Draw two circles with centers $B$ and $C$ and radius $2$. Consider the following sets of points: [asy][asy]usepackage("contour", "outline"); texpreamble("\contourlength{1.7pt}");import graph;size(8cm); pair B=(-1,0),C=(1,0),D=(0,sqrt(3)); filldraw((0,sqrt(3))--(0,0)--C--arc(B,C,(0,sqrt(3)))--cycle,cyan); filldraw(C--(3,0)--arc(C,(3,0),(0,sqrt(3)))--arc(B,(0,sqrt(3)),C,direction=CW)--cycle,magenta); xaxis(xmin=-4,xmax=4,arrow=Arrows); yaxis(ymin=-2.5,ymax=2.5,arrow=Arrows); draw(circle(B,2)^^circle(C,2)); dot(B^^C^^D);MP("B",B,SW);MP("C",C,SE);MP("D",D,2*W); label("\contour{white}{$S$}",(1/2,1/2),NW); label("\contour{white}{$S'$}",(2,1/2),NW); [/asy][/asy] We claim that these work. Clearly these are nonempty, bounded and respect (i) (the corresponding $T$'s can be $(0,0)$ and $(3,0)$). Note that since $S$ and $S'$ are the inverses of each other with respect to $\odot(B,1)$, it's enough to prove (ii) for $S$, and then (ii) for $S'$ and the fact that $BA\cdot BA'$ is constant would follow. Call the point $(0,\sqrt 3)$ as $D$. So take a triangle $P_1P_2P_3$; we'll try to fit some triangle $ABC$ similar to that in some order. Suppose $A$ is any point inside $S$; clearly $BC=1>BA>CA$. So $\angle A$ has to be the largest angle of $P_1P_2P_3$. Of course, the largest angle in $P_1P_2P_3$, say $\alpha$, is at least $60^\circ$, so we can find an arc $\mathcal C$ (shown in green) inside $S$ so that $\angle BAC=\alpha$ for all $A$ on that arc. [asy][asy]size(10cm);pair B=(-1,0),C=(1,0),D=(0,sqrt(3)),E=(0,1.2),F; filldraw((0,sqrt(3))--(0,0)--C--arc(B,C,(0,sqrt(3)))--cycle,cyan); xaxis(xmin=-4,xmax=2.5,arrow=Arrows); yaxis(ymin=-2.5,ymax=2.5,arrow=Arrows); draw(circle(B,2));draw(Arc(C,B,D,direction=CW)); pair[] x=intersectionpoints(circumcircle(B,C,E),circle(B,2)); F=x[0]; draw(arc(circumcenter(B,C,E),F,E),heavygreen); pair A=relpoint(arc(circumcenter(B,C,E),F,E),0.7); D(B--MP("A",A,N)--C);draw(B--E--C^^B--F--C,dotted); dot(A^^B^^C^^D^^E^^F);MP("B",B,SW);MP("C",C,SE);MP("D",D,2*W);[/asy][/asy] Of course, we need $\angle ABC=$smallest angle in $P_1P_2P_3$, say $\gamma$, and then $\angle ACB=\beta$. The largest possible value of $\gamma$ is given by $$\gamma=180^\circ-\alpha-\beta\le 180^\circ-\alpha-\gamma\implies \gamma\le 90^\circ-\tfrac12\alpha,$$which is attained at the leftmost point of $\mathcal C$, and the smallest possible value is given by $$\gamma=180^\circ-\alpha-\beta\ge 180^\circ-2\alpha,$$which is attained at the rightmost position on $\mathcal C$. Since the actual value of $\gamma$ is somewhere in between, and $\angle ABC$ strictly increases when $A$ moves on $\mathcal C$ right to left, at some unique intermediate point $\angle ABC$ is exactly $\gamma$, which is what we need. Since all these steps of locating $A$ were forced, its position is unique, which finishes the proof. $\blacksquare$

And after years of tackling IMO shortlist geometry problems and failing all of them past G2, my very first G3: The two nice subsets that satisfy the problem conditions in my solution are exactly the same as those provided in the solution above mine. But to address how one can actually find these subsets, what I shall do is partition these nice subsets into smaller regions. [asy][asy]usepackage("contour", "outline"); texpreamble("\contourlength{1.7pt}");import graph;size(8cm); pair B=(-1,0),C=(1,0),D=(0,sqrt(3)),E=(0,1),F=(1,2); filldraw(C--arc(B, C, D)--(0,1)--arc((0,0),E,C,direction=CW)--cycle,purple); filldraw((0,0)--C--arc((0,0),C,E)--cycle,fuchsia); filldraw(C--(3,0)--arc(C,(3,0),F)--cycle,green); filldraw(C--F--arc(C,F,D)--arc(B,(0,sqrt(3)),C,direction = CW)--cycle,yellow); draw(C--F); draw(arc((0,0),E,C,direction=CW)); xaxis(xmin=-4,xmax=4,arrow=Arrows); yaxis(ymin=-2.5,ymax=2.5,arrow=Arrows); draw(circle(B,2)^^circle(C,2)); dot(B^^C^^D^^E^^F);MP("B",B,SW);MP("C",C,SE);MP("D",D,2*W);MP("E",E,W);MP("F",F,N); [/asy][/asy] We denote $R = \{(x, y) : x \geq 0, y \geq 0\}$ for brevity, and we let each colored region include its boundary points. The union of the pink region and the green region is the set of points $A \in R$ where $\triangle{ABC}$ is obtuse and $BC$ is not the shortest side of $\triangle{ABC}$. The union of arc $EC$ and segment $CF$ is the set of points $A \in R$ such that $\triangle{ABC}$ is right and $BC$ is not the shortest side of $\triangle{ABC}$. And finally, the union of the yellow region and the purple region is the set of points $A \in R$ such that $\triangle{ABC}$ is acute and $BC$ is not the shortest side of $\triangle{ABC}$. Let $S$ be the union of the purple region, the pink region, and the arc $EC$. Let $S'$ be the union of the yellow region, the segment $CF$, and the green quarter circle. Observe that all points $A$ in $S$ are such that $BC$ is the longest side of $\triangle{ABC}$. All points $A$ in $S'$ are such that $BC$ is the second longest side of $\triangle{ABC}$. We verify that these subsets are indeed nice. $S$ and $S'$ are obviously nonempty and bounded, and condition (i) holds by taking $(0, 0) \in S$ and $(3, 0) \in S'$. Now, we show that condition (ii) holds. Denote $\angle{P_1}, \angle{P_2}, \angle{P_3}$ the angles of triangle $P_1P_2P_3$ at $P_1, P_2, P_3$ respectively. WLOG we have $\angle{P_1} \le \angle{P_2} \le \angle{P_3}$. For $P_1P_2P_3$, the point $A \in S$ in (ii) must be the intersection point of the ray emanating from $B$ with an angle of $\angle{P_1}$ above the positive x-axis and the ray emanating from $C$ with an angle of $\angle{P_2}$ above the negative x-axis since $BC$ is the longest side of $\triangle{ABC}$ and $A$ must lie in $R$. So, this point $A$ must be unique. The justification that condition (ii) holds for $S'$ is analogous to that for $S$, except that the point $A$ will be the intersection of the ray emanating from $B$ with an angle of $\angle{P_1}$ above the positive x-axis and the ray emanating from $C$ with an angle of $\angle{P_3}$ above the negative x-axis. Thus for any triangle $P_1P_2P_3$, upon the unique choices of $A$ and $A'$, $\triangle{ABC} \sim \triangle{CBA'}$. Thus $$\tfrac{BA}{BC} = \tfrac{BC}{BA'} \rightarrow 4 = BC^2 = BA \cdot BA'$$independent of $P_1P_2P_3$, and we're done.

This problem was proposed by C.R. Pranesachar from India, if I remember correct.

Is the spelling correct? Is it Pranesacher or Pranesachar? (Or are both spellings in Latin transcription equivalent?)

$S = \{(x, y): (x + 1)^2 + y^2 \le 4, x \ge 0, y \ge 0 \}$ $S' = \{(x, y): (x + 1)^2 + y^2 \ge 4, (x - 1)^2 + y^2 \le 4, x \ge 0, y \ge 0\}$

Finally, the point $T(0, 0)$ satisfies (i) for $S$ and the point $T(3, 0)$ satisfies (i) for $S'$.

Here's a sketch of the soluton which gives an intuitive argument. cjquines0 wrote: Let $B = (-1, 0)$ and $C = (1, 0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be nice if $\text{(i)}$ there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $TQ$ lies entirely in $S$; and $\text{(ii)}$ for any triangle $P_1P_2P_3$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1, 2, 3\}$ for which triangles $ABC$ and $P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$ are similar. Prove that there exist two distinct nice subsets $S$ and $S'$ of the set $\{(x, y) : x \geq 0, y \geq 0\}$ such that if $A \in S$ and $A' \in S'$ are the unique choices of points in $\text{(ii)}$, then the product $BA \cdot BA'$ is a constant independent of the triangle $P_1P_2P_3$. The following construction works: [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 = -3.880326658628674, xmax = 5.563410177753953, ymin = -2.903403066257509, ymax = 3.0713879444850263; /* image dimensions */ Label laxis; laxis.p = fontsize(10); string blank(real x) {return "";} xaxis(xmin, xmax, Ticks(laxis, blank, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, blank, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(shift((1,0))*xscale(2)*yscale(2)*arc((0,0),1,0,120), linewidth(0.5)); draw(shift((-1,0))*xscale(2)*yscale(2)*arc((0,0),1,0,60), linewidth(0.5)); draw((0,1.7320508075688774)--(0,0), linewidth(0.5)); draw((0,0)--(1,0), linewidth(0.5)); draw((1,0)--(3,0), linewidth(0.5)); draw((0,1.4484714474309524)--(0.2077478579274845,1.5941596882594833), linewidth(0.5)); draw((0,1.1359524501849139)--(0.4176104928340292,1.4108084528407323), linewidth(0.5)); draw((0,0.6775912542240576)--(0.5911836946149307,1.2116659811976151), linewidth(0.5)); draw((0,0.30256845752881123)--(0.7730640019377426,0.9253345584341466), linewidth(0.5)); draw((0.1428356262204876,0)--(0.8839341572096762,0.6714105236728624), linewidth(0.5)); draw((0.8527281548556536,0.7532585108746965)--(2.873603903382004,0.6997202392611769), linewidth(0.5)); draw((0.7255016159045673,1.0112587075079884)--(2.7368424984130044,0.9916542420150651), linewidth(0.5)); draw((0.5702341254247569,1.2386947934626789)--(2.5949198390889654,1.2067438447659178), linewidth(0.5)); draw((0.4176104928340292,1.4108084528407323)--(2.39330859765985,1.434813978077696), linewidth(0.5)); draw((0.11353062245363632,1.661339686174389)--(2.0585353274290394,1.6969098268867135), linewidth(0.5)); draw((0.221328959232708,1.8421920123240092)--(1.690084065398108,1.8771744678328703), linewidth(0.5)); draw((0.7845239922432099,1.988358642217544)--(1.2200689189867662,1.9878555457819354), linewidth(0.5)); draw((0.9300710363180413,0.5242382995987629)--(2.9402850002906638,0.48507125007266594), linewidth(0.5)); draw((0.611226872533924,0)--(0.9692641863171296,0.34928292899701413), linewidth(0.5)); draw((0.9907924495055433,0.19169095698993968)--(2.9895812872292615,0.2038781535799126), linewidth(0.5)); draw(circle((-1,0), 2), linewidth(0.5) + linetype("4 4")); draw(circle((1,0), 2), linewidth(0.5) + linetype("4 4")); /* dots and labels */ dot((-1,0),linewidth(4pt) + dotstyle); label("$B$", (-1.2472119045327783,-0.27812350920861235), NE * labelscalefactor); dot((1,0),linewidth(4pt) + dotstyle); label("$C$", (1.0421877263413073,-0.27271150182609394), NE * labelscalefactor); dot((3,0),linewidth(4pt) + dotstyle); dot((0,1.7320508075688774),linewidth(4pt) + dotstyle); dot((0,0),linewidth(4pt) + dotstyle); label("$O$", (-0.23302886251866956,-0.23704110773210868), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Firstly ignore the isosceles cases, in which the unique position of the point $A$ are on the boundaries of our regions. (even though these motivate the solution) Otherwise, note that $$\alpha+\beta=2\theta \implies min\{\alpha,\beta\} \le \theta, max\{\alpha,\beta\} \ge \theta$$The following diagram shows the only $3$ possible positions in the first quadrant. By the fact above we can reason out the following: On the ray $CA_1,$ why does $A_2$ lie inside the region while $A_1$ lie outside. On the ray $BA_2,$ why do the points $A_2, A_3$ lie in different regions. Now, $A_2, A_3$ are the unique points in the two regions. Since $BA_2 \cdot BA_3=BC^2=4 \text{ unit}^2$ is independent of $\triangle P_1P_2P_3,$ we are done. $\square$ [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 = -3.619602909462244, xmax = 6.201255331689688, ymin = -2.3313788332925762, ymax = 3.88200645619919; /* image dimensions */ Label laxis; laxis.p = fontsize(10); string blank(real x) {return "";} xaxis(xmin, xmax, Ticks(laxis, blank, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, blank, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(circle((-1,0), 2), linewidth(0.4) + linetype("2 2")); draw(circle((1,0), 2), linewidth(0.4) + linetype("2 2")); draw(shift((-1,0))*xscale(2)*yscale(2)*arc((0,0),1,0,60), linewidth(0.4)); draw(shift((1,0))*xscale(2)*yscale(2)*arc((0,0),1,0,120), linewidth(0.4)); draw((-2.8793852415718164,3.255190725397494)--(-1,0), linewidth(0.4)); draw((-1.532088886237956,0.9216049851068763)--(1,0), linewidth(0.4)); draw((-2.8793852415718164,3.255190725397494)--(1,0), linewidth(0.4)); draw((-1,0)--(2.8793852415718146,3.255190725397493), linewidth(0.4)); draw((2.8793852415718146,3.255190725397493)--(1,0), linewidth(0.4)); draw((1.5320888862379558,0.9216049851068762)--(-1,0), linewidth(0.4)); /* dots and labels */ dot((-1,0),linewidth(4pt) + dotstyle); label("$B$", (-1.2640807250877486,-0.3375313150228304), NE * labelscalefactor); dot((1,0),linewidth(4pt) + dotstyle); label("$C$", (1.1470837156105538,-0.3468050244101315), NE * labelscalefactor); dot((2.8793852415718146,3.255190725397493),dotstyle); label("$A_1$", (2.918362208585076,3.344131311735724), NE * labelscalefactor); dot((1.5320888862379558,0.9216049851068762),dotstyle); label("$A_2$", (1.7236743474264072,0.9164302555231366), NE * labelscalefactor); dot((-2.8793852415718164,3.255190725397494),dotstyle); dot((-1.532088886237956,0.9216049851068763),dotstyle); dot((-0.39493084363469844,0.5077133059428725),dotstyle); dot((0.3949308436346985,0.5077133059428725),dotstyle); label("$A_3$", (0.3830080927883642,0.5991133330945851), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $\omega_B$ and $\omega_C$ denote the circles centered at $B$ and $C$ with radius $BC$. Let $S$ be the region of the first quadrant in $\omega_B$ and let $S'$ be the region of the first quadrant in $\omega_C$ but not $\omega_B$, where both $S$ and $S'$ includes their boundaries. To prove that both of them are nice: if we have a triangle $P$ with angles $\alpha \le \beta\le \gamma$ then draw a ray from $B$ above $y=0$ such that the angle between segment $BC$ and that ray is $\alpha$. Draw a ray from $C$ above $y=0$ such that the angle between segment $BC$ and that ray is $\beta$. The intersection is $A$. Clearly, since $\angle ABC \le \angle ACB$, $A$ lies in the first quadrant, and since $\angle BAC \ge \angle ACB$, $A$ lies in $\omega_B$, and so it lies in $S$. Conversely, if anything lies in $S$ then we must have $\angle BAC \ge \angle ACB \ge \angle ABC$ so the method above is the only way to find $A$. Thus, $S$ is nice. Similarly, the only way to construct $S'$ is by drawing the ray from $B$ above $y=0$ such that the angle between segment $BC$ and that ray is $\alpha$ and the ray from $C$ above $y=0$ such that the angle between segment $BC$ and that ray is $\gamma$. Clearly, $S'$ is also nice and $BA\cdot BA'=BC^2$ which is constant.