm » |
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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?
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Sir Col
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Re: Equilateral triangle
« Reply #2 on: Jun 21st, 2004, 12:42am » |
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Is it really that obvious how it follows from the formula? 
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ThudnBlunder
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Re: Equilateral triangle
« Reply #3 on: Jun 21st, 2004, 3:11am » |
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on Jun 21st, 2004, 12:42am, Sir Col wrote:| Is it really that obvious how it follows from the formula? |
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By symmetry, the maximum must be when a = b = c
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Eigenray
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Re: Equilateral triangle
« Reply #4 on: Jun 21st, 2004, 6:12am » |
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Perhaps more elementarily, consider the ellipse.
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Barukh
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Re: Equilateral triangle
« Reply #5 on: Jun 21st, 2004, 9:03am » |
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Like it always happens with Nick's problems, this one is very elegant.
on Jun 20th, 2004, 10:46pm, 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? |
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If Nick doesn't want, I do! Would you? 
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Leo Broukhis
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Re: Equilateral triangle
« Reply #6 on: Jun 22nd, 2004, 1:52pm » |
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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
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Barukh
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on Jun 22nd, 2004, 1:52pm, 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 |
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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.
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Grimbal
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Re: Equilateral triangle
« Reply #8 on: Jun 25th, 2004, 4:58am » |
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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.
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Sir Col
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Re: Equilateral triangle
« Reply #9 on: Jun 29th, 2004, 5:02am » |
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on Jun 25th, 2004, 3:49am, Barukh wrote:
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?
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Eigenray
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Re: Equilateral triangle
« Reply #10 on: Jun 29th, 2004, 5:30am » |
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on Jun 29th, 2004, 5:02am, Sir Col wrote:Good question!
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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?) |
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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? |
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It demonstrates that no matter which two sides we pick, they must be equal.
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| « Last Edit: Jun 29th, 2004, 5:33am by Eigenray » |
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Sir Col
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Re: Equilateral triangle
« Reply #11 on: Jun 29th, 2004, 10:32am » |
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Aha! Thanks, Eigenray.
on Jun 29th, 2004, 5:30am, Eigenray wrote:
I'll tell you why AY = p-a, if you tell me why AYa = p. |
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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
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rmsgrey
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Re: Equilateral triangle
« Reply #12 on: Jun 30th, 2004, 12:35pm » |
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on Jun 29th, 2004, 5:30am, Eigenray wrote:
It demonstrates that no matter which two sides we pick, they must be equal. |
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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.
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Eigenray
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Re: Equilateral triangle
« Reply #13 on: Jun 30th, 2004, 1:25pm » |
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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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