Found problems: 33
1985 Traian Lălescu, 2.2
We are given the line $ d, $ and a point $ A $ which is not on $ d. $ Two points $ B $ and $ C $ move on $ d $ such that the angle $ \angle BAC $ is constant. Prove that the circumcircle of $ ABC $ is tangent to a fixed circle.
1985 Traian Lălescu, 1.2
For the triangles of fixed perimeter, find the maximum value of the product of the radius of the incircle with the radius of the excircle.
1985 Traian Lălescu, 1.4
Let $ ABC $ a right triangle in $ A. $ Let $ D $ a point on the segment $ AC, $ and $ E,F $ the projections of $ A $ upon the lines $ BD, $ respectively, $ BC. $ Show that the quadrilateral $ CDEF $ is concyclic.
2016 Romania National Olympiad, 1
The orthocenter $ H $ of a triangle $ ABC $ is distinct from its vertices and its circumcenter $ O. $ $ M,N,P $ are the circumcenters of the triangles $ HBC,HCA, $ respectively, $ HAB. $ Prove that $ AM,BN,CP $ and $ OH $ are concurrent.
1978 Romania Team Selection Test, 1
In a convex quadrilateral $ ABCD, $ let $ A’,B’ $ be the orthogonal projections to $ CD $ of $ A, $ respectively, $ B. $
[b]a)[/b] Assuming that $ BB’\le AA’ $ and that the perimeter of $ ABCD $ is $ (AB+CD)\cdot BB’, $ is $ ABCD $ necessarily a trapezoid?
[b]b)[/b] The same question with the addition that $ \angle BAD $ is obtuse.
2000 Junior Balkan Team Selection Tests - Romania, 3
Let $ D,E,F $ be the feet of the interior bisectors from $ A,B, $ respectively $ C, $ and let $ A',B',C' $ be the symmetric points of $ A,B, $ respectively, $ C, $ to $ D,E, $ respectively $ F, $ such that $ A,B,C $ lie on $ B'C',A'C', $ respectively, $ A'B'. $
Show that the $ ABC $ is equilateral.
[i]Marius Beceanu[/i]
1985 Traian Lălescu, 1.3
Let $ H $ be the orthocenter of $ ABC $ and $ A',B',C', $ the symmetric points of $ A,B,C $ with respect to $ H. $ The intersection of the segments $ BC,CA, AB $ with the circles of diameter $ A'H,B'H, $ respectively, $ C'H, $ consists of $ 6 $ points. Prove that these are concyclic.
2016 District Olympiad, 4
Consider the triangle $ ABC $ with $ \angle BAC>60^{\circ } $ and $ \angle BCA>30^{\circ } . $ On the other semiplane than that determined by $ BC $ and $ A $ we have the points $ D $ and $ E $ so that
$$ \angle ABE =\angle CBD =\angle BAE +30^{\circ } =\angle BCD +30^{\circ } =90^{\circ } . $$
Note by $ F,H $ the midpoints of $ AE, $ respectively, $ CD, $ and with $ G $ the intersection of $ AC $ and $ DE. $ Show:
[b]a)[/b] $ EBD\sim ABC $
[b]b)[/b] $ FGH\equiv ABC $