Point $ P$ is taken inside a right angle $ KLM$. A circle $ S_1$ with center $ O_1$ is tangent to the rays $ LK,LP$ of angle $ KLP$ at $ A,D$ respectively. A circle $ S_2$ with center $ O_2$ is tangent to the rays of angle $ MLP$, touching $ LP$ at $ B$. Suppose $ A,B,O_1$ are collinear. Let $ O_2D,KL$ meet at $ C$. Prove that $ BC$ bisects angle $ ABD$.
Problem
Source: Russian MO 1994
Tags: geometry, parallelogram, geometry unsolved
28121941
12.06.2008 12:58
In the number 5 of Kvant 1994, the following outlined solution is published: The quadrilateral O_1DO_2B is a paralelogram (O_1D and O_2B are perpendicular to LP), and if E is the point in which circle O_2 touches LM, the quadrilateral CLEO_2 is a rectangle. The triangles DCL and BDO_1 are equal, and therefore the angles DCB and DBC are equal.
stergiu
12.06.2008 21:37
28121941 wrote: In the number 5 of Kvant 1994, the following outlined solution is published: The quadrilateral O_1DO_2B is a paralelogram (O_1D and O_2B are perpendicular to LP), and if E is the point in which circle O_2 touches LM, the quadrilateral CLEO_2 is a rectangle. The triangles DCL and BDO_1 are equal, and therefore the angles DCB and DBC are equal. Probably , something in the solution is different from the wording. For example - in my figure - $ O_1D$ is parallel to $ O_2B$ but these radii are not equal, so the four points do not form a parallelogram. If you can see the problem in the original wording , please correct me !May be I have missunderstood the problem. Babis
Attachments:
28121941
12.06.2008 22:10
I am trying of solve the problem in another way; I will check the statement of the problem to see if both circles is supposed to be of the same radius. I just translated the sketched solution of Kvant, in which no pictures are included.
28121941
12.06.2008 22:28
I have checked the original version of the problem in Kvant, 1994, issue 5, and in fact both circles is supposed to be of the same radius.
discredit
13.06.2008 08:51
So, when the radii are not equal, does the problem still hold?
28121941
13.06.2008 13:03
I think the picture by Stergiou provides a clear example that the problem does not work if the radiuses are different. Maybe it is possible to think a bit more about the conditions for that.