i don't think it's too hard, so please check my solution carefully: In fact, we just have to prove that $\frac{AB}{AM}=\frac{BP}{PM}$. By the law of sine, $\frac{AB}{AM}=\frac{sin\widehat{AMB}}{sin\widehat{ABM}}$ and $\frac{BP}{PM}=\frac{sin\widehat{BMP}}{sin\widehat{PBM}}$ Since $\widehat{PBM}+\widehat{ABM}=180$ so $sinPBM=sinABM$ Since the two triangles $ABM$ and $NCM$ are similar, we get $\widehat{BMP}=\widehat{CMN}=\widehat{AMB}$ and as a result, $sin\widehat{BMP}=sin\widehat{AMB}$ So we get the desired property. (Q.E.D)

My solution: Notice that: $\triangle{CNM} \sim \triangle{BAM}$ (1) We have: $\frac{AB}{BP}.\frac{PM}{MN}.\frac{NC}{CA} = 1$ (Menelaus) $\Rightarrow \frac{BP}{PM} = \frac{CN}{NM}$ $\Rightarrow \frac{BP}{PM} = \frac{BA}{AM}$ (because of (1)) $\Rightarrow$ Bisectors of $\angle{BAM}, \angle{BPM}$ intersect on $BC$ Q.E.D

I think that it is unnecessary to use Menelaus Theorem. Just observe that $\triangle$$PBM$$\sim$$\triangle$$ACM$ Hence $\implies$ $PB$$/$$PM$$=$$AM$$/$$AC$ But since $AB$$=$$AC$ from the converse of the angle bisector theorem we are done. It's too easy

Well, it´s G1, that´s usually the easiest geometry problem in the shortlist.

(See attachment) Internal bisector of $ \angle BAM $, external bisector of $ \angle MPA $, external bisector $ BC $ of $ \angle AMP $ are concurrent at A-excenter of $ \triangle APM $ .

Hey, where can we get Balkan Mo Shortlists???

I will post them

Nice & Beautiful Balkan MO SL 2014 G1 wrote: Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$ Solution: The condition $\angle BAM=\angle MNC$ $\implies$ $\angle AMB=\angle NMC$. Hence, $BM$ bisects $\angle AMN$ $\implies$ bisectors of $\angle BAM$, $\angle BPM$ concur at the $P-$excenter WRT $\Delta APM$

It suffices to show that line $BC$ bisects $\angle{AMP}$, since this implies the incenter of $\triangle{APM}$ lies on $BC$ as desired. Since $\angle{MNC}=\angle{ABM}$ and $\angle{ABM}=\angle{NCM}$ (since $\triangle{ABC}$ is isosceles), $\angle{NMC}=\angle{AMB}=\angle{BMP}$, as desired.

dizzy wrote: Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$ Let $T=$ bisector of $\angle BAC \cap BC$ The problem is the same as proving that $TP$ is the bisector of $\angle BPM$ Let $NC=a, MN=c$ and $MC=b$ Since $\triangle BAM \cong \triangle CNM$ $\Rightarrow AB=ak, AM=ck$ Since $AT$ is a bisector of $\angle BAM$ $\Rightarrow \frac{BT}{TM} = \frac {a}{c}$ $\Rightarrow PT$ is bisector of $\angle BPM \Leftrightarrow \frac{BP}{PM} = \frac{a}{c}$ Let $2 \alpha = \angle MNC$ and $\beta=\angle MCN$ $\Rightarrow \angle PBM = \beta$ and $\angle PMB=2 \alpha + \beta$ In $\triangle MNC:$ $\frac{sin \beta}{sin( 2 \alpha + \beta )}=\frac{c}{a}$ In $\triangle BPM:$ $\frac{BP}{PM}=\frac{sin( 2 \alpha + \beta )}{sin \beta} = \frac{a}{c} _\blacksquare$

Let $\angle MBA = \beta$, $\angle BAM = \alpha$. Then, $\angle BMA = 180 - \alpha - \beta$ and $\angle BMN = \alpha + \beta \Rightarrow \frac{BA}{AM} = \frac{\sin(180 - \alpha - \beta)}{\sin(\beta)} = \frac{\sin(\alpha + \beta)}{\sin(\beta)} = \frac{BP}{PM}$. Therefore, the two bisectors meet on $BC$.

let $\angle MAC=2\beta$ , $\angle BAM = 2\alpha$ so $\angle MNC= 2\alpha$ we can see that $\triangle ABM \sim\triangle NCM$ so $\angle AMB =\angle NMC =90-\alpha+\beta$ Then it is easy to see that $BM$ is external angle bisector of $\angle AMP$. angle bisector of $\angle BAM$ is also external angle bisector of $\angle MAP$ Let $X$ is intersection of $BM$ and angle bisector of $\angle BAM$ so X is excenter of $\triangle AMP$ Then $PT$ is angle bisector of $\angle APM$.