$AB < AC$ on $\triangle ABC$. The midpoint of arc $BC$ which doesn't include $A$ is $T$ and which includes $A$ is $S$. On segment $AB,AC$, $D,E$ exist so that $DE$ and $BC$ are parallel. The outer angle bisector of $\angle ABE$ and $\angle ACD$ meets $AS$ at $P$ and $Q$. Prove that the circumcircle of $\triangle PBE$ and $\triangle QCD$ meets on $AT$.
Problem
Source: Korean Summer Program Practice Test 2023 #6
Tags: geometry, angle bisector, circumcircle, power of a point, radical axis
Seungjun_Lee
04.08.2023 05:23
My solution Let $CD \cap AS = A_1$, $BE \cap AS = A_2$, the inner angle bisector of $\angle DCA, \angle EBA$ meets $AS$ at $Q', P'$. Also, let $CQ', BP'$ meets $AT$ at $X,Y$ First, $C(A_1, A;Q', Q) = -1$ and $A(A_1, T ; B, C) = -1$ Hence, $AT, Q'D, QC$ are concurrent (at point $Z$) In the same way, $AT, P'E, PB$ are concurrent at point $W$ $Q$ is the $D$ - excenter of $\triangle ADC$ and $P$ is the $E$ - excenter fo $\triangle AEB$ Hence, $D,X,Q$ and $E,Y,P$ are collinear Since $X$ is the orthocenter of $\triangle QQ'Z$, $\angle A'BQ = \frac{\pi}{2}$ so $Q',D,C,Q$ are concyclic (= $\omega_1$) In the same way, $P', E,B,P$ are concyclic (= $\omega_2$) Here, $\omega_1$ is the circle of Apollonios about $\triangle ACA_1$ so $AD:A_1D = AQ':A_1Q'$ which means that $\angle ADQ' = \angle Q'DA_1$ Hence, $\angle ADQ' = \angle Q'DA_1 = \angle AQC$ and since $\angle Q'AD = \angle CAQ$, $\triangle AQ'D \sim \triangle ACQ$ Therefore, $AQ' : AD = AC : AQ \Leftrightarrow AQ' \times AQ = AD \times AC$ In the same way, $AP' \times AP = AE \times AB$ Since $DE \parallel BC$, $AD \times AC = AE \times AB$ Hence, $P_{\omega_1}(A) = AQ' \times AQ = AP' \times AP = P_{\omega_2}(A)$ So $A$ is on the radical axis of $\omega_1$ and $\omega_2$ But since $QQ'$ and $PP'$ is the diagonal of $\omega_1$ and $\omega_2$, line $AS$ is the line that connects the center of two circles Therefore, $AT$ is the radical axis of $\omega_1$ and $\omega_2$ which means that the two circles meet on the line $AT$
MathLuis
04.08.2023 05:28
wait what, i like this problem but it actually doesnt take too long to realice something...(prolly 20 mins at most, idk, maybe i got lucky, maybe it is true) Sketch: Reflect $E,B,C,D$ over $AS$ and let the points be $E', B', C', D'$ respecitivily, by noting that $P,Q$ are excenters in $\triangle ABE, \triangle ACD$ respectivily, its trivial by one line angle chase (for each) that $B'E'EBP$ and $D'C'QCD$ are cyclic, note that $AB \cdot AE=AC \cdot AD$ by thales so we have $B'A \cdot AE=DA \cdot AC'$ so $A$ lies in the radax of $(PBE), (QCD)$, the previous cyclics we found also prove that $(PBE)$ and $(QCD)$ are self-reflective when reflection over $AS$ meaning that their radax is also perpendicular to $AS$, but $A$ lies on their radax so their radax must be $AT$, clearly these circles do intersect so we are done
wassupevery1
04.08.2023 05:50
Lemma. Let $ABC$ be a triangle with $A$-excenter $I_a$. Then
(a) The circumcenter of triangle $I_aAB$ lies on $I_aC$,
(b) If $D$ is the intersection, different from $A$, of line $BC$ and $(I_aAB)$, then $CA=CD$.
ProofLet $I_b$ be the $B$-excenter of triangle $ABC$. Then $(I_aAB)$ has diameter $I_aI_b$, but $C$ lies on $I_aI_b$, so the circumcenter of triangle $I_aAB$ (namely the midpoint of $I_aI_b)$ lies on $I_aC$.
On the other hand, $BI_b$ is the internal angle bisector of $\angle ABD$, so $I_bA=I_bD.$ But we have shown that $I_a, I_b, C$ and the circumcenter of triangle $I_aAB$ are collinear, hence $CA=CD$.
Returning to our problem, let $X$ be the intersection, different from $C$, of line $CA$ and $(QCD)$. Let $Y$ be the intersection, different from $B$, of line $BA$ and $(PBE)$.
Note that $P$ is the $E$-excenter of triangle $AEB$, $Q$ is the $D$-excenter of triangle $ADC$, applying the lemma (part b), we get that $AX=AD, AY=AE$. This means $AX \cdot AC = AX \cdot AE \cdot \frac{AC}{AE} = AY \cdot AD \cdot \frac{AB}{AD} = AY \cdot AB$, hence $A$ lies on the radical axis of $(PBE)$ and $(QCD)$.
Now, applying the lemma, but with part a for triangles $AEB, ADC$, we get that the circumcenters of triangles $PBE, QCD$ lie on $AS$. This means the radical axis of $(PBE)$ and $(QCD)$ must be the line from $A$, perpendicular to $AS$, which is $AT$. Finally, we can see that $A$ has negative power wrt $(PBE)$ and $(QCD)$, so the intersection of the two circles lies on $AT$, and we are done.
SQTHUSH
04.08.2023 06:05
Let $\odot(ABE)\cap AF=O_{1}$ Note that $P$ is the circumcenter of $A-$excircle So $\angle PEB=\dfrac{1}{2}\angle AEB=\dfrac{1}{2}\angle PO_{1}B$ Note that $O_{1}B=O_{1}E$ hence $O_{1}$is the circumcenter of $\triangle PEB$ Similarly,let $O_{2}$ be the circumcenter of $\triangle DCQ$ Note that $O_{2}\in AS$ Hence the root axis of $\odot(PEB)$ and $\odot(QCD)$ is parallel to $AT$ Let $CA\cap \odot(QCD)=Y,BA\cap \odot(PEB)=X$ Note that $\angle BYE=\dfrac{1}{2}\angle BO_{1}E=\dfrac{1}{2}\angle BAC=\dfrac{1}{2}\angle DO_{2}C=\angle DXE$ Which means that $X,Y,D,E$ are cyclic Hence $\angle XYE=\angle XDE=\angle XBC$ So $X,Y,B,C$ are cyclic Notice $AY\cdot AC=AX\cdot AB$ Which means that $A$ is on the root axis of $\odot(PEB)$ and $\odot(QCD)$ Thus $AT$ is the root axis of $\odot(PEB)$ and $\odot(QCD)$
pokmui9909
04.08.2023 06:51
Let angle bisector of $\angle ABE$ meets $AS$ at $F$. Since $P, F$ are excenters of $ABE$, we get $\angle PBF = \angle PEF = 90$. Let $X, Y$ be the circumcenter of $PBE, QCD$, then $X, A, S, Y$ are collinear. $\triangle PAB \sim \triangle EAF$, so $PA \cdot AF = AB \cdot AE$. Note that $AT \perp XY$ and $AB \cdot AE = AD \cdot AC$. Thus, $P_{(PBE)}(T) = P_{(QCD)}(T)$ and we are done!