Orthocentre

[ins-]

It is given an acute-angled triangle ABC (AC>BC) with orthocentre H. M is the middle point of the segment AB. A line through H, perpendicular to HM intersects AC and BC in the points E and F respectively. Prove that EH=HF.

What do you think is the source of this problem?

I think the problem may be solved by using trygonomethry. Is there some pure geometric solution?

[Профессор Снэйп]

ins- писал(а):

A line, perpendicular to HM intersects AC and BC in the points E and F respectively.

Что-то я не совсем понял. Вот эта прямая, перпендикулярная $HM$ --- она сама через точку $H$ проходит? Или это просто какая-то прямая, перпендикулярная прямой $HM$?

Добавлено спустя 1 минуту 27 секунд:

Пока писал вопрос, автор успел поправиться. Проходит, оказывается.

[нг]

Do you have a proof? Or, as usual you leave others to prove your statements?

Why do you think it is so?

[ins-]

This problem is not mine and is not so symetrical as usualy. Mine problems are more symmetrical as you observed

Solution:
Let O=circumcenter of ABC and let C1 a point on its circumcircle such that C, O and C1 are collinear. It's well know that C1, M and H are collinears too. Now, we can see C1HEA and C1HFB are cyclics, then i just need to prove C1E=C1F.
Doing an easy angles translation we can observe that. (Using that C1AHB is paralelogram)

The solution is not mine. I think it is good problem. It was the reason to share with you.

If I should solve the problem I will use trygonometry ... Sine law for the triangles: CHE, CHF, definition for sine for triangle MHD where D is the feet of the altitude through C. And maybe some other trygonometric formulaes and the reader of my solution will have a headache...

What about the source of this problem do you know it?
Is it some russian problem?

[Leader171]

Это задача с Петербурских отборов - 1995. Авторское решение.
Пусть L - основание высоты, опущенной из А. Тогда AHM=90-AHE=90-LHF=LFH=BFH. Кроме того, HAM=FBH=90-C. Значит, AHM подобен BFH. Аналогично доказываем, что CHM подобен ВЕН. Значит, EH/FH=EH/BH*BH/FH=HM/CM*AM/HM=AM/CM=1.

[ins-]

Thank you! I supposed so - it is russian problem (or spanish/portuguese). There is some specific "sign" in the problems from the russian competitions. I think the "autor" is the person proposed the problem. For example in Bulgarian Math Olympiad I saw - there are some problems that appears in Chinese math olympiad, UK MO, Moskow Math Olympiad, as well as problems proposed but not chosen for IMO. The situation with Canadian, UK, Indian, Polish, Iranian, Irish math olympiads is the same.