Apollonius Circle.

I have received a solution via PM from user caoquyetthang ,Told me it's belong to his/her friend , who is also a AoPSer but can't post her solution for some reasons . ladykillah : Circle $ \mathcal C $ is a circle that is tangent to $AB,AC$ at points $Q,R$ resp and tangent to $ \omega $ at $ F' $ from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=16265&sid=35f37c614c0547728dd1676e6da0e524#p16265 $ F' \equiv F $ from http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=403068225&t=211949 $ P $ is on $ RQ $ we have $ \frac{QP}{PR}=\frac{sin\angle BAF}{sin\angle CAF}=\frac{BF}{CF} $ $ BF,CF $ cuts $ \mathcal C $ at $ X,Y $ resp.easy to prove $ XY $ parallel to $ BC \Longrightarrow \frac{BF}{CF}=\frac{BX}{CY} $ $ \frac{BQ^{2}}{CR^{2}}=\frac{BX.BF}{CY.CF}=\frac{BF^{2}}{CF^{2}} \Longrightarrow \frac{BQ}{CR}=\frac{BF}{CF} $ $ \Longrightarrow \frac{QP}{PR}=\frac{BQ}{CR}$ and $ \angle BQP=\angle CRP $

Let $M$ be the midpoint of $BC$ and $AI$ meet the cirumcircle at $K$. The points $D', M, K$ are collinear and the line passing through them passes through the cirumcentre $O$ of triangle $ABC$. If $AK$ meets $BC$ at $R$, then $ARMD'$ is cyclic Let $FK$ extended meet $BC$ at $X$. We have $\angle XFA=\angle AD'K=\angle ARX$ and so, $ARFX$ is cyclic. This implies that $\angle FAK=\angle FXR$. But since $AK$ is parallel to $PE$, we have that $\angle FPE=\angle FXR$ which in turn implies that $FXPE$ is cyclic. Now, since tangents at $D', K'$ are parallel to $BC$, we have $FB, FD', FC, FK$ to be harmonic and hence, $X, B, E, C$ are harmonic. But then since $\angle XPE=\angle D'FK=90^{\circ}$, we have that $PE$ is the angle bisector of $BPC$ as required.

goodar2006 wrote:

let $N$ be a midpoint of arc $BC$ we have : $\angle FPE=\angle FAN=\angle FDN $ let $M$ be a midpoint of $BC$ then $\angle DME= 90=\angle DFS $ (1) $ \angle SEF= \angle DEM $ (2) (1),(2)give us that $\angle FSE=\angle FDN =\angle FPE $ so $ FSPE $ is cyclic . now use harmonic as Goutham solution's and get the result .

siavosh wrote: goodar2006 wrote:

let $N$ be a midpoint of arc $BC$ we have : $\angle FPE=\angle FAN=\angle FDN $ let $M$ be a midpoint of $BC$ then $\angle DME= 90=\angle DFS $ (1) $ \angle SEF= \angle DEM $ (2) (1),(2)give us that $\angle FSE=\angle FDN =\angle FPE $ so $ FSPE $ is cycle . now use harmonic as Goutham solution's and get the result . Nice! You could also observe that $\angle SFA=\angle SEP=\frac{A}{2}+C$, Hence $FSPE$ is cyclic

Proposed by Mehdi E'tesami Fard

Nice problem ! Let $ D' $ be the antipole of $ D $ w.r.t. the circumcircle , $ T $ the intersection of $ PE $ and $ DD' $ . Then $ \angle PFD = \angle ID'D = \angle PTD $ , so $ P , F , T , D $ are concyclic . Therefore , $ PE \cdot TE = FE \cdot DE = BE \cdot CE $ and $ P , B , T , C $ are concyclic . We can conclude that $ PE $ bisects $ \angle BPC $ because $ BT = CT $

PE and (PDF) intersect at K. We have KCPB is concylic. AI intersect (w) at D'. We have angleFKD=angleFPK=angleFAD=angleFDD' so K,D,D' are collinear so we have K is on DD' so KB=KC- done

Let $AI$ meet $\omega$ at $Q$. Let $QF$ intersect $CB$ at $R$. Let $DQ$ intersect $BC$ at $M$. Since $\angle DMB = \angle DFR = 90$, $\angle EPF = \angle QAF = \angle QDF = \angle ERF$, so $EPRF$ is cyclic. Then $\angle EPR = 90$. But since $EF$ is the bisector of $\angle CFB$, $(C, B; E, R) = -1$, so $PE$ bisects $\angle CPB$.

Harder problem: Prove that $\angle IPE=90^{o}$

Swistak wrote: Harder problem: Prove that $\angle IPE=90^{o}$ You can see here, as in mahanmath's post.

Goutham wrote: Let $M$ be the midpoint of $BC$ and $AI$ meet the cirumcircle at $K$. The points $D', M, K$ are collinear and the line passing through them passes through the cirumcentre $O$ of triangle $ABC$. If $AK$ meets $BC$ at $R$, then $ARMD'$ is cyclic Let $FK$ extended meet $BC$ at $X$. We have $\angle XFA=\angle AD'K=\angle ARX$ and so, $ARFX$ is cyclic. This implies that $\angle FAK=\angle FXR$. But since $AK$ is parallel to $PE$, we have that $\angle FPE=\angle FXR$ which in turn implies that $FXPE$ is cyclic. Now, since tangents at $D', K'$ are parallel to $BC$, we have $FB, FD', FC, FK$ to be harmonic and hence, $X, B, E, C$ are harmonic. But then since $\angle XPE=\angle D'FK=90^{\circ}$, we have that $PE$ is the angle bisector of $BPC$ as required.

I found an extension for this Let $ABC$ be a triangle with circumcircle $(O)$ and a point $P$. $AP$ cuts $(O)$ again at $D$. $E$ is on $(O)$ such that $DE\perp BC$. $EP$ cuts $BC$ at $F$ and $(O)$ second time at $G$. $Q$ is a point on $AG$ such that $FQ\parallel AP$. Prove that $\dfrac{\tan\widehat{PAB}}{\tan\widehat{PAC}} =\dfrac{\tan\widehat{FQB}}{\tan\widehat{FQC}} $.

For proving $\angle IPE=90$, let $\ell_1$ be the line through $I$ perpendicular to $AI$.Let it intersect $BC$ at $A'$.Then it can be proven that $(A',B,E,C)=-1$.(Hint:-$F$ is the point where the A-mixtilinear incircle touches $\odot ABC$ and if $\ell_1\cap AB=X, \ell_1\cap AC=Y$ then $FX$ bisects $\angle AFB$ and similar for $FY$.) Since $BP$ bisects $\angle BPC$, $(A',B,E,C)=-1\Longrightarrow PA'\perp EP$ and hence $P\in \ell_1$.Hence $PI\perp EP$.

I also have a very nice solution to $ \angle IPE =90^{o}$. $\angle DAI=90^{o}$, so we have to show that $FE \cdot FD=FI^2$ Let $o_1$ be a circle with centre $N$ and $o_2$ be a circle with centre $F$, both passing through $I$. $\angle NFI=90^{o}$, so $o_1, o_2$ and $DF$ concur, so $DF$ is these circles radical axis. So by theorem about radical axis (applied to circles $o_1, o_2, \omega$) we know that $L, M, E$ are collinear and so our thesis follows.

Let $M$ be the midpoint of $\triangle ABC$, and let the line through $F$ perpendicular to $EF$ hit $BC$ at $X$. Note that $\angle BFE=\angle CFE$, so the circle $(XEF)$ is an $E-$Apollonious circle of $\overline{BC}$. Note that $AI, DM$ intersect at the midpoint $Y$ of arc $BC$ (not containing $A$), and thus we have: \[\angle FPE=\angle FAI=\angle FDM=90^\circ-\angle DEC=90^\circ-\angle BEF=angle FXE\] And so $P$ lies on $(XEF)$, and thus $PE$ bisects $\angle BPC$ as desired.

Swistak wrote: I also have a very nice solution to $ \angle IPE =90^{o}$. $\angle DAI=90^{o}$, so we have to show that $FE \cdot FD=FI^2$ Let $o_1$ be a circle with centre $N$ and $o_2$ be a circle with centre $F$, both passing through $I$. $\angle NFI=90^{o}$, so $o_1, o_2$ and $DF$ concur, so $DF$ is these circles radical axis. So by theorem about radical axis (applied to circles $o_1, o_2, \omega$) we know that $L, M, E$ are collinear and so our thesis follows. Nice solution. But We can also prove $FE \cdot FD=FI^2$ without using so much circles. See ($M$ is the midpoint of $BC$) \[FE \cdot FD=FI^2 \iff NI^2-NF^2=FE^2- FE\cdot DE\] \[\iff NB^2=FE^2+NF^2-FE \cdot FD\iff NB^2=DN^2-DM\cdot DN=ND\cdot NM\] Which is obvious.

Meh, you messed so many letters, it became pretty hard to understand xd. \[ FE\cdot FD=FI^{2}\iff NI^{2}-NF^{2}=FD^{2}-FD\cdot DE \] \[ \iff NB^{2}=FD^{2}+NF^{2}-DE\cdot FD\iff NB^{2}=DN^{2}-DM\cdot DN=ND\cdot NM \]

goldeneagle wrote: Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$. Proposed by Mr.Etesami I think there is no need for $DF$ to go through $I$, rather the statement is also true when $I$ is replaced with any arbitrary point $J$.

Mahi wrote: goldeneagle wrote: Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$. Proposed by Mr.Etesami I think there is no need for $DF$ to go through $I$, rather the statement is also true when $I$ is replaced with any arbitrary point $J$. And why not representing a proof?

goodar2006 wrote: And why not representing a proof? Of course! Let $N$ be the arbitrary point, $DN$ meets $BC$ and $\omega$ at $E$ and $F$. Let $G$ be the diametrically opposite point of $D$ on $\omega$. Let $M$ be the midpoint of $BC$(not in the image) Let $SF \perp FD$ meets $BC$ at $S$, then $\angle SFD = \angle SMD = 90^\circ$ implies $SFMD$ cyclic or $\angle FSE = \angle FDG$ (As $D,G,M$ collinear). So $\angle FSE = \angle FDG = \angle FAG = \angle FPE$ So $FSEP$ cyclic. $\angle EPF=\angle ESF= 90 ^\circ$. The rest of the solution is like goutham's post (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2668107#p2668107 ) using the idea of Apollonian circles.

Let $G$ be the midpoint of the arc $BC$ not containing $A$. So, $DG$ is the diameter of $\omega$. Let $PE \cap DG = H$ and as $PE \parallel AG$ and $ADGF$ concyclic, we must have $PDHF$ concyclic. So, $EP.EH=ED.EF=EB.EC \implies PBHC$ concyclic and as $H$ lies on the perpendicular bisector of $BC$, we must have $HB=HC \implies \angle BPE = \angle EPC$. comment: $I$ doesn't need to be the incenter of $\triangle ABC$, the problem is true for any $I \in AG$.

goldeneagle wrote: Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$. Proposed by Mr.Etesami We prove that tangent to $\odot(BIC)$ at $I$ meets $\odot(IE)$ at $P$ then $P$ lies on line $\overline{AF}$. Then $\overline{PE} \parallel \overline{AI}$ and $(IB, IC; IP, IE)=-1 \implies \angle BPE=\angle CPE$ as desired. Now apply inversion at center $I$ with radius $\sqrt{IB \cdot IC}$ followed by reflection in the bisector of angle $BIC$. Note that $B \leftrightarrow C$ are swapped; $A \leftrightarrow H$ with $H$ the orthocenter of $\triangle BIC$; $F \leftrightarrow K$ with $IBKC$ a parallelogram; $E \leftrightarrow G$ with $\overline{IG}$ a median and $G \in \odot(BIC)$; $P \leftrightarrow Q$ with $\overline{IQ} \parallel \overline{BC}$ and $\angle IGQ=90^{\circ}$; and we need to prove $Q \in \odot(IKH)$. Let $R$ be the midpoint of $\overline{IQ}$, $M$ the midpoint of $\overline{BC}$, $O$ the circumcenter of $\triangle BIC$, $X$ the projection of $O$ on $\overline{IM}$ and $Y$ the projection of $M$ on $\overline{IO}$. Notice that $XOYM$ is cyclic with diameter $\overline{MO}$. As $I=\overline{MX} \cap \overline{YO}$ and $R'=\overline{XO} \cap \overline{YM}$ then $\overline{R'I}$ is the polar of $S=\overline{MO} \cap \overline{XY}$ hence $\overline{R'I} \perp \overline{MO}$. Consequently, $R'=R$ and the claim is proved. $\blacksquare$

Dear Mathlinkers https://artofproblemsolving.com/community/c6t48f6h1599839_geometry_with_incircle Sincerely Jean-Louis

Let $M$ be the midpoint of minor arc $BC$ in $\odot(ABC)$; by radical axes on $\odot(BIC)$, $\odot(IFM)$ and $\odot(ABC)$ we obtain that $\overline{MF}, \overline{BC}$, and the line through $I$ perpendicular to $\overline{AI}$ are concurrent. Projecting through $F$ onto $\odot(ABC)$ implies this concurrency point is the harmonic conjugate of $E$ in $\overline{BC}$, so it suffices to show by Apollonius circles that $\overline{AI} \perp \overline{IP}$. Let $P' \in \overline{AF}$ with $\overline{P'I} \perp \overline{AI}$; it suffices to show that $$\frac{FP'}{FA} = \frac{FI}{FD} = \frac{FE}{FI} \iff FD \cdot FE = FI^2 $$but if $\{I, J\} \equiv \overline{DI} \cap \odot(BIC)$ then $(DF;IJ)$ is harmonic, and since $\overline{MF} \perp \overline{FI}$, $F$ is the midpoint of segment $\overline{IJ}$ which implies the result.

Note that $F$ is the touchpoint of the $A$-mixtilinear incircle on $\omega$ (this is a well-known lemma). Let $T_B$ and $T_C$ be the touchpoints of the $A$-mixtilinear incircle on $\overline{AC}$, $\overline{AB}$, respectively. We will prove by barycentric coordinates that $\overline{AE}, \overline{BT_B}, \overline{CT_C}$ concur, and that $P$ lies on $\overline{T_BT_C}$. Note that $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$. It is well known that $I = (a : b : c)$. We now compute $F$. It is a well known lemma that $\overline{AF}$ is isogonal to the $A$-Nagel cevian (this is proved by noticing that the $A$-mixtilinear incircle and the $A$-excircle are inverses under $\sqrt{bc}$ inversion). The Nagel point has barycentric coordinates $(s - a : s - b : s - c)$. Hence, its isogonal conjugate has coordinates $\left(\frac{a^2}{s - a} : \frac{b^2}{s - b} : \frac{c^2}{s - c}\right)$. Since $F$ lies on $\overline{AF}$, we have $F = \left(t : \frac{b^2}{s - b} : \frac{c^2}{s - c}\right)$, for some $t$. Note that $(ABC)$ has equation \begin{align*} a^2yz + b^2zx + c^2xy &= 0. \end{align*}Since $F$ lies on $(ABC)$, we have \begin{align*} \frac{a^2b^2c^2}{(s - b)(s - c)} + \frac{tb^2c^2}{s - c} + \frac{tb^2c^2}{s - b} &= 0 \\ a^2 + t(s - b) + t(s - c) &= 0 \\ a^2 + at &= 0 \\ t &= -a. \end{align*}Therefore, the coordinates of $F$ are $\left(-a: \frac{b^2}{s - b} : \frac{c^2}{s - c}\right)$. Next, we compute $E$. Note that $E = \overline{IE} \cap \overline{BC}$. Since $E$ lies on $\overline{BC}$, its coordinates are of the form $(0, v, w)$ for some $v, w$. Since $F, E, I$ are collinear, we have \begin{align*} \begin{vmatrix} 0 & v & w \\ -a & \frac{b^2}{s - b} & \frac{c^2}{s - c} \\ a & b & c \end{vmatrix} &= 0 \\ \begin{vmatrix} 0 & v & w \\ -1 & \frac{b^2}{s - b} & \frac{c^2}{s - c} \\ 1 & b & c \end{vmatrix} &= 0. \end{align*}Expanding, \begin{align*} -v\left(-c - \frac{c^2}{s - c}\right) + w\left(-b - \frac{b^2}{s - b}\right) &= 0 \\ v\left(\frac{sc}{s - c}\right) &= w\left(\frac{sb}{s - b}\right) \\ v(c(s - b)) &= w(b(s - c)). \end{align*}Hence, $E = (0 : b(s - c) : c(s - b))$. Let $M_B$ be the midpoint of $\widehat{AC}$. By the Shooting Lemma, $F, T_B, M_B$ are collinear. We now compute $M_B$. Since $M_B$ lies on $\overline{BI}$, its coordinates are of the form $(a : t : c)$ for some $t$. Since $M_B$ lies on $(ABC)$, we have \begin{align*} a^2tc + b^2ac + c^2at &= 0 \\ at + b^2 + ct &= 0 \\ t &= -\frac{b^2}{a + c}. \end{align*}Hence, $M_B = \left(a : -\frac{b^2}{a + c} : c\right)$. Now, we compute $T_B$. Since $T_B$ lies on $\overline{AC}$, we have $T_B = (u, 0, w)$ for some $u, w$. Since $T_B, F, M_B$ are collinear, we have \begin{align*} \begin{vmatrix} u & 0 & w \\ -a & \frac{b^2}{s - b} & \frac{c^2}{s - c} \\ a & -\frac{b^2}{a + c} & c \end{vmatrix} &= 0 \\ \begin{vmatrix} u & 0 & w \\ -a & \frac{1}{s - b} & \frac{c^2}{s - c} \\ a & -\frac{1}{a + c} & c \end{vmatrix} &= 0. \end{align*}Expanding, \begin{align*} u\left(-\frac{c^2}{(a + c)(s - c)} - \frac{c}{s - b}\right) + w\left(\frac{a}{s - b} - \frac{a}{a + c}\right) &= 0 \\ u\left(\frac{c^2}{(a + c)(s - c)} + \frac{c}{s - b}\right) &= w\left(\frac{a}{s - b} - \frac{a}{a + c}\right) \\ u\left(\frac{c^2s - c^2b + cas + c^2s - c^2a - c^3}{(a + c)(s - b)(s - c)}\right) &= w\left(\frac{a^2 + ac - as + ab}{(a + c)(s - b)}\right) \\ u\left(\frac{c^2(2s - a - b - c) + cas}{(a + c)(s - b)(s - c)}\right) &= w\left(\frac{a(a + b + c - s)}{(a + c)(s - b)}\right) \\ u\left(\frac{cas}{(a + c)(s - b)(s - c)}\right) &= w\left(\frac{as}{(a + c)(s - b)}\right) \\ u(c) &= w(s - c). \end{align*}Hence, $T_B = (s - c : 0 : c)$. Similarly, $T_C = (s - b : b : 0)$. Recall that $E = (0 : b(s - c) : c(s - b))$. Then, \begin{align*} \frac{BE}{EC} \cdot \frac{CT_B}{T_BA} \cdot \frac{AT_C}{T_CB} &= \frac{c(s - b)}{b(s - c)} \cdot \frac{s - c}{c} \cdot \frac{b}{s - b} \\ &= 1. \end{align*}Therefore, by Ceva's theorem, $\overline{AE}, \overline{BT_B}, \overline{CT_C}$ are concurrent, proving that $\overline{AE}, \overline{BT_B}, \overline{CT_C}$ concur. Now, redefine $P$ as $\overline{T_BT_C} \cap \overline{AF}$. We will show that $\overline{PE} \parallel \overline{AI}$, which is enough to prove (b). Since $P$ lies on $\overline{AF}$, we have $P = \left(t' : \frac{b^2}{s - b} : \frac{c^2}{s - c}\right) = \left(t : b^2(s - c) : c^2(s - c)\right)$, for some $t$. Since $P, T_B, T_C$ are collinear, we have \begin{align*} \begin{vmatrix} t & b^2(s - c) & c^2(s - b) \\ s - c & 0 & c \\ s - b & b & 0 \end{vmatrix} &= 0 \\ \begin{vmatrix} t & b(s - c) & c(s - b) \\ s - c & 0 & 1 \\ s - b & 1 & 0 \end{vmatrix} &= 0. \end{align*}Expanding, \begin{align*} t(-1) - b(s - c)(-(s - b)) + c(s - b)(s - c) &= 0 \\ t &= (b + c)(s - b)(s - c). \end{align*}Hence, $P = ((b + c)(s - b)(s - c) : b^2(s - c) : c^2(s - b))$. Also, we have $E = (0 : b(s - c) : c(s - b)) = (0 : sb(s - c) : sc(s - b))$. Written this way, it is not difficult to verify that the sum of the coordinates of $P$ and the sum of the coordinates of $E$ are both $s^2(b + c) - 2sbc$. Therefore, we have \begin{align*} \vec{PE} &= ((b + c)(s - b)(s - c), (s - c)(b^2 - sb), (s - b)(c^2 - sc)) \\ &= ((b + c)(s - b)(s - c), -b(s - b)(s - c), -c(s - b)(s - c)) \\ &= (b + c, -b, -c), \end{align*}where $\vec{PE}$ is the displacement vector of $\overline{PE}$. Also, recall that $A = (1, 0, 0) = (a + b + c : 0 : 0), I = (a : b : c)$. Hence, \begin{align*} \vec{AI} &= (b + c, -b, -c). \end{align*}Therefore, $\vec{PE} = \vec{AI}$, implying that $\overline{PE} \parallel \overline{AI}$, proving that $P$ lies on $\overline{T_BT_C}$. Now, let $T = \overline{T_BT_C} \cap \overline{BC}$. By the Ceva/Menelaus configuration (recall that $\overline{AE}, \overline{BT_B}, \overline{CT_C}$ concur), we have $(BC;ET) = -1$. It is well known that $\angle AIT_B = 90^\circ$, and that $T_B, I, T_C$ are collinear. Since $\overline{PE} \parallel \overline{AI}$, we have $\angle AIP = \angle IPE = \angle EPT = 90^\circ$. Thus, by the right angles and bisector lemma, $\overline{PE}$ bisects $\angle BPC$, as desired. $\Box$

let's beat the clock to this.

Too easy for TST level. Iran TST 2012 D1 P2 wrote: Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$. Proposed by Mr.Etesami [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.37, xmax = 11.37, ymin = -6.65, ymax = 6.65; /* image dimensions */ pen qqttcc = rgb(0,0.2,0.8); pen ttffqq = rgb(0.2,1,0); pen zzttqq = rgb(0.6,0.2,0); /* draw figures */ draw(circle((-0.017678952177886148,0.641303685653366), 3.7374969131495366), linewidth(0.1) + red); draw((-1.83,3.91)--(-3.33,-1.09), linewidth(0.4) + qqttcc); draw((-3.33,-1.09)--(3.17,-1.31), linewidth(0.4) + qqttcc); draw((3.17,-1.31)--(-1.83,3.91), linewidth(0.4) + qqttcc); draw((-1.83,3.91)--(-0.18827728060230695,-3.092297710604824), linewidth(0.4) + ttffqq); draw((0.07065620525783442,4.377756559136868)--(-1.9813271893911304,-2.5387857433236922), linewidth(0.4)); draw((-1.83,3.91)--(-1.9813271893911304,-2.5387857433236922), linewidth(0.4)); draw((-0.18827728060230695,-3.092297710604824)--(-7.19,-0.97), linewidth(0.4)); draw((-7.19,-0.97)--(-3.33,-1.09), linewidth(0.4)); draw((-7.19,-0.97)--(-1.9149551950022576,0.2896469586452608), linewidth(0.4)); draw((-3.33,-1.09)--(-1.9813271893911304,-2.5387857433236922), linewidth(0.4)); draw((-1.9813271893911304,-2.5387857433236922)--(3.17,-1.31), linewidth(0.4)); draw((-1.83,3.91)--(1.0497212187695046,-1.238236718173737), linewidth(0.4)); draw((-1.9149551950022576,0.2896469586452608)--(-1.5691858982817963,-1.1495967849812316), linewidth(0.4) + zzttqq); draw((-1.9149551950022576,0.2896469586452608)--(-3.33,-1.09), linewidth(0.4) + zzttqq); draw((-1.9149551950022576,0.2896469586452608)--(3.17,-1.31), linewidth(0.4) + zzttqq); /* dots and labels */ dot((-1.83,3.91),dotstyle); label("$A$", (-1.97,4.21), NE * labelscalefactor); dot((-3.33,-1.09),dotstyle); label("$B$", (-3.59,-1.53), NE * labelscalefactor); dot((3.17,-1.31),dotstyle); label("$C$", (3.33,-1.59), NE * labelscalefactor); dot((0.07065620525783442,4.377756559136868),dotstyle); label("$D$", (0.15,4.57), NE * labelscalefactor); dot((-0.18827728060230695,-3.092297710604824),dotstyle); label("$K$", (-0.17,-3.49), NE * labelscalefactor); dot((-1.0536101842688363,0.5985317691006009),dotstyle); label("$I$", (-0.89,0.43), NE * labelscalefactor); dot((-1.9813271893911304,-2.5387857433236922),dotstyle); label("$F$", (-2.17,-3.03), NE * labelscalefactor); dot((-7.19,-0.97),dotstyle); label("$T$", (-7.11,-0.77), NE * labelscalefactor); dot((-1.9149551950022576,0.2896469586452608),dotstyle); label("$P$", (-1.83,0.49), NE * labelscalefactor); dot((1.0497212187695046,-1.238236718173737),dotstyle); label("$Z$", (0.75,-1.59), NE * labelscalefactor); dot((-1.5691858982817963,-1.1495967849812316),dotstyle); label("$E$", (-1.43,-1.61), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $AI\cap\omega=K$. Let $Z$ be the $A-\text{Excircle Touch point}$ with $BC$ and let $KF\cap BC=T$ and also let $AI\cap BC=X$. As $D$ is the midpoint of $\widehat{BAC}$ we get that $\angle BFE=\angle EFC$ and as $\angle EFT=90^\circ$. So, $(TE;BC)=-1$. So it just suffices to show that $\angle TPE$ or $TPEF$ is a cyclic quadrilateral. It's well known that $F$ is the $A-\text{ Mixtillinear Incircle Touch Point}$ with $\odot(ABC)$ and $\angle BAF=\angle CAZ$. Because $\sqrt{bc}$ Inverting with a flip around the angle bisector of $\angle BAC$ swaps $\{F,Z\}$. So we get that $\angle KAF=\frac{\angle A}{2}-\angle BAF$. Also we get that \begin{align*} \angle XTF &=\angle CBF-\angle BFT \\ &=\angle FAC-\angle BFT \\ &=2(\frac{\angle A}{2}-\angle BAF)+\angle BAF-\angle BFT \\ &=\angle A-\angle BAF-\angle BFT \\ &=\frac{\angle A}{2}-\angle BAF \\ &=\angle KAF \end{align*} So we get that $FTAX$ is a cyclic quarilateral and as $PE\|AK$ we get by Reim's Theorem that $ETPE$ is also cyclic $\implies \angle TPE=90^\circ$ and as $(TE;BC)=-1$ we get that $PE$ bisects $\angle BPC$. $\blacksquare$

Since $\angle BEP = 90 - \angle AFD = 90 - \angle PFE$, the circumcenter of $(EFP)$ is on $BC$. Since $\frac{BE}{CE}=\frac{BF}{CF}$ by the angle bisector theorem, the circle through $E$ and $F$ centered on $BC$ is an appolonius circle, so we are done by another application of the angle bisector theorem.

We present two approaches. In both approaches $M$ is the antipode of $D$, i.e.\ the other arc midpoint of $\widehat{BC}$.

. Let $T = \overline{FM} \cap \overline{BC}$. Then we make three observations: We have $(TB;EC) = -1$, since $\angle TFE = 90^{\circ}$ and $\overline{EF}$ bisects $\angle BFC$. We have $\overline{TI} \perp \overline{AM}$ (actually line $TI$ is the extension of the $A$-mixtilinear touch chord). By angle chasing $TPEF$ is cyclic, since $\angle (PE, BC) = \angle (AI, BC) = 180^{\circ} - \angle AFM =\angle TFP$. Thus $\angle TPE = 90^{\circ}$ as well. The second and third observations together imply that $T$, $P$, $I$ are collinear. Since $\angle TPE = 90^{\circ}$, the desired conclusion follows. Second one-liner by Eric Shen from Canada. Let line $ME$ meet $\omega$ again at $G$; by Reim's theorem $GPEF$ is cyclic. The three points $E$, $F$, $G$ are already on the Apollonian circle of $E$ wrt $\overline{BC}$; thus so is $P$.

Let $T$ be the midpoint of $ \overarc{BAC} $ not containing $A$. $ Then $ $ AI\cap \omega=T $ $ DT\cap BC=M(midpoint-BC) $ $ DT\cap PE=K $ Claim 1: $ FPDK-cyclic $ Proof: $ \angle FPK=\angle FAT=\angle FDT\implies FPDK-cyclic $ Claim 2: $ BPCK-cyclic $ Proof: $ PE*EK=FE*ED=BE*EC\implies BPCK-cyclic $ $ BM=MC,\angle DMC=90\implies \angle BCK=\angle CBK $ $ \angle BPK=\angle BCK=\angle CBK=\angle CPK $

What was the morality of this act?

The same as #25, #35. Let $DM$ - diameter of $(ABC)$ and let $PE\cap DM = Q$. By Reim $DPFQ$ is cyclic. Then by Radical Axis $BPCQ$ is cyclic. Since $Q$ is midpoint of the arc $BC$ result follows.

Clearly doing $\sqrt{bc}$ inversion we get that $F$ is the intouch point of the $A$-mixtilinear incircle with $(ABC)$. Let $G$ to be the midpoint of the minor arc $BC$ and then we let $GF \cap BC=J$ then since $JE \perp DG$ and $JG \perp DF$ we have that $E$ is ortocenter on $\triangle DJG$ thus if we let $GE \cap DJ=K$ then $K$ lies on $(ABC)$ and also since $\angle DKG=\angle EFJ=90$ we have $JKEF$ cyclic. And now by reims on $(ABC),(JEF)$ we have $P$ lying on $(JKEF)$ thus $\angle JPE=90$ and now since $-1=(B,C;D,G) \overset{F}{=}(B,C;E,J)$ meaning that $PE$ bisects $\angle BPC$ thus we are done .

Let $M_a$ be the midpoint of arc $BC$ (which clearly lies on $AI$), $Q = FM_a \cap BC$, and $R = AM_a \cap BC$. It's easy to see that $(EFQ)$ is the $E$-Apollonian Circle wrt $BC$. The Shooting Lemma yields $$M_aA \cdot M_aR = M_aB^2 = M_aF \cdot M_aQ$$which implies $ARFQ$ is cyclic. Hence, we have $$\angle FQE = \angle FQR = \angle FAR = \angle FPE$$so $EFQP$ is cyclic, which clearly suffices. $\blacksquare$ Remark: Once we spot and construct the $E$-Apollonian Circle, angle chasing in reverse becomes extremely natural. Edit: Because $ARFQ$ is cyclic and $AR \parallel PE$, Reim's implies $PEFQ$ is cyclic.

Let circumcircle of PBC meet PE at S. we need to prove S lies on perpendicular bisector of BC. Let AI meet ABC at K. we know DK is diameter of ABC so in fact we need to prove D,K,S are collinear. Let DK meet PE at S'. ∠S'DF = ∠KAF = ∠S'PF so S'PDF is cyclic. PE.ES' = DE.EF = BE.EC = PE.ES so S and S' are same so S lies on DK which is perpendicular bisector of BC. we're Done.

yoink Let $PE$ intersect perp bis of $BC$ again at $X$ Suffices to show that $PBCX$ concyclic, trivial by GOOD DIAGRAM Basically let $M$ be the midpoint of arc $BC$ not with $A$, note $DMAF$ then Reim's to $DXPF$ then $ED\cdot EF=EB\cdot EC=EP\cdot EX$

Let $M$ be the arc midpoint of $BC$, and let $MF$ intersect $BC$ at $Q$. Note that $$(QE;BC)=^F(MD;BC)=-1.$$so it suffices to show that $\angle QPE=90$. However, since $\angle QFE=90$, this is equivalent to $QPEF$ cyclic. Let $R$ be the intersection of $AI$ and $BC$. We have $$\angle PEQ=\angle ARQ,$$which intercepts arc $AB$ plus arc $MC$ on $(ABC)$. However, $$\angle PFQ=180-\angle AFM,$$and $\angle AFM$ intercepts arc $AM$ containing $C$, so $\angle PFQ$ intercepts arc $AB$ plus $BM$, and since $M$ is the arc midpoint, it is equal to $\angle PEQ$, done.

Redefine $P=MN \cap AF$ where $MN$ is $A$ mixtilinear touch chord. $G$ is the other midpoint of arc $BC$. $X =(AMN) \cap (ABC)$ and $H=DX \cap BC \cap FG$(It is well known that these concur). Now since $AI \perp MN$ at $I$, some trivial angle chasing yields that $(XHFP)$ is cyclic and it is easy to see that $E$ is orthocenter of $DHG$ so $(XEFH)$ is cyclic implying $(XHFEP)$ is cyclic so this implies that $HP \perp PE$. It is also well known that $(HE;BC)=-1$ so this implies that $PE$ bisects $\angle BPC$ by the famous lemma about harmonic bundles, angle bisectors and perpendiculars. Now it remains to show that $PE \parallel AI$ . Note the following rare angle chase I somehow chose to write:- $$ \angle FPE = \angle FXE = \angle FXG = \angle FAG = \angle FAI$$which proves that $PE \parallel AI$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8.9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.598165822680246, xmax = 6.920592775478876, ymin = -4.50164554803081, ymax = 7.050025886446059; /* image dimensions */ draw((-2,5)--(-3,0)--(3,0)--cycle, linewidth(0.6)); /* draw figures */ draw(circle((0,2), 3.6055512754639896), linewidth(0.6)); draw((-2,5)--(-3,0), linewidth(0.6)); draw((-3,0)--(3,0), linewidth(0.6)); draw((3,0)--(-2,5), linewidth(0.6)); draw((-1.8210582127866786,1.3982372913692824)--(-1.3977060251916877,0), linewidth(0.6)); draw((-2,5)--(-1.692794816346267,-1.183464419425983), linewidth(0.6)); draw((-1.692794816346267,-1.183464419425983)--(0,5.6055512754639905), linewidth(0.6)); draw((-3,0)--(-1.8210582127866786,1.3982372913692824), linewidth(0.6)); draw((-1.8210582127866786,1.3982372913692824)--(3,0), linewidth(0.6)); draw((-6.439122274489518,0)--(3,0), linewidth(0.6)); draw((-6.439122274489518,0)--(0,-1.6055512754639896), linewidth(0.6)); draw((-2,5)--(0,5.6055512754639905), linewidth(0.6)); draw(circle((-0.9860241491363453,1.6510652625189504), 1.6510652625189512), linewidth(0.6)); draw((-6.439122274489518,0)--(0.8062713305997746,2.193728669400225), linewidth(0.6)); draw(circle((-0.6955995074477237,0.6918578313527275), 2.123966108099418), linewidth(0.6)); draw(circle((-1.3477997537238617,2.8459289156763634), 2.250641552438296), linewidth(0.6)); draw((-6.439122274489518,0)--(0,5.6055512754639905), linewidth(0.6)); draw((-4.832960731148251,0.4863065855857564)--(-2,5), linewidth(0.6)); draw((-4.832960731148251,0.4863065855857564)--(-1.692794816346267,-1.183464419425983), linewidth(0.6)); draw(circle((-3.918414149840598,0), 2.5207081246489103), linewidth(0.6)); draw((0,5.6055512754639905)--(0,-1.6055512754639896), linewidth(0.6)); draw((0,-1.6055512754639896)--(-2,5), linewidth(0.6)); /* dots and labels */ dot((-2,5),linewidth(3pt) + dotstyle); label("$A$", (-2.229546966040998,5.104481223797324), NE * labelscalefactor); dot((-3,0),linewidth(3pt) + dotstyle); label("$B$", (-2.9439266468573337,0.08862389040605122), NE * labelscalefactor); dot((3,0),linewidth(3pt) + dotstyle); label("$C$", (3.0751021532122182,-0.2153674631328138), NE * labelscalefactor); dot((0,5.6055512754639905),linewidth(3pt) + dotstyle); label("$D$", (-0.01041008520727439,5.727663498551997), NE * labelscalefactor); dot((-0.9860241491363453,1.6510652625189506),linewidth(3pt) + dotstyle); label("$I$", (-0.9223841458238732,1.7453767671928655), NE * labelscalefactor); dot((-1.3977060251916877,0),linewidth(3pt) + dotstyle); label("$E$", (-1.256774634716626,0.14942216111382423), NE * labelscalefactor); dot((-1.692794816346267,-1.183464419425983),linewidth(3pt) + dotstyle); label("$F$", (-1.6823625296710387,-1.5225302833499332), NE * labelscalefactor); dot((-1.8210582127866786,1.3982372913692824),linewidth(3pt) + dotstyle); label("$P$", (-1.7583603680557554,1.4869841166848303), NE * labelscalefactor); dot((0,-1.6055512754639896),linewidth(3pt) + dotstyle); label("$M_A$", (0.05038818550049886,-1.9025194752735146), NE * labelscalefactor); dot((-6.439122274489518,0),linewidth(3pt) + dotstyle); label("$X$", (-6.728618998416219,0.07342432272910795), NE * labelscalefactor); dot((-2.778319628872465,1.1084018556376751),linewidth(3pt) + dotstyle); label("$K$", (-3.004724917565107,1.228591466176795), NE * labelscalefactor); dot((0.8062713305997746,2.193728669400225),linewidth(3pt) + dotstyle); label("$L$", (0.977561813794041,2.201363797501163), NE * labelscalefactor); dot((-3.571177939257481,2.496677084392556),linewidth(3pt) + dotstyle); label("$U$", (-3.8711002751508756,2.687749963163347), NE * labelscalefactor); dot((-1.873763924386207,2.459104456760453),linewidth(3pt) + dotstyle); label("$N$", (-1.8191586387635286,2.550953854070858), NE * labelscalefactor); dot((-4.832960731148251,0.4863065855857564),linewidth(3pt) + dotstyle); label("$Y$", (-5.041466986275511,0.5750100560682352), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $M_A$ be the arc-midpoint of minor-arc $\widehat{BC}$, $K,L$ be the touch-points of $A-$Mixtillinear Incircle with $AB$ and $AC$, it is also clear that $F$ is the $A-$ Mixtillinear touch-point. Now define $U=(AKL)\cap(ABC)$ and $N=AF\cap (FKL)$. It is known that $DU, BC, M_AF$ concur and $AU, FF, KL$ concur, let the concurrence points be $X,Y$ respectively. We have $UFBC$ is harmonic (motivated from ISL 2016 G2) because $(FN;KL)\stackrel{F}{=}(PY;KL)=(UF;BC)=-1$. Also note that $\odot(XUPEF)$ because $\angle FUM_A = \angle FAM_A = \angle FPE=\angle FAM_A=\angle FDM_A=\angle FXE$, here we used the fact that $\angle DM_AX=\angle DFX=90^{\circ}$. Now because $XI\perp AI$ and $\angle XPE=90^{\circ}$ we have $\overline{X-P-I}$ and since $(XE;BC)=-1$ (kinda obvious finishing) we are done!!!

Really nice problem! Claim: $\frac{FI}{ID} = \frac{FE}{EI}$ Proof: Let $DI \cap (BIC) = K$, and $M$ midpoint of minor arc $\widehat{BC}$. $BD$ and $CD$ are tangent to $(BIC) \Rightarrow BICK$ is harmonic. So $(D, E ; I, K) = -1$. Therefore, the claim is equivalent to: $\frac{KD}{KE} = \frac{IF}{FE}$. Perform inversion wrt $(BIC)$ and denote by $X’$ the inverse of point $X$ for every point $X$ in the plane. $K’ \equiv K, B’ \equiv B, C’ \equiv C, I’ \equiv I$ and $D’$ is the midpoint of $BC$. $E \in (BC) \Rightarrow E’ \in (MBC)$ $F \in (BMC) \Rightarrow F’ \in (BC)$, moreover, since $MFED’$ is cyclic, $D - E’ - F’$ collinear. It follows that $MD’IE’F’K$ is cyclic, $E’$ is the orthocenter of $\triangle DF’M$ and $K$ is the reflection of $I$ wrt $F’M$. By the inversion distance formula, we get that the conclusion is equivalent to: $KD’ \cdot MI \cdot F’E’ = MD’ \cdot KE’ \cdot IF’$ which is true by the following lemma: Lemma: In a cyclic hexagon $ABCDEF$, the main diagonals concur iff $AB\cdot CD\cdot EF = BC \cdot DE \cdot FA$ Therefore, the claim is proved $\square$ By Thales, we get that $\angle AIP = 90^{\circ}$ Let $R, S$ be the intersections of $IP$ with $AB$, $AC$ respectively, and let $RS \cap BC = T$. We know that $IE$ is the $I$-symmedian in $ \triangle BIC$, so, by Steiner, $\frac{BE}{EC} =\frac{BI^2}{CI^2} = (\frac{\sin(\frac{B}{2})}{\sin(\frac{C}{2})})^2$, by Law of Sines. By Menelaus in $\triangle ABC$, we get that: $\frac{AS}{SC} \cdot \frac{BT}{CT} \cdot \frac{BR}{AR} = 1$ By Ratio Lemma in $\triangle AIB$ and $\triangle AIC$ we get that $\frac{CT}{BT} = \frac{CE}{BE}$ so $(C, B ; E, T) = -1$. We are done from EGMO Lemma 9.18 $\blacksquare$