Let $ABC$ be a triangle, let its $A$-symmedian cross the circle $ABC$ again at $D$, and let $Q$ and $R$ be the feet of the perpendiculars from $D$ on the lines $AC$ and $AB$, respectively. Consider a variable point $X$ on the line $QR$, different from both $Q$ and $R$. The line through $X$ and perpendicular to $DX$ crosses the lines $AC$ and $AB$ at $V$ and $W$, respectively. Determine the geometric locus of the midpoint of the segment $VW$. Adapted from American Mathematical Monthly
Problem
Source: Spring Stars of Mathematics 2021 P3 (senior & junior level)
Tags: geometry, Locus, romania
27.03.2021 19:13
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.
27.03.2021 19:27
Anyway, this problem wasn't at "senior level" in the contest(as stated above).It was the third problem for juniors at SM2021.
27.03.2021 19:33
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.
27.03.2021 19:40
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.
27.03.2021 20:59
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.