Let {$E$}$=VW\cap BC$. I will prove that E is the midpoint of $VW$. $ABDC$ is a harmonic quadrilateral so $AB\cdot CD=BD\cdot AC$ and then $\frac{AB}{AC}=\frac{BD}{CD}$ $B-E-C$ is a transversal in the triangle $AVW$ so from Menelaus's theorem we have that $\frac{AB}{BW}\cdot \frac{EW}{EV}\cdot \frac{VC}{AC}=1$. To prove that $EV=EW$ we have to prove that $\frac{AB}{BW}\cdot \frac{VC}{AC}=1$ or $\frac{AB}{AC}=\frac{BW}{VC}$. But $\frac{AB}{AC}=\frac{BD}{CD}$ so it is enough to prove that $\frac{BD}{CD}=\frac{BW}{VC}$. $\sphericalangle DBW = \sphericalangle DCA$ because ABDC is a cyclic quadrilateral. On the other hand $WRXD$ is cyclic so $\sphericalangle QXD = \sphericalangle RWD$.Also $\sphericalangle QXD = \sphericalangle DVQ$ because $VXDQ$ is cyclic. Then $\sphericalangle DVQ = \sphericalangle RWD$. $\triangle BDW \sim \triangle CDV$ ($\sphericalangle DVQ = \sphericalangle BWD$ and $\sphericalangle DBW = \sphericalangle DCV$ ) and from this we get $\frac{BD}{CD}=\frac{BW}{VC}$ and the problem is done.

Anyway, this problem wasn't at "senior level" in the contest(as stated above).It was the third problem for juniors at SM2021.

The problem was on both tests, the only difference being that the juniors' one asked to prove that the midpoint of $VW$ lies on $BC$ (so the locus was already given), while the seniors' one asked to identify the locus yourself. However, I do think this was misplaced (on both tests). I should also mention that this was my exact solution.

trigadd123 wrote: The problem was on both tests, the only difference being that the juniors' one asked to prove that the midpoint of $VW$ lies on $BC$ (so the locus was already given), while the seniors' one asked to identify the locus yourself. However, I do think this was misplaced (on both tests). I should also mention that this was my exact solution. Interesting...I didn't know.

Let the foot from $D$ to $BC$ be $F$. Then $BFDR$ and $QFDC$ are cyclic. We have $\angle ACD = \angle RBD$ so $\angle QFC = \angle QDC = \angle BDR = \angle BFR \implies R,F,G$ are collinear. Also from $\triangle BRD \sim \triangle CQD$ we have $\frac{\sin{C}}{\sin{B}} = \frac{AB}{AC} = \frac{BD}{DC} = \frac{DR}{DQ}$. Let $VW\cap BC = E$. $\angle QVD = \angle RXD = \angle RWD \implies \triangle WRD \sim \triangle VQD \implies \frac{DW}{DV} = \frac{DR}{DQ} = \frac{\sin{C}}{\sin{B}}$. Also $\angle VCD = \angle FXD = \angle FED \implies DECV$ is cyclic. Also $\angle WDE = \angle WDX + \angle XDE = \angle BRF + \angle BFR = \angle ABC$. Finally by the ratio lemma we get $\frac{WE}{EV} = \frac{DW}{DV} \cdot \frac{\sin{\angle WDE}}{\sin{\angle EDV}} = \frac{\sin{C}}{\sin{B}} \cdot \frac{\sin{B}}{\sin{C}} = 1$. Thus $E$ is the midpoint of $VW$ and so we are done.