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Source: Iranian TST 2022 problem 8

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#1 h Apr 2, 2022, 10:05 AM • 2 Y

Y by ImSh95, PHSH

In triangle $ABC$ , with $AB<AC$ , $I$ is the incenter, $E$ is the intersection of $A$ -excircle and $BC$ . Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$ . Point $Q$ lies on $BC$ such that $\angle AIQ=90$ . Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$ , circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$ . if $M$ is the Midpoint od the arc $BC$ , which does not contain $A$ , prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$ .

Proposed by Amirmahdi Mohseni

This post has been edited 2 times. Last edited by TheBarioBario, May 7, 2022, 3:57 PM

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#2 h Apr 6, 2022, 3:34 PM • 4 Y

Y by ImSh95, PHSH, PRMOisTheHardestExam, MehmetBurak

$[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 = -3.64, xmax = 19.02, ymin = -10.32, ymax = 5.14; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((5.017079622156162,-0.0702900807401852),0.6,42.858436552730765,96.22504243845378)--(5.017079622156162,-0.0702900807401852)--cycle, linewidth(1) + qqwuqq); draw(arc((9.287124516792435,-2.6616991692150336),0.6,129.03759007102846,182.40419595675158)--(9.287124516792435,-2.6616991692150336)--cycle, linewidth(1) + qqwuqq); draw(arc((6.977847723380699,3.270642374059604),0.6,-173.77495756154622,-120.4083516758231)--(6.977847723380699,3.270642374059604)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw(circle((6.22,-0.44), 3.7872417403698964), linewidth(1)); draw(circle((5.462152276619301,-4.150642374059604), 4.104554117726785), linewidth(1)); draw((5.017079622156162,-0.0702900807401852)--(8.83601638261342,3.4733258675802405), linewidth(1)); draw((3.3499012003832087,-2.9109781222095554)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((3.7807714588156975,1.223303560456394)--(9.51817699151674,-4.779949943924331), linewidth(1)); draw((4.68,3.02)--(2.5273396127849685,-1.2811059770878679), linewidth(1)); draw((4.68,3.02)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(4.68,3.02), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(5.512055018726986,1.9938748559082409), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(5.017079622156162,-0.0702900807401852), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((2.5273396127849685,-1.2811059770878679)--(9.51817699151674,-4.779949943924331), linewidth(1)); draw((5.512055018726986,1.9938748559082409)--(3.2149945745387947,-7.585412638186182), linewidth(1)); draw((3.3499012003832087,-2.9109781222095554)--(6.977847723380699,3.270642374059604), linewidth(1)); draw((5.462152276619301,-4.150642374059604)--(-0.4704346677023634,-0.6688517675314166), linewidth(1)); draw((4.68,3.02)--(8.83601638261342,3.4733258675802405), linewidth(1)); draw((4.68,3.02)--(5.462152276619301,-4.150642374059604), linewidth(1)); draw((3.2149945745387947,-7.585412638186182)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((5.512055018726986,1.9938748559082409)--(8.83601638261342,3.4733258675802405), linewidth(1)); /* dots and labels */ dot((4.68,3.02),dotstyle); label("$A$", (4.54,3.36), NE * labelscalefactor); dot((2.5273396127849685,-1.2811059770878679),dotstyle); label("$B$", (1.96,-1.6), NE * labelscalefactor); dot((9.287124516792435,-2.6616991692150336),dotstyle); label("$C$", (9.5,-2.78), NE * labelscalefactor); dot((5.462152276619301,-4.150642374059604),linewidth(4pt) + dotstyle); label("$M$", (5.66,-3.98), NE * labelscalefactor); dot((5.017079622156162,-0.0702900807401852),linewidth(4pt) + dotstyle); label("$I$", (5.38,-0.18), NE * labelscalefactor); dot((-0.4704346677023634,-0.6688517675314166),linewidth(4pt) + dotstyle); label("$Q$", (-1.04,-0.8), NE * labelscalefactor); dot((3.4398904867702496,2.1318069706821787),linewidth(4pt) + dotstyle); label("$G$", (3.1,2.42), NE * labelscalefactor); dot((6.977847723380699,3.270642374059604),linewidth(4pt) + dotstyle); label("$N$", (7,3.54), NE * labelscalefactor); dot((8.83601638261342,3.4733258675802405),linewidth(4pt) + dotstyle); label("$F$", (8.92,3.64), NE * labelscalefactor); dot((3.3499012003832087,-2.9109781222095554),linewidth(4pt) + dotstyle); label("$K$", (3.08,-3.36), NE * labelscalefactor); dot((3.7807714588156975,1.223303560456394),linewidth(4pt) + dotstyle); label("$X$", (3.46,1.4), NE * labelscalefactor); dot((5.512055018726986,1.9938748559082409),linewidth(4pt) + dotstyle); label("$Y$", (5.6,2.16), NE * labelscalefactor); dot((9.51817699151674,-4.779949943924331),linewidth(4pt) + dotstyle); label("$B_1$", (9.8,-4.96), NE * labelscalefactor); dot((3.2149945745387947,-7.585412638186182),linewidth(4pt) + dotstyle); label("$C_1$", (3.74,-7.6), NE * labelscalefactor); dot((7.76,-3.9),linewidth(4pt) + dotstyle); label("$D$", (7.7,-4.44), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $D$ be the antipode of $A$ in $\odot(ABC)$ , $N$ be the antipode of $M$ in $\odot(ABC)$ , $K$ be the touchpoint of $A$ -mixtilinear incircle of $\bigtriangleup ABC$ with $\odot(ABC)$ , and $G$ be the point on $\odot(ABC)$ such that $\angle AGI = 90^{\circ}$ (i.e. the so-called Sharky-Devil point). Let $QF$ cut $AB, AC$ at $X, Y$ respectively.

Claim 01. $Q, G, N, F$ are concyclic.
Proof. It's known that $A, G, Q$ are collinear (e.g. by considering the radical center of three circles $\odot(AI)$ , $\odot(ABC)$ , and $\odot(BIC)$ ). Since $\overline{IG} \perp \overline{AQ}$ , we obtain $AG \times AQ = AI^2$ . A property of mixtilinear touchpoint tells us that $K, I, N$ are collinear, and $\overline{AK}, \overline{AE}$ are isogonal with respect to $\angle BAC$ . Therefore, $\angle ANK = \angle ACK = \angle AEB = \angle AIF \implies \angle ANI = \angle AIF$ . Therefore, $\bigtriangleup AIN \sim \bigtriangleup AFI$ , i.e. $AI^2 = AN \times AF$ . Therefore, $AG \times AQ = AN \times AF \implies Q, G, N, F$ are concyclic. $\square$

Claim 02. $\overline{XI}$ is the external angle bisector of $\angle AXY$ , and similarly $\overline{YI}$ is the external angle bisector of $\angle AYX$ .
Proof. The inversion with center $A$ and radius $AI$ swaps $\{G, Q\}$ and $\{N, F\}$ . Therefore, $\odot(AGN) \equiv \odot(ABC)$ maps to $\overline{QF}$ under this inversion, i.e. it swaps $\{X, B\}$ and $\{Y, C\}$ . So, $AX \times AB = AI^2 \implies \bigtriangleup AXI \sim \bigtriangleup AIB$ . We also know that $\angle AXY = \angle ACB$ , since $AX \times AB = AY \times AC$ . By angle chasing, we obtain $\angle BXI = 180^{\circ} - \angle AXI = 180^{\circ} - \angle AIB = 180^{\circ} - (90^{\circ} + \frac{\angle ACB}{2}) = 90^{\circ} - \frac{\angle ACB}{2}$ . Therefore, $\overline{XI}$ is the external angle bisector of $\angle AXY$ . Similarly for $\overline{YI}$ . $\square$

Now, we will locate $B_1$ , the center of $\omega_b$ , and $C_1$ , defined similarly. Since $\omega_b$ is tangent to $\overline{AB}$ at $B$ , we must have $\overline{B_1B} \perp \overline{AB}$ . Moreover, since $\omega_b$ is also tangent to $\overline{FQ}$ , it must lie on the bisector of the angle formed by $\overline{FQ}$ and $\overline{AB}$ , and since it intersects the interior of $\bigtriangleup ABC$ , $B_1 \in \overline{XI}$ . This means $\angle IB_1B = \angle XB_1B = 180^{\circ} - 90^{\circ} - (90^{\circ} - \frac{\angle ACB}{2}) = \frac{\angle ACB}{2} = \angle ICB \implies B_1 \in \odot(BIC)$ . Similarly, $C_1 \in \odot(BIC)$ . Since $\angle B_1MC_1 = 2 \times \angle B_1IC_1 = 2 \times \angle XIY = 2 \times (90^{\circ} - \frac{\angle XAY}{2}) = 180^{\circ} - \angle BAC = \angle BMC$ , so $B_1C_1 = BC$ , i.e. $BB_1 = CC_1$ . Therefore,
$\text{Pow}(M, \omega_b) = B_1M^2 - R_{\omega_b}^2 = B_1M^2 - BB_1^2 = C_1M^2 - CC_1^2 = C_1M^2 - R_{\omega_c}^2 = \text{Pow}(M, \omega_c).$ So, $M$ lies on the radical axis of $\omega_b$ and $\omega_c$ , as desired. $\blacksquare$ .

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#3 h Apr 30, 2022, 3:55 PM • 3 Y

Y by ImSh95, PHSH, PRMOisTheHardestExam

Following three steps we can actually easily prove M lies on the radical axis of $\omega_b$ and $\omega_c$ .

Step 1

Step 2

Last Step

This post has been edited 9 times. Last edited by ashraful7525, Apr 30, 2022, 4:11 PM

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PHSH

#6 h Aug 11, 2022, 8:18 PM • 2 Y

Y by PRMOisTheHardestExam, GeoKing

Great problem!

The key observation was that line $FQ$ is tangent to the incircle. After I managed to prove this, the finish was natural!

Key Lemma: Line $FQ$ is tangent to the incircle of $ABC$
proof

Now let $U$ and $S$ be intersections of line $FQ$ with line $AB$ and $AC,$ respectively. Further, let $O_b$ and $O_c$ be the centers of $w_b$ and $w_c,$ respectively. From the Key Lemma it follows that $O_b$ is the intersection of line $UI$ with the perpendicular from $B$ to $AB,$ similarly $O_c$ is the intersection of line $VI$ with the perpedicular from $C$ to $AC.$

$[asy] import graph; import geometry; import olympiad; size(300); pair A,B,C,I,D,E,S,K,L,Lr,Ls,T,W,F,Q,H,P,J,Z,Y,G; A=dir(134); B=dir(204.87); C=dir(-24.87); I=incenter(A,B,C); D=foot(I,B,C); S=B*0.5+C*0.5; E=2*S-D; L=extension(A,I,B,C); Lr=2*I-L; Ls=bisectorpoint(L,Lr); Q=extension(I,Ls,B,C); J=foot(D,Q,I); K=2*J-D; H=2*I-D; Z=foot(D,S,I); G=2*Z-D; F=extension(S,G,Q,K); T=extension(F,I,A,E); W=extension(Q,K,A,I); Y=2*I-A; path w,e,f,g; w=incircle(A,B,C); e=circumcircle(I,L,E); f=circumcircle(A,I,G); g=circumcircle(A,W,T); pair U,V,O,Ar,Ob,Oc,By,Cy,Bx,Cx,M; U=extension(A,B,F,Q); V=extension(A,C,F,Q); O=(0,0); Ar=2*O-A; Ob=extension(B,Ar,U,I); Oc=extension(C,Ar,V,I); By=foot(C,A,I); Cy=foot(B,A,I); Bx=2*By-C; Cx=2*Cy-B; M=extension(O,S,A,I); path a,t; a=circumcircle(U,B,C); t=circumcircle(B,I,C); fill(A--B--C--cycle,0.1*cyan+0.9*white); draw(A--Bx); draw(A--C); draw(U--Ob,red); draw(V--Oc,red); draw(Cx--Oc,green); draw(Bx--Ob,green); draw(Q--C); draw(Q--F); draw(w,blue); draw(a,dotted); draw(t,magenta); dot("$A$",A,dir(A)); dot("$B$",B,dir(140)); dot("$C$",C,dir(0)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$I$",I,dir(65)); dot("$B'$",Bx,dir(Bx)); dot("$C'$",Cx,dir(35)); dot("$U$",U,dir(U)); dot("$V$",V,dir(90)); dot("$O_b$",Ob,dir(Ob)); dot("$O_c$",Oc,dir(Oc)); dot("$M$",M,dir(-35)); [/asy]$

Claim 3: Quadrilateral $UVBC$ is cyclic.
proof
Claim 4: Points $O_b,O_c$ lies on the circumcircle of $BIC,$ which has center $M.$
proof

To finish let $B'$ and $C'$ be the second intersections of $(BIC)$ with $AB$ and $AC,$ respectively. Combining $O_bB \perp AB, O_cC \perp AC$ with Claim 4, we get that $O_b$ and $O_c$ are the reflections of $B'$ and $C'$ with respect to $M,$ respectively. So we have by Pythagorean Theorem that the power of $M$ with respect to $w_b$ is
${MO_b}^2 - {O_bB}^2 = r^2 - (4r^2 - {BB'}^2) = {BB'}^2 -3r^2$ where $r$ is the radius of $(BIC).$ We have an analogous expresion to the power of $M$ with respect to $w_c;$ so that it is sufficient to show that $BB' = CC'.$ Indeed, we have trivially that $AB' = AC$ and $AC' = AB$ (say from $\sqrt{bc}$ inversion), so that $BB'=CC'=|AC-AB|$ . So done!

This post has been edited 4 times. Last edited by PHSH, Aug 12, 2022, 3:45 PM

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Shayan-TayefehIR

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Shayan-TayefehIR

#7 h Nov 7, 2022, 5:42 PM • 1 Y

Y by Mahdi_Mashayekhi

Firstly , we'll show that we have $\triangle FIQ \sim \triangle AEI_{a}$ which $I_{a}$ is $A$ -excenter of triangle $\triangle ABC$ . Suppose that $H$ is the foot of altitude from $A$ to $BC$ and let $AI$ meet line $BC$ at point $D$ , so we have $\triangle AIF \sim \triangle HEA$ and since quadrilateral $QIEI_{a}$ is cyclic , triangles $\triangle QID$ and $\triangle EI_{a}D$ are similar too. So one can see that :
$\frac{AE}{FI}=\frac{HE}{AI} , \frac{QI}{EI_{a}}=\frac{ID}{DE} (I)$ Now since by homothety of incircle and $A$ -excircle of triangle $\triangle ABC$ , $AE$ passes trough the antipode of $K$ ( touch point of incircle to $BC$ ) in the incircle , so if we put $H_{a}$ as the midpoint of the altitude $AH$ , then points $E , I , H_{a}$ are collinear and applying Menelaus theorem for these point in triangle $\triangle ADH$ , one can see that :
$\frac{AI}{ID}=\frac{EH}{DE} \implies \frac{AE}{EI_{a}}=\frac{FI}{IQ} , \angle FIQ=\angle AEI_{a}$ So we have $\angle FQI=\angle EI_{a}A=\angle IQD$ and the distance of the point $I$ from lines $BC$ and $FQ$ are equal. As the result , the incircle of triangle $\triangle ABC$ is tangent to $FQ$ and if $FQ$ intersects segments $AB$ and $AC$ at points $X$ , $Y$ respectively , then quadrilateral $BXYC$ is circumscribed and $I$ lies on the bisectors of angles $\angle YXB$ and $\angle XYC$ .
Now let $O_b$ and $O_c$ be points on the lines $IX$ , $IY$ such that we have $\angle XBO_b=\angle YCO_c=90$ , then obviously these points are centers of circles $\omega_b$ and $\omega_c$ . So we have $\angle XQB=2\angle IQD=\angle B-\angle C$ and also $\angle AXY=\angle C$ , so one can see that $\triangle AXY \sim \triangle ABC$ and while $I$ is the $A$ -excenter of triangle $\triangle AXY$ we have :
$\angle O_bII_a=\angle AIX=\frac{\angle AYX}{2}=\frac{\angle B}{2}=\angle O_bBI_a$ So quadrilateral $IBI_aO_B$ and similarly $ICI_aO_c$ are cyclic and both points $O_b$ , $O_c$ lies on the circumcircle of triangle $\triangle BIC$ with center $M$ , as the result we have $MO_b=MO_c$ and it's enough to show that equality $BO_b=CO_c$ holds.
Let $W_a$ be the touchpoint of $A$ -mixtilinear incircle with circumcircle of triangle $\triangle ABC$ and suppose that $QI$ intersects lines $AB$ , $AC$ at points $X'$ , $Y'$ respectively. Now since we know that $\angle CAE =\angle BAW_a$ , then one can see that :
$\frac{EC}{BE}=\frac{BW_{a}}{CW_{a}}.\frac{AC}{AB} (II)$ Now applying Menelaus theorem again for points $Q , X' , Y'$ and $Q , X , Y$ in triangle $\triangle ABC$ , while $AX'=AY'$ we can get :
$\frac{QB}{QC}=\frac{BX'}{CY'} \implies \frac{BX}{CY}=\frac{QB}{QC}.\frac{AX}{AY}=\frac{BX'}{CY'}.\frac{AC}{AB}$ Also a property of mixtilinear shows us that quadrilaterals $BX'IW_a$ and $CY'IW_a$ are cyclic which implies $\triangle BX'W_a \sim \triangle IY'W_a$ and $\triangle CY'W_a \sim \triangle IX'W_a$ , so since $IX'=IY'$ , according to $(II)$ one can see that :
$\frac{BX'}{CY'}=\frac{BW_a}{CW_a} \implies \frac{BX}{CY}=\frac{EC}{BE}$ $\triangle BXO_b \sim \triangle CEI_a , \triangle CYO_c \sim \triangle BEI_a \implies BO_b=CO_c$ So we're done.

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levifb

#8 h Dec 19, 2023, 2:31 PM

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Solved with @levimpcbranco

Let $T$ be the contact point of $(ABC)$ and the $A$ -Mixtilinear incircle and $N$ the midpoint of arc $BC$ that contains $A$ . It's well known ( $\sqrt{bc}$ inversion) that $\angle BAT = \angle BCT = \angle CAE \Rightarrow AEB = \angle ACT = \angle ANT = \angle ANI$ since $T$ , $I$ , $N$ are collinears. Hence $\angle AIF = \angle AEB = \angle ANI \Rightarrow AI^2 = AN \cdot AF$ .

Let $S$ be the $A$ -Chianca Point, i.e, intersection of $(ABC)$ with the circle of diameter $AI$ . Since $Q$ is the radical center of $(BIC)$ , $(ABC)$ and such circle, $Q$ , $S$ , $A$ are collinears and by the metric relationships, $AS \cdot AQ = AI^2 = AN \cdot AF \Rightarrow SNFQ$ is cyclic. Hence, $\frac{1}{2} \cdot (B-C) = \angle ASN = \angle AFQ = \angle FQI$ .

Let $X$ , $Y$ be the points where $\omega_b$ and $\omega_c$ touch $FQ$ and $R_b$ , $R_c$ the radius of $\omega_b$ and $\omega_c$ .

Main Claim: $\triangle IBX \sim \triangle ICY$

Proof

So $\frac{BX}{CY} = \frac{BI}{CI} = \frac{\sin \frac{C}{2}}{\sin \frac{B}{2}} \Rightarrow 2R_b = \frac{BX}{\sin \frac{C}{2} } = \frac{CY}{\sin \frac{B}{2}} = 2R_c$ .

Since the radius of the two circles are equal, their radical axis is the perpendicular bisector of $XY$ . Let $Z$ be the midpoint of $XY$ and $P$ the midpoint of $BC$ . For $MX=MY$ , it's suffice that $\angle MZQ = 90^\circ$ . In the current spiral similarity, we have $P \rightarrow Z$ , so $QZIP$ is cyclic, but also $QIPM$ due to $\angle MPQ = 90^\circ$ . Hence, $QZIPM$ is cyclic and $\angle QZM = \angle QPM = 90^\circ$ and we are done.

This post has been edited 1 time. Last edited by levifb, Dec 19, 2023, 2:33 PM

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662 posts

#9 h Mar 28, 2024, 3:44 PM

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Config soln:
Let the tangent from $Q$ to incircle meet $AB,AC,AM_{BC}$ at $U,V,F'$ , it's well known that $BCUV$ is bicentric, so we have $F \equiv F'$ by PoP at $A$ , and therefore we also have $FQ$ tangent to incircle, now clearly the $O_b \equiv UI \cap (BIC),O_c \equiv VI \cap (ABC)$ , for finishing just let $(BIC)\cap AB,AC = C',B'$ notice $B',M,O_B$ are collinear and by PoP, we're done.

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730 posts

#10 h Apr 28, 2025, 10:59 AM

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Let $T$ be $A-$ mixtilinear touch point, $N$ be the midpoint of arc $BAC$ . Let $QH$ be the tangent to the incircle which meets $AB,AC,AN$ at $Y,Z,W$ . Let incircle touch $BC$ at $D$ . Let $U,V$ be the circumcenters of $\omega_B,\omega_C$ .

Claim: $QF$ is tangent to the incircle (which is equavilent to $F=W$ ).
Proof: Let $K\in (ABC)$ such that $AK\parallel BC$ . Note that $T,D,K$ are collinear. Let $TL_1,TL_2$ be tangent to the incircle and $L_1,L_2\in (ABC)$ . By DDIT at $ABDC$ we get $(\overline{TA},\overline{TD}),(\overline{TB},\overline{TC}),(\overline{TL_1},\overline{TL_2})$ is an involution. Project onto $(ABC)$ to see that $L_1L_2\parallel BC$ . DDIT gives $(\overline{TH},\overline{TD}),(\overline{TQ},\overline{TQ}),(\overline{TL_1},\overline{TL_2})$ is an involution. Projecting onto $(ABC)$ implies $(TH\cap (ABC),K),(M,M),(L_1,L_2)$ is an involution hence $T,H,A$ are collinear. Notice that $W,A,H,I$ are concyclic.
$\measuredangle AIW=\measuredangle AHW=\measuredangle THQ=\measuredangle TIQ=\measuredangle TMA=\measuredangle BEA=\measuredangle AIF$ Which proves the claim.
Claim: $U,V\in (BIC)$ .
Proof: Let $E',F'$ be the tangency points of the incircle with $AC,AB$ . Since $U$ lies on $YI$ and $UB\perp AB$ ,
$\measuredangle BUI=\measuredangle BUY=90-\measuredangle IYB=\measuredangle YIH=\measuredangle F'DH=\frac{\measuredangle C}{2}=\measuredangle BCI$ Thus, $U$ lies on $(BIC)$ . Similarily $V\in (BIC)$ .
Claim: $UB=VC$ .
Proof:
$\measuredangle UCV=180-\measuredangle CBU-\measuredangle VCB-(90-\frac{\measuredangle A}{2})=180-\measuredangle A-90+\frac{\measuredangle A}{2}=90-\frac{\measuredangle A}{2}=\measuredangle BVC$ Hence $UB=VC$ .

We observe $Pow(M,\omega_B)=MU^2-UB^2=MV^2-VC^2=Pow(M,\omega_C)$ as desired. $\blacksquare$

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