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Trignometry

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3 Mar 2008 19:11:46 IST

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Triangle Prob

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In a triangle ABC, with sides a, b and c, prove that iff cosA = a/b+c, cosB = b/a+c and cosC = c/a+b, the triangle is equilateral.

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sreeraman nagasubramaniyan

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3 Mar 2008 19:39:27 IST

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First part:

that is if cosA = a/c+b , cosB = b/a+c , cosC = c/a+b to prove triangle is equilateral

writing a =2RsinA ,b = 2RsinB , c=2RsinC

cosA = sinA/ sinB+sinC

=> tanA = sinB+sinC

similarly

tanB = sinC+sinA

tanC = sinA+sinB

now tanA+tanB+tanC = tanAtanBtanC

2(sinA+sinB+sinC) = 2sin((B+C)/2)cos((B-C)/2) 2sin((A+C)/2)cos((C-A)/2) 2sin((A+B)/2)cos((A-B)/2)

using sinA+sinB+sinC = 4cosA/2cosB/2cosC/2

8cosA/2cosB/2cosC/2[1-cos((B-C)/2)cos((C-A)/2)cos((A-B)/2)] = 0

but cosA/2cosB/2cosC/2 cannot be equal to zero because they r angles of a triangle

so cos((B-C)/2)cos((C-A)/2)cos((A-B)/2) = 1

prod of three numbers whose max value individually is 1 is found to be 1

hence they r individually 1

so B=C=A

hence equilateral

Hari Shankar

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3 Mar 2008 19:40:21 IST

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great work sboosy, but more solutions awaited. In fact, there is a one-line solution!

Hari Shankar

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3 Mar 2008 20:03:17 IST

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Ok, here it is

Basically you have to put these two facts together.

For a,b and c +ve, a/b+c + b/c+a + c/a+b >=3/2 with equality when a=b=c (Ask anchit for the proof or use Rearrangement Inequality)

In a triangle, cosA + cosB + cosC<=3/2 with equality when A=B=C

Hence, the triangle must be equilateral.

anchit saini

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3 Mar 2008 20:06:37 IST

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i think this is the basis for soln

a/b+c + b/c+a + c/a+b >=3/2 ( equality when a=b=c)

through this we get the ans in one line

i had given a similar qn also , i think

anchit saini

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3 Mar 2008 20:07:12 IST

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oh bhatt sir
u have already done it

DAMN!!

Hari Shankar

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3 Mar 2008 20:08:34 IST

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hey sorry anchit. But let me acknowledge your solution. Good job.

And I was under the illusion all this while that I had made up this problem Seconds before u posted I was bragging to sboosy about it!

While you are at it, can you also post the proof of this inequality. It is called Nesbitt's Inequality and is very useful.

anchit saini

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3 Mar 2008 20:19:16 IST

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wait sir
i'll post the proof

anchit saini

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3 Mar 2008 20:26:13 IST

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TO PROVE--

a/(b+c) + b/(a+c) + c/(a+b)>=3/2

or
a^2/(ab + ac) + b^2/(ab +bc) + c^2/(ac+ab) >=3/2
also
a^2/(ab + ac) + b^2/(ab +bc) + c^2/(ac+ab)>=(a+b+c)^2 / 2(ab+bc+ca)

(used LEMMA above)

so its sufficient to prove that
(a+b+c)^2 / 2(ab+bc+ca)>=3/2
or
(a+b+c)^2>=3(ab+bc+ca)
or
a^2 + b^2 + c^2>=ab + bc + ca
or
(a-b)^2 + (b-c)^2 + (c-a)^2>=0

which is true cos it is sum of squares
hence

PROVED

also above i had written that i had posted such a qn
i am sorry for that
actually i had done such a qn before(where this inequality was used not the same qn)

Brad

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3 Mar 2008 20:32:36 IST

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Well sir,

cos A + cos B + cos C
= 2 cos (A+B)/2 .cos(A-B)/2 +1 - 2sin ² C/2
= 2 sin C/2 [ (cos [A-B]/2 - sin C/2 )] + 1

2 sin C/2 [ 1 - sin C/2 ] + 1

as cos A-B/2 has maximum value = 1

So L.H.S

1 - 2(sin ² C/2 - sin C/2)
= 1 - 2 ( sin ²C/2 - sin C/2 + 1/4) + 2.1/4
=3/2 - 2(sin C/2 - 1/2) ²
So cos A + cos B + cos C

3/2

Equality holds when sin C/2 = 1/2 and cos A-B/2 = 1

so cos A + cos B + cos C = 3/2

when cos A-B/2 = 0
A-B =0
i,e A = B & C/2 = 30 deg or C =60

thus A+B+C =180
so A = B = C =60
thus triangle is equilateral

Hari Shankar

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3 Mar 2008 22:06:51 IST

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Guys, for many of these triangular inequalities, there is a fundamental inequality you can learn from which it is easy to derive many of these inequalities. It is called Jensen's inequality.

Defn: (simplistic, working defn)
1. A function is said to be convex in an interval (a,b) if f"(x)>0 for all x

(a,b)
More simply, it is bowl-shaped. e.g. f(x) = x² for all R, x³ for R⁺ etc.

2. A function is said to be concave in an interval (a,b) if f"(x)<0 for all x

(a,b).
More simply, it is an inverted bowl. e.g. f(x) = sinx for all [0,

] cos x for [0,

/2] .

The result is:

let w₁+w₂+...+w_n = 1 with w_i>0

(1) If f(x) is convex then f(w₁x₁+w₂x₂+...+w_nx_n)

w₁f(x₁)+w₂f(x₂)+...+w_nf(x_n)

(2) If f(x) is concave then f(w₁x₁+w₂x₂+...+w_nx_n)

w₁f(x₁)+w₂f(x₂)+...+w_nf(x_n)

So, lets look at the function f(x) = sinx. f'"(x) = -sinx<0 for [0,

]

So, in a triangle, we have sin(A+B+C/3)

(sinA + sinB + sinC)/3

Hence sin

(sinA + sinB + sinC)/3

which gives sinA + sinB + sinc

3/2

Now you can use AM-GM to find the maximum value of sinA sinB sinC

I'll post more 2moro. It's past bedtime for me now.

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