hi hui jack, soarer and "me"!!! it's so great that you can online in the mainland!!! what are your result today?

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 $D_1,D_2;E_1,E_2;,F_1,F_2$ in turn. Then, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ 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 $X = MN\cap BC$, $Y = NL\cap CA$, $Z = LM\cap AB$ are collinear. Applying the Pascal theorem to the cyclic hexagon $F_1D_1D_2E_2F_2E_1$, we see that the points $F_1D_1\cap E_2F_2$, $D_1D_2\cap F_2E_1$ and $D_2E_2\cap E_1F_1$ are collinear. But $F_1D_1\cap E_2F_2 = N$, $D_1D_2\cap F_2E_1 = BC\cap F_2E_1$ and $D_2E_2\cap E_1F_1 = M$; hence, it follows that the points N, $BC\cap F_2E_1$ and M are collinear. In other words, the point $BC\cap F_2E_1$ lies on the line MN. Equivalently, the lines BC, $F_2E_1$ and MN concur; in other words, the point $X = MN\cap BC$ can be also considered as the point $X = BC\cap F_2E_1$. In other words, $X = D_1D_2\cap F_2E_1$. Similarly, $Y = E_1E_2\cap D_2F_1$ and $Z = F_1F_2\cap E_2D_1$. But applying the Pascal theorem to the cyclic hexagon $D_1D_2F_1F_2E_1E_2$, we see that the points $D_1D_2\cap F_2E_1$, $D_2F_1\cap E_1E_2$ and $F_1F_2\cap E_2D_1$ are collinear; in other words, the points X, Y, Z are collinear. $\blacksquare$ EDIT: Looks like I have forgotten that I already solved exactly the same problem some time ago - see Hyacinthos message #9873! Darij

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 $D_1,D_2;E_1,E_2;,F_1,F_2$ in turn. Then, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ 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 $X = MN\cap BC$, $Y = NL\cap CA$, $Z = LM\cap AB$ are collinear. Applying the Pascal theorem to the cyclic hexagon $F_1D_1D_2E_2F_2E_1$, we see that the points $F_1D_1\cap E_2F_2$, $D_1D_2\cap F_2E_1$ and $D_2E_2\cap E_1F_1$ are collinear. But $F_1D_1\cap E_2F_2 = N$, $D_1D_2\cap F_2E_1 = BC\cap F_2E_1$ and $D_2E_2\cap E_1F_1 = M$; hence, it follows that the points N, $BC\cap F_2E_1$ and M are collinear. In other words, the point $BC\cap F_2E_1$ lies on the line MN. Equivalently, the lines BC, $F_2E_1$ and MN concur; in other words, the point $X = MN\cap BC$ can be also considered as the point $X = BC\cap F_2E_1$. In other words, $X = D_1D_2\cap F_2E_1$. Similarly, $Y = E_1E_2\cap D_2F_1$ and $Z = F_1F_2\cap E_2D_1$. But applying the Pascal theorem to the cyclic hexagon $D_1D_2F_1F_2E_1E_2$, we see that the points $D_1D_2\cap F_2E_1$, $D_2F_1\cap E_1E_2$ and $F_1F_2\cap E_2D_1$ are collinear; in other words, the points X, Y, Z are collinear. $\blacksquare$ 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

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.

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.)

Believe or not but I have solved this similar problem before: $D_1E_2$ meets $D_2F_1$ at P. $D_1E_2$ meets $E_1F_2$ at Q. $E_1F_2$ meets $D_2F_1$ 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).

minhkhoa wrote: Believe or not but I have solved this similar problem before: $D_1E_2$ meets $D_2F_1$ at P. $D_1E_2$ meets $E_1F_2$ at Q. $E_1F_2$ meets $D_2F_1$ 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 $QR\cap BC$, $RP\cap CA$ and $PQ\cap AB$ are collinear. But we have $QR\cap BC = F_2E_1\cap D_1D_2$, $RP\cap CA = E_1E_2\cap D_2F_1$ and $PQ\cap AB = E_2D_1\cap F_1F_2$, and the collinearity of the points $F_2E_1\cap D_1D_2$, $E_1E_2\cap D_2F_1$ and $E_2D_1\cap F_1F_2$ follows from the Pascal theorem applied to the cyclic hexagon $F_2E_1E_2D_1D_2F_1$. This solves the problem. Darij

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.

i didn't knew that desargue's theorem works in both directions.Could S.O. please explain?

How to do it using trig form of Ceva?

indybar wrote: How to do it using trig form of Ceva? Trig Ceva proof: Construct $LL_1 \perp AB$, $LL_2 \perp AC$ where $L_1, L_2 \in AB, AC$. Thus we have: $$\frac{\sin{\angle{BAL}}}{\sin{\angle{CAL}}}=\frac{LL_1}{LL_2}=\frac{LF_2 \cdot \sin{\angle{LF_2L_1}}}{LE_1 \cdot \sin{\angle{LE_1L_2}}}$$ Since $\triangle{LF_2D_1} \sim \triangle{LE_1D_2}$, so: $$\frac{LF_2}{LE_1}=\frac{F_2D_1}{E_1D_2}$$ Connect $D_2F_1$ and $D_1E_2$, we can see that: $$\frac{D_2F_1}{D_1E_2}= \frac{2R \cdot \sin{\angle{LF_2L_1}}}{2R \cdot \sin{\angle{LE_1L_2}}}=\frac{\sin{\angle{LF_2L_1}}}{\sin{\angle{LE_1L_2}}}$$ Put it back into the first equation we got: $$\frac{\sin{\angle{BAL}}}{\sin{\angle{CAL}}}=\frac{D_1F_2}{D_2E_1}\cdot\frac{D_2F_1}{D_1E_2}$$ Similarly we can get other two trig relations, multiply them togather and finishes the problem.

Attachments:

Sorry for bumping an 11 old thread, but is this problem is purely projective?