A circle meets the three sides BC,CA,AB of a triangle ABC at points D1,D2;E1,E2;F1,F2 respectively. Furthermore, line segments D1E1 and D2F2 intersect at point L, line segments E1F1 and E2D2 intersect at point M, line segments F1D1 and F2E2 intersect at point N. Prove that the lines AL,BM,CN are concurrent.
Problem
Source: china mathematical olympiad cmo 2005 final round - Problem 2
Tags: geometry, inequalities, conics, trigonometry, China
22.01.2005 14:53
hi hui jack, soarer and "me"!!! it's so great that you can online in the mainland!!! what are your result today?
22.01.2005 15:27
I'm pretty busy now, so I usually don't open a thread unless it is in the "geometry" or in the "inequalities" forum. Please post problems into separate threads in respective forums. mecrazywong wrote: A circle meets three sides BC,CA,AB of triangle ABC at points D1,D2;E1,E2;,F1,F2 in turn. Then, line segments D1E1 and D2F2 intersect at point L, line segments E1F1 and E2D2 intersect at point M, line segments F1D1 and F2E2 intersect at point N. Prove that the lines AL,BM,CN are concurrent. This one is pretty simple: The Desargues theorem, applied to triangles ABC and LMN, shows that instead of proving that the lines AL, BM, CN are concurrent, it is enough to prove that the points X=MN∩BC, Y=NL∩CA, Z=LM∩AB are collinear. Applying the Pascal theorem to the cyclic hexagon F1D1D2E2F2E1, we see that the points F1D1∩E2F2, D1D2∩F2E1 and D2E2∩E1F1 are collinear. But F1D1∩E2F2=N, D1D2∩F2E1=BC∩F2E1 and D2E2∩E1F1=M; hence, it follows that the points N, BC∩F2E1 and M are collinear. In other words, the point BC∩F2E1 lies on the line MN. Equivalently, the lines BC, F2E1 and MN concur; in other words, the point X=MN∩BC can be also considered as the point X=BC∩F2E1. In other words, X=D1D2∩F2E1. Similarly, Y=E1E2∩D2F1 and Z=F1F2∩E2D1. But applying the Pascal theorem to the cyclic hexagon D1D2F1F2E1E2, we see that the points D1D2∩F2E1, D2F1∩E1E2 and F1F2∩E2D1 are collinear; in other words, the points X, Y, Z are collinear. ◼ EDIT: Looks like I have forgotten that I already solved exactly the same problem some time ago - see Hyacinthos message #9873! Darij
25.01.2005 17:37
darij grinberg wrote: I'm pretty busy now, so I usually don't open a thread unless it is in the "geometry" or in the "inequalities" forum. Please post problems into separate threads in respective forums. mecrazywong wrote: A circle meets three sides BC,CA,AB of triangle ABC at points D1,D2;E1,E2;,F1,F2 in turn. Then, line segments D1E1 and D2F2 intersect at point L, line segments E1F1 and E2D2 intersect at point M, line segments F1D1 and F2E2 intersect at point N. Prove that the lines AL,BM,CN are concurrent. This one is pretty simple: The Desargues theorem, applied to triangles ABC and LMN, shows that instead of proving that the lines AL, BM, CN are concurrent, it is enough to prove that the points X=MN∩BC, Y=NL∩CA, Z=LM∩AB are collinear. Applying the Pascal theorem to the cyclic hexagon F1D1D2E2F2E1, we see that the points F1D1∩E2F2, D1D2∩F2E1 and D2E2∩E1F1 are collinear. But F1D1∩E2F2=N, D1D2∩F2E1=BC∩F2E1 and D2E2∩E1F1=M; hence, it follows that the points N, BC∩F2E1 and M are collinear. In other words, the point BC∩F2E1 lies on the line MN. Equivalently, the lines BC, F2E1 and MN concur; in other words, the point X=MN∩BC can be also considered as the point X=BC∩F2E1. In other words, X=D1D2∩F2E1. Similarly, Y=E1E2∩D2F1 and Z=F1F2∩E2D1. But applying the Pascal theorem to the cyclic hexagon D1D2F1F2E1E2, we see that the points D1D2∩F2E1, D2F1∩E1E2 and F1F2∩E2D1 are collinear; in other words, the points X, Y, Z are collinear. ◼ Darij Darij, nice proof! In fact, all the contestants who got this problem in cmo used the trigo form of Ceva's Theorem(including me), but except 2, they worked out the same solutions as you, which is nice and simple
25.01.2005 22:13
There are two theorems related to this theme. Steiner's Theorem Let ABCDEF is an arbitrary inscribed hexagon. Then Pascal's lines of ABCDEF, ADECF, ADCFEB are concurrent. Kirkman's theorem Let ABCDEF is an arbitrary inscribed hexagon. Then Pascal's lines of ABFDCE, AEFBDC, ABDFEC are concurrent. There are 60 Pascal's lines for ABCDEF. Each such line is included into one Steiner's triple, and it is inluded into three Kirkman's triples.
27.01.2005 00:44
There is another reduction to Pascal's theorem by taking the points dual to the lines AL, BM, CN with respect to the circle (or conic) through the 6 points on the sides. Are the CMO contestants are expected to know projective geometry? If yes, the problem can be solved in a few minutes, but if not then it could be a difficult and time consuming problem (at least, I did not see how to solve it without projective geometry). Question: is the problem statement true if we replace the 3 sides of the triangle by a smooth cubic curve? (Like the proof of Pascal using addition law on an elliptic curve.)
28.01.2005 04:13
Believe or not but I have solved this similar problem before: D1E2 meets D2F1 at P. D1E2 meets E1F2 at Q. E1F2 meets D2F1 at R. AP,BQ,CR are also concurrent. And this problem is also solved by similar proof (use the trigo form of Ceva's Theorem).
28.01.2005 09:48
minhkhoa wrote: Believe or not but I have solved this similar problem before: D1E2 meets D2F1 at P. D1E2 meets E1F2 at Q. E1F2 meets D2F1 at R. AP,BQ,CR are also concurrent. This problem is actually much simpler: In order to prove that the lines AP, BQ and CR are concurrent, by the Desargues theorem, it is enough to show that the points QR∩BC, RP∩CA and PQ∩AB are collinear. But we have QR∩BC=F2E1∩D1D2, RP∩CA=E1E2∩D2F1 and PQ∩AB=E2D1∩F1F2, and the collinearity of the points F2E1∩D1D2, E1E2∩D2F1 and E2D1∩F1F2 follows from the Pascal theorem applied to the cyclic hexagon F2E1E2D1D2F1. This solves the problem. Darij
07.02.2005 11:25
eh this is also seem to be right for the pappus lines. A,B,C on d1 D,E,F on d2 we denote (ABCDEF) the papus line through AE :cap: BD BF :cap: CE and AF :cap: CD then (ABCDEF) (ABCEFD) (ABCFDE) are concurent. Does anyone know any theorems about this. These kind of problems are really beautiful and interesting.
21.02.2005 15:49
i didn't knew that desargue's theorem works in both directions.Could S.O. please explain?
29.01.2006 14:32
How to do it using trig form of Ceva?
26.02.2006 11:12
indybar wrote: How to do it using trig form of Ceva? Trig Ceva proof: Construct LL1⊥AB, LL2⊥AC where L1,L2∈AB,AC. Thus we have: sin∠BALsin∠CAL=LL1LL2=LF2⋅sin∠LF2L1LE1⋅sin∠LE1L2 Since △LF2D1∼△LE1D2, so: LF2LE1=F2D1E1D2 Connect D2F1 and D1E2, we can see that: D2F1D1E2=2R⋅sin∠LF2L12R⋅sin∠LE1L2=sin∠LF2L1sin∠LE1L2 Put it back into the first equation we got: sin∠BALsin∠CAL=D1F2D2E1⋅D2F1D1E2 Similarly we can get other two trig relations, multiply them togather and finishes the problem.
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20.07.2017 13:05
Sorry for bumping an 11 old thread, but is this problem is purely projective?