Difficult and beautiful - /Open problem/

[juna]

My answer - other triangle is not exists. Prove it.

[ins-]

I tried to solve your problem. Unfortunately I've no enough time recently . I suppose your problem may be solved by using inequalities between sidelengths or some great application of trygonometry. I understood my thinking is more problem creating oriented than problem solving oriented. I'll be happy to see your solution I believe all people here will enjoy it.

I have a gift for all people here. I created an open problem - it is inspired by your problem - I believe it may be even more difficult than the initial problem.

It is given triangle ABC with points M, N, P on its sides (internally). U, V, W are intersection points of AN and CM, BP and AN, CM and BP. Triangle UVW is equaliteral. If AM=BN=CP is it true that ABC is equaliteral?

The problem is edited (b) was too easy. I'm sorry

I believe it is very good problem, but it isn't easy at all.

P.S. You may refer for future my first problem as "Теорема Боби".

[ins-]

"Артамонов Ю.Н."'s problem were far simpler than I expected. It was a very good excercise at a middle level, if I'm not wrong. My idea for solution is the following:
1. It is easy to see that equaliteral triangle have desired properties.
2. It is easy to see that if two of the sides of the triangle are equal - the triangle with these properties is equaliteral.
3. We will prove that the case all sides of ABC are different is impossible.
4. Without loss of generality (checked) we may assume that a>b>c.
5. Let Angles AMP, MNB, NPC are M, N, P. AM=BN=CP=x. Side of MNP is y.
6. From 4 it is easy to see A>B>C, sinA>sinB>sinC, P>M>N.
7. sin(120-P)=(x/y)sinA, sin(120-M)=(x/y)sinB, sin(120-N)=(x/y)sinC (standard notations - all angles are in degrees).
8. It is easy to see that M, N, P are all less than 120 degrees.
9. It is easy to see that at least two of M, N, P are greater than 30 degrees (proved by contradiction). Without loss of generality (checked) we may assume M, N > 30 degrees.
10. From 6, 7, 8, 9 follows that N>M.
11. From 6 and 10 we get a contradiction and it follows the only triangle with required properties is the equaliteral triangle.
Q.E.D

[juna]

My opinion - your solution is correct and nice. But only in point 7 the misprint was happened : sin(120-N)=(x/y)sinC

ins- писал(а):

It is easy to see that if two of the sides of the triangle are equal - the triangle with these properties is equaliteral.

I don't understand this thesis.

[ins-]

Thank you for valuable remarks.
I have no much time and I tried to solve the problem for short time (it took me 30 minutes to 1 hour intensive thinking) it is the reason for my mistake.
I tried to say if two of the sides of ABC are equal a=b for example - it is easy to prove ABC with required properties is equaliteral.
Proof:
1. AMP is same as CNP (by three equal sides).
2. From 1 follows for angles A, C: A=C.
3. But a=b => A=B.
4. From 2), 3) follows A=B=C => ABC, equaliteral.
Q.E.D.

What about my second problem? I think it is very diffcult but I believe we all together may solve it.

[нг]

ins-
Your last post removed as duplicating the one in the topic you dedicated to this problem.

Please read forum’s rules and follow them. This is your second double post.

[ins-]

If you want to comment my problem - please see: http://dxdy.ru/viewtopic.php?t=9676 it is a topic dedicated to it. I'm sorry for double posting.