Let O1 and O2 two exterior circles. Let A, B, C be points on O1 and D, E, F points on O1 such that AD and BE are the common exterior tangents to these two circles and CF is one of the interior tangents to these two circles, and such that C, F are in the interior of the quadrilateral ABED. If CO1∩AB={M} and FO2∩DE={N} then prove that MN passes through the middle of CF.
Problem
Source: Romanian I TST 2007
Tags: function, geometry, geometric transformation, reflection, geometry proposed
13.04.2007 18:28
It can be easily computed trigonometrically. Denote L=BD∩CF, K=CF∩AE and prove that BL=LC=KE=KF. Now if you denote ∠ABC=α and ∠CAB=γ you can express everything needed in terms of trigonometric functions with α and γ. That's a boring solution, anything more interesting?
14.04.2007 19:32
Hello! Thanks for the trigo solution, I need to learn to do trigo like that Just thought I'd point out that this problem seems quite related to a lemma that appeared in the forums before: Lemma: Let the incircle of triangle ABC have centre I and meet BC,CA,AB at D,E,F, and M be the midpoint of BC. Then ID,EF and AM are concurrent. (Similarly, if IA is the A−excentre and is tangent to BC,CA,AB at DA,EA,FA then IADA,EAFA, and AM are concurrent. ) For the incircle case, refer to Virgil Nicula's proof in http://www.mathlinks.ro/Forum/viewtopic.php?t=80872. The excircle case can be done in the same way. For this problem, just extend AD and BE to intersect, say at X, and let FC meet AD and BE at Y and Z respectively. The two circles are an incircle and excircle of triangle XYZ. Using the lemma , the median of the triangle XYZ passes through N and M. Also FY=CZ=s−y, so so the median also passes through the midpoint of CF.
16.04.2007 16:36
Megus wrote: and prove that BL=LC=KE=KF. Are you sure? In my diagram it is clear that BL≠LC, KE≠KF!
22.04.2007 16:27
Here is my complete solution: Invalid URL X,Y are the points where internal and external common tangents meet. The parallel to XY through N meets PX,PY at X′,Y′. The quadrilaterals O2NX′D and O2NEY′, having two opposite right angles, are cyclic, and the triangle O2DE, two of its sides being the radii, is isosceles. Therefore ∠O2X′N=∠O2DN=∠O2EN=∠O2Y′N⇒O2X′Y′isosceles. O2N, being height, is also median: NX′=NY′, but then K, the intersection of the line PN and XY, is the midpoint of XY. It is known that the midpoint of the side of a triangle is the midpoint of the points of contact of incircle and excircle with that side. Therefore KF=KC. Now consider the triangles KFN,KCM: they have both a right angle and ∠FKN=∠CKM, and KF=KC. Then they are congruent. KM=KN.
11.12.2014 08:17
Oh-hoo...oo, my solution same as @bryan_hooi's solution.
11.12.2014 10:39
My solution: Let X=AD∩BE,Y=CF∩AD,Z=BE∩CF . Let T be the reflection of F in O2 . It's well known that CY=ZF . Since AC∥DT,BC∥ET , so figure △ABC∩O1∩M and figure △DET∩O2∩N are homothetic with center X , hence we get X,M,N are collinear . ... (&) Since XM pass through the midpoint of YZ (see lemma 7 ) , so combine with (&) we get MN pass through the midpoint of CF . Q.E.D
08.07.2015 19:47
A little typo in question, D, E, F are on O2, not O1.
06.05.2018 13:59
Firsly, denote L=AD∩BE. Let X be the second intersection of the diameter CM with the circle (O1). Y is analogous. Because CF is the common internal tangent, CX⊥CF and YF⊥CF,so CX||YF. We know that the homotethy that takes (O1) to (O2) is centered at L, and because diameters CX and YF are parallel, then that homotethy takes CX to YF.This shows that lines CY, XF and MN all pas through L . By intersecting CF with LA and LB at S and T, a well known lemma gives that LM bisects ST, and because in the triangle LST , O1 and O2 are the incircle and excircle,the conclusion follows.