# Triangle problem - Evaluating the given expression

## Homework Statement

In a triangle ABC, with usual notation, if ##a^2b^2c^2 (\sin 2A + \sin 2B + \sin 2C) = λ(∆)^x## where ∆ is the area of the triangle and x ##\in## Q, find (λx).

## The Attempt at a Solution

The usual notation is:
a,b,c are three sides of the triangle opposite to the angles A,B and C respectively.

I remember a formula relating ∆ and the three sides i.e
$$\Delta=\frac{abc}{4R} \Rightarrow abc=4R \Delta$$
Also, from the law of sines,
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
where R is the circumradius of triangle.

The given expression can be written as:
$$2a^2b^2c^2 (\sin A \cos A + \sin B \cos B + \sin C \cos C)$$
I can substitute abc and sines from the above two relations but what should I replace cosines with? Law of cosines doesn't seem to be of much help.

Any help is appreciated. Thanks!

ehild
Homework Helper

## Homework Statement

In a triangle ABC, with usual notation, if ##a^2b^2c^2 (\sin 2A + \sin 2B + \sin 2C) = λ(∆)^x## where ∆ is the area of the triangle and x ##\in## Q, find (λx).

## The Attempt at a Solution

The usual notation is:
a,b,c are three sides of the triangle opposite to the angles A,B and C respectively.

I remember a formula relating ∆ and the three sides i.e
$$\Delta=\frac{abc}{4R} \Rightarrow abc=4R \Delta$$
Also, from the law of sines,
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
where R is the circumradius of triangle.

The given expression can be written as:
$$2a^2b^2c^2 (\sin A \cos A + \sin B \cos B + \sin C \cos C)$$
I can substitute abc and sines from the above two relations but what should I replace cosines with? Law of cosines doesn't seem to be of much help.

Any help is appreciated. Thanks!

Do you remember that sin(2θ)= 2 sinθ cosθ??

ehild

Hi ehild! Do you remember that sin(2θ)= 2 sinθ cosθ??

ehild

Yes, and I have even used it. I wrote ##\sin(2A)=2\sin A\cos A## and similarly the others to get the final expression I wrote.

ehild
Homework Helper
Sorry, it was stupid of me. How can you get the area of the triangle, using R and the central angles?

ehild

Sorry, it was stupid of me. How can you get the area of the triangle, using R and the central angles?

ehild

Only this comes to my mind:
$$\Delta = 2R^2\sin A \sin B \sin C$$

How do I use this?

ehild
Homework Helper
Do you remember that the area is 0.5absin(C)? See the coloured triangles in the picture. How do you get their area in terms of R and the central angles? How are the central angles related to the angles of the big triangle?

ehild

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Do you remember that the area is 0.5absin(C)? See the coloured triangles in the picture. How do you get their area in terms of R and the central angles? How are the central angles related to the angles of the big triangle?

ehild

Yes, I remember that. I misunderstood what you meant by the central angle, sorry. The areas of three triangles formed are 0.5R2sin(2A), 0.5R2sin(2B) and 0.5R2sin(2C). Am I correct?

ehild
Homework Helper
Yes, it is correct. So what is the area of the big triangle?

ehild

• 1 person
Yes, it is correct. So what is the area of the big triangle?

ehild

$$\Delta = \frac{1}{2}R^2\sin 2A +\frac{1}{2}R^2\sin(2B) +\frac{1}{2}R^2\sin(2C)$$

Also,
$$R=\frac{abc}{4\Delta}$$

$$\Rightarrow \Delta=\frac{1}{2}R^2(\sin 2A+\sin 2B+\sin 2C) \Rightarrow \Delta=\frac{1}{2}\frac{a^2b^2c^2}{16\Delta^2}(\sin 2A+\sin 2B+\sin 2C)$$

$$\Rightarrow a^2b^2c^2(\sin 2A+\sin 2B+\sin 2C)=32\Delta^3$$

Hence, ##\lambda=32## and ##x=3## and therefore ##\lambda x=96##.

Thanks a lot ehild! How did you think of those triangles?

ehild
Homework Helper
I just drew a picture with triangle and circumcircle. It is obvious, is it not, seeing the double angles in the equation. The central angle is twice the inscribed angle....

ehild

It is obvious, is it not, seeing the double angles in the equation.

It wasn't for me at least. Thank you once again ehild! ehild
Homework Helper
I forgot to tell you that you gave the idea of using circumcircle... I have never seen the equation ##\Delta=\frac{abc}{4R}##. So thank you, Pranav ehild