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One day I had a question.

In an acute angled triangle,what is maximum and minimum of $I$?

$$ I=\frac{\sin(A) + \sin(B) + \sin(C)}{\cos(A) + \cos(B) + \cos(C)}$$

Attempt As law of cosines,

$$\cos(A) + \cos(B) + \cos(C) = \frac{r}{R} + 1$$

$R$ is radius of the circumcircle

$r$ is radius of the incircle

so $ I=\frac{a+b+c}{2(r+R)}$

But I can't evaluate this, could not solve further.

Would you mind solving my question?

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    $\begingroup$ put $A = \pi - (B+C)$. then, $sin(A) = sin(B+C)$ and $ cos(A) = -cos(B+C)$. $0< B < \pi/2$, $0< A< \pi/2$ , and $ 0< C+B < \pi/2$. Now, you can use partial derivatives to find the extrema in the constraints. $\endgroup$
    – Arfin
    Commented Sep 7, 2023 at 12:54
  • $\begingroup$ thanks to see my question Mr.Aria Can I just do a partial differentiation of the substituted function with b or c? But I think it is difficult to find b and c whose derivatives are 0 Is it difficult to think geometry(consider triangles)? I'm thinking I can express I in terms of a, b, c and the inequality(e.g. AM-GM Cauthy) can be applied. But I haven't been able to do it yet. $\endgroup$
    – ykk
    Commented Sep 7, 2023 at 13:47
  • $\begingroup$ This question should be useful to you. $\endgroup$
    – Jam
    Commented Sep 7, 2023 at 15:48
  • $\begingroup$ Thanks Jam. After I look it, I think that this problem should be solve to use partial differentiation now. $\endgroup$
    – ykk
    Commented Sep 7, 2023 at 16:10

1 Answer 1

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Remarks: The maximum of $I$ does not exist for acute triangles. The supremum of $I$ is $2$. The following gives the minimum of $I$. Similarly, we can prove that $I \le 2$.


The minimum of $I$ is $1 + \frac{1}{\sqrt 2}$ when e.g. $A = B = \pi/4$ and $C = \pi/2$.

Proof.

WLOG, assume that $A \le B \le C$. Then $C \ge \pi/3$.

Letting $u = \cos \frac{B - A}{2} \le 1$ and using partial fraction decomposition, we have \begin{align*} I &= \frac{2\cos \frac{C}{2}\,\cos \frac{B - A}{2} + \sin C}{2\sin\frac{C}{2}\, \cos \frac{B - A}{2} + \cos C}\\ &= \frac{2u\cos \frac{C}{2} + \sin C}{2u\sin\frac{C}{2} + \cos C}\\ &= \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}} + \frac{\sin C-\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}}}{2u\sin\frac{C}{2} + \cos C}\\ &\ge \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}} + \frac{\sin C-\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}} }{2\sin\frac{C}{2} + \cos C}\\ & = \frac{2\cos \frac{C}{2} + \sin C}{2\sin\frac{C}{2} + \cos C}\tag{1} \end{align*} where we use $$\sin C -\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}} = \frac{- \cos \frac{3C}{2}}{\sin \frac{C}{2}} \ge 0.$$

Let $f(C) := \frac{2\cos \frac{C}{2} + \sin C}{2\sin\frac{C}{2} + \cos C}$. We have $$f'(C) = - \frac{(1 + \sin \frac{C}{2})(1 - 2\sin\frac{C}{2})^2}{(2\sin\frac{C}{2} + \cos C)^2}\le 0.$$ Thus, we have $$f(C) \ge f(\pi/2) = 1 + \frac{1}{\sqrt 2}. \tag{2}$$

From (1) and (2), we have $I \ge 1 + \frac{1}{\sqrt 2}$.

Also, when $A = B = \pi/4$ and $C=\pi/2$, we have $I = 1 + \frac{1}{\sqrt 2}$.

We are done.

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  • $\begingroup$ Mr River Li, Your idea was fantastic! I understand it that make it a one-variable function of c instead of from the bivariate function of b and c with inequality evaluation. I see. thank you. $\endgroup$
    – ykk
    Commented Sep 9, 2023 at 7:51
  • $\begingroup$ @Taro Yes, it is. $\endgroup$
    – River Li
    Commented Sep 9, 2023 at 9:48

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