Here's a sketch. Let $O_B,O_C$ be the centers of $(PAB),(PAC)$; then $\triangle{AO_BB},\triangle{AO_CC}$ are directly similar, and thus $\triangle{AO_BO_C},\triangle{ABC}$ are as well. Use POP at the one of $X,Y$ opposite side $BC$ to $A$ and Pythagorean to compute $PA/XY$ (which is independent of $P$). (Alternatively, invert about $A$ to compute all of this.) Thus there are either 0 or 2 solutions (reflections about midpoint $M$ of $BC$), and it's not hard to see that the foot of the $A$-angle bisector is one of these, completing the problem.

Sketch of my solution: Bash out $PA$, $XY$, using stuff about common tangents, radical axis, heron's formula, and mid-line of a trapezoid. It turns out that \[ (\frac{PA}{XY})^2 = 1- \frac{l^2}{(R+r)^2} \] where $l$ is the distance between the two centers and $R$, $r$ are the radii of the two circles. Afterwards use spiral similarity to get \[ \frac{l}{R+r} = \frac{BC}{AB+BC} \] then you are pretty much done. Also, @math154 couldn't it be also one solution for an isosceles triangle?

Invader_2011 wrote: Also, @math154 couldn't it be also one solution for an isosceles triangle? Yeah, of course, I was just being lazy.

Attached is a diagram for this problem. Notice the spiral similarity.

math154 wrote: Use POP at the one of $X,Y$ opposite side $BC$ to $A$ and Pythagorean to compute $PA/XY$. You've lost me here.

Okay, If I said how far the two possible p's were from B to C, without realizing that one was the angle bisector, do you think that is still full points?

iglenini wrote: Okay, If I said how far the two possible p's were from B to C, without realizing that one was the angle bisector, do you think that is still full points? If your description leads to the complete solution you should get full points. (such as $BP=\frac{ab}{b+c}$ or $\frac{ac}{b+c}$ etc)

Hello, a slightly no-pythag way to do it is to notice that $1 - \displaystyle \left (\dfrac{PA}{XY}\right)^2 = \dfrac{4XA\cdot XP}{XY^2}$, which letting $MN$ be the length of the tangent, gives $\left ( \frac{MN}{XY}\right )^2$, which is $\frac{O_1O_2 \sin \theta}{(r_1+r_2) \sin \theta} = \frac{O_1 O_2}{r_1+r_2} = \frac{BC}{AB+AC}$.

Let $X$-nearer $A$, $Y$-nearer $P$ and $l$ - leight of the common external tangent. We see not difficult : \[ XY^2=AP^2+l^2 .\] Hence, \[ \frac{PB\cdot PC}{AB\cdot AC}=\frac{l^2}{l^2+AP^2} .\] So, from $l^2=d^2-(R_1-R_2)^2=4R_1 R_2 \sin^2 \frac{A}{2}$ we have more facts: for example $ \frac{AP^2}{l^2}=\frac{\sin B \sin C}{\sin^2 \frac{A}{2}} $ $ \Longrightarrow $ \[ \frac{PB\cdot PC}{AB\cdot AC}=\frac{\sin^2 \frac{A}{2}}{\cos^2 \frac{B-C}{2}} .\] Hence, to be \[ PB\cdot PC \cdot \cos^2 \frac{B-C}{2}= (s-b)(s-c) \] here, $s$-semiperimetr of the triangle $ABC.$ Let $s-b=x$, $s-c=y$ and $BP=u$, $CP=v$. We have two facts: \[ uv\cdot \cos^2 \frac{B-C}{2}=xy (1)\] and \[ u+v=x+y=BC \] System has positive roots. Suppose that $u=x+\alpha$ and $v= y - \alpha $, then from (1), $ \alpha^2 +\alpha (x-y) +xy \tan^2 \frac{B-C}{2}=0 .$ Also, solving the kvadratic equation $ \Rightarrow $ locus $P$ $\rightarrow $ foot bisector of angle $BAC$ or it's reflection point to regard midpoint $BC.$

So I derived $PB*PC=bc \frac{1-\cos A}{1 + \cos (B-C)}$, but didn't bother to check the bisector... how many points should this be?

I would give a 7, but probably at least a 6? In fact, I guessed the bisector through trig first---it becomes clear if you let $\alpha,\beta,\gamma$ be half-angles and rewrite the trig expression as $\sin^2\alpha/\cos^2(\beta-\gamma) = \cos^2(\beta+\gamma)/\cos^2(\beta-\gamma)$, and rewrite $PB\cdot PC/AB\cdot AC$ as a sine product.

to 2# how to see that the foot of the -angle bisector is one of these?

If $D$ is the foot, you just want to show $DB\cdot DC/AB\cdot AC$ equals whatever expression you get for $1 - (PA/XY)^2$, and this is just a (boring) standard computation. (You can get everything in terms of sides and angles of $\triangle{ABC}$, but some ways are nicer than others.)

An 'easy, short, and straightforward' calculation: -Parameters: Let $r_1$, $r_2$ be the circumradii of circle $\omega_1$, $\omega_2$, respectively. Let $d = O_1O_2$. -Variable: Let $x=PB/BC$. Then $PC/BC=1-x$. -Calculation of $1-(AP/XY)^2 = d^2/(r_1+r_2)^2$:

-Proof of $AB\cdot AC/BC^2 = r_1r_2/d^2$:

-An equation of $x$ is then $x(1-x)=\frac{r_1r_2}{(r_1+r_2)^2}$.

-Now it is easy to see that

Here's a solution that does the calculations with $a,b,c$ alone. As before, let $O_B$ and $O_C$ be the circumcenters of $ABP$ and $ACP$, and let $M$ be the midpoint of $AP$ (on $O_BO_C$). Assume $X$ is closer to $A$ than $B$. Since $\triangle AO_BB \sim \triangle AO_CC$, by spiral similarity we have that $AO_BO_C \sim \triangle ABC$. Therefore $\tfrac{MA}{MX} = \tfrac{PA}{XY}$ depends only on (the ratios of the sides of) $ABC$, since these are the same ratios in $\triangle AO_BO_C$. We'll perform the main calculation with the convenient scaling $O_BO_C = a$, $AO_C = b$, and $AO_B = c$. Let $B_1$ and $C_1$ be the tangency points of $X$, and let $h = AM$ be the height of $AO_BO_C$. Note that by Power of a Point, we have $XB_1^2 = XC_1^2 = XM^2 - h^2$. Also, by Pythag we easily obtain $B_1C_1 = a^2 - (b-c)^2$. Then \begin{align*} 1 - \left( \frac{MA}{MX} \right)^2 &= 1 - \frac{h^2}{h^2 + XB_1^2} \\ &= \frac{XB_1^2}{h^2+XB_1^2} \\ &= \frac{B_1C_1^2}{(2h)^2 + B_1C_1^2} \\ &= \frac{a^2 - (b-c)^2}{\left( \frac{4[AO_BO_C]}{a} \right)^2 + a^2-(b-c)^2} \\ &= \frac{a^2(a^2-(b-c)^2)}{(a+b+c)(-a+b+c)(a-b+c)(a+b-c) + a^2(a^2-(b-c)^2)} \\ &= \frac{a^2}{(a+b+c)(-a+b+c) + a^2} \\ &= \left( \frac{a}{b+c} \right)^2. \end{align*}Now, for $P$ the foot of the angle bisector or its reflection, we have \[ \frac{PB \cdot PC}{AB \cdot AC} = \frac{\left( \frac{b}{b+c} a \right)\left( \frac{c}{b+c} a \right)}{bc} = \left( \frac{a}{b+c} \right)^2. \]Since there are clearly at most two solutions as $\tfrac{PA}{XY}$ is fixed, these are the only two solutions.

[asy][asy] /* USAMO 2013 Problem 6, free script by liberator, 23 December 2014 */ /* Manually converted from GeoGebra file, see original tube.geogebra.org/student/m424825 */ unitsize(1.5cm); pointpen=black; /* Initialize objects */ pair A = (2.1,3.4); pair B = origin; pair C = (5.5,0); pair P = (3.17,0); pair X = (1.95,3.88); pair Y = (3.32, -0.48); pair Ob = (1.58,1.37); pair Oc = (4.33,2.23); pair D = (0.67,3.25); pair E = (3.23,4.5); pair K = (1.91,-0.7); pair L = (4.73,-0.26); pair Q = foot(D,Ob,Oc); pair R = foot(E,Ob,Oc); pair S = foot(Ob,Oc,E); /* Draw objects */ draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(A--Ob--Oc--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(D--Q--Ob--cycle, rgb(0.4,0.6,0.8)); draw(E--R--Oc--cycle, rgb(0.4,0.6,0.8)); draw(Ob--S--Oc--cycle, rgb(0.4,0.6,0.8)); draw(D--E, rgb(0.4,0.6,0.8)); draw(X--Y, rgb(0.4,0.6,0.8)); draw(K--L, rgb(0.4,0.6,0.8)); draw(Q--K, rgb(0.4,0.6,0.8)+dashed); draw(R--L, rgb(0.4,0.6,0.8)+dashed); draw(circumcircle(A,B,P), rgb(0.9,0,0)); draw(circumcircle(A,C,P), rgb(0,0.6,0)); /* Place dots on and label points */ Drawing("A", A, dir(40)); Drawing("B", B, dir(220)); Drawing("C", C, dir(-40)); Drawing("D", D, dir(90)); Drawing("E", E, dir(90)); Drawing("O_B", Ob, dir(-90)); Drawing("O_C", Oc, dir(0)); Drawing("X", X, dir(90)); Drawing("Y", Y, dir(-90)); Drawing("P", P, 2*dir(75)); Drawing("Q", Q, dir(200)); Drawing("R", R, dir(240)); Drawing("S", S, dir(70)); [/asy][/asy] Suppose $X$ is closer to $A$, and $Y$ is closer to $P$. Let $O_B$ be the center of $(ABP)$ and $O_C$ be the center $(ACP)$. Let the tangent through $X$ touch $(ABP)$ at $D$ and $(ACP)$ at $E$. Note that by symmetry, we have $XA=PY$. $XD^2 = XA \cdot XP = XE^2 \implies XD = XE$. Now \begin{align*}1 - (\frac{PA}{XY} )^2 = \frac{XY^2 - PA^2}{XY^2} = \frac{(XY+PA)(XY-PA)}{XY^2} = \frac{4XP \cdot XA}{XY^2} = \frac{4XD^2}{XY^2} = (\frac{DE}{XY} )^2. \end{align*} Let $Q,R$ be the projections of $D,E$ to $O_BO_C$. Let $S$ be the projection of $Q$ onto $O_CE$. Denote the radii of $(ABP), (ACP)$ by $r_B, r_C$. We have $\triangle O_BQD \sim \triangle O_CRE \sim \triangle O_CSO_B$. Hence \begin{align*} \frac{DE}{O_BO_C} = \frac{O_BS}{O_BO_C} = \sin{\angle O_BO_CE} = \frac{DQ+ER}{O_BD + O_CE} = \frac{XY}{r_B + r_C} \implies \frac{DE}{XY} = \frac{O_BO_C}{r_B+r_C}. \end{align*} $A$ is the center of the spiral similarity $\overline{O_BO_C} \to \overline{BC} \implies \triangle AO_BO_C \sim \triangle ABC$. Hence \[\boxed{( \frac{DE}{XY} ) ^2 = ( \frac{O_BO_C}{r_B + r_C} ) ^2 = ( \frac{a}{b+c} ) ^2}\, ,\] where $a,b,c$ denote the lengths of $BC, CA, AB$. If $P$ is the foot of the angle bisector of $\angle A$ or its isotomically conjugate line, we have \[PB \cdot PC = \frac{ac}{b+c} \cdot \frac{ab}{b+c} = \frac{a^2bc}{(b+c)^2} \implies \boxed{\frac{PB \cdot PC}{AB \cdot AC} = ( \frac{a}{b+c} )^2} \, ,\] which satisfies the conditions of the problem. Noting that $PC = a - PB$, we may rearrange the last equation to get \[PB^2 - aPB + \frac{a^2bc}{(b+c)^2} = 0, \] which is a quadratic equation in $PB$, so has at most two solutions. Hence we have \[\boxed{P = \{ \begin{gathered} (0,b,c) \\ (0,c,b) \end{gathered}}\] in barycentric coordinates.

Let $O_c$ be the center of $\odot(ABP)$ and $O_b$ be the center of $\odot(APC)$. It is a well-known fact that $\triangle AO_cO_b$ is spirally similar to $\triangle ABC$. Let the tangent from $X$ hit $\odot(ABP)$ at $E$ and $\odot(APC)$ at $H$. Then we have that $XA\cdot XP=XE^2$, so that $$1-\bigg(\frac{PA}{XY}\bigg)^2=\frac{(XY+PA)(XY-PA)}{XY^2}=\frac{(2XY-2XA)(2XA)}{XY^2}=\frac{4XP\cdot XA}{XY^2}=\bigg(\frac{HE}{XY}\bigg)^2$$ Our goal is now to compute $\frac{HE}{XY}$; let $G,I,K$ be the projections of $X,E,H$ onto line $O_CO_B$, respectively. Project $O_B$ onto line $EO_c$ to obtain another point $L$. Then it follows that $\triangle O_BO_CL\sim \triangle EO_CI\sim HKO_B$. Additionally, $HE=LO_B$, hence $$HE=LO_B=\frac{O_BO_C}{2}\bigg(\frac{EI}{EO_C}+\frac{HK}{HO_B}\bigg)=2\frac{O_BO_C}{2}\bigg(\frac{EI+HK}{EO_C+HO_B}\bigg)=O_BO_C\frac{XY}{EO_C+HO_B}$$from which we derive that $$\frac{HE}{XY}=\frac{O_BO_C}{EO_C+HO_B}=\frac{O_BO_C}{AO_C+AO_B}=\frac{BC}{AB+AC}$$Our original condition now simplifies to $$PB\cdot PC=\frac{AB\cdot AC\cdot BC^2}{(AB+AC)^2}$$Note that if we let $PB$ be a variable, then the above equation reduces to a quadratic, hence it suffices to find one solution, and the other solution is simply the reflection of the first about the midpoint of $\overline{BC}$. If we let $D$ be the foot of the angle bisector from $A$, then by the angle bisector theorem, $$DB=\frac{BC\cdot AB}{AB+AC}, DC=\frac{BC\cdot AC}{AB+AC}$$hence $D$ is our desired point$.\:\blacksquare\:$

Not too bad for a 3/6, I guess. Let $\odot(PAB) = \omega_1$, with center $O_1$ and $\odot(PAC) = \omega_2$, with center $O_2$. Let the radii of $\omega_1$ and $\omega_2$ be $r_1$ and $r_2$ respectively. WLOG, $A$ lies between $X$ and $P$, and $P$ lies between $Y$ and $A$. Let $t_1$ be the line segment formed by the tangent through $X$ and the points it touches $\omega_1$ and $\omega_2$ and $t_2$ be the line segment defined similarly, but for $Y$ instead. By a well-known lemma, $X$ and $Y$ are the midpoints of $t_1$ and $t_2$ respectively. Observe that \[ 1 - \left(\frac{PA}{XY}\right)^2 = \frac{(XY-PA)(XY+PA)}{XY^2} = \frac{4XA \cdot XP}{XY^2} = \frac{t^2}{XY^2} \] Using $\sin$, compute $\frac{t}{XY} = \frac{O_1O_2}{r_1+r_2}$. Let $AO_1$ hit $\omega_1$ again at $O_1'$ and similarly define $O_2'$. $O_1O_2 \perp AP$ by radical axis, so homothety tells us that $O_1'O_2' \perp AP$. Thus, $O_1', P, O_2'$ collinear, so $\triangle AO_1O_2 \sim \triangle AO_1'O_2' \sim \triangle ABC$. Then from our spiral similarity, we have $\frac{O_1O_2}{r_1+r_2} = \frac{O_1O_2}{O_1A+O_2A}=\frac{a}{b+c}$. Now $PB \cdot PC = \frac{a^2bc}{(b+c)^2}$. Using the quadratic formula, $PB = \frac{ab}{b+c}$ or $PB = \frac{ac}{b+c}$. Call the foot of the $A$-angle bisector on side $BC$ as $D$, and its reflection about the midpoint $M$ of side $BC$ as $D'$. By angle-bisector theorem, we now know that the solutions are $D$ and $D'$, and we are done.

v_Enhance wrote: Here's a solution that does the calculations with $a,b,c$ alone. We'll perform the main calculation with the convenient scaling $O_BO_C = a$, $AO_C = b$, and $AO_B = c$. Let $B_1$ and $C_1$ be the tangency points of $X$, and let $h = AM$ be the height of $AO_BO_C$. Note that by Power of a Point, we have $X_1B^2 = X_1C^2 = XM^2 - h^2$. Also, by Pythag we easily obtain $B_1C_1 = a^2 - (b-c)^2$. What's $X_1$?

Eek, sorry, $XB_1$ and $XC_1$ were meant here. I'll amend the post, thanks for pointing that out.

v_Enhance wrote: Now, for $P$ the foot of the angle bisector or its reflection, we have \[ \frac{PB \cdot PC}{AB \cdot AC} = \frac{\left( \frac{b}{b+c} a \right)\left( \frac{c}{b+c} a \right)}{bc} = \left( \frac{a}{b+c} \right)^2. \]Since there are clearly at most two solutions as $\tfrac{PA}{XY}$ is fixed, these are the only two solutions. For which tringle $P$ be the foot of angle bisecto?

P is the foot of angle bisector in $\triangle$ ABC

tenniskidperson3 wrote: Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\] Let $O_1, O_2$ be the circumcentre of $\triangle APB,$ $\triangle APC$ and let $\omega_1$, $\omega_2$ be their circumcircles. Let $\ell_1, \ell_2$ be the common tangents. Let $\ell_1 \cap \omega_1 = Q$, $\ell_1 \cap \omega_2 = R$. Claim 1 : $\triangle AO_1O_2 \sim \triangle ABC$. Proof. Note that $\angle{AO_1B} = 2.\angle{APB} = \angle{AO_2B}$ and $\tfrac{AO_1}{BO_1} = \tfrac{AO_2}{BO_2} = 1$ yields the desired result. $\square$ Claim 2 : $\tfrac{{QR}^2}{{XY}^2} = \tfrac{PB \cdot PC}{AB \cdot AC}$ Proof. Note that $ \tfrac{{QR}^2}{{XY}^2} = \tfrac{(XY - PA) \cdot (XY + PA)}{{XY}^2} = \tfrac{PB \cdot PC}{AB \cdot AC}$. $\square$ Claim 3 : $\tfrac{PB \cdot PC}{AB \cdot AC} = \tfrac{a^2}{(b+c)^2}$ Proof. $\tfrac{PB \cdot PC}{AB \cdot AC} = {(\tfrac{O_1O_2}{O_1Q + O_1R})}^2 = \tfrac{a^2}{(b+c)^2}$ from Claim 1 and Claim 2. $\square$. From here the problem is reduced to solving a quadratic which yields $PB = \tfrac{ab}{b+c}$ which gives $P$ to be the foot the $A$- angle bisector or the reflection over its midpoint. $\blacksquare$

[asy][asy] unitsize(2inches); pair A=dir(125); pair B=dir(215-40); pair C=dir(-35+40); pair I=incenter(A,B,C); pair P=extension(A,I,B,C); pair O1=circumcenter(A,P,B); pair O2=circumcenter(A,P,C); pair I1=incenter(A,P,B); pair I2=incenter(A,P,C); pair UUU=2*O1-(2*foot(O1,A,I1)-A); pair VVV=2*O2-(2*foot(O2,A,I2)-A); pair T=extension(UUU,VVV,O1,O2); pair U=intersectionpoints(circle((O1+T)/2,abs(O1-T)/2),circumcircle(A,B,P))[0]; pair UU=intersectionpoints(circle((O1+T)/2,abs(O1-T)/2),circumcircle(A,B,P))[1]; pair V=intersectionpoints(circle((O2+T)/2,abs(O2-T)/2),circumcircle(A,C,P))[0]; pair VV=intersectionpoints(circle((O2+T)/2,abs(O2-T)/2),circumcircle(A,C,P))[1]; pair X=extension(A,P,T,V); pair Y=extension(A,P,T,VV); pair S=extension(O1,O2,X,Y); draw(A--B--C--cycle); draw(circumcircle(A,P,B)); draw(circumcircle(A,P,C)); draw(T--V); draw(T--VV); draw(X--Y); draw(A--O1--O2--cycle,dotted); draw(T--O1,dotted); dot("$A$",A,dir(45)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$P$",P,2*dir(30)); dot("$T$",T,dir(220)); dot("$X$",X,dir(120)); dot("$Y$",Y,dir(260)); dot("$U$",U,dir(100)); dot("$U'$",UU,dir(260)); dot("$V$",V,dir(120)); dot("$V'$",VV,dir(250)); dot("$O_1$",O1,dir(270)); dot("$O_2$",O2,dir(0)); dot("$S$",S,dir(55)); [/asy][/asy] Let $O_1$ and $O_2$ be the circumcenters of $(ABP)$ and $(ACP)$ with circumradii $R_1$ and $R_2$, and let $S$ be the midpoint of $XY$. By symmetry, we see that $S$ is the midpoint of $AP$ as well and $S$ passes through $O_1O_2$. Furthermore, let $\theta=\frac{1}{2}\angle UO_1U'=\frac{1}{2}\angle VO_2V'$. Note that \begin{align*} 1-\left(\frac{PA}{XY}\right)^2 &= \frac{(XY-PA)(XY+PA)}{XY^2} \\ &= \frac{(2\cdot XA)(2\cdot XP)}{XY^2} \\ &= \frac{4\cdot XU^2}{XY^2} \\ &= \frac{UV^2}{XY^2} \\ &= \frac{O_1O_2^2\sin^2\theta}{[(R_1+R_2)\sin\theta]^2} \\ &= \left(\frac{O_1O_2}{R_1+R_2}\right)^2. \end{align*}Note that by the well known spiral similarity lemma, there is a spiral similarity at $A$ sending $B\mapsto C$ and sending circles $(ABP)\mapsto (APC)$. In particular, we see that there is a spiral similarity at $A$ sending $BO_1\mapsto CO_2$, so there is also a spiral similarity at $A$ sending $O_1O_2$ to $BC$. Thus, \[\frac{O_1O_2}{R_1+R_2}=\frac{BC}{AB+AC}.\]Thus, \[\frac{PB\cdot PC}{AB\cdot AC}=-\left(\frac{PA}{XY}\right)^2=\left(\frac{BC}{AB+AC}\right)^2,\]so \[PB\cdot PC=\frac{bca^2}{(b+c)^2}.\]Also, our steps are reversible, the condition for $P$ given in the problem is equivalent to the above condition. This is a quadratic equation in $P$ that is not satisfied everywhere (say at $P=B$ and $P=C$), so it has at most two solutions. We can name two solutions, namely the angle bisector foot and its isotomic point, so those are the only two solutions.

edit: i dont agree with my previous remarks

mathlogician wrote:

@above Is the isotomic point equivalent to reflection of foot of angle bisector across midpoint? Yes

Let $(PAB)$ be denoted $\omega_1$, $(PAC)$ be $\omega_2$, with centers $O_1,O_2$ respectively. Note that \[O_1P=\frac{AP\cdot AB\cdot PB}{4[PAB]} = \frac{AP\cdot AB\cdot PB}{4\cdot \frac{PB}{BC}[ABC]} = \frac{AP\cdot AB\cdot BC}{4[ABC]}.\]Hence, the ratio \[O_1P:AP:O_2P = \frac{AB\cdot BC}{4[ABC]}:1:\frac{AC\cdot BC}{4[ABC]}\]is fixed. This implies that lengths $O_1P,O_1A,AP,O_2P,O_2A$ are in a fixed ratio to each other as $P$ varies, so $\frac{PA}{XY}$ is fixed. Moreover, because \[\measuredangle O_1AO_2=\measuredangle O_1AP+\measuredangle PAO_2=90^\circ-\measuredangle PBA+90^\circ-\measuredangle ACP=180^\circ-\measuredangle CBA-\measuredangle ACB=\measuredangle BAC,\]and we have the prior obtained ratios, $\triangle O_1AO_2\sim\triangle BAC$. Now, we compute $\tfrac{PA}{XY}$. Let $\ell_1$ denote one common external tangent of $\omega_1$ and $\omega_2$ and $\ell_2$ denote the other. Let $\ell_1\cap\omega_1=Q,\ell_1\cap\omega_2=R,\ell_2\cap\omega_1=S,\ell_2\cap\omega_2=T$. Let $QR\cap ST=K$. WLOG, $AB>AC$. Since \[\frac{AB}{AC}=\frac{O_1Q}{O_2R} = \frac{O_1K}{O_2K} = \frac{O_1O_2+O_2K}{O_2K}=\frac{O_1O_2}{O_2K}+1,\]we have $O_2K=O_1O_2\cdot \frac{AC}{AB-AC}$. Hence, we have \[RT=2\cdot O_2R\cdot\frac{RK}{O_2K} = 2\cdot \frac{O_2R}{O_1O_2}\cdot \frac{AB-AC}{AC}\cdot \sqrt{O_2K^2-O_2R^2} = \]\[2\cdot\frac{AC}{BC}\cdot\frac{AB-AC}{AC}\cdot\sqrt{O_1^2O_2^2\cdot\frac{AC^2}{(AB-AC)^2}-O_1^2O_2^2\cdot\frac{AC^2}{BC^2}}=\]\[2\cdot \frac{O_1O_2\cdot AC}{BC^2}\cdot\sqrt{BC^2-(AB-AC)^2}=2\cdot \frac{O_1O_2\cdot AC}{BC^2}\cdot\sqrt{(BC+AC-AB)(BC+AB-AC)}.\]Thus, we have \[XY=(RT+QS)/2 = \frac{O_1O_2\cdot (AC+AB)}{BC^2}\cdot\sqrt{(BC+AC-AB)(BC+AB-AC)}.\]Note also that \[AP=\frac{4[O_1AO_2]}{O_1O_2} = \frac{O_1O_2}{BC^2}\sqrt{(AB+AC+BC)(AB+AC-BC)(BC+AC-AB)(BC+AB-AC)}.\]Hence, we have \[\left(\frac{AP}{XY}\right)^2 = \frac{(AB+AC+BC)(AB+AC-BC)}{(AC+AB)^2}=\frac{(AB+AC)^2-BC^2}{(AC+AB)^2},\]so \[1-\left(\frac{AP}{XY}\right)^2=\frac{BC^2}{(AC+AB)^2}.\]Then, $P$ is any point on segment $BC$ such that \[BC\cdot PB-PB^2 = PB\cdot PC = BC^2 \cdot \frac{AB\cdot AC}{(AC+AB)^2}.\]This is a quadratic in $PB$, so we rearrange it to yield \[(2PB-BC)^2 = 4PB^2-4PB\cdot BC+BC^2 = BC^2 \cdot \frac{(AC-AB)^2}{(AC+AB)^2}.\]Finally, we get that \[PB = BC/2 \pm \frac{1}{2}BC\cdot \frac{AC-AB}{AC+AB}.\]

Can someone please tell the motivation behind all these solutions. Like, how could one suddenly think of constructing the centers of the two circles (which eventually works quite well for us), since at first look, it seems quite tough to compute the length $XY$ (and even if think of constructing the center of the circles, its not yet over since we have to another clever thing to apply power of a point wisely to get the length of $XY$ somehow into the expressions).

This ability is not built overnight. We practice many problems and slowly we develop the "feel" for a particular type of problem.

But is there some particular thing which motivated us to do these constructions (since I think there is always some motivation of even a ingenious solution too). Like, a motivation could even be seen a similar problem before or finding some obvious reasons for not going in some other direction and in some particular direction only.

Let $O_1$ and $O_2$ be the circumcenters of $PAB$ and $PAC$. Let $E$ and $F$ be points on the circle such that $XE$ and $XF$ are tangent to the circles, and $E'$ and $F'$ are the reflections of $E$ and $F$ across $O_1O_2$ respectiely. Let $D$ be the intersection of the two tangents. Claim: $AO_1O_2 \sim ABC$. Proof: First notice that $AP \perp O_1O_2$ since $AP$ is the Radical Axis of the two circles. We have $\angle{ABP} = \frac{\angle{AO_1P}}{2} = \angle{AO_1O_2}$, and similarly $\angle{ACP} = \frac{\angle{AO_2P}}{2} = \angle{AO_2O_1}$, which follows by AA similarity. Claim: $\left(\frac{AP}{XY}\right)^2$ is constant. Proof: By Power of a Point, $$XE^2 = XA\cdot XP = XA(XA+AP) \implies EF^2 = 4XA(XA+AP).$$Next, $$XY^2 = (AP+XA+YP)^2 = (AP+2XA)^2 = AP^2+4AP\cdot XA + 4XA^2 = AP^2 + 4XA(XA+AP) = AP^2+EF^2. $$ Let $\alpha = \angle{EO_1D}$. We have $EF = O_1O_2\sin\alpha$, and $$XY = \frac{EE'+FF'}{2} = \frac{EE'}{2}+\frac{FF'}{2} = EO_1\sin\alpha + FO_2\sin\alpha = (EO_1+FO_2)\sin\alpha$$$$\implies \left(\frac{EF}{XY}\right)^2 = \left(\frac{O_1O_2}{EO_1+EO_2}\right)^2 = \left(\frac{a}{b+c}\right)^2$$$$\implies \left(\frac{AP}{XY}\right)^2 = 1-\left(\frac{EF}{XY}\right)^2 = 1-\left(\frac{a}{b+c}\right)^2.$$ Therefore, $\left(\frac{AP}{XY}\right)^2$ is constant as desired. Because there are at most two points, they are isotomic to each other. Checking shows that the foot of the angle bisector is a solution because that means $$\frac{(BP)(CP)}{(AB)(AC)} = \frac{(\frac{ac}{b+c})(\frac{ab}{b+c})}{bc} = \left(\frac{a}{b+c}\right)^2.$$ Finally, this means the two solutions are the foot of the angle bisector and its isotomic point. Remarks: mathlogician wrote:

@above Is the isotomic point equivalent to reflection of foot of angle bisector across midpoint? I used to think that geometry that involved a lot of algebra were bad also, but after attempting many of them, I will disagree and these problems are not as simple as just "bashing", otherwise everyone would get it. It is actually quite similar to synthetic geo problems because where need to try extending different lines or angle chasing in different ways until you solve the problem, except here you need to find a nice way to express lengths. In fact, this problem took me longer than it should've because I kept trying to bash $EF$ using the Pythagorean Theorem, which is a way to do it, but I didn't have enough insight. Even though the condition is contrived, you can break it apart and it is a nice problem imo.

Here's a synthetic way to get $\frac{AP}{XY}$ constant: The center of homothety of the 2 circles is precisely the circumcenter $O$ of $\triangle APZ$ where $Z$ is the intersection of the $A$-external bisector and $BC$. Now we have $\angle AOP=2\angle AZP$, which is constant.

Lemma 1: Suppose two circles centered at $O_B$ and $O_C$ intersect at two points, $A$ and $P$. Then, if $B$ is on the first circle and $C$ is on the second circle such that $B,P,C$ are collinear, then $$\triangle AO_BO_C\sim \triangle ABC.$$ Reflect $A$ over $O_B$ to get $O_B'$, and define $O_C'$ similarly. Then, since $\angle O_B'PA=\angle O_C'PA=90,$ $O_B',P,O_C'$ are collinear. Thus, $\triangle AO_B'O_C'\sim\triangle ABC$ by spiral similarity lemma. However, clearly $\triangle AO_BO_C\sim \triangle AO_B'O_C'$ which shows the lemma. Thus, if we let $O_B$ denote the center of $(ABP)$ and $O_C$ the center of $(ACP)$, then $\triangle AO_BO_C\sim \triangle ABC.$ Thus, we can "discard" points $B$ and $C$, and construct the diagram starting from $\triangle AO_BO_C$ in the following manner: Start with $\triangle AO_BO_C.$ Let $\omega_B$ be the circle centered at $O_B$ passing through $A$, and let $\omega_C$ be the circle centered at $O_C$ passing through $A$. The second intersection of $\omega_B$ and $\omega_C$ is $P$, and let $AP$ meet the external tangents at $X$ and $Y$. Clearly, $AP/XY$ only depends on the angles of the original triangle $\triangle AO_BO_C,$ which is of course determined by $\triangle ABC$ already by Lemma 1. Thus, $AP/XY$ does not depend on $P$. Let $M$ be the midpoint of $BC$. Note that $$PB\cdot PC=BM^2-PM^2,$$so there are at most two possible locations of $P$ that satisfy the condition, and they are reflections of each other about $M$. It is easy to see that $P$ being the foot from $A$ to $BC$ works, so the solution is either the foot or the reflection of the foot across $M$.

Define $O_B$, $O_C$ and $r_B$, $r_C$ as the centers and radii of $(PAB)$, $(PAC)$, respectively. The only noncomputational idea of this question to note the spiral similarity $\triangle AO_BO_C \sim \triangle ABC$. If we let the length of the mutual external tangent be $k$, we have \begin{align*} 1-\left(\frac{PA}{XY}\right)^2 &= \frac{(XY+PA)(XY-PA)}{XY^2} = \frac{4 \cdot XA \cdot XB}{XY^2} \\ &= \frac{k^2}{XY^2} = \left(\frac{O_BO_C}{r_B+r_C}\right)^2 = \left(\frac{a}{b+c}\right)^2. \end{align*} Thus we have the system of equations \[PB \cdot PC = \frac{a^2bc}{(b+c)^2}, PB+PC = a \implies \{PB,PC\} = \left\{\frac{ab}{b+c}, \frac{ac}{b+c}\right\}.\] Hence the angle bisector theorem tells us $P$ is the foot of the $A$-angle bisector or the reflection of this point over the midpoint of $BC$. $\blacksquare$