I haven't managed to solve it, but here's an equivalent statement, which might be helpful (well, they're equivalent in the setting in which I'm looking at it, $BC\le AB\le CA$): Take points $X,Y$ on $(AB,(AC$ s.t. $AX=AY=a=BC$. Then $A,N,X,Y$ are concyclic.

Using the law of sine Let A>B>C . Then $NA+NC=NB <=>sin \measuredangle{BDN} = sin\measuredangle{ADN} + sin\measuredangle{CDN} <=> \frac{1}{2} BD.DPsin \measuredangle{BDN} = 1/2 DC.DP sin\measuredangle{CDN}+1/2 IM.MPsin\measuredangle{CDN} <=> S[BDP] = S[CDQ]+S[IPD] <=> S[BNE] = S[CNE] + S[INE] <=> S[BNC] = S[INC]$ (obvious) where E is the intersection of DP and BC.

Here's something extra: it seems that the Euler lines of $BCI_a, CAI_b, ABI_c$ concur at $N$ ($I_a$ and the others are the excenters).

Really? I may try it.

I don't think $S[BNC]=S[INC]$ Maybe I'm wrong, though.

Here's what's going on (I think I have proofs of these things): Let $\omega_a$ be the circle passing through $A$ and $X\in(AB,Y\in(AC$ s.t. $AX=AY=a$. Define $\omega_b,\omega_c$ similarly. Then the circles $\omega_a,\omega_b,\omega_c$ and the Euler lines of $BCI_a, CI_b, ABI_c$ all pass through a point on the circumcircle of $ABC$, and this is the point $N$.

I'm taking advantage a few minutes at school. It's very easy. I 'll post my solution right after school.

Here's a question: $N=X(?)$ in the ETC database. I bet Darij could be of some help here .

Here is my solution for the concurence Let ABC be an acute triangle and orthic triangle (the pedal triangle of the orthocenter H). We 'll prove that the 3 Euler line of AB'C', BC'A', CA'B' are concurent. Let $O_a, O_b, O_c$ and $H_a, H_b, H_c$ be the circumcenters and orthocenters of AB'C', BC'A', CA'B'.

Noticing the 3 triangles are similar: AB'C' ~ A'BC'~ A'CB', we obtain The Euler line of of AB'C' intersects AB', AC', B'C' and forming respectively the acute angles $a, b, c$. The Euler line of of A'BC' intersects A'B, A'C', BC' and forming respectively the acute angles $a, b, c$. The Euler line of of A'CB' intersects A'C, A'B', B'C and forming respectively the acute angles $a, b, c$. We also have $O_aO_b // AB, O_cO_b // CB, O_aO_c // AC$. Hence applying the trig Ceva to triangle $O_aO_bO_c$, we 'll obtain our claim. Grobber 's problem is equivalent to mine by showing that A'B'C' have A, B, C its' excenters. Therefore, if you call $K_a, K_b, K_c$ such that there exists m, n, p such that $ mK_aA + nK_aB' + pK_aC' = 0 , mK_bA' + nK_bB + pK_bC' = 0, mK_cA' + nK_cC + pK_cB' = 0$ (they are all vectors) then $O_aK_a, O_bK_b, O_cK_c$ are concurent.

I did something very similar (angle-chase). Now try the rest of it, that thing with the $\omega$s. In fact, if I'm not wrong, I showed that the initial problem (the one posted by lqdschool) was equivalent to the fact that $N$ lies on $\omega_a$, so the problem is a consequence of this problem.

grobber wrote: Here's a question: $N=X(?)$ in the ETC database. N is the triangle center X(100), defined as the anticomplement of the Feuerbach point of triangle ABC. It is actually well-known that this point X(100) lies on the circumcircle of triangle ABC and on the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$. What concerns the circles $\omega_a$, $\omega_b$, $\omega_c$, they were known to me under a different definition. In fact, applying Theorem 2.8 of my paper "Generalization of the Feuerbach point" to triangle $I_aI_bI_c$, you see that if E and F are the orthocenters of the triangles $CAI_b$ and $ABI_c$, then the points E, F, A and N lie on one circle touching the line $I_bI_c$. (Hereby, the point N has to be defined as the point of intersection of the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$ and not as the second intersection of the line DP with the circumcircle!) Now it remains to show that this circle through the points E, F, A and N is the circle $\omega_a$. In other words, you have to prove that this circle intercepts on the sides AB and AC of the triangle ABC two chords of length a = BC. Well, this is not so obvious, but I leave it as an exercise. Darij

I think this thread went off an a tangent, no pun intended. Has anyone actually solved the original problem?

I think treegoner has, but there's something wrong with his notations (I don't know what exactly ), and I have by proving those things in my previous messages: I first showed by Ptolemy and by using the condition we must prove that the problem is equivalent to the fact that the Euler line of $BCI_a$, $\omega_a$ (see one of my previous messages) and the circumcircle of $ABC$ are concurrent (at $N$, of course). Then I showed this last fact, but it's sort of long, so maybe I'll post it later (can't make any promises ).

Actually, Grobber, I wonder how you got in the circles $\omega_a$, $\omega_b$, $\omega_c$ ! They are very nice but you really don't need them... The line DP is the Euler line of triangle $BCI_a$. This is because the point D is the circumcenter and the point P is the orthocenter of triangle $BCI_a$. [Why this? Well, the fact that D is the circumcenter of triangle $BCI_a$ is well-known - actually, the point D is the center of a circle through the points B, C, I and $I_a$. The fact that P is the orthocenter of triangle $BCI_a$ can be shown as follows: Since the point P is the reflection of the point I in the midpoint M of the segment BC, the point M is the midpoint of the segment IP. In other words, the segments BC and IP have the same midpoint M. But these two segments BC and IP are the diagonals of the quadrilateral BICP; hence, the diagonals of the quadrilateral BICP bisect each other, and this quadrilateral is a parallelogram. Thus, BP || CI. Now, $CI \perp CI_a$ (since the internal and the external angle bisector of an angle are perpendicular). Hence, $BP \perp CI_a$, and consequently, our point P lies on the B-altitude of triangle $BCI_a$. Similarly, the same point P lies on the C-altitude of triangle $BCI_a$. Hence, P is the orthocenter of triangle $BCI_a$, qed..] Now, from Theorems 2.1, 2.5, 2.7 of my paper "Generalization of the Feuerbach point", applied to the triangle $I_a I_b I_c$, we see that the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$ concur at one point Q on the circumcircle of triangle ABC, and that one of the segments AQ, BQ, CQ equals the sum of the two others. Now, the Euler line of triangle $BCI_a$ is the line DP, and thus we see that the point Q is the point of intersection of the line DP with the circumcircle of triangle ABC different from the point D. But this is exactly the way we defined the point N. Hence, Q = N, and we see that one of the segments AN, BN, CN equals the sum of the two others. And the problem is dead! Darij

treegoner wrote: Using the law of sine Let A>B>C . Then $NA+NC=NB <=>sin \measuredangle{BDN} = sin\measuredangle{ADN} + sin\measuredangle{CDN} <=> \frac{1}{2} BD.DPsin \measuredangle{BDN} = 1/2 DC.DP sin\measuredangle{CDN}+1/2 IM.MPsin\measuredangle{CDN} <=> S[BDP] = S[CDQ]+S[IPD] <=> S[BNE] = S[CNE] + S[INE] <=> S[BNC] = S[INC]$ (obvious) where E is the intersection of DP and BC. <=> S[BDP] = S[CDP]+S[IPD] <=> is obvious