# Triangle problem - Evaluating the given expression

1. Oct 19, 2013

### Saitama

1. The problem statement, all variables and given/known data
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).

2. Relevant equations

3. 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!

2. Oct 20, 2013

### ehild

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

ehild

3. Oct 20, 2013

### Saitama

Hi 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.

4. Oct 20, 2013

### ehild

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

ehild

5. Oct 20, 2013

### Saitama

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

How do I use this?

6. Oct 20, 2013

### 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

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Last edited: Oct 20, 2013
7. Oct 20, 2013

### Saitama

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?

8. Oct 20, 2013

### ehild

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

ehild

9. Oct 20, 2013

### Saitama

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

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?

10. Oct 20, 2013

### 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

11. Oct 20, 2013

### Saitama

It wasn't for me at least.

Thank you once again ehild!

12. Oct 20, 2013

### 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