My solution: Lemma : Let $ P,P' $ be the isogonal conjugate of $ \triangle ABC $ . Let $ \ell $ be the isogonal conjugate of $ AP $ WRT $ \angle BPC $ . Let $ \ell ' $ be the isogonal conjugate of $ AP' $ WRT $ \angle BP'C $ . Then $ \ell, \ell' $ are symmetry with respect to $ BC $. Proof of the lemma : Let $ X $ be the intersection of $ BP' $ and $ CP $ . Let $ Y\in BC $ be a point such that $ XA,XY $ are isogonal conjugate WRT $ \angle BXC $ . Since $ CA,CY $ are isogonal conjugate WRT $ \angle P'CP $ , so $ A $ and $ Y $ are isogonal conjugate WRT $ \triangle CXP' \Longrightarrow P'Y, P'A $ are isogonal conjugate WRT $ \angle BP'C $ . Similarly, we can prove $ PY, PA $ are isogonal conjugate WRT $ \angle BPC $ . By easy angle chasing we get $ \angle PYB=\angle CYP' $ , so $ PY \equiv \ell $ and $ P'Y\equiv \ell' $ are symmetry with respect to $ BC $ . ____________________________________________________________ Back to the main problem : Let $ M^* $ be the isogonal conjugate of $ M $ WRT $ \triangle ABC $ . From the lemma we get $ PK $ is the isogonal conjugate of $ BM^* $ WRT $ \angle AM^*C $ , so from $ B \in PK $ we get $ PK $ is the bisector of $ \angle AM^*C $ and $ M^* \in BK \Longrightarrow \angle ABP=\angle CBN $ . Q.E.D

I think it's old problem and well known resut

Isn't the isogonal conjugate $P'$ of $P$ with respect to $\triangle ABC$ the point $B$ (if we may even speak of it, since $P$ lies on a side)? And what is then $\ell'$? EDIT. Ahh, I have received by private communication the information that the pair $P,P'$ in the lemma is generic (it has nothing to do with the actual point $P$ on $AC$). Makes sense now.

Observe that the conclusion is equivalent (by Steiner's theorem) to $\dfrac{BC}{BA}=\dfrac{MC}{MA}$, or otherwise said, $B$ is on the $M$-Apollonius Circle of $\triangle{MCA}$.

Let $F$ be the foot of the bisector of $\widehat{CMA}$. By the previous observation, it is enough to prove that $B\in (MFK)$, which is trivial by a simple angle chasing

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Here my solution: We have $ \angle AKP=\angle CKN$ and the bisectors of the angles $ \angle PBN $, $ \angle PKN $, $ \angle AKC$ are intersect on the side $AC$. So the Appalonian circle of the triangles are same...

Write sin. To triangles BAK BAM CNB and BCK.

Use the lemma for solution: LEMMA: if $\frac{\sin A}{\sin (K-A)}=\frac{\sin B}{\sin (K-B)}$ then we have $K=180^\circ$ or $\angle A = \angle B$. PROOF: obviously! in this problem given $K=\angle B$

Let l be a isogonal conjucate of AP WRT angleBPC. What does it mean? Can anyone explain it please

incenters and apollonian circles OP

Sardor wrote: Inside the triangle $ ABC $ a point $ M $ is given. The line $ BM $ meets the side $ AC $ at $ N $. The point $ K $ is symmetrical to $ M $ with respect to $ AC $. The line $ BK $ meets $ AC $ at $ P $. If $ \angle AMP = \angle CMN $, prove that $ \angle ABP=\angle CBN $. My solution: $\angle{AMP} = \angle{CMN}$ $\Rightarrow \angle{AKP} = \angle{CKN}$ $\Rightarrow A, C$ are isogonal conjugates in $\triangle{NBK} (\because \overline{A, P, N, C}$ bisects $\angle{BNK})$ $\Rightarrow \angle{ABP} = \angle{CBN}$ Q.E.D

Mirali wrote: Write sin. To triangles BAK BAM CNB and BCK. You wanna say Write Sines Law to triangles $BAK,BAM$,CMB and $BCK$ .

TelvCohl wrote: My solution: Lemma : Let $ P,P' $ be the isogonal conjugate of $ \triangle ABC $ . Let $ \ell $ be the isogonal conjugate of $ AP $ WRT $ \angle BPC $ . Let $ \ell ' $ be the isogonal conjugate of $ AP' $ WRT $ \angle BP'C $ . Then $ \ell, \ell' $ are symmetry with respect to $ BC $. Nice Lemma Another proof Denote $Q$ the reflection of $P$ on $BC$ Then is easy notice that $A$ and $Q$ are isogonal conjugate WRT $\triangle BP'C$ then $P'Q \equiv \ell'$ is sufficient.

Reflect $B$ in $AC$ to construct isoceles trapezoid $BMKD.$ Define $(X) \equiv \odot(BMKD)$ and note that $X \in AC$ by symmetry. By Brokard's Theorem, $P \equiv BK \cap DM$ and $N \equiv BM \cap DK$ are conjugate WRT $(X).$ Thus, the inversion about $(X)$ fixes $M$ and swaps $\{P, N\}.$ Then if $C^*$ is the image of $C$, we have $\angle AMP = \angle CMN = \angle C^*MP.$ Thus, $C^* \equiv A$, i.e. $\{A, C\}$ are swapped under inversion. Hence, $\angle ABP = \angle CBN$, as desired. [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 = 895.6852661373015, xmax = 1025.8593320711423, ymin = 961.7607726188968, ymax = 1047.8836477262905; /* image dimensions */ pen qqzzqq = rgb(0.,0.6,0.); draw(arc((974.7317865761167,1019.1066129485729),4.439323459143672,-135.17068440153136,-109.50018009608581)--(974.7317865761167,1019.1066129485729)--cycle, qqzzqq); draw(arc((974.7317865761167,1019.1066129485729),4.439323459143672,-46.26536898749708,-20.59486468204402)--(974.7317865761167,1019.1066129485729)--cycle, qqzzqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((xmin, 0.*xmin + 1003.0464807889357)--(xmax, 0.*xmax + 1003.0464807889357)); /* line */ draw(circle((939.0537626361713,1003.0464807889358), 39.12606851249346), linetype("2 2") + blue); draw((956.6736065886997,1037.980566383565)--(990.0977647830306,1003.0464807889357), red); draw((956.6736065886997,1037.980566383565)--(974.7317865761167,986.9863486292985), red); draw((974.7317865761167,1019.1066129485729)--(956.6736065886997,968.1123951943064), red); draw((956.6736065886997,968.1123951943064)--(990.0977647830306,1003.0464807889357), red); draw((958.5756818085035,1003.0464807889362)--(974.7317865761167,1019.1066129485729)); draw((1017.4707063664763,1003.0464807889362)--(974.7317865761167,1019.1066129485729)); /* dots and labels */ dot((958.5756818085035,1003.0464807889362),linewidth(3.pt) + dotstyle); label("$A$", (958.2797269112273,1000.3828867134497), S * labelscalefactor); dot((956.6736065886997,1037.980566383565),linewidth(3.pt) + dotstyle); label("$B$", (956.651974976208,1040.0408429484694), NE * labelscalefactor); dot((1017.4707063664763,1003.0464807889362),linewidth(3.pt) + dotstyle); label("$C$", (1017.4707063664763,999.4950220216209), S * labelscalefactor); dot((939.0537626361713,1003.0464807889358),linewidth(3.pt) + dotstyle); label("$X$", (937.8588389991664,999.7909769188972), S * labelscalefactor); dot((974.7317865761167,1019.1066129485729),linewidth(3.pt) + dotstyle); label("$M$", (975.0011786073352,1020.0638873823215), NE * labelscalefactor); dot((990.0977647830306,1003.0464807889357),linewidth(3.pt) + dotstyle); label("$N$", (990.3908332656998,1000.3828867134497), SE * labelscalefactor); dot((974.7317865761167,986.9863486292985),linewidth(3.pt) + dotstyle); label("$K$", (975.0011786073352,984.2533448118932), SE * labelscalefactor); dot((969.0445386893457,1003.0464807889357),linewidth(3.pt) + dotstyle); label("$P$", (968.4901708672578,999.4950220216209), S * labelscalefactor); dot((956.6736065886997,968.1123951943064),linewidth(3.pt) + dotstyle); label("$D$", (956.0600651816554,964.8682990402976), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $MX$ be bisector of $\angle PMN$ in $\Delta PMN$. Since $PN$ is axis of symmetry for $M,K$ we have $\angle NMX=\angle PMX=\angle PKX=\angle BKX$ $\implies$ $BMXK$ is cyclic. From $\odot(BMXK)$ and $MX=XK$ we clearly have that $BX$ bisects $\angle PBN$. Let $Y'$ be on $\overline{AC}$ such that $\angle Y'MX=\frac{\pi}{2}$. From $\angle PMX=\angle XMN$ $\implies$ $(Y',X;P,N)=-1...(*)$. From $\angle AMX=\angle XMC$ $\implies$ $(Y',X;A,C)=-1...(**)$. Also $(*)$ combined with $\angle PBX=\angle XBN$ clearly leads to the fact that $\angle Y'BX=\frac{\pi}{2}$. This combined with $(**)$ yields to a result, hence we are done.

Let $\gamma$ be the apollonian circle of ABC, and D be intersection of B angle bisector of ABC with AC. From angle chasing, we have that it is also apollonian circle of AMC. But center of apollonian circle is perpendicular bisector of BD $\cap$ line AC. Since angle PBN = DBN, it is also apollonian circle of PBN. But M lies on $\gamma$, so it is also apollonian circle of PMN and we're done... (Or you could just show BMDK is cyclic for all 4 triangles to share same apollonian circle)

Sardor wrote: Inside the triangle $ ABC $ a point $ M $ is given. The line $ BM $ meets the side $ AC $ at $ N $. The point $ K $ is symmetrical to $ M $ with respect to $ AC $. The line $ BK $ meets $ AC $ at $ P $. If $ \angle AMP = \angle CMN $, prove that $ \angle ABP=\angle CBN $. Lemma: Let $ABC$ be a triangle with $P,Q$ two points that are symmetric in line $BC$. Then lines $AP, AQ$ are isogonal in angle $BAC$ if and only if $\odot(APQ)$ is the $A$-appollonian circle of triangle $ABC$. (Proof) Suppose isogonality is true. Let the internal/external bisector of $\angle A$ meet $BC$ at point $D$. Then $AD$ is an angle bisector of $\triangle APQ$ and $DP=DQ$; thus proving $D$ lies on $\odot(APQ)$ as desired. Suppose it is the appollonian circle. Then $APQD$ is cyclic and $DP=DQ$ forces isogonality. Now let $B'$ be the reflection of $B$ in $AC$. The condition states that $MB, MB'$ are isogonal in angle $AMC$. Thus, $\odot(BMKB')$ is an appollonian circle in triangle $AMC$. Now it is also an appollonian circle in triangle $ABC$ so the other half of the lemma shows $BM, BK$ are isogonal in angle $ABC$.

how does proving $D$ lies on $(APQ)$ prove that it is the $A$ apollonious circle?

So far i don't understand why nobody just realize $A$ and $C$ are isogonal conjugate WRT $\triangle MBP$ makes the problem trivial LoL

drmzjoseph wrote: So far i don't understand why nobody just realize $A$ and $C$ are isogonal conjugate WRT $\triangle MBP$ makes the problem trivial LoL Its quiet similar the solving of tranquanghuy above

When you don't know/like basic isogonality, trig can always save you. Btw, the proof provided here is similar to the trig proof of the so-called First Isogonality Lemma. Denote $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$, $\angle AMP = \angle CMN = x$, $\angle PMN = \varphi$ $\angle ABP = y$, $\angle CBN = z$, with the aim to show $y = z$. We also have $\angle APB = \angle NPK = \angle NPM = a$ by symmetry. The Sine Law gives $\frac{AP}{\sin x} = \frac{AM}{\sin a}$ from $AMP$, $\frac{AM}{\sin(\beta - z)} = \frac{AB}{\sin \angle AMB} = \frac{AB}{\sin (\varphi+x)}$ from $AMB$ and $\frac{AP}{\sin y} = \frac{AB}{\sin a}$ from $ABP$. Hence $\sin y = \frac{\sin(\beta-z)\sin x}{\sin(\varphi + x)}$. Analogously, we obtain $\sin z = \frac{\sin(\beta-y)\sin x}{\sin(\varphi + x)}$. Hence $\sin z \sin(\beta - z) = \sin y \sin(\beta -y)$, so $\cos (\beta - 2z) - \cos \beta = \cos (\beta - 2y) - \cos beta$, i.e. $\cos (\beta - 2z) = \cos (\beta - 2y)$. Hence either $\beta - 2z = \beta - 2y$, which yields $y=z$ (and we are done), or $\beta - 2y = 2z - \beta$, i.e. $\beta = y + z$, thus $P \equiv N$, $ABC$ is isosceles and the desired statement again holds.

Here's a solution with inversion: First, we apply Steiner's Theorem two times to get that it suffices to prove that $$\left(\frac{AB}{BC} \right)^2 = \frac{AP}{PC} \cdot \frac{AN}{NC} = \left(\frac{AM}{MC}\right)^2 \iff \frac{AB}{BC} = \frac{AM}{MC}.$$ This indicates that inversion at $M$ will be helpful, since then the condition to prove will be $A'B' = B'C'$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.867176196192233cm); 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 = -8.428173232593108, xmax = 13.739767257887475, ymin = -1.3323343708626068, ymax = 8.751962455287543; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-1.5515876110562743,2.2402211472872597), 2.91571299059628), linewidth(0.8) + sexdts); draw(circle((3.7965862420573786,5.269487736531138), 6.146496533134772), linewidth(0.8) + wvvxds); draw((-1.6593466180870384,8.100073587098848)--(-3.4698820623953255,0.044423196315059865), linewidth(0.8) + dbwrru); draw((-1.6593466180870384,8.100073587098848)--(0.4461412537696319,0.11643640627104741), linewidth(0.8) + dbwrru); draw((-3.4698820623953255,0.044423196315059865)--(0.4461412537696319,0.11643640627104741), linewidth(0.8) + yqqqyq); draw((-1.5515876110562743,2.2402211472872597)--(0.4461412537696319,0.11643640627104741), linewidth(0.8) + wrwrwr); draw((-2.3458339800350014,5.045672082323553)--(-1.5515876110562743,2.2402211472872597), linewidth(0.8) + wrwrwr); draw((-4.163767174046186,0.9448832862526423)--(1.1064506274868002,1.0417992825106144), linewidth(0.8) + yqqqyq); /* dots and labels */ dot((-1.5515876110562743,2.2402211472872597),linewidth(3.pt) + dotstyle); label("$K'$", (-1.4231624126914098,2.491234035000401), NE * labelscalefactor); dot((-4.163767174046186,0.9448832862526423),linewidth(3.pt) + dotstyle); label("$A'$", (-4.831635964022788,0.6670124673176675), NE * labelscalefactor); dot((-2.3458339800350014,5.045672082323553),linewidth(3.pt) + dotstyle); label("$M$", (-3.0387583492132116,5.103519319922075), NE * labelscalefactor); dot((1.1064506274868002,1.0417992825106144),linewidth(3.pt) + dotstyle); label("$C'$", (1.4080294603521932,0.8567315103566717), NE * labelscalefactor); dot((-3.4698820623953255,0.044423196315059865),linewidth(3.pt) + dotstyle); label("$N'$", (-4.304103744739928,-0.2961765204188158), NE * labelscalefactor); dot((-1.6593466180870384,8.100073587098848),linewidth(3.pt) + dotstyle); label("$B'$", (-2.3674448123059653,7.905523647882754), NE * labelscalefactor); dot((0.4461412537696319,0.11643640627104741),linewidth(3.pt) + dotstyle); label("$P'$", (0.3280902922840147,-0.5253640655017396), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Indeed let us invert at $M$, let $A,B,C,N,P, K \mapsto A',B',C', N', P', K'$ respectively. Let the circumcircle of $\displaystyle \triangle A'MC'$ be called $\omega$. Then $K'$ is the center of $\omega$, $MB'$ intersects $\omega$ for the second time at $N'$, and $P'$ is a point on $\omega$ such that $M,P',B',K'$ are concyclic. The angle condition translates to $\angle A'MP' = \angle C'MN'$, which implies that $A'C' \parallel N'P'$ since we get an isosceles trapezium $A'C'P'N'$. Note that $A'B' = B'C'$ is the same as proving that $B'$ lies on the perpendicular bisector of $A'C'$ which happens if and only if $B'$ lies on the perpendicular bisector of $P'N'$. This is easy to see since $$\angle M'B'P' = 180^\circ - \angle MK'P' = 180^\circ - 2 \cdot \angle BN'P' \implies \angle BN'P' = \angle BP'N'. \quad \square$$