X,Y the intersections of AP with BD,CE respectively. Bisector theorem=> XA/XP=BA/BP, YA/YP=CA/CP AP,BD,CE meet at a point <=> X=Y <=> BP/BA=CP/AC (*) Consider B',C',P' the images of B,C,P under the inversion centered A with coefficient r radius Show that ABC is similar to AC'B' ABP similar to AP'B' ACP similar to AC'P' The angle condition => B'C'P' is isoceles at P' And get (*)

reflect BC in BD. reflect BC in CE. reflections will intersect in T on AP (because of given equal angles). the concurrcency point of BD,CE and AP is the incenter of TBC. Col. Pestich

what is inversion?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=14958.

Since I went through the trouble of finding a solution, I'm posting it . Let $S=AP\cap BC$ and $T\ne P$ the intersection between $AP$ and the circumcircle $(BPC)$. We have $\angle ABT=\angle ACT$, so $\frac{[ABT]}{[ACT]}=\frac{AB\cdot BT}{AC\cdot CT}\ (*)$ ($[XYZ]$ denotes the area of the triangle $XYZ$). At the same time, we have $\frac{[ABT]}{[ACT]}=\frac{BS}{CS}=\frac{[PBT]}{[PCT]}=\frac{PB\cdot BT}{PC\cdot CT}$. This, together with $(*)$, shows that $\frac{AB}{AC}=\frac{PB}{PC}$, or, in other words, the bisectors from $B,C$ in the triangles $ABP,ACP$ respectively cut $AP$ in the same point, dividing $AP$ into the ratio $\frac{AB}{PB}=\frac{AC}{PC}$.

Same approach as grobber's, but alternative way to achieve AB/BP = AC/CP: Consider the spiral similarity that takes AP to AC. Let it take B to Q. Then by angle condition, $\angle QBC = \angle APC-\angle ABC = \angle APB-\angle ACB = \angle QCB$. Hence $QB = QC$. Then $\frac{AB}{BP}= \frac{AQ}{QC}= \frac{AQ}{QB}= \frac{AC}{CP}$.

I looked through previous posts and i didn't notice my solution... Solution: It is enough to prove that $ \frac {AB}{BP} = \frac {AC}{CP}$. Let $ \omega$ be the circumcirlce of $ \triangle ABC$.Denote $ D\in\omega\cap AP$.Since $ \angle BDA = \angle BCA$ and $ \angle ADC = \angle ABC$,thus we conclude that $ \angle PBD = \angle PCD$,so from sine law: $ \frac {BP}{PC} = \frac {\sin{\angle BDA}}{\sin{\angle ADC}} = \frac {AB}{AC}$.Thus we are done.

nice one; i haven't read the solutions yet, but i am trying to come up one myself ;-)

Hmm okay I was doing this problem and if you drop perpendiculars from P to like X on BC, Y on AC, and Z on AB, you get cyclic quads AZPY , BXPZ, CYPX. The solution manual dude is like ∠XZY = ∠APB − ∠C and XY = PC sin∠C. How is this?

Here is my solution: Let D is on the side AB,E is on BC and F is on CA such that PD is perpendicular to AB,PE is perpendicular to BC and PF is perpendicular to CA then the triangle DEF is isosceles and the proof is completed by angle bisector theorem.

From the given equality we conclude that P is on the Apollonius Circle. Therefore the bisectors of BAC and BPC meet at the same point on BC. Now with the simple bisectors theoreom we get the conclusion.

Arne wrote: Let $ P$ be a point inside a triangle $ ABC$ such that \[ \angle APB - \angle ACB = \angle APC - \angle ABC. \] Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point. Let $\Delta ACP\sim\Delta AF_1B$ and $\Delta ABP\sim\Delta AF_2C$, such that $AF_1$ and $AF_2$ are the same straight line. (it's possible because $\measuredangle BAP+\measuredangle CAP=\measuredangle BAC$). Hence, $\frac{AB}{AP}=\frac{AF_1}{AC}$ and $\frac{AC}{AP}=\frac{AF_2}{AB}$, which gives $AF_1=AF_2$ and $F_1\equiv F_2$. Let $F_1\equiv F_2\equiv F$. Hence, $\measuredangle CBF=\measuredangle ABF - \measuredangle ABC = \measuredangle APC - \measuredangle ABC= $ $=\measuredangle APB - \measuredangle ACB= \measuredangle ACF - \measuredangle ACB=\measuredangle BCF$. Hence, $CF=BF$. Since $\frac{AB}{AP}=\frac{BF}{CP}$ and $\frac{AC}{AP}=\frac{CF}{BP}$, we obtain $\frac{AB}{BP}=\frac{AC}{CP}$, which ends the proof.

I have 13 written solutions of this problem 4 are of inversion from A,B,C,P one of them is cut and paste method let $AP=x$ $CP=z$ $BP=y$ draw triangles having sides $yx,yz,by$ $zx,zy,zc$ $xy,xz,xa$ (in short multiply the sides of traingles $APC,APB,BPC$ by $y,z,x$) where as usual $a,b,c$ are the sides the 3 triangles fit together again two sides will come congruent (base angles equal) which will be $by=cz$ thus $P$ lies on appolonious circle

$1)$ The given condition is equivalent to assuming that $\angle PAC + \angle PBC=\angle PAB +\angle PCB$ The to prove that condition also can be converted to prove that $\frac{BP}{PC}=\frac{AB}{AC}$ $2)$ After doing inversion from $A/B/C/P$, mark the points $A^',B^',C^',P^'$ $3)$ Then do some angle chasing using the modified to prove that condition and deduce that $P^'B^'=P^'C^'/P^'C^'=A^'C^'/P^'B^'=A^'B^'/A^'B^'=A^'C^'$ (according to whether the inversion is done from $A/B/C/P$). $4)$ Use the fact that if $O$ is the center of inversion,$r$ is the radius and $A,B$ are any two points then $A^'B^'=\frac{AB*r^2}{OA*OB}$ $5)$ Deduce the to prove that condition $\frac{BP}{PC}=\frac{AB}{AC}$ The solution of inversion from $A$ is given by Moubinool

Apply an inversion with center $A$ and arbitrary radius $r$ and let $X'$ be the image of $X$.Then note that $\angle{P'B'C'}=\angle{P'B'A}-\angle{C'B'A}=\angle{APB}-\angle{ACB}$ $\angle{P'C'B'}=\angle{P'C'A}-\angle{B'C'A}=\angle{APC}-\angle{ABC}$ Thus $\triangle{B'P'C'}$ is $P'$-isosceles so $B'P'=\frac{r^2 \cdot BP}{AB \cdot AP}=\frac{r^2 \cdot CP}{AC \cdot AP} \implies \frac{AB}{BP}=\frac{AC}{CP}$ ot the bisectors concur with $AP$ as desired//

Arne wrote: Let $ P$ be a point inside a triangle $ ABC$ such that \[ \angle APB - \angle ACB = \angle APC - \angle ABC. \]Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point. It's easy to prove that $P$ lies on the Apollonius circle of the point $A$ and so we immediately get $\frac{AB}{AC}=\frac{PB}{PC}$

If $PB$ intersect the circumcircle at $N$ and $CP$ intersect it at $M$ we can apply inversion with center $A$ and radius $\sqrt {AM.AN}$ too

I have a different solution.

It's enough to prove that CP/AC=PB/AB. Let the circlecumcircles of triangles APC and APB intersect AC in Q and G respectively. Then it’s easy to see that Q and G are isogonal conjugate. Note that angles CPQ and BPG are equal . So by Shteiner theorem we have (AQ/QB)(AG/GB)=AB^2/AC^2=BP^2/CP^2 . So AB/AC=BP/CP and result follows.

Here is a solution using Normal Inversion Quote: Let $ P$ be a point inside a triangle $ ABC$ such that \[ \angle APB - \angle ACB = \angle APC - \angle ABC. \]Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point. $BD \cap AP = C$ . Let $\angle APB = x$ and $\angle APC = y$ Invert about $A$ with radius of inversion as $r$ and let primes/dashes denote inverses. $BPP'B'$ is a cyclic quadrilateral $\Longrightarrow \angle BPP' = 180 - x$ and $\angle BB'P' = x$ $CPP'C'$ is a cyclic quadrilateral $\Longrightarrow \angle CPP' =180 - y$ and $\angle CC'B' = y$ $BB'C'C$ is a cyclic quadrilateral $\Longrightarrow \angle BB'C' = C$ and $\angle CC'B' = B$ Thus, $\angle P'B'C' = C-x$ and $\angle P'C'B' = B-y$ $\Longrightarrow P'B' = P'C'$ By, Inversion Distance Formula $P'B' = P'C' = \frac{PB \cdot r^2}{AP \cdot AB} = \frac{PC \cdot r^2}{AP \cdot AC} \Longrightarrow$ $\frac{AB}{PB} = \frac{AC}{PC}$ By, Angle Bisector Theorem $\frac{AB}{PB} = \frac{AC}{PD} = \frac{AX}{XP}. $ Thus, $X = CE \cap AP$. Hence, we are done $ $ $\blacksquare$

Let $\Psi$ refer to the inversion around the $A$ with radius $1$, $\Psi(X)$ be the image of $X$ under $\Psi$, $X'$ be $\Psi(X)$ where $X$ is a point. Claim: $P'B' = P'C'$. Proof: $\angle P'B'C' = \angle P'B'A' - \angle C'B'A' = \angle APB - \angle ACB = \angle P'C'A' - \angle B'C'A' = \angle P'C'B'$. Now, by the inversion distance formula, $P'B' = \frac{PB}{AB \cdot AP}$. Similarly, $P'C' = \frac{PC}{AC \cdot AP}$. Setting the two equal gives $\frac{AB}{PB} = \frac{CA}{PC}$, so the angle bisectors of $\angle ABP$ and $\angle ACP$ concur with $AP$, as desired.

Let $Q$ be the intersection of $AP$ and $BD$, and let $\Omega_B$ and $\Omega_C$ be the circumcircles of $\triangle ABP$ and $\triangle APC$, respectively. Let $R$ be the intersection of the tangents from $B$ and $C$ to $\Omega_B$ and $\Omega_C$, respectively. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -8.310099767583345, xmax = 16.194800154907682, ymin = -11.11738676680911, ymax = 7.337675020204087; /* image dimensions */ /* draw figures *//* special point */ draw(circle((-3.238013249960513,-2.134293777766263), 4.1324647838105815), linewidth(1.6) + linetype("2 2")); draw(circle((5.12,0.24), 6.855569017729871), linewidth(1.6) + linetype("2 2")); draw((-5.420933946887003,-5.643157186388007)--(-1.5880573948301944,1.654493741421877), linewidth(1.6)); draw((-1.5880573948301944,1.654493741421877)--(8.845279347716643,-5.515095163273459), linewidth(1.6)); draw((8.845279347716643,-5.515095163273459)--(-5.420933946887003,-5.643157186388007), linewidth(1.6)); draw((-5.420933946887003,-5.643157186388007)--(0.15738427123569654,-4.489827870597285), linewidth(1.6)); draw((0.15738427123569654,-4.489827870597285)--(8.845279347716643,-5.515095163273459), linewidth(1.6)); draw((-5.420933946887003,-5.643157186388007)--(-0.5558921299286373,-1.9789449850664533), linewidth(1.6)); draw((-0.5558921299286373,-1.9789449850664533)--(8.845279347716643,-5.515095163273459), linewidth(1.6)); draw((1.7528091427582853,-10.106058827512621)--(-5.420933946887003,-5.643157186388007), linewidth(1.6) + linetype("2 2")); draw((1.7528091427582853,-10.106058827512621)--(8.845279347716643,-5.515095163273459), linewidth(1.6) + linetype("2 2")); draw((-1.5880573948301944,1.654493741421877)--(0.15738427123569654,-4.489827870597285), linewidth(1.6)); draw((0.15738427123569654,-4.489827870597285)--(1.7528091427582853,-10.106058827512621), linewidth(1.6) + linetype("2 2")); /* dots and labels */ label("$A$", (-1.4658384425983202,1.9600411220015657), NE * labelscalefactor); dot((0.15738427123569654,-4.489827870597285),dotstyle); label("$P$", (0.27578162670590495,-4.1814612276501775), NE * labelscalefactor); dot((1.7528091427582853,-10.106058827512621),dotstyle); label("$R$", (1.864628005720286,-9.80353303031645), NE * labelscalefactor); dot((-5.420933946887003,-5.643157186388007),dotstyle); label("$B$", (-5.285180699844428,-5.342541273852994), NE * labelscalefactor); dot((8.845279347716643,-5.515095163273459),dotstyle); label("$C$", (8.953327235169063,-5.2203223216211185), NE * labelscalefactor); dot((-2.084674828630707,-3.1303809703085426),dotstyle); label("$D$", (-1.954714251525822,-2.837052753099547), NE * labelscalefactor); dot((-0.5558921299286373,-1.9789449850664533),dotstyle); label("$Q$", (-0.4269773486273789,-1.6759727068967298), NE * labelscalefactor); dot((1.60455184832151,-2.791572898179365),dotstyle); label("$E$", (1.7118543154304415,-2.5009506344618893), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Keeping in mind the angle condition and the fact that $RB$, $RC$ are tangents, note that $$\angle RBC=180^{\circ}-\angle CBA-\angle APB=180^{\circ}-\angle ACB-\angle CPA=\angle BCR$$ Hence, $RB=RC$, so $R$ lies in the radical axis of $\Omega_B$ and $\Omega_C$, that is, $AP$. Also, we have that $\triangle ACR\sim \triangle PRC$ and $\triangle ABR\sim\triangle PRB$. Finally, because of these similarities and the Angle Bisector Theorem, $$\dfrac{AC}{CP}=\dfrac{RC}{RP}=\dfrac{RB}{RP}=\dfrac{AB}{PB}=\dfrac{AQ}{QP}$$Then, the angle bisector of $\angle ACP$ also passes through $Q$, so $AP$, $BD$, and $CE$ concur. We are done $\square$.

Construct points $X,Y$ such that $\triangle XBP, \triangle ABC, \triangle YPC$ are directly similar. Then, the angle condition just says that $\angle XPA=\angle APY$. We also have parallelogram $AXPY$ (This fact appears in Prasolovs geometry book as problem 1.49, but the complex numbers proof is also pretty intuitive). So, $YA=YP$ because $\angle YAP=\angle XPA=\angle APY$ and also, $\triangle AYC$ and $\triangle BPC$ are directly similar (by spiral similarity). Hence $$\frac{BP}{CP}=\frac{AY}{CY}=\frac{PY}{CY}=\frac{BA}{CA}$$which is enough.

Invert around $A$ with an arbitrary radius $r.$ Notice $$\angle C'B'P'=\angle AB'C'-\angle AB'P'=\angle AC'B'-\angle AC'P'=\angle P'C'B'$$so $$\frac{r^2}{AB\cdot AP}\cdot BP=B'P'=C'P'=\frac{r^2}{AC\cdot AP}\cdot CP.$$If $X=\overline{BD}\cap\overline{AP}$ and $Y=\overline{CE}\cap\overline{AP},$ we see $$\frac{AX}{XB}=\frac{AB}{BP}=\frac{AC}{PC}=\frac{AY}{BY}$$and $X=Y.$ $\square$

Let $Q,R$ be such points on $AC,AB$ respectively, that $\angle APQ=\angle ABC,\angle APR=\angle ACB.$ Clearly $ARPQ$ is cyclic, and $\angle ARQ=\angle APQ\implies QR\parallel BC.$ Also $\angle BRP+\angle CQP=\pi.$ By angle condition $\angle BPR=\angle CPQ,$ which yields (sine law) $$\frac{|BP|}{|CP|}=\frac{|BR|}{|CQ|}=\frac{|AB|}{|AC|}.$$Thus we are done.

sol without invertion: Take a point $Q$ such that $\triangle APB\sim \triangle AQC$…

why D and E are given to be incenters? I think if they are points on bisector of angles, this would still work

adorefunctionalequation wrote: why D and E are given to be incenters? I think if they are points on bisector of angles, this would still work can someone please answer this?

@above you're right, although letting them be incenters makes the statement a bit cleaner.

khina wrote: @above you're right, although letting them be incenters makes the statement a bit cleaner. Thanks for answer

Mogmog8 wrote: Invert around $A$ with an arbitrary radius $r.$ Notice $$\angle C'B'P'=\angle AB'C'-\angle AB'P'=\angle AC'B'-\angle AC'P'=\angle P'C'B'$$so $$\frac{r^2}{AB\cdot AP}\cdot BP=B'P'=C'P'=\frac{r^2}{AC\cdot AP}\cdot CP.$$If $X=\overline{BD}\cap\overline{AP}$ and $Y=\overline{CE}\cap\overline{AP},$ we see $$\frac{AX}{XB}=\frac{AB}{BP}=\frac{AC}{PC}=\frac{AY}{BY}$$and $X=Y.$ $\square$ In this solution, it'd probably make more sense to replace $\tfrac{AX}{XB}$ and $\tfrac{AY}{BY}$ in the display equation with $\tfrac{AX}{XP}$ and $\tfrac{AY}{YP}$, respectively; then the conclusion follows by the Angle Bisector Theorem.

I commented on the inversive angle structure in 2008 G3, but a bit more on it here: the fundamental structure of circles is of the form $\angle XAY = \angle XBY$ for some fixed points $A$ and $B$ (such a structure effectively defines a circle), while a line is defined by $\angle XYA = \angle XYB$. Thus as inversion turns circles into lines, it turns circular angle structures into linear angle structures. So in a sense, the motivation behind inverting angles is the exact same as the motivation behind inverting circles! Invert about $A$. The angle condition becomes $$\angle P'B'A - \angle C'B'A = \angle P'C'A - \angle B'C'A \iff \angle P'B'C' = \angle P'C'B',$$hence $P'C'B'$ is an isosceles triangle. The desired condition is equivalent to $$\frac{BP}{AB} = \frac{PC}{AC} \iff P'C' = P'B'$$using inversion distance formula, but this is evident.

Here is my solution. Construct outside $\Delta{ABC}$ point $D$ and $E$ such that ${\Delta{ADP}}\sim{\Delta{ABC}\sim{\Delta{APE}}}$(as image) Easy to proof that ${\Delta{ADB}}\sim{\Delta{APC}}$ and ${\Delta{AEC}}\sim{\Delta{APB}}$ Therfore, we have: $\widehat{BPD}=\widehat{APB}-\widehat{APD}=\widehat{APB}-\widehat{ACB}$$=\widehat{APC}-\widehat{ABC}=\widehat{APC}-\widehat{APE}$$=\widehat{CPE}$ Next, we have ${APBD}\sim{AECP}$ $\Rightarrow$ $\widehat{BDP}=\widehat{CPE}$, $\frac{DB}{PC}=\frac{AB}{AC}$ So: $\widehat{BPD}=\widehat{CPE}=\widehat{BDP}$ $\Rightarrow$ $\Delta{BDP}$ is issosilate $\Rightarrow$ $PB=DB$ $\Rightarrow$ $\frac{PB}{PC}=\frac{DB}{PC}=\frac{AB}{AC}$ $\Rightarrow$ $\frac{PB}{AB}=\frac{PC}{AC}$ Apply angle bisector theorem, we get $AP,BD,CE$ concur.

The desired concurrence is equivalent to $\frac{BA}{BP}=\frac{CA}{CP}$. Now invert about $A$ with radius $1$ and denote images with $\bullet'$; the angle condition translates to $\angle C'B'P=\angle AB'P'-\angle AB'C'=\angle AC'P'-\angle AC'B'=\angle B'C'P$, so $P'B'=P'C'$. The above length condition is equivalent to $$\frac{\frac{1}{AB'}}{B'P'\cdot \frac{1}{AB'\cdot AP'}}=\frac{\frac{1}{AC'}}{C'P'\cdot \frac{1}{AC'\cdot AP'}},$$which is now clearly true. $\blacksquare$

Suppose $BD$ and $CE$ meet $AP$ at $X, Y$ respectively. Then $PX/XA = BP/BA$ and $PY/YA = CP/CA$. It suffices to show $BP/BA = CP/CA$. Invert about $A$ with radius $1$. We make the following claim: Claim: $\triangle PB^*C^*$ is isosceles. Proof: In the inverted diagram \[\angle P^*C^*B^* = \angle P^*C^*A - \angle AC^*B^* = \angle AP^*C^* - \angle AB^*C^*,\]and \[\angle P^*B^*C^* = \angle P^*B^*A - \angle AB^*C^* = \angle AP^*B^* - \angle AC^*B^*. \square\] Therefore $\triangle PB^*C^*$ is isosceles, whence $B^*P^* = C^*P^*$, and in particular, \[\frac{BP}{AB \cdot AP} = \frac{CP}{AC \cdot AP}.\]We thus obtain the desired result. $\blacksquare$

Perform a classic overlay at $A$ with radius $\sqrt{AB \cdot AC}$. In our inverted diagram, our angle condition can be interpreted as \[\angle ABP - \angle ABC = \angle ACP - \angle ACB \iff \angle CBP = \angle BCP \iff BP = CP.\] Angle Bisector Theorem says our desired statement is equivalent to $\frac{AB}{BP} = \frac{AC}{CP}$, or in our inverted picture, \[\frac{AC}{BP \cdot \frac{AB \cdot AC}{AB \cdot AP}} = \frac{AB}{CP \cdot \frac{AB \cdot AC}{AC \cdot AP}} \iff AB \cdot AC = AB \cdot AC. \quad \blacksquare\]

Invert about $A$. Let the image of the object $X$ under inversion be $X'$. Then some angle chasing gives us $B'P' = P'C'$, after which the inversion distance formula gives us $AB/BP = AC/CP$, which is clearly sufficient by the angle bisector theorem. $\square$

For all points $X$ define $XP \cap (ABC) = X'$. Angle chase gives $A'$ is the midpoint of arc $B'C'$, so $AA'$, $B'E'$, $C'D'$ concur at the incenter of $AB'C'$. Using Pascal's on hexagon $BB'E'CC'D'$, we find $BD' \cap CE'$ must lie on the line connecting $BB' \cap CC' = P$ and $B'E' \cap C'D' \in AP$, which is $AP$ as desired. $\blacksquare$

Let $X = (APB) \cap BC$ and $Y = (APC) \cap BC$. Also let $U$ and $V$ be the internal and external angle bisector of $\angle A$, respectively. By the angle condition, we have that $AX$ and $AY$ are isogonal lines. Also since $(VU;XY)=-1$ and $(VU;BC)=-1$, by DDIT, we have $(U,V),(X,B),(Y,C)$ are part of the same involution, which implies $A \in (VUP)$ and the result follows by Apollonius circle.