Areas and triangle inequalities

[ins-]

Let $ABC$ is an acute-angled triangle and $M, N, P$ are points from the segments $AB, BC, CA$ respectively.
a) If $AM=BN=CP$ - prove that $S_{MNP} \geq \frac{1}{4}S_{ABC}$
b) If $\frac{AM}{BM}=\frac{BN}{CN}=\frac{CP}{AP}$ - prove that $S_{MNP} \geq \frac{1}{4}S_{ABC}$
c) If $AN, BP, CM$ intersects at a common point - prove that $S_{MNP} \leq \frac{1}{4}S_{ABC}$

(Оффтоп)

I hope you will like the problem. For a) I think it is a well known problem posted by some Vietnamese guys but I cannot find the link. It inspired me for b) and c).

[ins-]

Any ideas?

[ins-]

(Оффтоп)

I'm sorry - the problem I had in mind (my source of inspiration) and I saw on some sites is a different problem. It is the text:
Suppose that $M, N, P$ are three points lying respectively on the edges $AB,BC,CA$ of a triangle $ABC$ such that $AM+BN+CP=MB+NC+PA$. Prove that $S_{MNP} \leq \frac{1}{4}S_{ABC}$

[ins-]

I have an idea for a), b) and c).
If $AM:BM=m,BN:CN=n,CP:AP=p$
$\frac{S_{MNP}}{S_{ABC}} = 1-\frac{m}{(m+1)(p+1)} - \frac{p}{(p+1)(n+1)}-\frac{n}{(n+1)(m+1)} $
so b) is trivial.
For c) the problem can be rewriten as:

For positive reals $m, n, p$ such that $mnp=1$. Prove the inequality:
$\frac{m}{(m+1)(p+1)} + \frac{p}{(p+1)(n+1)} + \frac{n}{(n+1)(m+1)} \geq \frac{3}{4}$

a) Can be rewritten as:

For positive reals $a,b,c,d$ such that $a,b,c>d$ prove the following inequality:
$\frac{a-d}{ab}+\frac{b-d}{bc}+\frac{c-d}{ca} \leq \frac{3}{4d}$

[ins-]

After expanding of b) it becomes trivial by using A.M.-G.M. It remains to prove only a). Does someone have an idea how to be proved?

[vvsss]

Существует проективное преобразование, переводящее данный треугольник в правильный.
Для правильного Вы легко докажете. Затем, поскольку b,c - проективные обобщите на случай произвольного треугольника.

[ins-]

"Существует проективное преобразование, переводящее данный треугольник в правильный." - thank you for the valuable opinion. It is a very powerful method but unfortunately I cannot use it. Probably it makes the problem very easy. Can you send some link describing the method?

The following inequality was born while i made a research on the problem.

For positive reals $a,b,c,d$ such that $a+b>c, c+a>b, b+c>a$ and $a,b,c>d$ prove that:
$\frac{a-d}{ab}+\frac{b-d}{bc}+\frac{c-d}{ca} \leq \frac{3}{4d}$.

You can try to prove it.

Also I think the original problem is very interesting. I tried to solve but it is not a very easy. You can try it. I think it is very beautiful one.

Suppose that $M, N, P$ are three points lying respectively on the edges $AB,BC,CA$ of a triangle $ABC$ such that $AM+BN+CP=MB+NC+PA$. Prove that $S_{MNP} \leq \frac{1}{4}S_{ABC}$

[vvsss]

Изображено решение задачи о построении отрезков, пересекающихся
на вписанной окружности.
У меня в библиотеке несколько построений и доказательств этим методом.
Возьмите на сайте deoma-cmd.ru программу по геометрии GInMA.
Вы публикуете интересные задачи, но не даете время для их решения.
Я обычно Ваши задачи решаю примерно за неделю.
Мне они нравятся.
С уважением Владимир
Не видели ли Вы, чтобы теорему Стокса применяли для поиска площади или объема?

[ins-]

Thank you dear Vladimir,
It is interesting to see how an algebraic inequality can be proved in pure geometric way.
You are correct. Sometimes I'm impatient to see the solutions because I have already "discovered" new problems.

[vvsss]

Не видели ли Вы, чтобы теорему Стокса применяли для поиска площади или объема?

[ins-]

"Не видели ли Вы, чтобы теорему Стокса применяли для поиска площади или объема?" - I'm not a mathematician and most of the problems I'm posting are based on existing ones and based on my "intuition". It is the first time I'm understanding about the existense of this therorem.

[ins-]

a=2, b=4, c=3, d=1.5 is a counter-example for my inequality. So a) is a false statement.

It remains to prove the following:

$M, N, P$ are three points lying respectively on the edges $AB,BC,CA$ of a triangle $ABC$ such that $AM+BN+CP=MB+NC+PA$. Prove that $S_{MNP} \leq \frac{1}{4}S_{ABC}$