Circles C(O1) and C(O2) intersect at points A, B. CD passing through point O1 intersects C(O1) at point D and tangents C(O2) at point C. AC tangents C(O1) at A. Draw AE⊥CD, and AE intersects C(O1) at E. Draw AF⊥DE, and AF intersects DE at F. Prove that BD bisects AF.
Problem
Source: CWMO 2005-5
Tags: conics, hyperbola, geometry, geometric transformation, reflection, circumcircle, perpendicular bisector
25.11.2005 23:45
Comment: It comes as a pleasant surprise to me that in Western part of China they do know about Apollonious theorem concerned with a tangent to hyperbola! Thank you (Chairman Mao?). M.T.
03.12.2005 12:19
I solved it,but I use at least 4 hours!!! Oh,my God.
18.12.2005 17:28
What's that?? I mean, stealing problems from other olympiads is OK if one has no phantasy; proposing ugly but original problems is also halfways OK, as a matter of honor, but stealing ugly problems from other sources sux, particularly if the other source is the ISL 1998. To make it short: The problem is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 . Finding out why is not that easy, since the proposers at the CWMO apparently tried everything to make the situation as confusing as possible, so here are the details: First let me rewrite the problem in a constructive way: Problem. Let k1 be a circle with center O1, and let D and A be two points on this circle. The tangent to the circle k1 at the point A intersects the line O1D at a point C. The perpendicular to the line O1D through the point A meets the circle k1 at a point E (apart from A). Let F be the feet of the perpendicular from the point A to the line DE. Let k2 be the circle through the points A and C which touches the line O1D at the point C. (You can easily see that there is only one such circle: The center of this circle must lie on the perpendicular bisector of the segment AC - since the segment AC is a chord of this circle - and on the perpendicular to the line O1D at the point C - since the circle touches the line O1D at the point C; hence, the center of this circle is uniquely determined as the point of intersection of these two lines; moreover, since we know a point on the circle - namely, the point A -, the radius is uniquely determined, too, so there is only one such circle.) Let B be the point of intersection of the circles k1 and k2 (different from A). Prove that the line BD bisects the segment AF. Solution. Let Y be the midpoint of the segment AF. Then, we want to show that the line BD passes through this point Y. Let G be the point diametrically opposite to the point D on the circle k1, i. e. the reflection of the point D in the center O1 of this circle. Then, the segment DG is a diameter of the circle k1, and thus < DAG = 90°. Since the segment AE is a chord of the circle k1, the perpendicular bisector of this segment AE must pass through the center O1 of this circle k1. In other words, the perpendicular bisector of the segment AE is the perpendicular to the line AE through the point O1. But the perpendicular to the line AE through the point O1 is the line DG (since the line DG coincides with the line O1D, and AE⊥O1D). Hence, the perpendicular bisector of the segment AE is the line DG. In other words, the point E is the reflection of the point A in the line DG. Now, let the line DY intersect the circle k1 at a point B' (apart from D). In order to solve the problem, it will be enough to prove that B' = B. In fact, the point B' is known to lie on the line DY; in other words, the line B'D passes through the point Y; hence, once it will be shown that B' = B, we will conclude that the line BD passes through the point Y, and the problem will be solved. Now, consider our situation from another viewpoint: We have a triangle ADG with < DAG = 90°. Its circumcircle is k1, and the tangent at the point A to this circumcircle k1 intersects the line DG at the point C. The point E is the reflection of the point A in the line DG. The point F is the foot of the perpendicular from the point A to the line DE. The point Y is the midpoint of the segment AF. The line DY intersects the circle k1 at a point B' (apart from D). Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 , the line DC (i. e., the line O1D) is tangent to the circumcircle of triangle ACB'. Thus, the circumcircle of triangle ACB' is the circle through the points A and C which touches the line O1D at the point C. Hence, the circumcircle of triangle ACB' is simply the circle k2. Consequently, the point B' lies on the circle k2. Thus, the point B' is the point of intersection of the circles k1 and k2 (different from A), and this yields B' = B, so the problem is solved. Darij
20.11.2014 03:23
My solution: Let M=AE∩CD,G=AF∩BD . Easy to see M is the midpoint of AE . Since ∠BCM=∠BAC=∠BEM , so we get B,C,E,M are concyclic , hence ∠BMA=90−∠CMB=90−∠CEB=90−∠EDB=∠BGA . ie. A,B,G,M are concyclic Since ∠AMG=∠ABD=∠AED , so we get MG∥DE . ie. AG=GF Q.E.D