Let $ABC$ be a triangle with $AB > AC$. The internal angle bisector of $\angle BAC$ intersects the side $BC$ at $D$. The circles with diameters $BD$ and $CD$ intersect the circumcircle of $\triangle ABC$ a second time at $P \not= B$ and $Q \not= C$, respectively. The lines $PQ$ and $BC$ intersect at $X$. Prove that $AX$ is tangent to the circumcircle of $\triangle ABC$.
Problem
Source: Baltic Way 2020, Problem 11
Tags: geometry, geometry proposed
MarkBcc168
15.11.2020 11:37
First, we will show that $XB\cdot XC = XD^2$. By radical center theorem on $\odot(BPD)$, $\odot(CQD)$, and $\odot(ABC)$, we get that if $T=BP\cap CQ$, then $DT\perp BC$. In addition, $TPDQ$ are concyclic with diameter $TD$, so $BC$ is tangent to this circle, and thus $XP\cdot XQ = XD^2$.
To finish, let $E$ be the reflection of $D$ across $X$, so that $(BC;DE)=-1$ or that $AE$ is the external bisector. This means that $X$ is the center of the $A$-Apollonius circle of $\triangle ABC$, implying the conclusion.
mitokondrio
15.11.2020 18:24
Let $PD$ and $QD$ intersect $(ABC)$ again at $B'$ and $C'$, respectively, and let $M$ be the midpoint of arc $BC$ not containing $A$. Also let $D'$ be the reflection of $D$ over the midpoint of $BC$. Clearly $B'$ and $C'$ are the antipodes of $B$ and $C$ and $M$ lies on $AD$. Now negative inversion at $D$ fixing $(ABC)$ swaps $(B,C),(P,B'),(Q,C'),(A,M)$ and thus maps $BC\to BC$, $PQ\to (B'DD'C')$ and $AA\to (DMD')$. Since these all intersect at $D'$, inverting back we get that $BC$, $PQ$ and $AA$ are concurrent.
Gryphos
18.11.2020 11:42
I had a similar solution: After inversion at $D$ we have that $\angle DB'A' = \angle A'C'D$, hence the triangle $A'B'C'$ is $A'$-isosceles. Moreover, the images of the circles with diameters $BD$ and $CD$ are the lines through $B'$ and $C'$ perpendicular to $B'C'$, and thus $B'C'Q'P'$ is a rectangle. This implies that $DX'Q'P'$ is symmetric wrt the perpendicular bisector of $P'Q'$. By symmetry, the circumcircles of $\triangle A'B'C'$ and $A'DX'$ are thus tangent at $A'$. (In fact, the problem was created by inversion at $D$.)
iman007
01.04.2021 03:14
trivial. $X$ is the anti-homologous point of the circles with diameter $BD$ and diameter $CD$ so then: \[XC \times XB=XP \times XQ =XD^2\]so X is the center of the $A$-appolonian circle of $ABC$ hence we are done.
PROA200
01.04.2021 04:15
Let $B'$ be the $B$-antipode and $C'$ be the $C$-antipode. Let $M$ be the midpoint of arc $BC$ not containing $A$. Trivially, $MM$, $BC$, $B'C'$ are all parallel so $(M,M)$, $(B,C)$, $(B',C')$ are the pairs of an involution. Projecting this involution at $D$ onto from $(ABC)$ to itself, $(A,A)$, $(C,B)$, and $(P,Q)$ are the pairs of an involution (since $D$ lies on both $PB'$ and $QC'$). Now, we have that $AA$, $BC$, $PQ$ concur so we're done.
WinterSecret
17.08.2022 17:17
Wlog for simplicity $AC=1$, $AB=a$, $DC=k$, $BD=ak $ and $XC=b.$ Notice that $\angle DPB=\angle DQC=90$ so, we have $\angle PDB=90-\angle PBD=\angle PQC-90=\angle PQD$ meaning that $(PDQ)$ is tangent to BC. Therefore, $XD^2=XQ\cdot XP=XC\cdot XB$ $\rightarrow (b+k)^2=b(b+ak+k)$ which leads to $b=\frac{k}{a-1}$ Now, applying steward to $\triangle ABX$ we have; \[\frac{a^2k}{a-1}+AX^2\cdot(ak+k)-(ak+k)\cdot\frac{k}{a-1}\cdot\frac{ka^2}{a-1}=\frac{ka^2}{a-1}\]Meaning that $AX=\frac{ak}{a-1}\rightarrow XA^2=\frac{a^2k^2}{a-1}=XC\cdot XB\rightarrow XA$ is tangent to $(ABC)$
bin_sherlo
23.03.2023 22:30
Let the tangent and $BC$ intersect at $Y$. $\angle DPC=90$ The intersection of $(ABC)$ and $YP$ be $Q$. We will prove that $\angle BQD=90$.
$\angle ABC=\angle CAY$ $\angle BAD=\angle DAC$ $\implies $$|AY|=|YD|$
$|YD|^2=|YP|.|YQ|$
$\implies \angle PQD=\angle PDY$
$\angle DCP=90-\angle PDC=90-\angle PQD$
$\implies \angle BQD=90$