apparently circles have two intersections :'(

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apparently circles have two intersections :'(

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Source: 2020 USOJMO Problem 2

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554183

#65 h Sep 13, 2021, 3:31 PM

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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.

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Reason: Latex blunder

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#66 h Sep 13, 2021, 3:46 PM

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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 :wacko:

:wacko:

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#67 h Nov 27, 2021, 7:23 PM

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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] /* 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$", (-0.8220863299424661,-3.45787850227881), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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.

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#68 h Jan 11, 2022, 2:03 AM

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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 $\omega^*$ .

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#69 h Mar 7, 2022, 5:46 PM

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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 $I$ .

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2963 posts

#70 h Mar 13, 2022, 3:40 AM

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How many points would missing the other case get?

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#71 h Mar 13, 2022, 10:24 PM • 1 Y

Y by centslordm

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 $B_1=\overline{BE}\cap\odot(I,2r).$ $\square$

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#72 h Mar 19, 2022, 5:29 AM

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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 $I$ .

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OronSH

#73 h Jan 25, 2024, 3:42 AM

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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.

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#74 h Mar 10, 2024, 3:29 AM

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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.

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#75 h Mar 16, 2024, 7:11 AM

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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.

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#76 h Jun 7, 2024, 9:38 PM

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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. $\blacksquare$

$[asy] size(300); defaultpen(linewidth(0.7)+fontsize(11)); pair I, A, B, C, M, N, X, P, Q, R1, R2; I = (0,0); A = dir(210); B = dir(330); C = dir(90); M = .5*B + .5*C; N = .5*C + .5*A; X = .5*dir(95); P = extension(B,C,X,rotate(90,X)*I); Q = extension(C,A,X,rotate(90,X)*I); R1 = 4*X; R2 = -2*X; draw(arc(I, 2, 30, 150)^^arc(I, 1, 210, 330), green+dashed+linewidth(1)); draw(A--B--C--cycle^^circle(I, .5), gray+linewidth(0.2)); draw(P--A--M--cycle^^Q--B--N--cycle^^P--Q^^R1--R2); draw(P--A^^P--R1^^P--R2^^Q--B^^Q--R1^^Q--R2); draw(P--A^^P--R1^^P--R2, red); draw(Q--B^^Q--R1^^Q--R2, blue); dot("$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$", R2, dir(270)); [/asy]$

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N3bula

#77 h Oct 17, 2024, 10:49 AM

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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 $A'$ .

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#78 h Feb 11, 2025, 5:52 AM

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flashsonic wrote:

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

Bruh, I definitely didn't just misread it in the exact same way :whistling:

:whistling:

:whistling:

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#79 h Mar 16, 2025, 6:05 PM

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The answer is the minor arc $AB$ in circumcircle $(ABC),$ and its $-2\times$ homothety centered at $C.$
Let $T$ be the tangency point of $\ell$ and $\omega.$
Let $D$ be the tangency point of $BC$ and $\omega.$
Let $R’$ be the intersection of $TI$ and $(ABC)$ along minor arc $AB.$
We claim $R’ = R.$
We have that $TQ = TD, AD = AI + ID = IR’ + IT = TR’.$
Using pythagorean theorem, we get $R$ ’s desired traits.
For the other point, do the same thing.

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