Prove-that-among-all-triangle-of-equal-perimeter-equilateral-triangle-has-the-largest-area-

Question Number 3324 by prakash jain last updated on 10/Dec/15

Prove that among all triangle of equal

perimeter, equilateral triangle has the

largest area .

Commented by 123456 last updated on 10/Dec/15

2s(x,y,z)=x+y+z A(x,y,z)=(√(s(s−a)(s−b)(s−c)))

2 s (x, y, z) = x + y + z

A (x, y, z) = \sqrt{s (s - a) (s - b) (s - c)}

Commented by Rasheed Soomro last updated on 11/Dec/15

Let x,x+l and x+l+m are measures of three sides where l,m≥0 Perimeter=3x+2l+m S=((3x+2l+m)/2) In Equilateral triangle S=((3x)/2) for l,m=0 A_△ =(√(((3x)/2)(((3x)/2)−x)^3 ))=(√(((3x)/2)×(x^3 /2^3 ))) =(√((3x^4 )/2^4 ))=(((√3) x^2 )/4) As S is constant (because perimeter is constant Hence for any triangle S S=((3x)/2) Area of any triangle A=(√(((3x)/2)(((3x)/2)−x)(((3x)/2)−x−l)(((3x)/2)−x−l−m))) =(√(((3x)/2)×(x/2)×((x−2l)/2)×((x−2l−2m)/2))) =((√(3x^2 (x−2l)(x−2l−2m)))/4) We have to prove A_△ ≥A (((√3) x^2 )/4)≥((√(3x^2 (x−2l)(x−2l−2m)))/4) for l,m≥0 C^(ONTINUE)

L e t x, x + l a n d x + l + m a r e m e a s u r e s o f t h r e e s i d e s

w h e r e l, m ⩾ 0

P e r i m e t e r = 3 x + 2 l + m

S = \frac{3 x + 2 l + m}{2}

I n E q u i l a t e r a l t r i a n g l e S = \frac{3 x}{2} f o r l, m = 0

A_{△} = \sqrt{\frac{3 x}{2} {(\frac{3 x}{2} - x)}^{3}} = \sqrt{\frac{3 x}{2} \times \frac{x^{3}}{2^{3}}}

= \sqrt{\frac{3 x^{4}}{2^{4}}} = \frac{\sqrt{3} x^{2}}{4}

A s S i s c o n s t a n t (b e c a u s e p e r i m e t e r i s c o n s t a n t

H e n c e f o r a n y t r i a n g l e S

S = \frac{3 x}{2}

A r e a o f a n y t r i a n g l e

A = \sqrt{\frac{3 x}{2} (\frac{3 x}{2} - x) (\frac{3 x}{2} - x - l) (\frac{3 x}{2} - x - l - m)}

= \sqrt{\frac{3 x}{2} \times \frac{x}{2} \times \frac{x - 2 l}{2} \times \frac{x - 2 l - 2 m}{2}}

= \frac{\sqrt{3 x^{2} (x - 2 l) (x - 2 l - 2 m)}}{4}

W e h a v e t o p r o v e

A_{△} ⩾ A

\frac{\sqrt{3} x^{2}}{4} ⩾ \frac{\sqrt{3 x^{2} (x - 2 l) (x - 2 l - 2 m)}}{4} f o r l, m ⩾ 0

C^{ONTINUE}

Commented by prakash jain last updated on 11/Dec/15

From your derivation In your formula for area a=x, b=x−l, c=x−l−m how s=((3x)/2)

From your derivation

In your formula for area

a = x, b = x - l, c = x - l - m

h o w s = \frac{3 x}{2}

Commented by Rasheed Soomro last updated on 12/Dec/15

For equilateral triangle of side x S=((3x)/2) and I have used this also in case of common triangle,because the perimeter is constant. But now I have changed my mind. I am working on different way. Pl see my answer below. I am always thankful to you for guidance.

F o r e q u i l a t e r a l t r i a n g l e o f s i d e x S = \frac{3 x}{2}

a n d I h a v e u s e d t h i s a l s o i n c a s e o f c o m m o n

t r i a n g l e, b e c a u s e t h e p e r i m e t e r i s c o n s t a n t .

B u t n o w I h a v e c h a n g e d m y m i n d . I a m w o r k i n g

o n d i f f e r e n t w a y . P l s e e m y a n s w e r b e l o w .

I a m a l w a y s thankful t o y o u f o r g u i d a n c e .

Answered by Rasheed Soomro last updated on 13/Dec/15

Area of Common Triangle Let the measures of sides of a common triangle are x,y,z Area(A) will be A=(√(S(S−x)(S−y)(S−x))) ; S=(1/2)(x+y+z) =(√(((x+y+z)/2)(((x+y+z)/2)−x)(((x+y+z)/2)−y)(((x+y+z)/2)−z))) =(√((((x+y+z)/2))(((x+y+z−2x)/2))(((x+y+z−2y)/2))(((x+y+z−2z)/2)))) =(√((((x+y+z)/2))(((−x+y+z)/2))(((x−y+z)/2))(((x+y−z)/2)))) =(1/4)(√((x+y+z)(−x+y+z)(x−y+z)(x+y−z))) Area of Equilateral Triangle ∴ Every side of of an equilateral triangle having same perimeter (x+y+z) is ((x+y+z)/3) ∴ A_△ =(√(S(S− ((x+y+z)/3))^3 )) =(√(((x+y+z)/2)×(((x+y+z)/2)−((x+y+z)/3))^3 )) =(√(((x+y+z)/2)×(((3x+3y+3z−2x−2y−2z)/6))^3 )) =(√(((x+y+z)/2)×(((x+y+z)/6))^3 )) =(√(((x+y+z)^4 )/(2^4 ×3^2 ×3)))=(((x+y+z)^2 )/(12))×(1/( (√3))) A_△ =(((x+y+z)^2 )/(12(√3))) −−−−−<><><><>−−−−− Now we have to prove: A_△ ≥A (((x+y+z)^2 )/(12(√3)))≥(1/4)(√((x+y+z)(−x+y+z)(x−y+z)(x+y−z))) for x,y,z>0 ⇒(((x+y+z)^2 )/(3(√3)))≥(√((x+y+z)(−x+y+z)(x−y+z)(x+y−z))) Yozzi′s contribution: Yozzi has proved that ⇒(((x+y+z)^2 )/4)≥(√((x+y+z)(−x+y+z)(x−y+z)(x+y−z))) (See answer by Yozzi. An approach I can′t even think of!) Now since 4<3(√3) ∴ (((x+y+z)^2 )/(3(√3)))≥(√((x+y+z)(−x+y+z)(x−y+z)(x+y−z))) ∴ A_△ ≥A QED

A r e a o f C o m m o n T r i a n g l e

L e t t h e m e a s u r e s o f s i d e s o f a c o m m o n t r i a n g l e a r e

x, y, z

A r e a (A) w i l l b e

A = \sqrt{S (S - x) (S - y) (S - x)}; S = \frac{1}{2} (x + y + z)

= \sqrt{\frac{x + y + z}{2} (\frac{x + y + z}{2} - x) (\frac{x + y + z}{2} - y) (\frac{x + y + z}{2} - z)}

= \sqrt{(\frac{x + y + z}{2}) (\frac{x + y + z - 2 x}{2}) (\frac{x + y + z - 2 y}{2}) (\frac{x + y + z - 2 z}{2})}

= \sqrt{(\frac{x + y + z}{2}) (\frac{- x + y + z}{2}) (\frac{x - y + z}{2}) (\frac{x + y - z}{2})}

= \frac{1}{4} \sqrt{(x + y + z) (- x + y + z) (x - y + z) (x + y - z)}

A r e a o f E q u i l a t e r a l T r i a n g l e

∴ E v e r y s i d e o f o f a n e q u i l a t e r a l t r i a n g l e h a v i n g

s a m e p e r i m e t e r (x + y + z) i s \frac{x + y + z}{3}

∴ A_{△} = \sqrt{S {(S - \frac{x + y + z}{3})}^{3}}

= \sqrt{\frac{x + y + z}{2} \times {(\frac{x + y + z}{2} - \frac{x + y + z}{3})}^{3}}

= \sqrt{\frac{x + y + z}{2} \times {(\frac{3 x + 3 y + 3 z - 2 x - 2 y - 2 z}{6})}^{3}}

= \sqrt{\frac{x + y + z}{2} \times {(\frac{x + y + z}{6})}^{3}}

= \sqrt{\frac{{(x + y + z)}^{4}}{2^{4} \times 3^{2} \times 3}} = \frac{{(x + y + z)}^{2}}{12} \times \frac{1}{\sqrt{3}}

A_{△} = \frac{{(x + y + z)}^{2}}{12 \sqrt{3}}

- - - - - <><><><> - - - - -

N o w w e h a v e t o p r o v e :

A_{△} ⩾ A

\frac{{(x + y + z)}^{2}}{12 \sqrt{3}} ⩾ \frac{1}{4} \sqrt{(x + y + z) (- x + y + z) (x - y + z) (x + y - z)}

f o r x, y, z > 0

\Rightarrow \frac{{(x + y + z)}^{2}}{3 \sqrt{3}} ⩾ \sqrt{(x + y + z) (- x + y + z) (x - y + z) (x + y - z)}

{Yozzi}^{'} s contribution :

Y o z z i h a s p r o v e d t h a t

\Rightarrow \frac{{(x + y + z)}^{2}}{4} ⩾ \sqrt{(x + y + z) (- x + y + z) (x - y + z) (x + y - z)}

(S e e a n s w e r b y Y o z z i . A n a p p r o a c h I {c a n}^{'} t e v e n

t h i n k o f!)

N o w s i n c e 4 < 3 \sqrt{3}

∴ \frac{{(x + y + z)}^{2}}{3 \sqrt{3}} ⩾ \sqrt{(x + y + z) (- x + y + z) (x - y + z) (x + y - z)}

∴ A_{△} ⩾ A

Q E D

Commented by Rasheed Soomro last updated on 14/Dec/15

(((x+y+z)^2 )/4)≥(√(Π_(r=1) ^4 a_r )) it cannot solely lead one to safely stating that (((x+y+z)^2 )/(3(√3)))≥(√(Π_(r=1) ^4 a_r )) . ⋮ Why transitive property of inequality doesn′t help ? If we have to prove that A>B and we have by some way (A/2)>B (A/2)>B ⇒ A>B ?

\frac{{(x + y + z)}^{2}}{4} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}}

i t c a n n o t s o l e l y l e a d o n e t o s a f e l y s t a t i n g

t h a t \frac{{(x + y + z)}^{2}}{3 \sqrt{3}} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}} .

⋮

W h y t r a n s i t i v e p r o p e r t y o f i n e q u a l i t y {d o e s n}^{'} t

h e l p ?

I f w e h a v e t o p r o v e t h a t A > B a n d w e h a v e

b y s o m e w a y \frac{A}{2} > B

\frac{A}{2} > B \Rightarrow A > B ?

Commented by Yozzii last updated on 13/Dec/15

If 4<3(√3)⇒(1/4)>(1/(3(√3)))⇒(((x+y+z)^2 )/4)>(((x+y+z)^2 )/(3(√3))) Now, we really weren′t given that A_△ ≥A. Even if we were to check that this is the case, we would have to decompose the inequality to a form that the information we have about the situation is satisfies. The AM−GM sadly didn′t help as I had hoped it would because while it shows that (((x+y+z)^2 )/4)≥(√(Π_(r=1) ^4 a_r )) it cannot solely lead one to safely stating that (((x+y+z)^2 )/(3(√3)))≥(√(Π_(r=1) ^4 a_r )) . The transitive property of inequalities wouldn′t have worked for us in this approach :( .

I f 4 < 3 \sqrt{3} \Rightarrow \frac{1}{4} > \frac{1}{3 \sqrt{3}} \Rightarrow \frac{{(x + y + z)}^{2}}{4} > \frac{{(x + y + z)}^{2}}{3 \sqrt{3}}

N o w, w e r e a l l y {w e r e n}^{'} t g i v e n t h a t

A_{△} ⩾ A . E v e n i f w e w e r e t o c h e c k

t h a t t h i s i s t h e c a s e, w e w o u l d h a v e

t o d e c o m p o s e t h e i n e q u a l i t y t o a

f o r m t h a t t h e i n f o r m a t i o n w e h a v e

a b o u t t h e s i t u a t i o n i s s a t i s f i e s . T h e

A M - G M s a d l y {d i d n}^{'} t h e l p a s I h a d

h o p e d i t w o u l d b e c a u s e w h i l e i t

s h o w s t h a t

\frac{{(x + y + z)}^{2}}{4} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}}

i t c a n n o t s o l e l y l e a d o n e t o s a f e l y s t a t i n g

t h a t \frac{{(x + y + z)}^{2}}{3 \sqrt{3}} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}} .

T h e t r a n s i t i v e p r o p e r t y o f i n e q u a l i t i e s

{w o u l d n}^{'} t h a v e w o r k e d f o r u s i n t h i s

a p p r o a c h : (.

Commented by Yozzii last updated on 14/Dec/15

Say we have the following problem. −−−−−−−−−−−−−−−−−−−−− We want to show p>q. We have that r>q and r=np where n,p,q∈R^+ ∴ np>q⇒p>(1/n)q. q=(√(Π_(r=1) ^4 a_r )). Let p=(u/(3(√3))) and r=(u/4) ,u=(x+y+z)^2 . u=3p(√3)⇒r=((3p(√3))/4)⇒ n=((3(√3))/4). ∴ p>(4/(3(√3)))q ((16)/(27))=((4/(3(√3))))^2 ⇒ 0<(4/(3(√3)))<1. ∴ The information only shows that p>(4/(3(√3)))q and not p>q since (4/(3(√3)))q<q. On a number line we have three cases due to this result: (i) 0^→ ...............(4/(3(√3)))q........p....q.... (ii) 0^→ ...............(4/(3(√3)))q........q....p..... (iii)0^→ ..............(4/(3(√3)))q.......p=q...... Which is correct?

S a y w e h a v e t h e f o l l o w i n g p r o b l e m .

- - - - - - - - - - - - - - - - - - - - -

W e w a n t t o s h o w p > q . W e h a v e t h a t

r > q a n d r = n p w h e r e n, p, q \in R^{+}

∴ n p > q \Rightarrow p > \frac{1}{n} q . q = \sqrt{\underset{r = 1}{\prod^{4}} a_{r}} .

L e t p = \frac{u}{3 \sqrt{3}} a n d r = \frac{u}{4}, u = {(x + y + z)}^{2} .

u = 3 p \sqrt{3} \Rightarrow r = \frac{3 p \sqrt{3}}{4} \Rightarrow n = \frac{3 \sqrt{3}}{4} .

∴ p > \frac{4}{3 \sqrt{3}} q

\frac{16}{27} = {(\frac{4}{3 \sqrt{3}})}^{2} \Rightarrow 0 < \frac{4}{3 \sqrt{3}} < 1 .

∴ T h e i n f o r m a t i o n o n l y s h o w s t h a t

p > \frac{4}{3 \sqrt{3}} q a n d n o t p > q s i n c e \frac{4}{3 \sqrt{3}} q < q .

O n a n u m b e r l i n e w e h a v e t h r e e c a s e s

d u e t o t h i s r e s u l t :

(i) \vec{0} \dots \dots \dots \dots \dots \frac{4}{3 \sqrt{3}} q \dots \dots . . p \dots . q \dots .

(i i) \vec{0} \dots \dots \dots \dots \dots \frac{4}{3 \sqrt{3}} q \dots \dots . . q \dots . p \dots . .

(i i i) \vec{0} \dots \dots \dots \dots . . \frac{4}{3 \sqrt{3}} q \dots \dots . p = q \dots \dots

W h i c h i s c o r r e c t ?

Commented by Rasheed Soomro last updated on 15/Dec/15

T^{H^{A} N} K_{S s s s}!

Commented by Yozzii last updated on 14/Dec/15

Let r>7 and r=1.2p. ∴1.2p>7⇒p>(7/(1.2)) but this doesn′t show that p>7. Similarly, (((z+y+x)^2 )/4)≥(√(Π_(r=1) ^4 a_r )) ⇒(x+y+z)^2 ≥4(√(Π_(r=1) ^4 a_r )) ⇒(((x+y+z)^2 )/(3(√3)))≥(4/(3(√3)))(√(Π_(r=1) ^4 a_r ))≱(√(Π_(r=1) ^4 a_r )) since (4/(3(√3)))≯1

L e t r > 7 a n d r = 1.2 p .

∴ 1.2 p > 7 \Rightarrow p > \frac{7}{1.2} b u t t h i s

{d o e s n}^{'} t s h o w t h a t p > 7 .

S i m i l a r l y,

\frac{{(z + y + x)}^{2}}{4} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}}

\Rightarrow {(x + y + z)}^{2} ⩾ 4 \sqrt{\underset{r = 1}{\prod^{4}} a_{r}}

\Rightarrow \frac{{(x + y + z)}^{2}}{3 \sqrt{3}} ⩾ \frac{4}{3 \sqrt{3}} \sqrt{\underset{r = 1}{\prod^{4}} a_{r}} ≱ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}} s i n c e \frac{4}{3 \sqrt{3}} ≯ 1

Commented by Rasheed Soomro last updated on 15/Dec/15

T H α n N k S^{SS S s s s s!}

Answered by Yozzi last updated on 12/Dec/15

For a triangle with side lengths x,y and z, the length of one side cannot exceed the sum of the lengths of the other two sides. Symbolically, x+y−z≥0 ∧ x+z−y≥0 ∧ z+y−x≥0. PROOF: Suppose for contradiction that one side of length a of a given triangle is more than the sum of the lengths of the other two sides, denoted by b and c. ∴ a>b+c with a,b,c>0. ⇒ a^2 >b^2 +c^2 +2bc ⇒ a^2 −b^2 −c^2 >2bc (∗) By the cosine rule, a^2 =b^2 +c^2 −2bccosθ where θ is the angle subtended by the meeting of the sides b and c, opposite to a. ⇒ −2bccosθ=a^2 −b^2 −c^2 ∴ −2bccosθ>2bc⇒cosθ<−1 ⇒ ∣cosθ∣>1. But, ∀θ∈R, ∣cosθ∣≤1. This is a contradiction, so that indeed a≤b+c in general and we can show that b≤a+c and c≤a+b likewise, for a,b,c>0. □ Hence, all terms under the square root of Mr. Rasheed′s rhs expression are non−negative. Since all terms are positive reals, by AM−GM, (1/4)({x+y+z}+{x+y−z}+{x+z−y}+{z+y−x})≥((x+y+z)(x+y−z)(z+y−x)(x+z−y))^(1/4) a_1 =x+y+z, a_2 =x+y−z a_3 =x+z−y, a_4 =z+y−x (1/4)(2(x+y+z))≥(Π_(r=1) ^4 a_r )^(1/4) ⇒(1/4)(x+y+z)^2 ≥(√(Π_(r=1) ^4 a_r ))

F o r a t r i a n g l e w i t h s i d e l e n g t h s

x, y a n d z, t h e l e n g t h o f o n e s i d e

c a n n o t e x c e e d t h e s u m o f t h e l e n g t h s

o f t h e o t h e r t w o s i d e s . S y m b o l i c a l l y,

x + y - z ⩾ 0 \land x + z - y ⩾ 0 \land z + y - x ⩾ 0 .

PROOF : S u p p o s e f o r c o n t r a d i c t i o n

t h a t o n e s i d e o f l e n g t h a o f a g i v e n

t r i a n g l e i s m o r e t h a n t h e s u m o f t h e

l e n g t h s o f t h e o t h e r t w o s i d e s, d e n o t e d

b y b a n d c .

∴ a > b + c

w i t h a, b, c > 0 .

\Rightarrow a^{2} > b^{2} + c^{2} + 2 b c

\Rightarrow a^{2} - b^{2} - c^{2} > 2 b c (*)

B y t h e c o s i n e r u l e,

a^{2} = b^{2} + c^{2} - 2 b c c o s θ

w h e r e θ i s t h e a n g l e s u b t e n d e d b y t h e

m e e t i n g o f t h e s i d e s b a n d c, o p p o s i t e

t o a .

\Rightarrow - 2 b c c o s θ = a^{2} - b^{2} - c^{2}

∴ - 2 b c c o s θ > 2 b c \Rightarrow c o s θ < - 1

\Rightarrow ∣ c o s θ ∣> 1 . B u t, \forall θ \in R, ∣ c o s θ ∣⩽ 1 .

T h i s i s a c o n t r a d i c t i o n, s o t h a t

i n d e e d a ⩽ b + c i n g e n e r a l a n d w e

c a n s h o w t h a t b ⩽ a + c a n d c ⩽ a + b

l i k e w i s e, f o r a, b, c > 0 .

H e n c e, a l l t e r m s u n d e r t h e s q u a r e

r o o t o f M r . {R a s h e e d}^{'} s r h s e x p r e s s i o n

a r e n o n - n e g a t i v e .

S i n c e a l l t e r m s a r e p o s i t i v e r e a l s, b y A M - G M,

\frac{1}{4} ({x + y + z} + {x + y - z} + {x + z - y} + {z + y - x}) ⩾ {((x + y + z) (x + y - z) (z + y - x) (x + z - y))}^{1 / 4}

a_{1} = x + y + z, a_{2} = x + y - z

a_{3} = x + z - y, a_{4} = z + y - x

\frac{1}{4} (2 (x + y + z)) ⩾ {(\underset{r = 1}{\prod^{4}} a_{r})}^{1 / 4}

\Rightarrow \frac{1}{4} {(x + y + z)}^{2} ⩾ \sqrt{\underset{r = 1}{\prod^{4}} a_{r}}

Commented by Rasheed Soomro last updated on 12/Dec/15

I didn′t understand last two steps. Where has a_r come from?

I {d i d n}^{'} t u n d e r s t a n d l a s t t w o s t e p s .

W h e r e h a s a_{r} c o m e f r o m ?

Commented by Yozzi last updated on 12/Dec/15

I^{'} v e a d d e d d e f i n i t i o n s f o r a_{r}, 1 ⩽ r ⩽ 4 .

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