Trigonometry

[prokhaled]

If Sin(a/2)*Sin(b/2)*Sin(c/2)=1/8
Prove That abc equilateral triangle

http://3kd.dk3.com

[thinktank2]

An Equilateral triangle , all three angles are 60degrees

A=60 B=60 and C=60

A/2 = 30
B/2 = 30
C/2 = 30

Sin(30) = 1/2

so...

we have Sin(A/2)*Sin(B/2)*Sin(C/2) = 1/8

[Guv]

Thinktank2: It is obvious that an equilateral triangle has this property. I think the point of the Thread is to prove that only an equilateral triangle satisfies the conditions.

I have not tried to analyze the problem, but it looks like an interesting one.

[Guv]

This is a very interesting problem.

A little numerical experimenting indicates the following.

• You can easily satisfy the condition without (A, B, C) being the angles of a triangle.
  
• If there is an additional requirement that (A, B, C) be the angles of a triangle, then it seems necessary for the triangle to be equilateral.

The proof of the latter statement does not seem easy.

When somebody comes up with a proof, it will probably be more obvious than it seems now.

[thinktank2]

Sin(A/2)*Sin(B/2)*Sin(C/2)=1/8

The problem is to find A, B and C that satisfy the following conditions

1) Sin(A/2) * Sin(B/2) * Sin(C/2) = 1/8

2) Sin(A/2) and Sin(B/2) and Sin(C/2) should be In the Range -1 to +1

3) A or B or C Should not be negative (Measurement Constraint)

4) A < 180 , B < 180 , C < 180 (The triangle Constraint)

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PHASE - I
__________________________________________________________________________

Let's just think of three numbers x,y,z such that

x*y*z = +(1/8) with x and y and z all in the range -1 to +1

Possible solutions are

a) x = 1/2 , y = 1/2 , z = 1/2

b) x = -1/2 , y = -1/2 , z = 1/2

c) x = 1/2 , y = -1/2 , z = -1/2

d) x = -1/2 , y = 1/2 , z = -1/2

__________________________________________________________________________

PHASE - II (Applying the Measurement Constraint)
__________________________________________________________________________

Angles whose sine is 1/2 = { 30^o , 150^o , 390^o ...}

Sine is an Odd function, So -Sin(theta) = Sin(-theta)

Therefore Angles whose sine is -(1/2) = { -30^o , -150^o , -390^o..}

now If x=Sin(A/2), y=Sin(B/2) and z=Sin(C/2)

the condition is... these angles should not be negative.

So the only possible solution for x,y,z where
x*y*z = 1/8 with x,y,z in range -1 to +1

is x=1/2 , y=1/2 and z=1/2

Which means A/2 can be 30 or 150 or 390 or....
or A can be 60,300,780....
Similarly,
B can be 60, 300, 780,....
C can be 60, 300, 780,....

__________________________________________________________________________

PHASE - III (Applying the Triangle Constraint)
__________________________________________________________________________

The values of A , B and C such that A<180, B<180, C<180
are A = 60 , B = 60 , C = 60

Which means ABC is an Equilateral triangle



__________________________________________________________________________



I created a probability problem from this.

There are 3 boxes, One contains Shoes, Another contains Caps, and Another
Contains Ties. Each of these are in two colors red and black.
That is.. there are red shoes and black shoes in the shoe box, red caps and
black caps in the cap box and Red ties and black ties in the tie box.

Suppose, If a person picks up one from each of the box
and if the Probability of the person wearing
a black shoe, red cap and red tie is 1/8

Prove that There are Equal number of red and black items in each box.

That is
Number of Red Shoes = Number of Black Shoes
Number of Red Caps = Number of Black Caps
Number of Red Ties = Number of Black Ties

[Guv]

Thinktank2: Knowing the solution, you start with the following analysis.

Let's just think of three numbers (x, y, z) such that

x*y*z = +(1/8) with x and y and z all in the range -1 to +1

Possible solutions are . . .

At this point you assume that the only possibilities are Abs(x) = Abs(y) = Abs(z) = 1/2

What about one of the following for (x, y, z)?

(1/3, 3/4, 1/2)
(9/20, 4/9, 5/8)

The above are only two out of an unbounded number of triplets whose product is 1/8, which also satisfy the conditions 0 < x < 1, 0 < y < 1, & 0 < z < 1

While numerical experiments have convinced me that the conditions can only be satisfied by an equilateral triangle, I do not see any obvious way to prove it. The numerical experiments do not constitute a proof.

Your argument is unconvincing because it dismisses consideration of so many triplets which satisfy the product = 1/8 condition.

BTW: the triangle condition is A + B + C < 180 and none of (A, B, C) is negative, which is a bit more restrictive than A < 180 * B < 180 & C < 180.

[thinktank2]

Ok, Here is a formal proof

Sin(A/2) * Sin(B/2) * Sin(C/2) = 1/8

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PHASE - I
_____________________________________

Sin(A/2) * Sin(B/2) * Sin(C/2) =

(1/2)[ Cos([A-B]/2) - Cos([A+B]/2) ] * Sin(C/2) ----- (1)

(or)

(1/2)[ Cos([B-C]/2) - Cos([B+C]/2) ] * Sin(A/2) ----- (2)

(or)

(1/2)[ Cos([C-A]/2) - Cos([C+A]/2) ] * Sin(B/2) ----- (3)

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PHASE - II
_____________________________________

Proceeding from equation (1)

(1/2)[ Cos([A-B]/2) - Cos([A+B]/2) ] * Sin(C/2)

= (1/2)[ Cos([A-B]/2) - Sin(C/2) ] * Sin(C/2)

// Since, A+B+C = 180 or A+B = 180 - C or [A+B]/2 = 90 - (C/2)
// Implies Cos([A+B]/2) = Cos(90 - [C/2]) = Sin(C/2)

= (1/2)[ Cos([A-B]/2) * Sin(C/2) - (Sin(C/2))^2 ]

// The maximum value of Cos is +1 so...

£ (1/2)[ 1 * Sin(C/2) - (Sin(C/2))^2 ] = -(1/2)[(Sin(C/2))^2 - Sin(C/2)] ---- (4)

Equation (4) = -(1/2)( [Sin(C/2) - (1/2)]^2 - 1/4 )

= 1/8 - (1/2)[ Sin(C/2) - 1/2 ]^2


Therefore,

Sin(A/2) * Sin(B/2) * Sin(C/2) <= 1/8 - (1/2)[ Sin(C/2) - 1/2 ]^2

1/8 £ 1/8 - (1/2)[ Sin(C/2) - 1/2 ]^2

Implies...

(1/2)[ Sin(C/2) - 1/2 ]^2 £ 0

The Square of a real number cannot be negative so...

[ Sin(C/2) - 1/2 ]^2 = 0 (or) [ Sin(C/2) - 1/2 ] = 0

(or)

Sin(C/2) = 1/2

Possible values of C/2 are 30,150,390,....
So C can be 60,300,780....

In a Triangle no Single angle can exceed 180 Degrees

So C must be 60 Degrees

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PHASE - III
_____________________________________

Repeating Phase II for Equations (2) and (3)

We get A = 60 , B = 60 , C = 60

Which means the triangle is an Equilateral triangle.

[prokhaled]

WOW! Thanks you too much M.R thinktank2

[thinktank2]

prokhaled :

Too much thanks to you too

BTW, Were you really after the proof that I gave latter or
do you feel the very first reply I gave you would have been sufficient ?

Guv :

Is the second attempt convincing ?

[NotLKH]

Originally posted by thinktank2
Ok, Here is a formal proof

.......
// Since, A+B+C = 180 or A+B = 180 - C or [A+B]/2 = 90 - (C/2)

..........

How do you know A+B+C = 180?
Where did prokhaled say that A, B, C are angles in a triangle?
All he said was, in essence,

Given: Sin(a/2)*Sin(b/2)*Sin(c/2)=1/8
Prove: A) Triangle, {or sum of the angles = 180} & B) A, B, C = 60.

[NotLKH]

Originally posted by thinktank2


Though I first suspected it to be sides of a traingle than angles,
it never made any sense.



He talks about the figure abc which gives a hint that it is a traingle
with vertex a,b,c

Point#1) Sorry, I wasn't clear enough, but I wasn't thinking that A,B,C could mean sides. When I say A,B,C or a,b,c, I meant only the angles themselves. I tend to Ucase and Lcase with no signifigence in meaning.

Point#2) He only talks about {if at all} a figure abc as part of what he wants proven, NOT that it is part of the given. And, when you say Hint, At least you see that he doesn't formally state that you are Given a triangle, that it is only a {possibly implied} assumption on the readers part. For a Formal proof, the Givins must be stated, or else the assumed properties cannot be used within the proof.

Anyways, enough nitpicking. I've got something more interesting to throw in the mix.

Ok. Assume that it is a given that a+b+c = 180, hence a triangle.

Now, I've never seen the following in trig, but surely it is mathematically presentable. Suppose a, b, c are complex angles, such that when you sum them together, the imaginary components sum to 0, and the real components add to 180. It still satisfies the requirements of a triangle. BUT, does your proof {BTW, Great job!} extend to complex angles such as these?

ie...

angle a = R+Si
angle b = T+Ui
angle c = V+Wi

where R+T+V=180, while S+U+W=0?


-Lou

[thinktank2]

I am not sure about that.

If at all by hook or crook I managed to get a value for
Sin(A/2) then there may be more than one value for A,B, C that are less than 180 degrees . That would effectively screw up the results.

It's my wild guess that

Sin(a/2)*Sin(b/2)*Sin(c/2)=1/8
for an equilateral triangle only for real values of angles.

Though, I may come up with a proof for that if I could grasp more about complex angles.

[Guv]

Thinktank2: While I did not study it to the very end, I suspect your proof is valid. What I studied looked correct, and seemed to be going in the right direction.

NotLKH: The original problem specified that the product of the sines of three angles was 1/8, and asked for a proof that double the angles formed an equilateral triangle.

It is obvious that an equilateral triangle satisfies the conditions. Some numerical experimenting easily found angles meeting the condition that did not form any triangle. Hence, it was decided to assume that the angles formed a triangle and prove the triangle could only be an equilateral triangle.

[NotLKH]

Originally posted by Guv
NotLKH: The original problem specified that the product of the sines of three angles was 1/8, and asked for a proof that double the angles formed an equilateral triangle.

Yes? Did I disagree about what the original Question was?

Originally posted by Guv

It is obvious that an equilateral triangle satisfies the conditions. Some numerical experimenting easily found angles meeting the condition that did not form any triangle. Hence, it was decided to assume that the angles formed a triangle and prove the triangle could only be an equilateral triangle.

And again, as I pointed out, Yes, it IS an assumption that the three angles are part of a triangle, ie... a+b+c = 180.

The question does NOT state ANYwhere that a, b, and c are the angles of a triangle, nor does it say a+b+c = 180. {Furthur, you also pointed that out earlier.}

it DOES, however, ask that you prove that a,b,c form an equilateral triangle. This means:

1) {Since this was not a Givin} Prove that it is a triangle, ie... a+b+c=180

and

2) Prove that a=b=c, hence = 60.


The request, if a+b+c=180 was supposed to be a given, should have been:

Given a Triangle, whose internal angles are a, b, and c, show that
if Sin(a/2)*Sin(b/2)*Sin(c/2)=1/8, then the triangle is an equilateral/equiangular triangle.

Propably, wherever he got the question from, there was a setup given, and this question was one of a series related to the setup.

But, again, I'm nitpicking. I still wonder if ThinkTank2's proof is adjustible to take into account the possibility of complex angles, whose real parts sum together to 180, and whose imaginary parts sum to 0.