Difficult and beautiful - /Open problem/

[TOTAL]

ins- писал(а):

I'm not sure.

Please, stop guessing.
Exclude the $\varphi$ from this
$\displaystyle \frac{\sin (\beta + \varphi)}{\sin \gamma}= \frac{\sin (\alpha + \varphi)}{\sin \beta} = \frac{\sin (\gamma + \varphi)}{\sin \alpha} \; \; (*)$
and see that for an acute triangle it follows that $\alpha=\beta=\gamma$
For an obtuse triangle the $(*)$ is impossible.

[ins-]

I cannot understand how this equation is got from: "незванны гост" i think there is something wrong
$\sin^2 \alpha \cos\gamma +\sin^2 \beta \cos\alpha +\sin^2\gamma \cos\beta = $
$\sin\alpha\cos\alpha\sin\gamma + \sin\beta\cos\beta\sin\alpha + \sin\gamma\cos\gamma\sin\beta$.
If we exclude $ \varphi $ from equations we probably will not get "good" equation ("third degree" equation)

[незваный гость]

TOTAL писал(а):

Что тут кажется неверным?

Виноват, не проснулся. Спасибо за пояснение, увидел, как Вы используете принадлежность стороне.

I think, your post completed the proof. Verbum sapienti sat est. Further comments were needed only for stupid like me.

ins- писал(а):

For triangle with M, N, P on the sides the following equalities may be written:

1. $ \frac{sin \alpha}{sin \beta}=\frac{(\gamma+ \varphi)}{(\alpha + \varphi)} $
2. $ \frac{sin \beta}{sin \gamma}=\frac{(\alpha+ \varphi)}{(\beta + \varphi)} $
3. $ \frac{sin \gamma}{sin \alpha}=\frac{(\beta + \varphi)}{(\gamma + \varphi)} $

No way! not for angles!

[ins-]

1. What does it mean: Verbum sapienti sat est.?
2. Is it possible to exclude $ \varphi $ from equalities to get a good equality? If you know way - please show it.

[незваный гость]

ins- писал(а):

1. What does it mean: Verbum sapienti sat est.?

You can google it

ins- писал(а):

2. Is it possible to exclude $ \varphi $ from equalities to get a good equality? If you know way - please show it.

Which part needs clarification — a set of equations for $\tg \varphi$ or how do we get from equations for the tangent to the $\varphi$-free equation?

[ins-]

How to get "good" equation after excluding tgFi.

[незваный гость]

$\tg \varphi = \frac{\sin\beta \sin \gamma - \sin^2 \alpha}{\sin \alpha \cos\alpha - \sin \gamma \cos\beta}$. Similar, $ \tg \varphi =\frac{\sin\alpha \sin \beta - \sin^2 \gamma}{\sin \gamma \cos\gamma - \sin \beta \cos\alpha} $, and therefore $\tg \varphi = \frac{\sin\beta \sin \gamma - \sin^2 \alpha}{\sin \alpha \cos\alpha - \sin \gamma \cos\beta} = $ $ \frac{\sin\alpha \sin \beta - \sin^2 \gamma}{\sin \gamma \cos\gamma - \sin \beta \cos\alpha} $.

Getting away from denominators: $(\sin\beta \sin \gamma - \sin^2 \alpha)$ $(\sin \gamma \cos\gamma - \sin \beta \cos\alpha) =$ $(\sin\alpha \sin \beta - \sin^2 \gamma)$ $(\sin \alpha \cos\alpha - \sin \gamma \cos\beta)$.

Now, we can open parentheses: $ \sin \beta \sin^2 \gamma \cos \gamma - $ $\sin^2 \alpha \sin \gamma \cos \gamma -$ $ \cos\alpha \sin^2\beta\sin \gamma + $ $\sin^2 \alpha \cos\alpha \sin\beta =$ $\sin^2\alpha \cos\alpha \sin\beta -$ $\sin\alpha\cos\alpha \sin^2\gamma-\sin \alpha \sin \beta \cos \beta \sin \gamma +$ $\cos\beta \sin^3\gamma$ .

Here we can reduce $\sin^2\alpha \cos\alpha \sin\beta$ and then divide both parts by $\sin \gamma$: $ \sin \beta \sin \gamma \cos \gamma - $ $\sin^2 \alpha \cos \gamma -$ $ \cos\alpha \sin^2\beta = $ $-\sin\alpha\cos\alpha \sin\gamma-\sin \alpha \sin \beta \cos \beta +$ $\cos\beta \sin^2\gamma$.

After regrouping we have: $ \sin \beta \sin \gamma \cos \gamma +$ $\sin\alpha\cos\alpha \sin\gamma + $ $\sin \alpha \sin \beta \cos \beta =$ $\cos\beta \sin^2\gamma+$ $\sin^2 \alpha \cos \gamma +$ $ \cos\alpha \sin^2\beta$

[ins-]

I think this part is not correct:

Цитата:

$\tg \varphi = \frac{\sin\beta \sin \gamma - \sin^2 \alpha}{\sin \alpha \cos\alpha - \sin \gamma \cos\beta}$. Similar, $ \tg \varphi =\frac{\sin\alpha \sin \beta - \sin^2 \gamma}{\sin \gamma \cos\gamma - \sin \beta \cos\alpha} $, and therefore $\tg \varphi = \frac{\sin\beta \sin \gamma - \sin^2 \alpha}{\sin \alpha \cos\alpha - \sin \gamma \cos\beta} = $ $ \frac{\sin\alpha \sin \beta - \sin^2 \gamma}{\sin \gamma \cos\gamma - \sin \beta \cos\alpha} $.

It should be:
$\tg \varphi = \frac{\sin\beta \sin \gamma - \sin^2 \alpha}{\sin \alpha \cos\alpha - \sin \beta \cos\gamma}$.

$\tg \varphi = \frac{\sin\alpha \sin \beta - \sin^2 \gamma}{\sin \gamma \cos \gamma - \sin \alpha \cos \beta}$.

[незваный гость]

I think, the time probably came for you to go beyond “I think”, “I do not believe” and to show your computations in support of your opinion.

It does not mean that what you are pointing to is incorrect: it is just irrelevant. The difference is in order of angles ( $\alpha, \beta, \gamma$ vs. $\gamma, \beta, \alpha$) and as such cannot reflect on the proof. But if we accept $\frac{\sin(\alpha + \varphi)}{\sin \gamma} = …$, my calculations are correct.

Добавлено спустя 2 минуты 16 секунд:

P.S. You'd better use tag [quote] for quotes (or button “Quote”) to avoid wrath of moderators. You can also correct your messages.

[ins-]

I think - idea for the tg is OK. It is great idea. I'm sorry for my stupid questions. I understood it. You are great matematicians!

Цитата:

Пусть теперь $\beta$ --- тупой угол. Тогда соотношение
$ \frac{\sin(\beta + \varphi)}{\sin \alpha} = \frac{\sin(\gamma + \varphi)}{\sin \beta}$
не выполняется, так как левая часть больше единицы, а правая меньше.

I don't understand why

Цитата:

правая часть

$ \frac{\sin(\gamma + \varphi )}{\sin \beta} $ part is smaller than 1. Indeed I think I know but I just want to be sure.
What do you think about this problem? Do you like it? Is it difficult enough?
It will be my last questions.

[незваный гость]

It is very similar to TOTAL’s explanation here:

We assume that $\beta$ is an obtuse angle and want to prove that $ \frac{\sin(\gamma + \varphi)}{\sin \beta} < 1$.

Indeed $\gamma + \varphi < $ $\gamma + \alpha $. $\beta$ is obtuse, therefore $\gamma + \alpha < \pi/2$, and $\sin(\gamma + \varphi) < $ $\sin(\gamma + \alpha) =$ $\sin\beta$. It remains to divide by positive $\sin \beta$ to conclude proof.

[ins-]

Your solution is far better than provided in mathlinks.ro. I think your forum is great I may use one russian word I know: "молодцы". Probably you haven't saw my questions because I edited my post too many times: "What do you think about this problem? Do you like it? Is it difficult enough? Do you want me to post more interesting problems here?" I asked them because this problem was invented by me ( but not proved ) and I was surprised I saw it in mathlinks.

[незваный гость]

This kind of questions is more difficult for me than geometry. I do not participate in competitions and do not prepare or judge them. Neither am I really involved in recreational mathematics.

The problem definitely has its charm. It is not pure technical calculations, as I initially thought.

I am not quite sure about your goals. Still I think you might enjoy couple of problems discussed here. Would you have problems with translation or understanding (the language is not good Russian), let me know (private message) and I will help you with translation.

P.S. It would be nice if you can post a reference to this topic on mathlinks.ro. It might benefit somebody there.

[juna]

We can change the condition of the problem and get a new problem (more simple given).
Whether there is a triangle ABC (except equaliteral) that AM=BN=CP and the triangle MNP is equaliteral.

[ins-]

Yes, you are right. This problem is very interesting too.