Let $\mathcal{R}$ be an inversion with center $A$ and arbitrary power. Let $X'$ denote the image of $X$ under $\mathcal{R}$. $\mathcal{R}$ takes the perpendicular bisector of $AC$ to the circle with center $C'$ and radius $AC'$, so $D'$ is the point on ray $C'B'$ satisfying $C'D'=C'A$. Similarly, $E$ is the point on ray $B'C'$ satisfying $B'E'=B'A$. $P'$ is the intersection of the line through $B'$ parallel to $A'C'$ and the line through $E'$ parallel to $A'D'$. It follows $\angle C'AD' = \angle E'P'B'$. But from parallel lines $AD'$ and $E'P'$ and $C'D'=C'A$, $\angle BE'P =\angle C'D'A = \angle C'AD'$ hence $\angle BE'P=\angle E'P'B'$ so $\triangle B'P'E'$ is isoceles and $B'P'=B'E'=B'A$. But now $\triangle B'AP'$ is isoceles and so $\angle B'AP'=\frac{1}{2}(180^{\circ}-\angle AB'P') = \frac{1}{2}(180^{\circ} - \angle AB'C'- \angle C'B'P')$ $ = \frac{1}{2}(180^{\circ} - \angle AB'C'- \angle B'C'A) = \frac{1}{2} \angle B'AC'$ as desired.

yunxiu wrote: Circle ${c_1}$ passes $A,B$ and tangent $AC$ at $A$，Circle ${c_2}$ passes $A,E$ and tangent $AD$ at $A$, ${c_1} \cap {c_2} = \left\{ {A,P} \right\}$. Do you mean the tangents to $c_1$ and $c_2$ are $AC$ and $BC$?

Lyub4o wrote: yunxiu wrote: Circle ${c_1}$ passes $A,B$ and tangent $AC$ at $A$，Circle ${c_2}$ passes $A,E$ and tangent $AD$ at $A$, ${c_1} \cap {c_2} = \left\{ {A,P} \right\}$. Do you mean the tangents to $c_1$ and $c_2$ are $AC$ and $BC$? I mean tangents to $c_1$ and $c_2$ are $AC$ and $AD$. Another solution:$BD \cap CE = I$, then we can prove $A,P,I$ are collinear.

Solution Let the center of $c_1$ be $O_1$ and let the center of $c_2$ be $O_2$. Now Let $P'$ denote the point on segment $[EO_1]$ such that $BO_1 = P'O_1$, which implies that $P'$ lies on $c_1$. Since $c_1$ is tangent to $AC$, it follows that $\angle{BO_1 A}=2\angle{BAC}$. Since $O_1$ and $E$ both lie on the perpendicular bisector of $AB$, it follows that $O_1 E$ bisects angle $\angle{BO_1 A}$ which implies that $\angle{BO_1 A}=\angle{BAC}$ and hence that $\angle{BP'E}=90^\circ + \frac{1}{2}\angle{BAC}$. However, since $P'$ lies on the perpencular bisector $EO_1$ of $AB$, $A$ is the reflection of $B$ about $EO_1$ and $\angle{AP'E}=\angle{BP'E}=90^\circ + \angle{BAC}$. Since $c_2$ is tangent to $AD$ and passes through $E$, it follows that $\angle{AO_2 E}=2\angle{DAE}=180^\circ - \angle{BAC}$. Combining this with the angle relation above yields that $P'$ lies on $c_2$. Hence $P'$ lies on both $c_1$ and $c_2$ and $P=P'$. Therefore $\angle{BAP}=\frac{1}{2} \angle{BO_1 P} = \frac{1}{2} \angle{BAC}$ which implies the result.

During the competition, it only took about 45 minutes to solve this problem using complex numbers. The expressions turn out to be surprisingly simple!

Let $ F = AB \cap DE $ , by angle chasing , we have $ \angle DEP = \angle AED - \angle AEP = \angle CAD - \angle DAP = \angle CAP = \angle ABP $ , so $B,F,E,P$ are concyclic . Consider $ \Delta AEF $ and $P$ , $ \angle EPF = \angle EBF = 180^o - \angle EAF $ . At the same time , $ \angle APF = \angle APB + \angle BPF = \angle APB + \angle BED = 180^o - ( \angle BAC - \angle BAD )$$ = 180^o - \angle CAD = 180^o - \angle ACD = 180^o - \angle AEF $ . It is easy to prove that $ P $ is exactly the orthocentre of $ \Delta AEF $ and therefore $ EP $ is perpendicular to $AB$ and $ \angle BAP = \angle ABP = \angle CAP $ .

Let $M,N$ be the midpoints of $AB, AC.$ Let the internal bisector of $\angle BAC$ meet the perpendicular bisector of $AB$ at the point $S.$ Note that $\angle SAC=\angle MAS=\angle MBS$ leads to the fact that $CA$ is tangent to $\odot(ASB).$ We will now show that $DA$ is tangent to $\odot(ASE).$ Obviously $E$ lies on the perpendicular bisector of $AB.$ Hence we get $\angle AES=\angle BES=\frac 12\angle AEB=\frac 12\angle ACB.$ Furthermore, $\angle DAC=\angle DCA=\frac{\pi}2-\frac{\angle ADC}{2}=\frac{\pi}{2}-\frac{\angle ABC}{2};$ which leads to $\angle DAS=\frac{\pi-\angle ABC-\angle BAC}{2}=\frac{\angle ABC}{2}.$ So, $\angle DAS=\angle AES,$ and $DA$ is tangent to $\odot(ASE).$ We are done.$\Box$

Another solution: Using tangencies and the equations $AD=DC, \: AE=EB$ we get the equation $\angle BPE = \angle APE.$ Since $AE=BE,$ by sine law we get $\angle PAE= \angle PBE$ which means the triangle $APB$ is isosceles. The rest is just angle calculation.

Let $\frac{\angle{CAB}}{2}=A,$ and so on. Easy to get $\angle{ABP}+\angle{AEP}=A+C$. \[\sin \angle{ABP}:\sin \angle{AEP}\] \[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2C}}{\sin {2A}}\] \[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2A+2B}}{\sin {2B+2C}}\] \[=\sin A:\sin C\] implies $\angle{ABP}=A$, since $A,B,C<\frac{\pi}{2}$ and the function $\sin x$ in increasing when $0<x<\frac{\pi}{2}$. So done.

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This seems rather easy for China. After a lookthrough my solution is relatively similar to Potla's. If $\omega$ is the circumcircle of $\triangle ABC$, then $D$ is the intersection of the perpendicular bisector of $AC$ and $\omega$ on the same side as $B$, and $E$ is defined similarly. The A-bisector meets $\omega$ again at $K$, and C-bisector meets $\omega$ at $L$ (they are just the midpoints of arcs.) Let $Q=EL\cap AK$, now we want $Q=P$. Notice $\angle ABQ=\angle QAC =\angle QAB$, so $(AQB)$ is tangent to $AC$. So we just need $DA$ tangent to $AQE$. But this happens iff $\angle DAK=\angle AEL =\angle LEB\iff BL=KD\iff BD\parallel LK$ and this is trivial, so we are done.

Dear Mathlinkers, For the first step of the Message 7 1. X the point of intersection of AC and DE is on C2 2. according to Pivot theorem, B, F, E, P are concyclic. Sincerely Jean-Louis

Dear Mathlinkers, For the second step… 3. Y, Z the points of intersection of AD and PF, BY and AE ; Y is on C1 and Z on the circle (BFEP) 4. then the angle chasing of message 7 is more clear. Sincerely Jean-Louis

My solution: Let $F=AB\cap DE,G=AC\cap DE,$ by angle chasing,we have $\angle PEF=\angle AED-\angle PEA=\angle CAD-\angle DAP=\angle CAP=\angle ABP\to FBPE$ cyclic,and $\angle PEG=\angle PEA+\angle DBA=180-\angle DAC+\angle DAP=180-\angle PAG\to APEG$ cyclic.$(1)$ If $\angle DAC=x,\angle BAE=y$ then we find $\triangle AFG$ isosceles. Then from $(1)$ $\angle BPE=\angle APE=x+y,$ from $EB=EA$ and using the Sine Law in $\triangle EPA,\triangle EPB$ we can find $\angle EAP=\angle PBE.$ Then we can find easily $AP=BP$ where using the Sine Law in $\triangle APE,\triangle BPE.$ Then $\angle BAP=\angle PBA=\angle PAC.$ As desired.

My solution Let $J$ be the ẽ-bisector of triangle $ABC$, we have $J, C, E$ and $J, B, D$ are collinear. Let $JC$ cut $(c_2)$ at R, we have $AR//DC$. Similarly, let $BD$ cut $(c_2)$ at $K$, we have $AK//CD$$\implies$ $A, K, R$ are collinear so we have $\angle AKC=\angle JBC$ $\implies$ $R, K, E, B$ are concyclic so $AP$ passes through $J$.

Joint solution with Naruto.D.Luffy: Note that $BD$ and $CE$ are the external angle bisectors of $\angle ABC$ and $\angle ACB$. So it suffices to show that $AP$ concurs with $BD$ and $CE$. Extend $DB$ so that it meets $\odot(ABP)$ at $I$ and extend $EC$ so that it meets $\odot(APE)$ at $J$. Now, $AJ\|DC$ and $AI\|DC$ by Reim's Theorem. So $A,I, J$ are collinear. Now, \begin{align*} \angle IJE&=\angle AJE\qquad[\text{Since }A,I,J\text{ are collinear}]\\ &=\angle DAE\\ &=\angle DBE, \end{align*}giving us that $I,B,E,J$ are concyclic. By Radical Axis Theorem on $\odot(IBEJ)$, $\odot(ABP)$ and $\odot(APE)$, we get $IB$, $AP$, and $JE$ to be concurrent, giving us what we need. NOTE: Darn, just saw that this is the same as the above solution...

The entire difficulty of this problem is identifying this common intersection point, after which phantom points kills. Let $M$ denote the $BC$ minor arc midpoint , $I$ be the incenter of $\triangle ABC$ and let $P'$ be the intersection of the internal $\angle A$-bisector and the $AB$ perpendicular bisector. Then, $AP=BP$ so, \[\measuredangle ABP' = \measuredangle P'AB = \measuredangle CAP' \]which implies that $AC$ is tangent to $(ABP)$ , i.e $P'$ lies on $c_1$. Further, since $P'$ lies on the $AB$ perpendicular bisector, \[\measuredangle P'AD = \measuredangle MAD = \measuredangle ICA = \measuredangle P'EA\]which implies that $AD$ is tangent to $(AEP')$ i.e $P'$ lies on $c_2$. But this means $P' = c_1 \cap c_2 \ne A$ and thus, $P' \equiv P$ which implies that $P$ lies on the internal $\angle BAC-$bisector as desired.