Quadratic equation and trigonometry

[Saitama]

Homework Statement

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
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Homework Equations

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!

 

[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 ...

 

[Saitama]

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 ...

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

 

[UltrafastPED]

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
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

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.

 

[Saitama]

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.

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.

 

[pasmith]

Homework Statement

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
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

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

Homework Equations

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!

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, $a + c \geq b$.

 

[Saitama]

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

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

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, $a + c \geq b$.

$$\frac ac +\frac ca=\frac{a^2 + c^2}{ac}=\frac{(a+c)^2-2ac}{ac}=\frac{(a+c)^2}{ac}-2$$
$$=4\left(\frac{(a+c)^2}{b^2}\right)-2$$
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?

 

[Saitama]

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

 

[sankalpmittal]

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

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.

 

[Saitama]

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.

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

 

[sankalpmittal]

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

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→∞S_n = 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 ?

 

[Saitama]

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,∞)...

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.

 

[sankalpmittal]

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.

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

 

[Saitama]

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

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

 

[sankalpmittal]

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

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.

 

[Saitama]

If A+B+C=pi,

then

cosA + cosB + cosC ≤ 3/2

You can prove this also.

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

 

[Saitama]

Has anyone got some idea about this? I have made zero progress on this problem so far.

As always, help is appreciated.

 

[Saitama]

Plugging in Wolfram Alpha also gives nothing.

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=sin^2(x+y)=4sin(x)sin(y)

I hope someone can help me out with this problem.

 

[Ray Vickson]

Has anyone got some idea about this? I have made zero progress on this problem so far.

As always, help is appreciated.

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!

 

[Saitama]

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!

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.

 

[Ray Vickson]

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.

No, lots of people have given you hints, but you seem to want more and more and more. Somewhere it has to stop.

 

[Saitama]

No, lots of people have given you hints, but you seem to want more and more and more. Somewhere it has to stop.

Okay. Then tell me what you can do with those "hints".

I tried what pasmith said and yes, I was able to determine the lower bound of range. [strike]Hints[/strike]Posts by sankalp are of no use. What about the upper bound? I haven't yet got a hint for that. Even if we have reached the second page, the only useful post for me was #7 and none other so your point about me asking for more and more hints is completely wrong. Please check the replies on the previous page and tell me if you can make any use of them. Simply saying that "A,B,C are the angles of the triangle...this provides additional constraints" doesn't count as a hint. I know enough English to understand the given problem. No one is fool enough that he/she won't try any trig identities after reading the problem statement.

 

[I like Serena]

The problem statement is unclear to me as far as the sum of integers in the range mentioned is concerned.
So I will limit myself to what I can say.

I can see that we have 2 equations:
\begin{aligned}
k&=\frac{\sin A}{\sin C} + \frac{\sin C}{\sin A} \\
b^2&=4ac
\end{aligned}
The cosine rule gives us:
$$b^2 = a^2 + c^2 - 2ac \cos B$$
You already had:
$$k=\frac a c + \frac c a = \frac{a^2+c^2}{ac}$$
Combining gives us:
$$k=\frac{b^2 + 2ac \cos B}{\frac 1 4 b^2}$$
$$k=\frac{b^2 + \frac 1 2 b^2 \cos B}{\frac 1 4 b^2}$$
$$k=4+2\cos B$$
The range for this is:
$$2 \le k \le 6$$

The question remains what the possible angles for B are.
Did you try to find actual triangles ABC that satisfy the criteria?

 

[Ray Vickson]

Homework Statement

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
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Homework Equations

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!

I might be wrong, but I am getting that the problem is impossible; that is, there is no such nontrivial triangle. Here is what I did: I started with the re-formulation by pasmith, which uses $$\frac{\sin(A)}{\sin(C)} +\frac{ \sin(C)}{\sin(A)} = \frac{a^2 + c^2}{ac} = f(a,c).$$ Presumably you want to find the smallest and largest feasible values of f(a,c), so that you can get lower and upper bounds on
$$S(a,c) \equiv \sum_{0 \leq n \leq f(a,c)} n.$$
Of course, the triangle can be scaled, because multiplying all sides by a constant k does not change f. We need some type of constraint on the size of the triangle, and we might as well choose the simplest one: ac = 1. Now we have $f(a,c) = a^2 + c^2.$ Also, we need $4ac = b^2$, so b = 2 (as -2 is not an allowed length). We also need the triangle inequality $a + c \geq 2$. The constraint set $$ac = 1, \; a+c \geq 2, \; a, c \geq 0$$
have just one feasible solution: $a = c = 1.$ That means that the triangle collapses into a line segment of length 2, so could not really be called a triangle at all.

As I said, I may be wrong, but that's my take on the problem.

 

[I like Serena]

I might be wrong, but I am getting that the problem is impossible; that is, there is no such nontrivial triangle. Here is what I did: I started with the re-formulation by pasmith, which uses $$\frac{\sin(A)}{\sin(C)} +\frac{ \sin(C)}{\sin(A)} = \frac{a^2 + c^2}{ac} = f(a,c).$$ Presumably you want to find the smallest and largest feasible values of f(a,c), so that you can get lower and upper bounds on
$$S(a,c) \equiv \sum_{0 \leq n \leq f(a,c)} n.$$
Of course, the triangle can be scaled, because multiplying all sides by a constant k does not change f. We need some type of constraint on the size of the triangle, and we might as well choose the simplest one: ac = 1. Now we have $f(a,c) = a^2 + c^2.$ Also, we need $4ac = b^2$, so b = 2 (as -2 is not an allowed length). We also need the triangle inequality $a + c \leq 2$. The constraint set $$ac = 1, \; a+c \leq 2, \; a, c \geq 0$$
have just one feasible solution: $a = c = 1.$ That means that the triangle collapses into a line segment of length 2, so could not really be called a triangle at all.

As I said, I may be wrong, but that's my take on the problem.

How about (a,b,c)=(1/2, 2, 2).
Seems to me that satisfies the criteria as well.

The triangle equalities should be:
$$ac = 1, \; |a-c| \le b = 2 \le a+c, \; a, c \geq 0$$

 

[I like Serena]

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

 

[Ray Vickson]

How about (a,b,c)=(1/2, 2, 2).
Seems to me that satisfies the criteria as well.

The triangle equalities should be:
$$ac = 1, \; |a-c| \le b = 2 \le a+c, \; a, c \geq 0$$

Yes, you are right. That means that the minimization problem has $f(a,c)_{\min} = 2$ with $a = c = 1$ (collapse into a line-segment), while the maximization problem has $f(a,c)_{\max} = 6$, with $(a,c) = (1/(1+\sqrt{2}),1+\sqrt{2}).$

Basically, we can use the constraint $c = 1/a$ to express $f(a,c)$ as $F(a) = a^2 + 1/a^2$, which we want to maximize or minimize on the set $\{ a > 0, a + 1/a \geq 2 \}$.

 

[Ray Vickson]

We already found b = 2.
I don't think infinity for either a or c is an option anymore.
The triangle inequality $|a-c| \le b = 2$ prohibits it.

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

I have deleted the message you cite above.

 

[Saitama]

Thank you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

 

[I like Serena]

Thanks you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.
I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

You're right. I guess a triangle should indeed have angles strictly between 0 and 180 degrees.

In the edge case A is a straight angle. That is correct.
If we exclude the edge case, angle A is slightly less than a straight angle.

 

[Saitama]

You're right. I guess a triangle should indeed have angles strictly between 0 and 180 degrees.

In the edge case A is a straight angle. That is correct.
If we exclude the edge case, angle A is slightly less than a straight angle.

So the final answer is 3+4+5=12? :)

 

[I like Serena]

Yep!

 

[Saitama]

Yep!

Thanks a lot I Like Serena!

 

[Ray Vickson]

Thank you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.



I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

The minimum of f is 2, and that corresponds to a triangle that has collapsed into a line segment, so is not really a triangle at all. However, if you take some (a,c) giving, say f = 2.00001, there will be a legitimate triangle giving that value of f. For f = 2 the (positive) integers <= f are 1 and 2. For f = 2.00001 the positive integers <= f are still just 1 and 2. That is why f_min = 2 gives a legitimate lower bound on the integer sum, even though no actual triangle achieves that value of f.

 

[skj91]

I believe everyone is overlooking the fact that the formula which provides the solutions to the quadratic equation in the form of: ax² + bx + c is actually not

b²-4ac

but in fact it is:x = (-b ± √(b² - 4ac)) / 2a The '±' in this particular formula is the most significant mathematical notation you'll ever meet as that is precisely how you can derive your upper and lower bounds. First substitute into that formula your values and record the result of subtracting the portion under the root from -b, then do the same thing again, only find the result of adding the value under the root sign to your -b.

That is assuming b represents your y-coordinate, a is your x-coordinate and c is your hypotenuse, which is the standard representation used.