This solution is super cute; the problem turns out to be far more symmetric than expected Solution with alifenix-: Let $O$ be the circumcenter of $ABC$. As $P$ is the orthocenter of $ADE$ and $AO = AP$, $\angle A = 60^{\circ}$; thus $EOPD$ is cyclic and $PO$ externally bisects $\angle EPD$. Let $DP$ and $EP$ meet $(ABC)$ again at $X$ and $Y$; then it's well-known that $P$ is the incenter of $BXY$. But $B$ and $C$ are reflections over $OP$, and by the external bisector condition $E$ and $X$; $D$ and $Y$ are also reflections over $OP$. So as $P$ is the reflection of $P$ in $OP$, $P$ is the incenter of $CDE$, as desired.

Man that was really fast.

franzliszt wrote: Man that was really fast. wu2481632 is really good at geometry though.

alifenix- wrote: franzliszt wrote: Man that was really fast. wu2481632 is really good at geometry though. And math in general. Don't doubt the Wu.

I heard he swept EGMO too. He's really good.

Anyway, here's the first solution I found: it's far longer and less nice, but I wanted to post it here for a solid comparison of how much easier it is with symmetry. First, note that $P$ is the orthocenter of $\triangle AED$, by orthocenter reflections. Let $O$ be the circumcenter of $\triangle ABC$; as $AO = AP$, it follows that $\angle A = 60^{\circ}$. Now note that $\angle AOC = 2\angle B$ and $\angle POC = \angle A$, so by isosceles $\triangle AOP$ we see that $\angle OPA = 2 \angle B + \angle A$. Letting $M$ be the midpoint of $\overline{BC}$, we also get $\angle MPB = 90^{\circ} - \angle B$, so therefore $3\angle B + \angle A = 90^{\circ}$. Next, we see that $\angle OAP = 90^{\circ} - \angle B - \angle A$, using the aforementioned fact in conjunction with $\triangle AOP$; thus $3 \angle OAP = 270^{\circ} - 3 \angle B - 3 \angle A = 180^{\circ} - 2 \angle A = 60^{\circ}$. Thus, as $\angle EAO = \angle PAD$, we know that $\overline{AO}$ and $\overline{AP}$ trisect $\angle EAD$. Thus $\angle BAD = \angle BCD = \tfrac{\angle A}{3}$. But $\angle PCB = \angle B$, so $\angle PCD = \angle B + \tfrac{\angle A}{3} = 30^{\circ}$, so $\overline{CP}$ bisects $\angle ECD$. Thus $P$ lies on the angle bisector of $\angle ECD$. Additionally, as $\angle A = 60^{\circ}$, it is well-known that $E, O, P,$ and $D$ are concyclic, with center at the midpoint of arc $\widehat{ED}$. It must then follow that $P$ is the incenter of $\triangle ECD$, as desired.

Sad im not as fast as wu Let $M$ be the midpoint of $AB$, note that $C$ is the reflection of $B$ over $OP$, let $A'$ be the reflection of $A$ over $OP$. Since $AOA'P$ is a parallelogram, $A'$ is the midpoint of minor arc $DE$. Then, $C$, $P$, $A'$ are parallel on the reflection of $AB$ over $OP$, and $d(O,DE)=d(M,DE)=\frac{MP+MB}{2}=\frac{MP+MA}{2}$, so $\angle DCE = 60^{\circ}$. Thus, $CP$ bisects $\angle DCE$ and $\angle DPE=\angle DBE = 120^{\circ}=90+\angle{DCE}/2$, implying the result.

Extend line $CP$ to meet $\Gamma$ at $X$. Since $PB = PC$, we have $PA = PX = R$. But $OX = R$ as well, so $APXO$ is a rhombus. Now, since $\overline{OX} \parallel \overline{PB}$, we have $\overline{OX} \perp \overline{DE}$, so $X$ is the midpoint of arc $DE$ not containing $C$. Now we are going to prove that $\overline{DE}$ is also the perpendicular bisector of $\overline{OX}$. This follows from $\overline{OX} \parallel \overline{PB}$ and $OB = PX = R$, so $OPBX$ is an isosceles trapezoid. Thus $\triangle OXD$ and $\triangle OXE$ are equilateral, so $\angle DCE = 60^\circ$ and $\angle DPE = \angle DBE = 120^\circ$. Since $\overline{CP}$ bisects $\angle DCE$, this is enough to show that $P$ is the incenter of $\triangle CDE$.

i think this is one of those problems where constructing the diagram basically kills it because it tells you how to solve the problem as well. i think there was some austria beginner contest that had this, but maybe my memory is bad. anyways:

edit: https://artofproblemsolving.com/community/c6h1657242p10506091 was the problem i was referencing

heh imagine having a purely synthetic solution Let $CP$ hit $(ABC)$ again at point $M$, and let the circumcenter be $O$. Because $\triangle CPB\sim \triangle APM$, $AP=PM=R$. However, note that $OM=OA=R$, so $APMO$ is a rhombus. Thus, $AP||OM$ which means that $OM\perp DE$. This is enough to imply that $M$ is the midpoint of arc $DE$. Let $\angle BCM=\theta$. Common arcs and the fact that $APMO$ is a rhombus gives $\angle PAM=\angle MAO=\theta$ as well. Now note that $$BM=2R\sin{\theta}=AP\sin\theta+AO\sin \theta=PO$$Combining this with the fact that $PB||MO$, $PBMO$ is an isosceles trapezoid. Because $DE$ perpendicularly bisects $PB$, it does the same to $MO$, so $MD=ME=EO=R=MP$, or $M$ is the center of $(PDE)$. Using incenter-excenter lemma gives the result.

khina wrote: i think this is one of those problems where constructing the diagram basically kills it because it tells you how to solve the problem as well. i think there was some austria beginner contest that had this, but maybe my memory is bad. Is IMO 2002 a beginner contest? Not exactly the same problem, but the same rhombus and incenter picture! [asy][asy]size(14cm); pair C = dir(180); pair K = dir(37); pair M = K*K; pair D = dir(60)*M; pair E = dir(-60)*M; pair O = origin; pair P = M+O-K; filldraw(unitcircle, invisible, heavycyan); filldraw(C--D--E--cycle, invisible, heavygreen); filldraw(M--D--O--E--cycle, invisible, blue); draw(D--E, blue); draw(M--O, blue); draw(C--P, deepcyan); draw(P--M, blue+dotted); pair A = P+O-M; pair B = -A+2*foot(O, P, A); draw(A--B--O, orange); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$O$", O, dir(-90)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); label("EGMO 2020", (0,-1), dir(-90)); add(shift(3,0)*CC()); pair B = dir(0); pair C = dir(180); pair D = dir(37); pair A = D*D; pair E = dir(60)*A; pair F = dir(-60)*A; pair O = origin; pair J = A+O-D; filldraw(unitcircle, invisible, heavycyan); filldraw(A--B--C--cycle, invisible, heavycyan); draw(A--D, red); draw(O--J, red); filldraw(C--E--F--cycle, invisible, heavygreen); filldraw(A--E--O--F--cycle, invisible, blue); draw(E--F, blue); draw(A--O, blue); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$O$", O, dir(-90)); dot("$J$", J, dir(J)); label("IMO 2002", (0,-1), dir(-90)); [/asy][/asy]

This feels really easy for a #5? Note that $P$ is the orthocenter of $\triangle ADE$ so $R = AP = 2R \cos \angle DAE \implies \angle DAE = 60^\circ$. Let $CP$ meet $\Gamma$ again at $M$ so $BP \cdot R = AP \cdot BP = CP \cdot MP \implies MP = R$ whence $AOMP$ is a rhombus so $AM$ bisects $\angle AOP$ and thus $M$ is the midpoint of $\widehat{DE}$ since $O, P$ are isogonal conjugates. So, $\angle CP$ bisects $\angle DCE$ and then by the converse of Fact 5 we are done.

I guess I'm slow Let $\omega=(ACEBD)$ and let $CP$ meet $\omega$ again at $M.$ Since $P$ is the reflection of $B$ over $DE,$ it must be the orthocenter of $\triangle ADE,$ so it follows from $AP=2R\cos \angle DAE=R$ that $\angle DAE=60^\circ$ (motivation is that $AO=AP$ which is suspicious). Notice that since $P,$ the intersection of the diagonals of $ACBM,$ lies on the perpendicular bisector of $BC,$ $AMBC$ must be an isosceles trapezoid and $APMO$ is a rhombus with side length $R.$ In particular, this implies that $AP||OM,$ meaning that $M$ is the midpoint of arc $DE$ not containing $A,C$ and that $P$ lies on the angle bisector of $\angle DCE.$ However, we also know that $(PEDO)$ passes through the incenter of $\triangle CDE$ because $\angle DCE=60^\circ,$ so we're done. $\blacksquare$

Anyways, here's my solution. Let $O$ be the center of $\Gamma$ and $M$ the arc midpoint of $\overline{DE}$. Claim: Quadrilateral $APMO$ is a rhombus. Proof. Since $PA = MO$ and both are perpendicular to $\overline{DE}$, it follows $APMO$ is a parallelogram. In fact though, because $AO = MO$, we get the rhombus. $\blacksquare$ Since $PC = PB$, and $PA = PM$, it follows that $C$, $P$, $M$ are collinear now. Claim: We have $MP = MD = ME$ equal to the radius of $\Gamma$. Proof. Note $PBMO$ is an isosceles trapezoid. Since $\overline{DE}$ is the perpendicular bisector of $\overline{PB}$, it is the perpendicular bisector of $\overline{OM}$ too. Hence $MDOE$ is a rhombus (with $60^{\circ}$ internal angles) $\blacksquare$ Since $MD = MP = ME$ we are done by Fact 5. [asy][asy]size(9cm); pair C = dir(180); pair K = dir(37); pair M = K*K; pair D = dir(60)*M; pair E = dir(-60)*M; pair O = origin; pair P = M+O-K; filldraw(unitcircle, invisible, heavycyan); filldraw(C--D--E--cycle, invisible, heavygreen); filldraw(M--D--O--E--cycle, invisible, blue); draw(D--E, blue); draw(M--O, blue); draw(C--P, deepcyan); draw(P--M, blue+dotted); pair A = P+O-M; pair B = -A+2*foot(O, P, A); draw(A--B--O, orange); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$O$", O, dir(-90)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B));[/asy][/asy]

slicc

Too easy for P5 Obviously, the orthocenter of $ADE$ lies on $AB$ and thus, $\angle ADB=\angle EAO$. Now note that by PoP we get $PM=R\implies APMO$ is rhombus $\implies \angle PAM=\angle OAM$ where $M=\Gamma\cap CP$ and thus, $MD=ME\implies \angle DCP=\angle ECP$ and thus $I\in CM$. Now, note that as $APMO$ is rhombus, we have $MO\perp DE\implies DOEM$ is rhombus and as $\angle DAE=60\implies \angle DME=120\implies DM=MO=ME=MP$ and we are done by fact 5.

Solved with nprime06. Let $F$ be the midpoint of minor arc $DE$ and $O$ the circumcenter. First, note that $B$ is the reflection of $P$ over line $DE$, $P$ is the orthocenter of $\triangle ADE$. Also, as $AP = AO,$ we have that $\angle DAE = 60^{\circ}$. Next, we have that $\angle DPE = 120^{\circ} = \angle DOE$, thus $DOPE$ is cyclic, and since $\angle DAE = 60^{\circ}$, its circumcenter is $F$. We now show that $CBFA$ is an isosceles trapezoid. We have that \[\angle CPA = 180^{\circ}- \angle CPB = 180^{\circ}-2\angle PBC = 180^{\circ}-\angle COA,\] and as $AP = R = CO$, we have that $CPOA$ is an isosceles trapezoid. Moreover, as $B$ and $F$ are the reflections of $P,O$ over line $BC$, $PBFO$ is cyclic. But since $FP = FD = DO$, $PFBO$ is an isosceles trapezoid. Hence, $AC = PO = BF$, and $CBFA$ is indeed an isosceles trapezoid, and thus $C,P,F$ are collinear. By fact $5$, we get that $P$ is the incenter of $\triangle CDE$, as needed.

alifenix- wrote: Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$. Prove $P$ is the incentre of triangle $CDE$. Let $K$ be the midpoint of arc $DE$ in $\Gamma$. We see that if $O$ is the circumcenter of $\triangle ABC$, then $OK || AP$ as both these lines are perpendicular to line $DE$ and $AP=AO=OK$, implying that $\square APKO$ is a rhombus. Now, it can be seen that $PB = PC$ and $PA = PK$, therefore points $P, K, C$ are collinear. However, we also see that in $\triangle ADE$, point $P$ is the orthocenter of $\triangle ADE$ and $AP = AO$, which can happen only if $\angle ADE = 90^\circ$ and it is well known (old USAMO problem iirc) that in this case, points $P, O, D, E$ are concyclic, forming a circle with center $K$. Looking at this properly, $P$ lies on the angle bisector of $\angle DCE$ and $KP = KD = KE$, which means that $P$ is the incenter of $\triangle CDE$ as desired.

Solved with mxlcv and jelena_ivanchic! Extend $CP$ to intersect $\Gamma$ again at $M$. We prove that $MD=ME=MP$. $\angle ABC=\angle PBC=\angle PCB=\angle MCB \implies BCAM$ is isosceles trapezoid. Hence $PA=PM$. As $PA=R$, we get $APMO$ rhombus. $PM=R=OB$ and $OM \parallel AP \implies OM \parallel PB$, together imply $PBMO$ is isosceles trapezoid. Hence $DE$, perpendicular bisector of $PB$ is also the perpendicular bisector of $MO$. Hence, $MD=DO=R=OE=EM$ and as $PM=PA=R$, we get $MD=ME=MP$. By Fact 5, we are done!

pad wrote: We restate the problem in terms of $\triangle ADE$. Restated wrote: Given $\triangle ADE$ inscribed in $\Gamma$, consider a line $AB\perp DE$, with $A,B\in \Gamma$. Let $N=AB\cap DE$. Let $P$ be the reflection of $B$ over $N$, and let $C\in \Gamma$ such that $PB=PC$. Given $PA=R$, prove $P$ is the incenter of $CDE$. Let $I,O$ be the incenter and circumcenter of $\triangle ADE$. By orthocenter reflections, $P$ is the orthocenter of $\triangle ADE$. So $PA=2R\cos \angle DAE=R$ implies $\cos \angle DAE=1/2$, or $\angle DAE=60$. Hence $\angle DCE=60$ too. A corollary of this is that $D,P,I,O,E$ are concyclic. Let $L$ be the arc midpoint of $DE$ not containing $C$. By Fact 5, it suffices to show $C,P,L$ collinear. Let $C'=PL\cap \Gamma$; we will show $\triangle PBC$ is iscocleles, which would finish. We have \begin{align*} \angle PCB &= \angle LCB = \angle LAB = \angle EAB-\angle EAL \\ &= 90-E-A/2 = 60-E. \\ \angle PBC &= \angle ABC = \angle ALC = \angle ALP \\ &=\angle ILP = 2\angle IDP = 2(D/2-30) = D-60. \end{align*}But $D-60=E-60$ since $D+E=120$, so we are done.

Wait why is $PA = 2R \cos{DAE}$? Edit: nvm

Assume that $AE > AD$ Consider the triangle $\triangle{DAE}$. Then, $P$ must be the orthocenter of $\triangle{DAE}$ because it lies on the altitude from $A$ to $DE$ and its reflection over $DE$ lies on the circumcircle. We claim that if $AP = AO$, then $\angle{DAE} = 60^{\circ}$. We can use trigonometry to find that \[AP = \frac{AD\cdot \cos{\angle{DAE}}}{\sin{\angle{AED}}} = 2AO\cdot \cos{\angle{DAE}} = AO\]Thus, $\angle{DAE} = 60^{\circ} \implies \angle{DBE} = 120^{\circ}$. Let $\angle{CBP} = \alpha$. Note that $\triangle{COB}$ is isosceles and so is $\triangle{CPB}$ so, \[\angle{PCO} = \angle{PBO} = \angle{PAO}\]which implies that $CAOP$ is cyclic. Further note that $CO = AP$ implying that $\angle{CPO} = \angle{ACP}$ so $CAOP$ is an isosceles trapezoid. Thus, $\angle{PAO} = \angle{COA} = 2\alpha$. This implies that \[\angle{CEB} = \frac{\angle{COB}}{2} = \frac{180-2(3\alpha)}{2} = 90-3\alpha\]Note that $\alpha{CED} = 90-\angle{EDB}-\alpha \implies \angle{DEB} = \angle{EDB}-2\alpha$. We can use this and the fact that $\angle{EDB}+\angle{DEB} = 60$ to find out that $\angle{DEB} = 30-\alpha$ and $\angle{EDB} = 30+\alpha$. Thus, $DP$ and $EP$ are angle bisectors, proving that $P$ is the incenter.

Cute question Angle chasing kills it(Skipping obvious things) A sketch- Let M be midpoint of Arc DE It is easy to Prove that PM is the angle bisector Then just prove that OB and MP intersect on DE We are done

Let $\alpha=\angle ABC=\angle PCB$, $\beta=\angle DCB=\angle DEB=\angle DEP$, and $\gamma=\angle PCE$. Then \[ AC=2AP\sin\alpha \Longrightarrow \frac{\sin\angle APC}{\sin\angle ACP}\cdot\cos\alpha=\sin(2\alpha)=\sin\angle APC \Longrightarrow \angle ACP+(90^{\circ}-\alpha)=180^{\circ} \Longrightarrow\]\[ \angle ABE+\gamma=90^{\circ}+\alpha \Longrightarrow \gamma=\alpha+\beta=\angle DCP, \]so $CP$ bisects angle $DCE$. Let $O$ be the center of $\Gamma$, and let $CP$ meet $\Gamma$ again at $Y$. Then $Y$ is the arc midpoint of minor arc $DE$, so $OY\perp DE$. We have \[ \angle YPE=\angle PCE+\angle PEC =180^{\circ}-\angle DCP-\angle DEP-\angle CDE =180^{\circ}-\gamma-\beta-\angle CBE\]\[ =180^{\circ}-\gamma-\beta-\alpha-(90^{\circ}-\beta) =90^{\circ}-\gamma-\alpha. \]However, we also have \[ \angle OBE=90^{\circ}-\angle BCE =90^{\circ}-\alpha-\gamma. \]Therefore, $\angle YPE=\angle OBE$, so the reflection of $CP$ over $DE$, call it $\ell$, goes through $O$. Since the line through $Y$ perpendicular to $DE$ is obviously not $\ell$, these two lines intersect at the reflection of $Y$ over $DE$, so $O$ and $Y$ are indeed reflections of each other over $DE$. Therefore, $DE$ is the perpendicular bisector of a radius of the circle, thus the arc it covers is exactly known: $\angle DCE=60^{\circ}$. But $\angle DPE=\angle DBE=120^{\circ}$. Therefore, if $I$ is the incenter of triangle $CDE$ and $P\ne I$, then $A,P,I$ are collinear and $D,P,I,E$ are concyclic. This is impossible as $P$ must be on the same side of $DE$ as $C$. $\blacksquare$

The length condition gives us that $\angle POB \cong \triangle POC$ by SSS congruence. Let $M$ be the midpoint of arc $DBE$. Then, from $\overline{MO} \parallel \overline{PA}$ and $MO = OA = AP$, it follows that $MO$ and $OA$ are symmetric WRT line $OP$. Altogether, we have that $OPBM \cong OPCA$; since $MP = BO = R$, it follows that both quadrilaterals are isosceles trapezoids. Since $B$, $P$ and $A$ are collinear, it follows that $C$, $P$ and $M$ are collinear; since line $DE$ is the perpendicular bisector of $\overline{BP}$, it is the perpendicular bisector of $\overline{MO}$ as well, and hence $\angle DCE = 60^{\circ}$ and $\angle DPE = \angle DBE = 120^{\circ}$. There is only one point $P'$ that lies on $\overline{CM}$ satisfying $\angle DP'E = 120^{\circ}$: the incenter of $\triangle CDE$.

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Groupsolved with a lot of people, including Debarchan, siddharthmaybe, rishon_f and pithon_with_an_i. Solution sketch: Note that $\angle DAE = 60^\circ$, so it suffices to show that, for example, $CP$ when extended meets the circumcircle $\omega$ of $\triangle CDE$ at the midpoint of arc $DE$. But this is true as letting this extension be $T$, we have $PA = PT$ and therefore $AT$ bisects $OP$ and also $\angle DAE$, and we're done. $\square$

First I will prove that $P$ is the orthocentre of $AED$. Note that $AP$ is clearly perpendicular to $DE$ and that $P$ is the reflection of $B$ over $BC\cap DE$. Thus this implies that $P$ is the orthocentre of $AED$. Now I will prove that $\angle DAE=60 \deg$. Let $EP\cap (ABC)$ be $K$ and $DP\cap (ABC)$ be $L$. As $AP=R$ we have from the orthocentre property that $AK=AL=R$, as $OL=OK=AO=R$. This means $\angle KAL=120 \deg$. As $angle DAE= \frac{\angle KAL}{2}$ this means that $\angle DAE=60\deg$. Thus this implies that $DEOP$ is cyclic. Let $M$ be the midarc of $DE$ not containing $A$. I will prove that $M$ is the centre of $(DEOP)$. Clearly we have that $OM=OB$. As $A$ reflects over $OP$ to $M$ because $AP$ and $AO$ are isogonal and $AO=AP$ and the reflection of $A$ over $OP$ lies on the circle we get that $MO=MP$, we also have that $BP$ is perpendicular to $DE$ and $OM$ is perpendicular to $DE$ that $OM$ is parallel to $BP$, so we get that $BPOM$ is an isocelles trapezium, and as $DE$ passes through the midpoint of $BP$ and is perpendicular to it, we get that $DE$ is the perpendicular bisector of $OM$, so $MD=OD=OE=ME$ thus $ME=MD=MO$, so $M$ is the centre of $(DEOP)$. Note that $B$ reflects to $C$ over $OP$ and that $POKL$ is cyclic with centre $O$, thus circle $POKl$ reflects over $OP$ to circle $DEOP$. Thus as it is well known that $P$ is the incentre of $BKL$ and $B$ reflects over $PO$ to $C$ we have that $P$ must be the incentre of $DEC$.

And then by this problem we have $\angle EAD=60^{\circ}.$ Now it’s just angle chasing.