Let $ABCD$ be a parallelogram. Let $\omega_1$ be the circle passing through $D$ tangent to $AB$ at $A$. Let $\omega_2$ be the circle passing through $A$ tangent to $CD$ at $D$. The tangents from $B$ to $\omega_1$ touch it at $A$ and $P$. The tangents from $C$ to $\omega_2$ touch it at $D$ and $Q$. Lines $AP$ and $DQ$ intersect at $X$. The perpendicular bisector of $BC$ intersects $AD$ at $R$. Show that the circumcircles of triangles $\triangle PQX$, $\triangle BCR$ are concentric.
Problem
Source: 2024 Israel TST Test 6 P3
Tags: geometry, parallelogram, concentric circles, TST, perpendicular bisector, circumcircle
v4913
20.03.2024 18:17
Claim: $R$ is the center of the spiral similarity taking $\triangle{ABP} \leftrightarrow \triangle{QCD}$. Proof: First note that $AB = CD$ and $\omega_1 \cong \omega_2$, so $\triangle{ABP} \cong \triangle{QCD}$. Let $R'$ be their spiral center. Then this spiral similarity is a rotation that also takes $\omega_1$ to $\omega_2$, so $R'$ is equidistant from the centers of $\omega_1, \omega_2$, which are reflections over $AD$, so $R' \in AD$. And since $R'B = R'C$, this implies $R' = R$. Now $\triangle{RBC} \sim \triangle{RAQ} \sim \triangle{RPD}$, which are all isosceles, and if $\angle{RDP} = \angle{APB} = \angle{PAB} = \theta$ then $\angle{PXQ} = 180^{\circ} - 2\theta$, so the measure of $\widehat{PQ}$ on $(PQX)$ and the measure of $\widehat{BC}$ on $(BCR)$ are both $180^{\circ} - 2\theta$. If $O$ is the spiral center of $PB$ and $QC$ then $OP = OQ, OB = OC$, and $\angle{BOC} = \angle{BP, CQ}$, which equals $360^{\circ} - 4\theta$ since $\angle{ABP} = \angle{QCD} = 180^{\circ} - 2\theta$, so $O$ is the circumcenter of both $(PQX)$ and $(BCR)$. $\square$
math_comb01
20.03.2024 18:20
Let $S$ be the intersection of perpendicular bisector of $AD$ with $BC$ Clearly, $ABRS,DCRS$ are cyclic Let $P' = (ABRS) \cap \overline{DS} $ Let $X = (DRP) \cap \overline{AP}$ Let $Q' = DX \cap (AS)$ Denote by $D_{ABD},D_{ACD}$ the dumpty points of $ABD,ACD$ Claim 1: $SPXQ,DRSCQ$ are cyclic
The first one is just miquel, for the second $\measuredangle SQD = \measuredangle APD = \measuredangle APS = \measuredangle ABS = \measuredangle SCD $
Claim 2: $P' \equiv P, Q' \equiv Q$
We have $\measuredangle APD = \measuredangle BAD = \measuredangle AD_{ABD}D$ so $P \equiv P'$ similarly we have $Q' \equiv Q$
Claim 3: $(SPX),(BCR)$ are concentric
Let $O$ be the circumcenter of $SPQX$ we claim that it is also the circumcenter of $BCR$, firstly the fact that $O$ lies on perpendicular bisector of $RC$ follows from $SQ \parallel RC$ similaarly it lies on perp bisector of $RB$.
Hence we're done
@above sniped me
atdaotlohbh
05.04.2024 19:10
I found this problem to be a lot easier, than #1 from this test... Obviously, $w_1$ and $w_2$ are symmetrical across $AD$, so they have equal radius, hence triangles $ABP$ and $DCQ$ are equal, thus and the angles $ABP$ and $DCQ$, call them $\phi$. Now let $M$ be the center of rotation that takes $BP$ to $CQ$. Obviously, the angle of rotation is $\angle BMC=\angle QMP=2\phi$. By angle chasing $\angle PXQ=180-\phi$, so $M$ is the center of $(XPQ)$. Now we only need to check that $\angle BRC = \phi$. Denote the midpoint of $BC$ by $N$, the center of $w_1$ by $O$. Then we will just check that triangles $RNB$ and $BAO$ are similar. Let $BC=a,AB=b,\angle BAD = \alpha$. Then $R=\frac{a}{sin\alpha}$, $RN=b sin\alpha$, so $\frac{BN}{RN}=\frac{AO}{BA}=\frac{b}{asin\alpha}$. Thus $M$ is the center for both circles.