# Homework Help: Quadratic equation and trigonometry

1. Oct 30, 2013

### Saitama

1. The problem statement, all variables and given/known data
If the quadratic equation ax2+bx+c=0 has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

2. Relevant equations

3. The attempt at a solution
Since the given quadratic has equal roots, its discriminant is zero.
$$b^2-4ac=0$$
The above is equivalent to:
$$\sin^2 B-4\sin A\sin C=0 \,\, (*)$$
Now I am thinking of writing (*) in terms of sinA and sinC. Substituting B with $\pi$-(A+C) doesn't look to me a good idea. I am honestly stuck here.

Any help is appreciated. Thanks!

Last edited: Oct 30, 2013
2. Oct 30, 2013

### UltrafastPED

Go back to your original condition from (1), and your result in (3), and eliminate the sin A * sin C in the denominator. This gives an expression in terms of sin^2 A, sin^2 B, sin^2 C ...

3. Oct 31, 2013

### Saitama

As per your suggestion, I get
$$4\left(\frac{\sin^2A+\sin^2C}{\sin^2B}\right)$$
How does this help?

4. Oct 31, 2013

### UltrafastPED

Note that A,B,C are the angles of the triangle ... this provides additional constraints, and you can use any law that pertains to arbitrary triangles.

5. Oct 31, 2013

### Saitama

And which constraints am I supposed to use?

I already said that using A+B+C=$\pi$ is not a good idea. I can't think of anything else.

6. Oct 31, 2013

### pasmith

Is that "the sum of those positive integers which are less than or equal to" the given quantity?

I would be tempted to use the law of sines to do away with trig functions entirely. Thus $a \sin C = c \sin A$ and hence
$$\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} = \frac ac + \frac ca = \frac{a^2 + c^2}{ac}$$
Also, by the triangle inequality, $a + c \geq b$.

7. Oct 31, 2013

### Saitama

No, I have copied down the question word by word. Its exactly what I have written.

$$\frac ac +\frac ca=\frac{a^2 + c^2}{ac}=\frac{(a+c)^2-2ac}{ac}=\frac{(a+c)^2}{ac}-2$$
$$=4\left(\frac{(a+c)^2}{b^2}\right)-2$$
From the triangle inequality, a+c>=b.
Hence,
$$4\left(\frac{(a+c)^2}{b^2}\right) \geq 4 \Rightarrow 4\left(\frac{(a+c)^2}{b^2}\right)-2\geq 2$$
Anyways, the above was obvious from AM-GM inequality.

How do I determine the upper bound for the range? Just a guess, Cauchy-Schwarz?

8. Nov 1, 2013

### Saitama

Can you please give me some more help? I am still stuck on this one.

9. Nov 1, 2013

### sankalpmittal

Well you can use weighted means.
If that does not help, you can use Tchebychef's inequality or Weierstress' inequality ?
I often see many books using the inequality to solve range you have to do.

However I am not an adept in using them though and its not in JEE syllabus also.

10. Nov 1, 2013

### Saitama

I doubt any of these inequalities would help. Can you show how would you even apply them here?

11. Nov 1, 2013

### sankalpmittal

No. Sorry just did not see the question properly.. :p

Ok, so by AM-GM inequality you already found out that the expression is greater than or equal to 2.
Hence its range is [2,∞)...

Now to find out sum of all integers, you just have to evaluate the term, I think :

lim n→∞Sn = 2+3+4+5+6+7.....+......+n

Edit:

Wait the above is not making sense.

Did you make use of b^2-4ac=0 and sin^2B−4sinAsinC=0
in using AM-GM inequality ?

Also make an intuitive guesswork or find the range of sinA/sinC + sinC/sinA.. Obviously this will find additional conditions.....And replace sinAsinC=sin^2B/4

Have you also tried cauchy-shwartz ?

Last edited: Nov 1, 2013
12. Nov 1, 2013

### Saitama

That isn't true. I have only proved that the lower bound for range is 2 but I still have to find the upper bound. I never stated that the upper bound is infinity.

13. Nov 1, 2013

### sankalpmittal

Is answer 9 ? I made two condition one from Pasmith's and other from UltrafastPED.

I also gave you hints in Edit in previous post.. I hope its 9.. :p

14. Nov 1, 2013

### Saitama

Sankalp, please come up with a proper proof for the answer you get. 9 is incorrect.

15. Nov 1, 2013

### sankalpmittal

Expression≥2 which you found using pasmith's hint.

Sorry yes 9 is incorrect. You made use of equal roots in finding Expression≥2 or not ?

You need to find the other inequality. I am already thinking of one:

If A+B+C=pi,

then

cosA + cosB + cosC ≤ 3/2

You can prove this also.

16. Nov 1, 2013

### Saitama

I honestly don't see that of any help here.

Last edited: Nov 1, 2013
17. Nov 1, 2013

### Saitama

As always, help is appreciated.

18. Nov 2, 2013

### Saitama

19. Nov 2, 2013

### Ray Vickson

Ultimately, it is your homework and you are responsible for solving the problem, or not. Sometimes the best thing to do is to say "I cannot do this problem" and just accept a less than perfect mark! I can guarantee that every one of the homework helpers here have encountered such situations themselves in the past!

20. Nov 2, 2013

### Saitama

This is not a homework. Had this been a homework style problem, I would never put it here. Also, I am no longer a high school student that my work is graded. I am only solving problems and problems for a specific purpose.

Instead of a paragraph long lecture, a few hints would have been great. I very well know its my responsibility to solve the problem and most of the times, I don't put it on forums until I have spent enough time on it.