I guess this is well known? Notice that the vector $\vec{MT}$ is half of $\vec{BP} + \vec{CQ}$, which must be parallel to the angle bisector since they have the same magnitude.

problems might not be original but I am posting them in order to get different solutions, out of respect of the proposers and the translator of that file in English, sometimes what is well known for a few is not obvious for everyone, afterall everyone should have the chance to try to solve a not known problem to him

anyone willing to post the rest problems of IMOC 2017 in sepearate threads, may start doing it so, do not expect any person to post all the problems from any contest / problem collection, when you can do it yourself since the source is public and also in English, I do not have so much time and I can post limited posts each days in HSO (to avoid getting banned for overposting again), therefore I prefer using them to post problems closer to my interests (which is geometry) PS. The IMOC geometry problems are collected inside aops here

Where can we find other years of IMOC

itslumi wrote: Where can we find other years of IMOC using search 2018, 2019

Sorry to revive an old topic, but how do we do this synthetically?

DrYouKnowWho wrote: Sorry to revive an old topic, but how do we do this synthetically? hint: use Sparrow`s lemma

Adilet160205 wrote: DrYouKnowWho wrote: Sorry to revive an old topic, but how do we do this synthetically? hint: use Sparrow`s lemma Can you describe it pls?

Aritra12 wrote: Adilet160205 wrote: DrYouKnowWho wrote: Sorry to revive an old topic, but how do we do this synthetically? hint: use Sparrow`s lemma Can you describe it pls? Please see below https://artofproblemsolving.com/community/c6h1698253p10898018 and below https://artofproblemsolving.com/community/c6h1116026p10828606

Let $(APQ)$ intersect $(ABC)$ at $N$. By Miquel N - center of rotation $BP\mapsto CQ$, so $N$ - midpoint of the arc $BAC$. $BNC\cup M \sim PNQ\cup T$. If $P$ moves along $AB$ then by gliding principle $T$ also moves on a line. It's a line through $M$, since there is case $T=M$ and in case $T\in AN$ we see that $TM$ is parallel to bisector of $\angle A$.

I feel like I've seen this configuration a few times before. Let $D$ be the point such that $ABDC$ is a parallelogram, and let $R$ and $S$ be on $BD$ and $CD$ so that $BR=CS=BP=CQ$. Then it is not hard to show $PQSR$ is a parallelogram. Indeed, $\triangle BPR$ and $\triangle CQS$ are both congruent isosceles triangles with the vertex angle equal to $180 - \angle BAC$, and so the other two angles equal $\frac{1}{2} \angle BAC$. Thus, $PR$ and $QS$ are both parallel to the bisector of $\angle BAC$. Since $PR=QS$, $PQSR$ is a parallelogram. By symmetry, it is easy to show that $TM$ is parallel to both $PR$ and $QS$, and thus also parallel to the bisector of $\angle BAC$.

Here is a solution from Barycentric Coordinates. Let \(BP=CQ=d\), then we have \(P=(d:c-d:0)\) and \(Q=(d:0:b-d)\). Thus \(T=(d(b+c):b(c-d):c(b-d))\). We have \(TM: (b-c)x+(b+c)y-(b+c)z=0\) putting \(y=b\), \(z=c\) gives \(x=-b-c\) which shows \(x+y+z=0\).

Zuss77 Very nice solution

Reflect $Q$ (or $P$) over $M$.

Menelaus theorem

I amusingly stumbled on this problem whilst looking for iranian/taiwanese geo and realised that there was a math exam that put a generalised version of this as an exercise in vector bashing somehow absolute respect for the exam producers in putting in some really hard questions (relatively speaking of course ). Of course since I don’t like bashing I’ll solve this synthetically So we generalise by saying $\frac{BM}{MC}=\frac{PT}{TQ}=\frac{BP}{CQ}$; we claim this problem still holds. Call the Miquel point of quadrilateral $BPQC$ $K$ ($M$ got taken up ), the arc midpoint of arc $DE$ on $(ADE)$ without $A$ $M_a’$, the arc midpoint of $BC$ on $(ABC)$ without $A$ $M_a$. So by the length condition $\triangle KTM\sim\triangle KPB\sim\triangle KQC$. Now clearly $A$, $M_a$, $M_a’$ collinear as they all lie on the angle bisector of $\angle CAB$, so it actually suffices to show that $\overline{TM}\parallel\overline{M_aM_a’}$. This is actually pretty easy; notice by the angle bisector theorem after computing a few ratios $K-T-M_a’$ and $K-M-M_a$, hence since it can be quite easily seen that $\triangle KM_aM_a’\sim\triangle KPB\sim\triangle KTM$, $\overline{TM}\to\overline{M_aM_a’}$ under a homothety through K, proving the claim