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mathaddiction wrote: [asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373926, xmax = 19.375392351983407, ymin = -27.00511038307099, ymax = 11.432112570887025; /* image dimensions */ /* draw figures */draw(circle((-2.380165385326512,-6.289736249880691), 8.494137272468013), linewidth(0.8)); draw((-5.664624261175962,1.5436983669642155)--(-10.449340811461191,-8.942805405744245), linewidth(0.8)); draw((-10.449340811461191,-8.942805405744245)--(5.4997143561562405,-9.461149698691813), linewidth(0.8)); draw((-10.449340811461191,-8.942805405744245)--(2.9757742635150985,-2.398688792590641), linewidth(0.8)); draw((-5.664624261175962,1.5436983669642155)--(2.9757742635150985,-2.398688792590641), linewidth(0.8)); draw((-5.664624261175962,1.5436983669642155)--(-2.656079168173235,-14.779391106702937), linewidth(0.8)); draw((-8.056982536318577,-3.699553519390015)--(-2.380165385326512,-6.28973624988069), linewidth(0.8)); draw((-2.380165385326512,-6.28973624988069)--(-2.656079168173235,-14.779391106702937), linewidth(0.8)); draw((-3.6913519691489567,-9.162440043119394)--(2.9757742635150985,-2.398688792590641), linewidth(0.8)); draw((-8.056982536318577,-3.699553519390015)--(-2.656079168173235,-14.779391106702937), linewidth(0.8)); draw(circle((-6.552709989817214,-11.861098256223594), 4.868281516238014), linewidth(0.8)); /* dots and labels */dot((-5.664624261175962,1.5436983669642155),dotstyle); label("$A$", (-5.505133709499788,1.9424247461546513), NE * labelscalefactor); dot((-10.449340811461191,-8.942805405744245),dotstyle); label("$B$", (-11.645519949032499,-9.501022336610855), NE * labelscalefactor); dot((5.4997143561562405,-9.461149698691813),dotstyle); label("$C$", (5.6592049078324145,-9.062423319501377), NE * labelscalefactor); dot((-2.656079168173235,-14.779391106702937),linewidth(4pt) + dotstyle); label("$P$", (-2.5146858655715194,-14.44522943857226), NE * labelscalefactor); dot((-8.056982536318577,-3.699553519390015),linewidth(4pt) + dotstyle); label("$M$", (-7.897491984642403,-3.4005087349971883), NE * labelscalefactor); dot((-3.6913519691489567,-9.162440043119394),linewidth(4pt) + dotstyle); label("$D$", (-4.548190399442742,-8.86306012990616), NE * labelscalefactor); dot((2.9757742635150985,-2.398688792590641),linewidth(4pt) + dotstyle); label("$E$", (3.147228718932669,-2.08471168366875), NE * labelscalefactor); dot((-6.451154121672622,-6.993876119625543),linewidth(4pt) + dotstyle); label("$X$", (-6.302586467880659,-6.670065044358762), NE * labelscalefactor); dot((-0.722053262074299,-6.1501080214494515),dotstyle); label("$F$", (-0.04258231459081731,-7.108664061468241), NE * labelscalefactor); dot((-2.380165385326512,-6.28973624988069),linewidth(4pt) + dotstyle); label("$O$", (-2.2355774001382143,-5.952357561815978), NE * labelscalefactor); dot((-2.4748132276524752,-9.201977552218032),linewidth(4pt) + dotstyle); label("$N$", (-2.3153226759763013,-8.902932767825202), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $l$ be the line through $D$ perpendicular to $AC$ and $F$ be a point on the line $l$. Let $X=BE\cap MP$ and $N$ be the midpoint of $BC$, let $O$ be the circumcenter of $\triangle ABC$. Obviously $B,X,N,P$ are concyclic. CLAIM. $AE,BX,l$ are concurrent. Proof. By Ceva's theorem on $\triangle ABD$ it suffices to show $$\frac{\sin\angle ABX}{\sin\angle DBX}\frac{\sin\angle BDF}{\sin\angle FDA}{\sin\angle DAE}{\sin\angle BAE}=1$$By easy angle chasing, $\angle ABX=\angle OMX=\angle OMP$ and $\angle BXD=\angle XPN=\angle MPO$, hence the left hand side equals $$\frac{\sin\angle OMP}{\sin\angle OPM}\frac{\sin(90^{\circ}+C)}{\sin(90^{\circ}-\frac{A}{2})}\frac{\sin(90^{\circ}-\frac{A}{2})}{\sin 90^{\circ}}=\frac{OP}{OM}\cos C=\frac{R}{R\cos C}\cos C=1$$hence we are done. $\blacksquare$ Now since $AE\cap BX=E$ we have $E$ lies on $l$, therefore $\angle ADE=\angle ADC-(90^{\circ}-C)=90^{\circ}-\frac{A}{2}=\angle EAD$, this completes the proof. wow!

Nice problem! First define some points: $N$ midpoint of $AD$ $D$ and $K$ the foots of the perpendiculars from $D$ to $NC$ and $AC$ respectively. $I$ the incenter of $\triangle ABC$ $D'$ a point on segment $AC$ such that $CD = CD'$ $R$ foot of the perpendicular from $B$ to $MP$ Notice that $\triangle ABP \sim \triangle ADC \rightarrow \triangle BMR \sim \triangle BAE \sim \triangle DKN$ $\frac{AE}{AB} = \frac{KN}{KD}$ $DKFC$ is cyclic with diameter $CD$. By power of a point: $KN\cdot NC = \frac{AC^2 - CD^2}{4}$ $[ADC] = 2\cdot [DNC]\rightarrow DK\cdot NC = \frac{AC\cdot DF}{2}$ If we divide both equations the result is: $\frac{KN}{KD} = \frac{AC^2 - CD^2}{2DF\cdot AC}$ $\frac{AE}{AB} = \frac{AC^2 - CD^2}{2DF\cdot AC}$ $2AE\cdot DF = \frac{AB}{AC}\cdot (AC - CD)(AC+CD)$ $2AE\cdot DF = \frac{AB}{AC} \cdot AD'\cdot (AC + CD)$ (*) Since $\angle DD'C = \angle BID = 90 - \frac{\angle C}{2}$ and $\angle BAI = \angle DAD'$, we have that $\triangle BAI \sim \triangle DAD'$ $\frac{BA}{AI} = \frac{DA}{AD'}$ If we substitute this in (*), we have that: $2AE\cdot DF = \frac{AI\cdot DA\cdot (AC + CD)}{AC}$(**) By angle bisector theorem: $\frac{CD}{DI} =\frac{CA}{AI} = \frac{AC + CD}{AD}$ Now substitute in (**) $2AE\cdot DF = AD^2\rightarrow AD\cdot AN = AE\cdot DF \rightarrow \frac{AN}{AE} = \frac{DF}{DA}$. We know that $\angle DAE = \angle AEF$, thus $\triangle ANE \sim \triangle DFA\rightarrow \angle ANE = \angle DFA = 90$, and since $N$ is the midpoint of $AD$, $AE = DE$. Done

See my solution to this excellent problem on my Youtube channel here: https://www.youtube.com/watch?v=cpWNkSJJO-0

Draw the line through $D$ perpendicular to $AC$, intersecting the line through $A$ perpendicular to $AB$ at $E'$. We claim that $E'B \perp MP$. Indeed, easy angle chasing shows that $E'A = E'D$. Let $\Omega$ be the circle with center $E'$ and radius $E'A$. Then \[ pow_{\Omega} (M) = AM^2, \quad pow_{\Omega} (P) = PD \cdot PA = BP^2\]Hence $pow_{\Omega} (M) - pow_{\Omega} (P) = BM^2 - BP^2 $, implying that $MP \perp E'B$, hence $E'=E$

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Solution. Let $Q$ be the point such that $QA\perp AP$ and $AQ=QC$. By APMO 2017, P2, we know that $AQPM$ is cyclic. Let $R=\overline{PQ}\cap \overline{BE}$. Since $PM\perp BE$ and $\angle EAM=90^\circ$, we have $\angle AEB= \angle BMP = \angle AQP=\angle AQR$, i.e. $AQRE$ is cyclic as well, implying that $\angle BRP=\angle EAQ=\angle BAP$, thus $R$ lies on $\Gamma$. Finally, notice that $\angle EAQ=\angle BAD=\angle DAC$ and $$\angle EQA=\angle ERA=\angle BRA=\angle DCA$$therefore $\bigtriangleup EAQ \sim \bigtriangleup DAC$; in turn, $A$ is the center of spiral similarity carrying $\overline{DC}$ to $\overline{EQ}$, then it also maps $\overline{DE}$ to $\overline{CQ}$, which immediately leads to the desired result. $\square$

Consider two circles: $\Omega$ - circle through $A,D$, tangent to $AB$ and point-circle $(B)$. Then $PM$ is their radical axis. Hence $E$ - center of $\Omega$. $\blacksquare$ @below Such solutions are never come naturally, despite simple apearance. You only need to search deliberatly for the way how it can be optimized and simplified.

zuss77 wrote: Consider two circles: $\Omega$ - circle through $A,D$, tangent to $AB$ and point-circle $(B)$. Then $PM$ is their radical axis. Hence $E$ - center of $\Omega$. $\blacksquare$ Nice solution! I wish I had found this, since I know you mentioned both the idea of using point-circles, and that $PB^2 = PD \cdot PA$ on my channel.

Cute angle chase. Let $E'$ be the point such that $AE'\perp AB$ and $DE'\perp AC$. Notice that $$\angle DE'A=90^\circ-\angle E'AC=90^\circ-(90^\circ-\angle BAC)=\angle BAC.$$Let $\omega$ be the circle through $A$, $M$ and $E'$, let $DE'$ intersect $\omega$ at $H$, let $N$ be the midpoint of $BC$. Since $\angle EAM=\angle EAB=90^\circ$, we have that $EM$ is the diameter of $\omega$. Let $O$ be the circumcentre of $(ABC)$. We have, $$\angle NMB=\angle BAC=\angle DE'A=\angle HE'A=\angle HMA=\angle HMB\implies \text{$M$, $H$ and $N$ are collinear.}$$Let $R$ be the foot of perpendicular to $MP$ from $B$. I claim that $E'$, $R$ and $B$ are collinear, which directly means that $E\equiv E'$. \begin{align*} \angle HE'B+ \angle DBE'&=\angle HDN\\&=90^\circ-\angle HND\\&=\angle ONM\\&=\angle NPM+\angle HMP\\&=\angle NBR+\angle HMP\\&=\angle NBR+\angle BAC-\angle AER\\&=\angle DE'A-\angle AE'R+\angle NBR\\&=\angle DE'B+\angle NBR\\&=\angle HE'B +\angle NBR, \end{align*} where we used facts we already proved above and the fact that $PHRB$ is cyclic. Hence, $\angle NBR=\angle DBE'$. By the things above, we have that $\angle DEA=\angle BAC$ and we have that $\angle EAD=90^\circ-\frac{\angle BAC}{2}$; those two imply that $\angle EDA=90^\circ-\frac{\angle BAC}{2}$ and hence $AE=DE$.

Really straightforward to trig bash, here is a sketch. Redefine $E$ to be such that $\angle EAB=90$ and $AE=ED$. We have to prove $BM^2+PE^2=ME^2+BP^2$. We have $AD=2AE.sin \frac{\alpha} {2}, BP^2=AB^2+AP^2-2AB.AP.cos \frac{\alpha} {2}, ME^2=AM^2+AE^2, PE^2=AP^2+AE^2-2AP.AE.sin \frac{\alpha} {2}$, so we now get to prove $AB^2+AP.AD=2AB.AP.cos \frac{\alpha} {2}$, which is equivalent to $BP^2=AB^2+AP^2-2AB.AP.cos \frac{\alpha} {2}$ by noticing $AP.AD=AP^2-PD.PA=PA^2-BP^2$, done (the former is LoC in $\triangle BAP$).

Dear Mathlinkers, here Problem 1 Sincerely Jean-Louis