Circles have two intersections...

*cires* I feel your pain alifenix

I can't be the only one who misread PR=RA, QR=RB for the entirety of the test...

the solution set for this problem was wacky

alifenix- wrote: Circles have two intersections... same

Nice problem but did not get this during the test.

flashsonic wrote: I can't be the only one who misread PR=RA, QR=RB for the entirety of the test... No. You were not.

I also did that and was wondering why the locus wasn't nice...

dchenmathcounts wrote: flashsonic wrote: I can't be the only one who misread PR=RA, QR=RB for the entirety of the test... No. You were not. I misread at first too.

I can't be the only one who didn't see the equilateral until staring at the problem for an hour...

Say the incircle is tangent to BC, AC at A', B'. Dilate arc A'B' with respect to the incenter by magnitudes -2 and 4. Those new arcs are the answers

I centered circles at $A$ and $B$ instead of $P$ and $R$ for a very long time -_-.

Let each $R$ be split into $R_1$ and $R_2$. The crux of the problem is prove the incenter lies on $R_1R_2$ by perpendicularity lemma. Then proceed to length chase.

I just saw congruent triangles and after some logic, AI=RI where I is the incenter for one side, for the other, flip A over BC to get A', and A'I=RI

Awesome_guy wrote: Let each $R$ be split into $R_1$ and $R_2$. The crux of the problem is prove the incenter lies on $R_1R_2$ by perpendicularity lemma. Then proceed to length chase. bruh I used radical center to do that

complexed

Would not going into too much detail about how every point on the locus is attainable be a 7-, or 0+ for this?

I tried to coord bash because I misread perpendicular bisectors like an idiot??

If the reflections of $A,B,C$ across opposite sides are $X,Y,Z,$ we show the locus is the union of minor arcs $\widehat{AB}$ and $\widehat{XY}$ on $(ABC),(XYZ)$ respectively. If $O$ is the center of $\triangle ABC$, note $$\angle (\overline{PO},\overline{QO}) = 180^\circ - \angle OPQ - \angle OQP = 180^\circ - \tfrac{1}{2}(\angle AQP+\angle BPQ) = 180^\circ - \tfrac{1}{2}(240^\circ) = 60^\circ$$hence the transformation $\Psi$ consisting of reflection in $\overline{PO}$ and then $\overline{QO}$ is a $120^\circ$ degree rotation about $O$. In particular, $\Psi(A)=B$ and the reflections of $A,B$ over $\overline{OP},\overline{OQ}$ respectively must coincide at some point $R_1 \in \widehat{AB}$. Similarly, the reflections of $X,Y$ over $\overline{OP},\overline{OQ}$ coincide at some point $R_2 \in \widehat{XY}$. These are exactly the set of possible locations of $R$ (since circles intersect at most twice), as claimed. Showing every point on the arcs is attainable is left as an exercise to the reader.

The only computation here is $2+1+1=4$. I'll assume in this solution that the segments are closed (i.e. contain the endpoints), so we can let $Q=C$ and $P$ be the tangency point from $\omega$ to $\overline{BC}$ for instance. Also let $\Omega$ be the circumcircle of $\triangle ABC$, and let $I$ be its center (incenter, circumcenter, etc.). Then, the answer is the closed minor arc $\widehat{AB}$ of $\Omega$, as well as its image under a homothety with scale factor $-2$ centered at $I$. [asy][asy] size(300); pair A,B,C,D,E,P,Q,R,S,T,T1,T2,A1,B1,I; C=dir(90); B=dir(210); A=dir(330); draw(A--B--C--cycle); D=midpoint(B--C); E=midpoint(A--C); I=incenter(A,B,C); draw(incircle(A,B,C)); draw(circumcircle(A,B,C)); T1=dir(260); P=extension(midpoint(T1--A),I,B,C); Q=extension(midpoint(T1--B),I,A,C); T=T1*(-0.5); draw(arc(I,B,A),red+linewidth(2)); A1=-2*A; B1=-2*B; draw(arc(I,B1,A1),red+linewidth(2)); T2=T*4; draw(T2--T1,heavycyan); draw(A--A1,blue); draw(B--B1,blue); draw(P--Q); draw(A--P--T1,orange); draw(B--Q--T1,purple); dot("$A$",A,dir(330)); dot("$B$",B,dir(210)); dot("$C$",C,dir(90)); dot("$T'$",T1,dir(260)); dot("$T''$",T2,dir(80)); dot("$I$",I,dir(235)*2); dot("$P$",P,dir(125)); dot("$Q$",Q,dir(65)); dot("$D$",D,dir(190)*1.3); dot("$E$",E,dir(-10)*1.3); dot("$T$",T,dir(115)); [/asy][/asy] Suppose we pick an arbitrary line $\ell$ satisfying the conditions. Let $T$ be the point where $\ell$ is tangent to $\omega$. I claim that for this choice of $\ell$, the possible locations of $R$ are exactly the image of $T$ upon a homothety of scale factor $-2$ centered at $I$, as well was its image upon a homothety of scale factor $4$ centered at $I$. Since the choices of $R$ are exactly the intersections of the circle centered at $P$ passing through $A$ and the circle centered at $Q$ passing through $B$, and these circles are different (they have different centers), they have at most two intersection points. Hence it suffices to only prove that these two possibilities both work, which implies no others work. Let rays $PI$ and $QI$ intersect $\Omega$ at $R,S$ respectively. Also let $D,E$ denote the points where $\omega$ is tangent to $\overline{CB}$ and $\overline{CA}$, respectively. Clearly, a homothety of scale factor $-2$ centered at $I$ sends $D$ to $A$, $E$ to $B$, and $T$ to some point $T'$ on minor arc $\widehat{AB}$ of $\Omega$. Observe that we have $\angle IDP=\angle ITP=90^\circ$ and $ID=IT$, so $\triangle IDP\cong \triangle ITP$. Hence $IDPT$ is a kite and $D$ is the reflection of $T$ over $\overline{PI}$. Similarly, $E$ is the reflection of $T$ over $\overline{PI}$. Since these are preserved by the homothety, we find that $T'$ is the reflection of $A$ over $\overline{PI}$, and the reflection of $B$ over $\overline{QI}$. Thus, line $\overline{PI}$ is the perpendicular bisector of $\overline{AT'}$ so $PA=PT'$. Similarly, $QB=QT'$, so we conclude that $T'$ is one admissible choice of $R$. The other intersection of the two circles previously mentioned is the reflection of $T'$ over $\overline{PQ}$—call this $T''$. If the radius of $\omega$ is $r$, then we have $TT'=3r$, since $\overline{T'T} \perp \overline{PQ}$ as $\overline{IT} \perp \overline{PQ}$. Since $IT=r$, it follows that $IT''=4r$, so it is the image of $T$ under a homothety of scale factor $4$ centered at $I$ and thus the image of $T'$ under a homothety of scale factor $-2$ centered at $I$, as desired. To see that every point on minor arc $\widehat{AB}$ and its corresponding image under the homothety of scale factor $-2$ work, and nothing else does, observe that we can pick $T$ to be any point on minor arc $\widehat{DE}$ on $\omega$, and nothing else. Hence we're done. $\blacksquare$

Nice problem, though I took a lot of time (~2 hours) because I was (stupidly) trying to angle chase a claim. Diagram added for reference. Claim: $I \in GH$. Proof. We will prove that it lies on the radical axes of the red and blue circles. Throughout this solution, $DP=PX=x, QX=QE=y$. \begin{align*} \iff IP^2-PA^2&=IQ^2-QB^2 \\ \iff (\frac{a^2}{12}+x^2)-(\frac{3a^2}{4}+x^2) = (\frac{a^2}{12}+y^2)-(\frac{3a^2}{4}+y^2) \end{align*}Which is trivially true $\Box$ Note that the radical axis of two circles must be perpendicular to the line connecting their centres. As a result, $X \in GH$ too. Claim: $G \in \odot{ABC}$ Proof. $$GI=GX-IX$$$$=GX -\frac{a}{2\sqrt{3}}$$However, $GX=\frac{\sqrt{3}a}{2}$ because $GX^2+XP^2=GP^2= PA^2=AD^2+DP^2$. . Therefore, $GI=\frac{a}{\sqrt{3}}=AI$, proving the claim $\Box$. Now a simple length chase proves that $\frac{GI}{IH}=\frac{1}{2}$. Therefore, the homothety of ratio $-2$ sends $G$ to $H$, and consequently, our answer is arc $AB$ and the arc made by the homothety we mentioned before.

solution set is points R such that RT perpendicular to PQ and RT=AD=BE which is easily proven (where T is point of tangency), the solution follows easily edit: dang your solutions are

Let $I$ be the center of the equilateral triangle $ABC$. Let the circles with radius $AP$ and $BQ$ intersect each other at $R_1$ and $R_2$. The locus of the all points of such $R$ is circles actually, the circles with radius $IR_1$ (this is the circumcircle of $ABC$) and $IR_2$. The point $D$ lies on $R_1R_2$, which follows by the PoP. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(32.04009188686031cm); 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 = -13.55596900292541, xmax = 18.4841228839349, ymin = -18.814372246868626, ymax = 6.0214386240112; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((1.3581702394607207,-4.918497679616305), 2.116571695140565), linewidth(0.8)); draw((xmin, 2.9836120376494693*xmin-15.63104086425135)--(xmax, 2.9836120376494693*xmax-15.63104086425135), linewidth(0.8)); /* line */ draw(circle((2.8849340119995452,-7.023517018225128), 6.526953344280609), linewidth(0.8) + red); draw(circle((3.6713070312575016,-4.677285011884334), 6.422444339501898), linewidth(0.8) + green); draw(circle((1.358407455868242,-4.918479292321957), 4.233082233658332), linewidth(0.8) + ffxfqq); draw((-2.6555321573673805,-3.5732482483295986)--(1.3259710422622277,-0.6854767519122333), linewidth(0.8)); draw((1.3259710422622277,-0.6854767519122333)--(1.3581702394607207,-4.918497679616305), linewidth(0.8)); draw((1.3581702394607207,-4.918497679616305)--(5.040173496202865,-7.007122820712979), linewidth(0.8)); draw((1.3581702394607207,-4.918497679616305)--(1.3742698380599672,-7.035008143468338), linewidth(0.8)); draw((-0.48283138891035104,-3.8741851090679673)--(1.3581702394607207,-4.918497679616305), linewidth(0.8)); draw((1.3259710422622277,-0.6854767519122333)--(-2.291633820082928,-7.062893466223699), linewidth(0.8)); draw((-2.291633820082928,-7.062893466223699)--(5.040173496202865,-7.007122820712979), linewidth(0.8)); draw((5.040173496202865,-7.007122820712979)--(1.3259710422622277,-0.6854767519122333), linewidth(0.8)); draw((1.3259710422622277,-0.6854767519122333)--(2.8849340119995452,-7.023517018225128), linewidth(0.8) + red); draw((2.8849340119995452,-7.023517018225128)--(-2.6555321573673805,-3.5732482483295986), linewidth(0.8) + red); draw((-2.6555321573673805,-3.5732482483295986)--(9.385575033116927,-7.608996542189704), linewidth(0.8)); draw(circle((1.3574812381604935,-4.918232986367356), 8.466908766081035), linewidth(0.8)); draw((2.8849340119995452,-7.023517018225128)--(9.385575033116927,-7.608996542189704), linewidth(0.8) + red); /* dots and labels */ dot((1.3259710422622277,-0.6854767519122333),dotstyle); label("$A$", (1.0105816478069403,0.4602392432543942), NE * labelscalefactor); dot((-2.291633820082928,-7.062893466223699),dotstyle); label("$B$", (-3.2867087827778745,-7.8183643803722145), NE * labelscalefactor); dot((5.040173496202865,-7.007122820712979),dotstyle); label("$C$", (5.813435658460557,-7.344398524057714), NE * labelscalefactor); dot((1.3581702394607207,-4.918497679616305),linewidth(4.pt) + dotstyle); label("$I$", (1.9269156366816433,-4.721787452450811), NE * labelscalefactor); dot((1.3742698380599672,-7.035008143468338),linewidth(4.pt) + dotstyle); label("$E$", (0.7577998577725393,-8.071146170406616), NE * labelscalefactor); dot((3.3650214378747707,-5.591122395259655),dotstyle); label("$D$", (3.5699972719052493,-6.459662258937313), NE * labelscalefactor); dot((3.6713070312575016,-4.677285011884334),linewidth(4.pt) + dotstyle); label("$Q$", (4.328342642008452,-4.721787452450811), NE * labelscalefactor); dot((2.8849340119995452,-7.023517018225128),linewidth(4.pt) + dotstyle); label("$P$", (3.001238244327847,-8.229134789178115), NE * labelscalefactor); dot((-2.6555321573673805,-3.5732482483295986),linewidth(4.pt) + dotstyle); label("$R_1$", (-3.760674639092376,-3.0787058172272097), NE * labelscalefactor); dot((9.385575033116927,-7.608996542189704),linewidth(4.pt) + dotstyle); label("$R_2$", (9.76315112774807,-8.450318855458216), NE * labelscalefactor); dot((-0.48283138891035104,-3.8741851090679673),linewidth(4.pt) + dotstyle); label("$M$", (-0.8220863299424661,-3.45787850227881), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $IA=IR_1$. And similarly $IB=IR_1$. Proof. The triangles $\triangle AEP$ and $\triangle AR_1D$ are congruent. Because $EP=DP, R_1P=AP$ and there have a right angle. Therefore $IE=R_1D$ and hence done. $\square$ Since we have $IA=IB=IR_1$, we can conclude that the point $R_1$ moves on the circumcircle of $ABC$. (Basically on minor arc $AB$). Now notice that the triangles $\triangle PR_1D$ and $\triangle PR_2D$ are congruent. Since the length $IR_1$ doesn't change therefore $DR_2$ also does not change. Since the line $R_1R_2$ are moving with respect to the fixed point $I$ with the same length, so the locus of the point of $R_2$ is a circle.

Let $E$, $F$, and $T$ be the tangency point of $AC$, $AB$, and $\ell$ with $\omega$ and let $TI$ intersect $(ABC)$ at $L$. We claim $BLET$ is cyclic. Indeed, since $BI=LI$ and $TI=EI$ it follows that $BLET$ is an isosceles trapezoid. Now, consider the inversion around $(ABC)$. $L \to L$, $B \to B$, and $E \to E^*$ where $E$ is the midpoint of $E^*B$. We claim $Q$ is the circumcenter of $LBE^*$. Indeed, $QB=QE^*$ by the midpoint property so let us show $QB=QL$. $LB=IL$ so by SAS it is sufficient to show $\angle QIL = \angle QIB$. Clearly, $\angle EIL = \angle TIB$ and $QE=QT$. Thus, by SSS, $\angle EIQ = \angle TIQ$. So, \[ \angle QIL = \angle EIL + \angle EIQ = \angle TIB + \angle TIQ = \angle QIB \]Proving the claim. Using the fact that $T$ lies on $(BLE)$ we have $T^*Q=BQ$. By symmetry, $T^*P=AP$ so $T^*=R$. Notice that a point $X$ is a valid choice of $T$ iff $X$ is on minor arc $EF$. Thus, all valid choices of $T^*$ is the arc $E^*F^*$ on $\omega^*$.

I did not notice that there was a second intersection point until reading the first few posts on the thread lol. Anyways, being able to directly construct $R$ is pretty cool. The two circles intersect at exactly two points. Let $PQ$ be tangent to the incircle at $T$. I claim that the two intersections are the points $R_1, R_2$ such that $RT\perp PQ$ and $RT = \frac{\sqrt3}{2} s$ where $s$ is the sidelength of $\triangle ABC$. It suffices to show that this definition of $R$ satisfies $QB=QR$ and $PA = PR$. Let $X,Y$ be the tangency points on $CB,CA$ respectively. Then \[PA^2 = PX^2 + AX^2 = PT^2 + (\frac{\sqrt3}{2} s)^2 = PT^2 + TR^2 = PR^2.\]A similar argument shows that $QB = QR$. Thus, we have described all points $R$ that lie on the intersection of the two circles. Now, note that we always have $IT\perp PQ$. Since $r= \frac{s\sqrt3}{6}$, the locus of points $R$ is the points $s(\frac{\sqrt3}{2} - \frac{\sqrt3}{6})= s\frac{\sqrt{3}}{3} = 2r$ in the direction opposite $T$ from the incenter, and $s(\frac{\sqrt3}{2} + \frac{\sqrt3}{6}) = \frac{s\cdot 2\sqrt3}{3} = 2R=4r$ in the direction of $T$ from $I$. $T$ must lie on the $120^{\circ}$ arc $XY$. Note that a homothety with factor $-2$ and the homothety with factor 4 take $T$ to $R_1$ and $R_2$. Thus, the locus of points $R$ is the arc $XY$ under homothety with factor $-2$ or $4$ centered at $I$.

How many points would missing the other case get?

Let $I$ be the incenter, $D,E,F$ be the contact points, $s$ a side length, $r$ the circumradius, $T=\omega\cap\ell,$ and $R_1,R_2=\odot(P,PA)\cap\odot(Q,QB).$ Claim: $\overline{R_1T}\perp\overline{PQ}$ so $I,T,R_1,$ and $R_2$ are collinear. Proof. Notice \begin{align*}(R_1P^2-TP^2)-(R_1Q^2-TQ^2)&=(AP^2-PD^2)-(BQ^2-QE^2)\\&=AD^2-BE^2\\&=0\end{align*}and the second part follows as $\overline{TI}\perp\overline{PQ}.$ $\blacksquare$ We also see $\triangle R_1TP\cong\triangle ADP$ so $$IR_1=IT+AD=\frac{s\sqrt{3}}{6}+\frac{s\sqrt{3}}{2}=2r.$$Also, $$IR_2=BE-IT=\frac{s\sqrt{3}}{6}-\frac{s\sqrt{3}}{2}=\frac{r}{2}.$$ Hence, all $R$ are on the circle with radius $\tfrac{1}{2}r$ or $2r.$ Notice $T$ must lie on minor arc $DE$ on $\omega,$ and so by homothety, $R$ must lie on the $120$ degree arcs bounded by $\overline{AD}$ and $\overline{BE}.$ That is, $R$ is on minor arc $\overline{AB}$ of $\odot(I,\tfrac{1}{2}r)$ or minor arc $\overline{A_1B_1}$ of $\odot(I,2r),$ where $A_1=\overline{AD}\cap\odot(I,2r)$ and $B_1=\overline{BE}\cap\odot(I,2r).$ $\square$

Very fun problem I would appreciate it if anyone could look over my writeup and give me some feedback. Let $I$ be the incenter of $\triangle ABC$, $\omega_1$ be the circle with center $P$ and radius $PA$, and $\omega_2$ be the circle with center $Q$ and radius $QB$. Denote $D, E, F$ as the tangency points of $\omega$ with $BC, CA$, and $AB$ respectively. Clearly, the intersections of $\omega_1$ and $\omega_2$ will be possible locations of the point $R$. The key claim is the following: Claim: $TI$ is the radical axis of $\omega_1$ and $\omega_2$. Proof. Since $PQ$ is the line connecting the centers of $\omega_1$ and $\omega_2$ and $TI$ is clearly perpendicular to $PQ$, it suffices to show that $T$ has equal power to $\omega_1$ and $\omega_2$. Let $PD=PT=x, QT=QE=y$, and $DB=BF=FA=EA=d$. By the Law of Cosines, we find \begin{align*} PA^2&=BP^2+BA^2-2(BP)(BA)\cos 60^\circ=(x+d)^2+(2d)^2-(x+d)(2d)=x^2+3d^2 \\ QB^2&=AQ^2+AB^2-2(AQ)(AB)\cos 60^\circ=(y+d)^2+(2d)^2-(y+d)(2d)=y^2+3d^2. \end{align*}Therefore, \begin{align*} \text{Pow}(T,\omega_1)=TP^2-PA^2=x^2-(x^2+3d^2)=-3d^2=y^2-(y^2+3d^2)=\text{Pow}(T,\omega_2), \end{align*}as desired. Let $R_1$ denote the intersection of $\omega_1$ and $\omega_2$ that lies on the opposite side of $C$ with respect to $AB$. We have $T, I, R_1$ collinear. By the Pythagorean Theorem, we have \begin{align*} R_1T=\sqrt{PR_1^2-PT^2}=\sqrt{PA^2-PT^2}=\sqrt{(x^2+3d^2)-x^2}=\sqrt{3}d. \end{align*}But as $TI=\tfrac{d}{\sqrt{3}}$ (inradius), we have $IR_1=\tfrac{2d}{\sqrt{3}}$. This implies that there is a negative homothety centered at $I$ with ratio $-2$ that sends $T$ to $R_1$. Letting $R_2$ denote the intersection of $\omega_1$ and $\omega_2$ that lies on the same side of $C$ with respect to $AB$, we find with similar methods that $IR_2=\tfrac{4d}{\sqrt{3}}$, so there is a positive homothety centered at $I$ with ratio $4$ that sends $T$ to $R_2$. Let $D'$ and $E'$ denote the images of $D$ and $E$, respectively, after a homothety of ratio $4$ centered at $I$. Then, as $T$ can be any point on minor arc $\widehat{DE}$, the desired locus is the union of two $120^\circ$ circular arcs $\widehat{D'E'}$ and $\widehat{AB}$ centered at $I$.

The answer is minor arc $AB$ on the circumcircle together with its image under a homothety at the center with scale factor $-2.$ Let $X$ be the tangency point of $PQ$ to the incircle and let $DEF$ be the contact triangle, with $D$ opposite $A,$ $E$ opposite $B$ and $F$ opposite $C.$ It is clear that $X$ is on minor arc $DE$ of the incircle. Let $I$ be the incenter. We claim that $X,I,R$ are collinear. To show this, let $AE=AF=BF=BD=c,$ $QE=QX=a$ and $PD=PX=b.$ Notice that $R$ is defined as the intersection of circles centered at $P,Q.$ Thus $R$ is on their radical axis, which is perpendicular to $PQ.$ Since $IX\perp PQ$ it suffices to show that $X$ lies on the radical axis. This is equivalent to showing that $QB^2-QX^2=PA^2-PX^2.$ By Law of Cosines we have $QB^2=a^2+3c^2$ so $QB^2-QX^2=3c^2.$ Similarly $PA^2-PX^2=3c^2$ so we finish. Now we show that $RX=3IX.$ Letting $r$ be the inradius, we have $RX=\sqrt{QR^2-QX^2}=\sqrt{QB^2-QX^2}=c\sqrt3=r$ so we finish. Now if $R,I$ are on the same side of $PQ$ we have that $R$ is the image of $X$ under a homothety at $I$ with scale factor $-2$ which is minor arc $AB.$ If $R,I$ are on opposite sides of $PQ$ we see that $R$ is the image of $X$ under a homothety at $I$ with scale factor $4,$ which is the aforementioned image of $AB.$ Thus we are done.

This is fake geometry The answer is the minor arc $\widehat{AB}$ along with its image under a homothety at $O$ with ratio $-2$. To see the first locus, let $E$ be the tangency point of $\ell$. I claim that $R = \ell \cap (ABC)$ satisfies the given conditions. Set $N = \overline{OP} \cap \overline{AR}$ and $A'$ the incircle touchpoint on $\overline{BC}$. As $OA = OR$ and $PE = PA'$, by symmetry, $N$ is the midpoint of $\overline{AR}$. In particular, we must have $PA=PR$. Similarly, $QB=QR$, hence $\widehat{AB}$ describes one set of the locus of $R$. To see the other locus, notice that $(P)$ and $(Q)$ have radical axis $\overline{OR}$. Their other intersection $R'$ is a point with $ER' = ER$ and hence $OR'=2OR$ fixed.

yea this is not such a beatiful problem Let $T$ be the intersection of $\ell$ and $\omega$. The loci are minor arc $AB$ in $(ABC)$ and its reflection over $T$. Let $R = (ABC) \cup \overline{IT}$ be the point on minor arc $AB$ in $(ABC)$. We claim that $R$ is a desired point $R$ and over all configurations of $\ell$, $R$ makes up the first locus. Suppose that $PD=PT=a$, $QD=QT=b$, and the inradius of $\omega$ is $r$. Notice that \begin{align*} AP^2 &= AB^2+BP^2-2 (AB) (BP) \cos 60^\circ \\ &=(2r\sqrt{3})^2+(a+r\sqrt{3})^2 - (2r\sqrt{3})(a+r\sqrt{3}) \\ &=12r^2+(a^2+2ar\sqrt{3}+3r^2) - (2ar\sqrt{3}+6r^2) \\ &= a^2+9r^2 = a^2+(3r)^2 \\ &= TP^2+TR^2 = RP^2. \end{align*} Hence, $AP=RP$, and analogously, $BQ=RQ$. Now that we know that $R$ is one of our solutions, we can reflect it about $T$; this preserves the length of $RP$ and $RQ$, so this produces the latter locus. The complete solution set has been proved, so we are done.

Define $X = \ell \cap \omega$, and $R_1$, $R_2$ as the images of $I$ under homotheties centered at $X$ with scale -3 and 3, respectively. We claim $R_1$, $R_2$ are the two desired points of $R$ for given $\ell$. Note we have at most two choices of $R$ for each $\ell$, so it suffices to show both satisfy the length condition. But this simply holds from \[\triangle PMA \cong \triangle PXR_1 \cong \triangle PXR_2, \quad \triangle QNB \cong \triangle QXR_1 \cong \triangle QXR_2,\] where $M$, $N$ are the midpoints of $BC$, $CA$. Hence our locus is the union of the images of minor arc $MN$ on $\omega$ under homotheties of scale -2 and 4. $\blacksquare$ [asy][asy] size(300); defaultpen(linewidth(0.7)+fontsize(11)); pair I, A, B, C, M, N, X, P, Q, R1, R2; I = (0,0); A = dir(210); B = dir(330); C = dir(90); M = .5*B + .5*C; N = .5*C + .5*A; X = .5*dir(95); P = extension(B,C,X,rotate(90,X)*I); Q = extension(C,A,X,rotate(90,X)*I); R1 = 4*X; R2 = -2*X; draw(arc(I, 2, 30, 150)^^arc(I, 1, 210, 330), green+dashed+linewidth(1)); draw(A--B--C--cycle^^circle(I, .5), gray+linewidth(0.2)); draw(P--A--M--cycle^^Q--B--N--cycle^^P--Q^^R1--R2); draw(P--A^^P--R1^^P--R2^^Q--B^^Q--R1^^Q--R2); draw(P--A^^P--R1^^P--R2, red); draw(Q--B^^Q--R1^^Q--R2, blue); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$C$", C, dir(90)); dot("$M$", M, dir(30)); dot("$N$", N, dir(150)); dot("$I$", I, dir(250)); dot("$X$", X, dir(50)); dot("$P$", P, dir(25)); dot("$Q$", Q, dir(155)); dot("$R_1$", R1, dir(90)); dot("$R_2$", R2, dir(270)); [/asy][/asy]

Let $I$ be the incentre, $R_1$ and $R_2$ be the two choices of $R$ for a specific line and $D$, $E$, $F$ the intouch points opposite $A$, $B$, $C$ respectively, reflect $A$ over $D$ to get $A'$ and reflect $B$ over $E$ to get $B'$, we clearly get that the circle centred at $P$ through $A$ also goes through $A'$, and the circle centred at $Q$ through $B$ also passes through $B'$, thus from the fact that its equilateral and equal lengths and radax we get that $R_1$, $I$ and $R_2$ are collinear, let $G$ be the point where $l$ meets $\omega$, form the fact that $R_1$ and $R_2$ are perpendicular to $l$ we get that $G$ lies on $R_1R_2$. Thus because $P$ is the centre of the circle through $R_1R_2AA'$ and from equal tangents length we get that the shape is infact an isoscelles trapezium which implies that $R_1$ lies on the circle centred at $I$ through $A$ and that $R_2$ lies on the circle centred at $I$ and through $A'$.