a) this is IMOSL 1998: http://www.mathlinks.ro/Forum/post-124397.html#p124397 this is my 1000's post

What about b)?

Nice problem. I will slightly rewrite it: Problem. Two points M and N are chosen inside a triangle ABC such that < NAC = < MAB and < NBC = < MBA. Assume that $AM\cdot AN\cdot BC=BM\cdot BN\cdot CA=CM\cdot CN\cdot AB=k$ for some real k. Prove that: a) We have $3k=BC\cdot CA\cdot AB$. b) The midpoint of the segment MN is the centroid of triangle ABC. Solution. We begin, as in http://www.mathlinks.ro/Forum/viewtopic.php?t=18479 post #3, by proving that the point N is the isogonal conjugate of the point M wrt triangle ABC. Then we continue along the line of the first proof of Theorem 1 in that post, showing that $\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$ and $\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$. Now, $\frac{k}{BC\cdot CA\cdot AB}=\frac{AM\cdot AN\cdot BC}{BC\cdot CA\cdot AB}=\frac{AM\cdot AN}{CA\cdot AB}=\frac{AM\cdot AN}{AB\cdot AC}$, and similarly $\frac{k}{BC\cdot CA\cdot AB}=\frac{BM\cdot BN}{BC\cdot BA}$ and $\frac{k}{BC\cdot CA\cdot AB}=\frac{CM\cdot CN}{CA\cdot CB}$. Hence, $\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$ becomes $\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=1$, so that $\frac{3k}{BC\cdot CA\cdot AB}=1$, and thus $3k=BC\cdot CA\cdot AB$. Thus, we solved part a) of our problem. (This was basically Amir.S's solution.) Now, let S be the midpoint of the segment MN. Then, part b) requires us to show that this point S is the centroid of triangle ABC. Using $\frac{k}{BC\cdot CA\cdot AB}=\frac{BM\cdot BN}{BC\cdot BA}$ and $\frac{k}{BC\cdot CA\cdot AB}=\frac{CM\cdot CN}{CA\cdot CB}$, we transform $\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$ into $\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$. Similarly, $\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=\frac{\left[CMA\right]+\left[CNA\right]}{\left[ABC\right]}$. Thus, $\left[BMC\right]+\left[BNC\right]=\left[CMA\right]+\left[CNA\right]$. Now we need two lemmata (basic olympiad knowledge; if you don't know them, prove them as exercises): Lemma 1. If P, Q, U and V are four points in the plane such that the points U and V lie in the same halfplane wrt the line PQ, and if M is the midpoint of the segment UV, then $\left[PMQ\right]=\frac12\left(\left[PUQ\right]+\left[PVQ\right]\right)$. Lemma 2. If T is a point inside a triangle ABC such that [BTC] = [CTA] = [ATB], then this point T is the centroid of triangle ABC. Now, applying Lemma 1 to the points B, C, M and N and the midpoint S of the segment MN, we get $\left[BSC\right]=\frac12\left(\left[BMC\right]+\left[BNC\right]\right)$. Similarly, $\left[CSA\right]=\frac12\left(\left[CMA\right]+\left[CNA\right]\right)$. Thus, $\left[BMC\right]+\left[BNC\right]=\left[CMA\right]+\left[CNA\right]$ yields [BSC] = [CSA]. Similarly, [CSA] = [ASB]. Hence, [BSC] = [CSA] = [ASB]. According to Lemma 2, this yields that the point S is the centroid of triangle ABC. Part b) of the problem is solved. Darij

Similar to above, will just post anyways. Let $\triangle M_aM_bM_c, \triangle N_aN_bN_c$ be the pedal triangles of $M, N,$ respectively. Observe that $M_bM_c \perp AN$, and so hence $[ AM_bNM_c] = \frac12 AN \cdot M_bM_c.$ However, observe by the Extended Law of Sines in $(AM)$ that $M_bM_c = AM \cdot \sin \angle A$. Therefore, we conclude that: $$[AM_bNM_c] = \frac12 AN \cdot AM \cdot \sin \angle A = \frac{AM \cdot AN}{AB \cdot AC} \cdot \Delta,$$where $\Delta$ denotes the area of $\triangle ABC.$ This implies that the area of $AM_bNM_c$ is one-third of the area of $\triangle ABC$, and we can obtain analogous results for $AN_bMN_c, BM_aNM_c, BN_aMN_c, CM_aNM_b, CN_aMN_b.$ Adding up $\frac{[AM_bNM_c]}{\Delta}, \frac{[BM_aNM_c]}{\Delta}, \frac{[CM_aNM_b]}{\Delta}$ gives us that: $$\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1,$$which is actually just 1998 G4 as pointed out above. Anyhow, this trivializes part (a) . Let's now turn to part (b) . Let $X$ be the midpoint of $MN.$ Observe that $$2[\triangle BXC] = [\triangle BMC] + [ \triangle BNC] = [\triangle BMN_a] + [\triangle CMN_a] + [\triangle BNM_a] + [\triangle CNM_a].$$Now, notice that $[\triangle BNM_a] = \frac12 BN \cdot BM_a \cdot \sin \angle NBM_a = \frac BM \cdot BN_c \cdot \sin \angle MBN_c = [\triangle BMN_c],$ where we used that $BM_a = BM \cos \angle MBM_a, BN_c = BN \cos \angle NBN_c.$ Therefore, we have that $$2 [\triangle BXC] = [\triangle BMN_a] + [\triangle CMN_a] + [\triangle BMN_c] + [\triangle CMN_b],$$and so combining these triangles in pairs gives us that $$2[\triangle BXC] = [BN_aMN_c] + [CN_aMN_b],$$which we know is $\frac23 \Delta.$ Therefore, we have that $[\triangle BXC] = \frac13 \Delta,$ with analogous results for $[\triangle CXA], [\triangle AXB].$ This is easily enough to imply that $X$ is the centroid of $\triangle ABC$, as desired. $\square$