Triangle problem - Evaluating the given expression

[Saitama]

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).

Homework Equations

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 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).

Homework Equations

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

 

[Saitama]

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]

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

ehild

 

[Saitama]

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]

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

 

[Saitama]

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]

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

ehild

 

[Saitama]

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]

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

 

[Saitama]

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]

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