## La ricerca ha trovato 8 risultati

19 set 2009, 09:34
Forum: Geometria
Argomento: Nice but not easy !
Risposte: 4
Visite : 2187

### Nice but not easy !

Let ABC an acute triangle. The circle with diameter the altitude BD meets sides

AB,BC at K,L respectively. The tangents of this circle at K,L meet at M.Prove that

line BM bisects segment AC.

Babis
15 set 2009, 09:53
Forum: Geometria
Argomento: triangolo
Risposte: 3
Visite : 1682
Have you any non metric solution to this problem ?

Of cource, it is easy to find that $3(c-a) = 2(m_c-m_a)$ .Then using the median formulae , we find a=c.

Μπάμπης
06 set 2009, 20:40
Forum: Geometria
Argomento: Isosceles triangle
Risposte: 2
Visite : 1469
Excellent !!!

Babis
04 set 2009, 23:05
Forum: Geometria
Argomento: Isosceles triangle
Risposte: 2
Visite : 1469

### Isosceles triangle

Well, a nice problem for beginers with olympiads is the following:

Problem

The incircle of a triangle ABC touches BC at D. Let E be the projection of B on the bisector of angle A. If M is the midpoint of side BC, prove that the triangle MDE is isosceles.

Enjoy it - Babis
04 set 2009, 10:12
Forum: Geometria
Argomento: Parallel segments - not difficult
Risposte: 5
Visite : 1982
Nice problem stergiu :wink: Let \alpha,\beta,\gamma be the angles \angle CAB,\angle ABC,\angle BCA . First case: M belongs to the arc AB wich not contains C. \angle MNA = \angle MBA because they lie on the same arc. Since M is on the bisector of the external angle B, \angle MBA = \dfrac{(\angle CAB...
02 set 2009, 23:01
Forum: Geometria
Argomento: Parallel segments - not difficult
Risposte: 5
Visite : 1982

### Parallel segments - not difficult

After having seen and loged in at your wonderfull site, I put a simple geometry problem for students of class 10-12. . Problem The bisector of the internal angle A and the bisector of the external angle B of a triangle ABC meet the circumcircle of the triangle at points M,N.If I is the incenter of ...
02 set 2009, 20:05
Risposte: 2
Visite : 2025
Thank you very much ! I visited these links.

To be more acurate, I was looking for the problem for year 2009, but nobody has them. Probably the competition did not appear this year.
If I learn something new, I will let you know !

Regards - Babis
01 set 2009, 18:53