[obm-l] Perimeter as a function of area and angles

>From: "ndergiades" <ndergiades@yahoo.gr>
>Reply-To: Hyacinthos@yahoogroups.com
>To: Hyacinthos@yahoogroups.com
>Subject: [EMHL] Re: Perimeter as a function of area and angles
>Date: Sun, 14 Apr 2002 12:18:28 -0000
>
>Dear Chris and Ignacio,
>
>[CP]
> > Now I have to prove that the equilateral triangle
>minimises the perimeter. What tactic do I use for this?
>[IL]
> >First, prove that of all triangles with equal basis an
>heights, the isosceles one have the minimum perimetre (inmediate).
> >
> >Second, prove that of all isosceles triangles with the same
>area, the equilateral one have the minimum perimetre. If you don't
>meet another better way, it leads to a one variable optimization
>problem.
>
>
>********
>If s = semiperimeter and E = area of ABC
>Using Cauchy's AM-GM inequality
>we have
>
>(s-a + s-b + s-c)^3 >= 27(s-a)(s-b)(s-c) or
>s^4 >= 27E^2
>
>the equality holds only when s-a = s-b = s-c
>or when the triangle is equilateral.
>
>If E is constant then s becomes minimum when
>s^4 = 27E^2 or the triangle is equilateral.
>
>If s is constant then E becomes maximum when
>s^4 = 27E^2 or the triangle is equilateral.
>
>Best regards
>Nikos Dergiades
>

>From: "ndergiades" <ndergiades@yahoo.gr>
>Reply-To: Hyacinthos@yahoogroups.com
>To: Hyacinthos@yahoogroups.com
>Subject: Re: [EMHL] Perimeter as a function of area and angles
>Date: Sun, 14 Apr 2002 20:58:57 -0000
>
>Dear Antreas, you wrote,
>
> >This problem and many similar ones are discussed in a good
> >monograph (Maxima and Minima [in Greek]) by George A.
>Strikis (Rhodos, 1953. 104 pages).
> >
> >Here are two problems from "Exercises and Problems" section
> >(with no solutions):
> >
> >- Let ABC be an equilateral triangle [of constant side].
> >Find on its incircle a point M such that MA * MB * MC = max
> >(p. 102, #23)
>
>
>***
>Let MA = x, MB = y, MC = z   then it is known that
>using e.g. Stewart's theorem we can prove that
>x^2 + y^2 + z^2 = k = constant.
>Also it is known that
>
>3(x^2 + y^2 + z^2) >= (x + y + z)^2    Schwartz inequality
>(x^2 + y^2 + z^2)^3 >= 27(xyz)^2         Cauchy inequality
>or by multiplication
>
>(x^2 + y^2 + z^2)^4 >= 9(x + y + z)^2(xyz)^2   or
>k^2 >= 3(x + y + z)xyz and hence xyz = max when x + y + z =
>min.
>
>This happens when M is e.g. the intersection of the incircle
>and the segment AI ( I = incenter) but I cannot give a proof.
>
> >- On each side of a right-angled triangle of constant
>perimeter we construct a square [outwardly].
> >Which is the minimum of the area of (triangle + three
>squares)? (p. 102, #26)
>
>
>*****
>Let p be the constant perimeter, and E be the required area
>then since a^2 = b^2 + c^2      p = sqrt(b^2 + c^2) + b + c
>and E = a^2 + b^2 + c^2 + bc/2 = 2(b^2 + c^2) + bc/2.
>We'll prove the inequality p^2 <= kE
>where k = (16sqrt(2) + 24 )/ 18 = 2.59
>and the equality when b = c
>or that minE = p^2/k
>
>It is known that 2bc <= b^2 + c^2 and
>b + c <= sqrt(2)sqrt(b^2 + c^2) or
>
>b + c <= (18k - 24)sqrt(b^2 + c^2)/16 or
>
>2(b + c)sqrt(b^2 + c^2) <= (2k-8)(b^2 + c^2)/8 + (2k -
>2)(b^2 + c^2) or
>            since 2k - 8 < 0
>2(b + c)sqrt(b^2 + c^2) <= (2k-8)bc/4 + (2k - 2)(b^2 + c^2)
>or
>
>(b+c)^2+2(b + c)sqrt(b^2 + c^2)+b^2+c^2 <= 2k(b^2+c^2) +
>kbc/2 or
>
>p^2 <= kE which finishes the proof
>because the equality holds only when b = c.
>
>Best regards
>Nikos Dergiades
>
>
> >PS: Another monograph by the same author:
> >Convex Regular Polyhedra [in Greek] (Rhodos, 1950. 38
>pages)
> >
> >Antreas
>