Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

I. for one, distinctly remember posting two or three of these on the forum, but what the heck! They're nice problems; we can solve them all over again. I remember giving an inversive proof, but I can't find it anymore, so here's another one: Let $M$ be the circumcenter of $BCD$. We have $\angle BMC=2\angle BDC$, and we want to show that $\angle BMC=\angle BAC$, so it's equivalent to showing that $\angle BAC=2\angle BDC$. Let $C'$ be the intersection of $DC$ with $S_1$. The tangent through $C'$ to $S_1$ is parallel to $AC$ because $C'$ is the image $C$ through the homothety centered at $D$ which turns $S_2$ into $S_1$. Now let $T$ be the intersection of the tangents to $S_1$ through $B,C'$. We now have $\angle BAC=\pi-\angle BTC'=2\angle TC'B=2\angle BDC'=2\angle BDC$, Q.E.D.

The problem can be simplified (no arrangement conditions are necessary): Let A, B and C be three non-collinear points, and $S_1$ and $S_2$ two circles such that the circle $S_1$ touches the line AB at the point B, the circle $S_2$ touches the line AC at the point C, and the circles $S_1$ and $S_2$ touch each other at a point D. Prove that the circumcenter of the triangle BCD lies on the circumcircle of the triangle ABC. Here is my solution, using directed angles modulo 180: If P is the circumcenter of triangle BCD, then we have to prove that this point P lies on the circumcircle of triangle ABC. In other words, we have to show that < BPC = < BAC. What we know is < BPC = 2 < BDC (since the point P is the circumcenter of triangle BCD). If t is the common tangent to the circles $S_1$ and $S_2$ at their point of tangency S, then, since the tangents to a circle at the endpoints of a chord make equal angles with the chord, we have < (BD; t) = < (AB; BD) and < (CD; t) = < (AC; CD), so that < BDC = < (BD; CD) = < (BD; t) - < (CD; t) = < (AB; BD) - < (AC; CD) = < (AB; AC) + < (AC; BD) - (AC; CD) = < (AB; AC) + < (CD; BD) = < BAC + < CDB = < BAC - < BDC. Thus, 2 < BDC = < BAC, and hence < BPC = < BAC, as desired. $\blacksquare$ Darij

grobber wrote: I. for one, distinctly remember posting two or three of these on the forum, but what the heck! They're nice problems; we can solve them all over again. Well, grobber, I told that you shall post the links for the ISL problems in the section where I and Peter VDD did it. I do not search the forum all the time. And from time to time these postings are not adapted to LaTeX. Rather than searching and adopting all threads I repost it. The same for Russian Olympiad and Kvant problems. I really appreciate it that you post these nice problems but please collect the links an extra thread for convenience and to prevent people from reposting them. Thank you. Other suggestions ?

Here is the first solution: Name ,(the tangent $AB$),(the common tangent of the two circles from $D$),and( the tangent $AC$), $T_1,T_2,T_3$ , respectively. Consider aline which is perpendicular, to $BD$, and also bisects it. The points which are on this line, have equal distance from $T_1,T_2$, Consider aline which is perpendicular, to $CD$, and also bisects it. The points which are on this line , have equal distance from $T_2,T_3$. Then the intersection of these two lines,$O$, have equal distance from $T_1,T_3$. And we conclude that $O$ lies on the bisector of $\angle BAC$. But $O$ lies on line which is perpendicular to $BC$, and also biscts it. Then it lies on the circum circle of triangle $BAC$.

The second solution: Consider aline which is perpendicular, to $BD$, and also bisects it. Consider aline which is perpendicular, to $CD$, and also bisects it. And name their intersection, $O$. We should say that: $180=\angle A+\angle BOC$. Ok…………………………………………………………………………………………………. $\angle BDC=X$ then : $\angle BOC=2\angle EOF=2(180-X)=360-2X$. we should know that if, $\angle BOC=360-2X+\angle A=180$, holds or not. Suppose it holds, then: We draw a tangent at $S$ which is parallel to $AC$. Then,$AC,A'M$ are parallel And, $\angle A=\angle A'= \frac{%Error. "arcBDM" is a bad command. -%Error. "arcBM" is a bad command. }{2}$. $%Error. "arcBDM" is a bad command. =360-%Error. "arcBM" is a bad command. $ $\angle A=180-%Error. "arcBM" is a bad command. +180=360-%Error. "arcBM" is a bad command. =2X$ $X=180-\frac{%Error. "arcBM" is a bad command. }{2}$ and this fact is true, because of the quality of homotechy,$F,D,M$ are collinear. (NOTE:$D$ is the inner homotechy center , and a tangent at$F$, to circle$S'$, is parallel to the tangent at $M$ , to the circle $S$. and finally: $\angle BDC=180-\angle BDM=180-\frac{%Error. "arcBM" is a bad command. }{2}$. And the second one is complete,too.

Now , I choose an $INVERSIVE$ proof for the third solution. Really, and absolutely more nice than the two recent solution, I think. Lets invert the shape round point $B$,and an arbitrary radius. Now every thing which I will say , is about the inversive form of the shape. It is clear that each point, for example $X$, turns into its inversive form,$X'$. I should prove that, the center of circumcircle of triangle $BDC$, lies on circle $CAB$. Or the other words, in the inversive shape, the inversion of the center of this circle, Lies on the inversion of circle $CAB$. The inversion of circle $CAB$, is $AC$, and the inversion of circle$BDC$, is $DC$. The inversion of the center of this circle is a point$B'$,with reflect point $B$, at line $CD$. Then I should prove that: $B'$ lies on $AC$ ………………………………………………………………………………………………………………………... and to do that, if $AC,BH$ meet at $O$, and if $X=Y$, the problem is solved. $X=180-\angle BCP=\frac{%Error. "arcBCP" is a bad command. }{2}$ $Y=\angle PCA=\frac{%Error. "arcAP" is a bad command. }{2}$ we know that $P$ is the midpoint of the arc.because with center $C$, $P,D$ are two homologues point. And finally: $\overarc{BCP}=%Error. "arcAP" is a bad command. $ and every thing is ok. It was really an enjoyable problem,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,. I will think about the fourth solution.

You are so COOL ashegh. Excelent Work

saeid wrote: You are so COOL ashegh. Excelent Work ure wellcome my darling

Notation: (see diagram as well) Let $ O$ be the circumcenter of $ \triangle BCD$. Let $ BD$ intersect $ S_2$ again at $ D$ and let $ CD$ intersect $ S_1$ again at $ F$. Let $ AC$ intersect $ BF$ at $ G$. Let $ \angle BDF=x$. (1) $ \angle BOC=2x$. Proof: $ 2x=2\angle BDF=2\pi-2\angle BDC=(\pi-2\angle BDO)+(\pi-2\angle CDO)$. Since $ BO=CO=DO$ $ =\angle BOD+\angle DOC=\angle BOC$. (2) $ BF\parallel CE$. Proof: Since $ S_1$ and $ S_2$ are tangent at $ D$, there is a homothety that maps $ S_1$ to $ S_2$ centered at $ D$. It is easy to see that $ BF$ goes to $ CE$. Since a homothety maps a line to another parallel line, we get the conclusion. (3) $ \angle ACE=\pi-x$. Proof: $ \angle ACE=\angle ACD+\angle DCE=\angle CED+\angle DCE$ by the tangent secant angle theorem in $ S_2$. $ =\pi-\angle CDE=\pi=\angle BDF=\pi-x$. (4) $ \angle BAC=\pi-2x$. Proof: $ \angle BAC=\angle AGF-\angle ABG=\angle ACE-\angle BDF=\pi-2x$ by the tangent secant angle theorem in $ S_1$, the parallel lines in (2), and (3). (5) $ ABOC$ is cyclic. Proof: $ \angle BAC+\angle BOC=\pi-2x+2x=\pi$ by (1) and (4) so $ ABOC$ is cyclic. In other words, $ O$ lies on the circumcircle of $ \triangle ABC$, as desired.

Another solution. Define $ M$ be $ CD \cap S_{1}$, and $ AC$ meet $ S_{1}$ at $ E, F$, $ AF$ and $ BM$ meet at $ X$. $ A,D,O$(circumcenter of $ \triangle BCD$)$ B$ are cyclic $ <=> \angle ABO=\angle OCE <=> \frac{\pi}{2}=\angle BXC+\angle OBC <=> X B C D cyclic.$ And using homothety, we easily know $ \angle XCD=\angle XBD$ so $ A D O B$ cyclic. Sorry for my poor english...

Somewhat wierd solution, but I thought it was interesting. Let $ O_1$ be the center of the circle containing $ B$, and $ O_2$ be the center of the circle containing $ C$. Let $ A'$ be the intersection between $ O_1B$ and $ O_2C$. $ A'BCA$ is cyclic, so it suffices to show the circumcenter lies on the circumcircle of $ A'BC$. The perpendicular bisectors of $ BD$ and $ CD$ are the angle bisectors of $ \angle{O_2O_1A'}$ and $ \angle{O_1O_2A'}$, so the incenter of $ O_1O_2A'$ is the circumcenter of $ BCD$. Let this point be $ I$. $ A'I$ is thus the angle bisector of $ \angle{O_1A'O_2}$, and $ BIC$ is isoceles. If $ A'I$ is not the perpendicular bisector of $ BC$, we are done. If $ A'I$ is the perpendicular bisector of $ BC$, then $ B,C,D$ are the points of tangency of the incircle to the sides since $ A'B=A'C$, and so $ \angle{A'BI}=\angle{A'CI}=90$, and so we are done.

I think all the solutions above just complicate the problem...

Let $O$ be the circumcentre of $\triangle BDC$ and let $P$ be the intersection of the common tangent to $S_1$ and $S_2$ (passing through $D$) and $AC$. Let the midpoints of $BD,CD$ be $M,N$ respectively. Let $PD$ and $BA$ intersect at $K$. Obviously $D$ is the reflection of $B$ through the line $OK$. Then $CD$ is the polar of $P$ wrt $S_2$ so $O,P,N$ are collinear (all lie on the perpendicular bisector of $CD$). So by trig ceva on $\triangle ODC$ obviously $\angle ODP=\angle PCO$ (lol user pco !). Now $\angle ACO=\angle PCO=\angle ODP=180^{\circ}- \angle KDO=180^{\circ}-\angle KBO=180^{\circ}-\angle ABO$, so $ABOC$ is cyclic.

The only solution without defining new lines or points. Let the centers of $~$ $S_1 \wedge S_2$ $~$ be $~$ $O_1 \wedge O_2$ $~$ respectively, and $~$ $O$ $~$ be the circumcenter of $~$ $\triangle{BDC}. $

Here's my solution Draw the common tangent to $S_1 $ and $S_2$ at the point $D$ . Let the tangent intersect $AC$ at $M$ and $AB$ at $N$. Now, \[\angle{AMN} = \angle{NBD}+\angle{NDB}\] Thus, \[\angle{AMN}=2\angle{NDB}\]. Similarly we can arrive at $\angle{ANM}=2\angle{CDN}$ Adding these two we get \[\angle{ANM}+\angle{AMN}=2(\angle{CDN}+\angle{NDB})\] Thus , $180^\circ-\angle{A}=2\angle{BDC}$ Therefore we can conclude that the angle subtended at the circumcentre of triangle BDC by the chord BC is $180^\circ - \angle{A}$ and thus we conclude that the circumcentre of triangle BDC lies on circumcircle of triangle ABC

Notations::$BC \cap C_1=J,BD \cap C_2=K,\angle{CBD}=p,\angle{BDJ}=x,\angle{CKD}=q$.Also let $O$ be the circumcenter of $\triangle{BCD}$. Proof:Easy to see that $\angle{JDC}=p+q$.So $\angle{BDC}=x+p+q \Rightarrow \angle{BOC}=360-2x-2p-2q$. Also note that $\angle{CBA}=180^{\circ}-x$. Now $\angle{BCA}=\angle{BCD}-\angle{DCA}=\angle{BCD}-\angle{DKC}=180-x-2p-q-q=180-x-2p-2q$. So we get that $\angle{BAC}=2x+2p+2q-180$.It readily follows that $A,B,O,C$ are concyclic and we are done.$\blacksquare$.

Let $AC$ intersect $S_1$ in M then N . $S_1$ and $S_2$ are tangeant externally , and MN is a chord of $S_1 $ tangeant to $S_2$ at C. thus DC is external bissector of $\angle MDN$ . now we may start angle chasing: \[\angle COB =2\angle CDB =2\angle CDM +2\angle MNB = \pi -\angle MDN+2\angle ANB =\angle MBN+ \angle ABM +\angle ANB = \pi -\angle CAB \] QED

Let $O$ be the circumcenter of $BDC$. Let $\angle BDC=x$, so $\angle BOC=360-2x$ and we want to show $\angle BAC=2x-180$. Let $E$ and $F$ the intersections of $AC$ with $S_1$, such that $E$ is closer to $A$ than $F$ is. Let $CD$ intersect $S_1$ again at $M$. Then $\angle BAC=\angle BAF=\angle BDF-\angle BDE$. Since $x=\angle BDC=\angle BDF+\angle FDC$. Combining the two, we need to show that $\angle BDE+\angle FDC+\angle BDC=180$, or that $\angle MDE=\angle FDC=\angle MEF$. Hence, we want to show that $M$ is the midpoint of arc $EF$ not containing $B$. Now, let lines $ED$ and $FD$ intersect $S_2$ again at $X$ and $Y$. Note that $EF||XY$ by homothety, so $C$ is the midpoint of arc $XY$ on $S_2$. Hence $M$ is the midpoint of arc $EF$ on $S_1$.

I am the new definition of overkill Let $O_1,O_2$ be the centers of $S_1,S_2$ respectively and $O$ be circumcenter of $\triangle BCD.$ It suffice to show that $ABCO$ is cyclic. Note that \[\angle BAC=180^\circ-\angle ABC-\angle ACB\]\[=180^\circ-\angle ABC-\angle ACB\]\[=180^\circ-\angle ABD-\angle CBD-\angle ACB\]\[=180^\circ-\angle CBD-90^\circ+\angle O_1DB-\angle ACB\]\[=90^\circ-\angle CBD+\angle O_1DB-\angle ACB\](Here comes the leap of faith step) \[=180^\circ-\angle DCA-\angle O_2CD-\angle CBD-\angle ACB+\angle O_1DB\]\[=180^\circ-\angle DCB-\angle CBD-\angle O_2CD-\angle O_1DB\]\[=\angle CDB-180^\circ+\angle CDB\]\[=-180^\circ+2\angle CDB\]\[=180^\circ-\angle COB\]so we are done.

Let $O,O_1,O_2$ be center of circles $(BCD),S_1,S_2$, respectively. Note that $O_1-D-O_2$ are collinear. Since $O_2C=O_2D$ and $OC=OD$, we have $\triangle O_2CO\cong \triangle O_2DO$. \begin{align*} \angle O_2DO &= \angle O_2CO \\ &= \angle O_2CA+\angle ACO \\ &= 90^{\circ}+\angle ACO \end{align*}Similarly, $O_1B=O_1D$ and $OB=OD$ so $\triangle O_1DO\cong \triangle O_1BO$. \begin{align*} \angle O_1DO &= \angle O_1BO \\ &= \angle ABO-\angle ABO_1 \\ &= \angle ABO-90^{\circ} \end{align*} Hence, we have \begin{align*} 180^{\circ} &=\angle O_2DO+\angle O_1DO \\ &= (90^{\circ}+\angle ACO)+(\angle ABO-90^{\circ}) \\ &= \angle ACO + \angle ABO. \end{align*}Thus, $A,B,C,O$ concyclic, as desired.

Let $O_1,O_2,O$ be centers of $S_1,S_2,\odot (BCD).$ Clearly $D\in O_1O_2,O_1B\perp AB,O_2C\perp AC,$ so $$\measuredangle BAC=\angle (BO_1,CO_2)=\measuredangle BO_1D+\measuredangle DO_2C=2\measuredangle BDC=\measuredangle BOC\implies O\in \odot (ABC).$$

[asy][asy] size(10cm); import geometry; draw(circle((-1,0),1)); draw(circle((2,0),2)); circle c1 = circle((point)(-1,0),1); circle c2 = circle((point)(2,0),2); point B = 2*dir(65)+2, C = dir(85)-1, D = (0,0); point E = intersectionpoint(tangent(c1,D),tangent(c1,C)); point F = intersectionpoint(tangent(c2,D),tangent(c2,B)); point A = intersectionpoint(tangent(c2,B),tangent(c1,C)); dot("$B$", B, dir(60)); dot("$C$", C, dir(60)); dot("$D$", D, dir(-15)); dot("$E$", E, dir(55)); dot("$F$", F, dir(60)); dot("$A$", A, dir(60)); draw(C--A--F--D); draw(C--D--B); [/asy][/asy] Let $\ell$ be the common internal tangent to both circles, and let $E = AC \cap \ell$ and $F = AB \cap \ell$. Let $O$ be the circumcenter of $BCD$. To prove that $O \in (ABC)$, we only need to show that $\measuredangle COB = \measuredangle CAB$ (all angles directed). However, $\measuredangle COB = 2 \measuredangle CDB$ (circumcenter angles). Furthermore, we have: \begin{align*} \measuredangle CAB &= \measuredangle EAF \\ &= -\measuredangle AFE - \measuredangle FEA \\ &= - \measuredangle BFD - \measuredangle DEC \\ &= \measuredangle FDB + \measuredangle DBF + \measuredangle ECD + \measuredangle CDE \\ &= 2(\measuredangle CDE + \measuredangle FDB) \\ &= 2\measuredangle CDB \end{align*}which finishes.

Does this work? (I think the part where $\measuredangle BO_1D=2\measuredangle BDO_1$ might be incorrect.) Let the center of $S_1$ be $O_1$ and let the center of $S_2$ be $O_2$. Also let the circumcenter of $BCD$ be $O$. Then let $BO_1$ and $CO_2$ intersect at $A'$. Then we have \begin{align*} \measuredangle BAC&=\measuredangle BA'C \\ &=\measuredangle A'O_1O_2 + \measuredangle O_1O_2A' \\ &=\measuredangle BO_1D+\measuredangle DO_2C \\ &=2(\measuredangle BDO_1+\measuredangle O_2DC) \\ &=2\measuredangle BDC \\ &=\measuredangle BOC \end{align*}and we are done. $\blacksquare$

OKAY so there is apparently config issues. aint no way im actually gonna fix the proof

Let $O_1$ denote the center of $S_1$ and $O_2$ be defined similarly. Let $O$ be the circumcenter of $BCD$. The key claim is that Claim: $\measuredangle OBO_1 = \measuredangle O_1DO$. Proof: We know that $\measuredangle O_1BD = \measuredangle BDO_1$ and $\measuredangle OBD =\measuredangle BDO$ by isosceles triangles. Subtracting gives the result. It is sufficient to prove that $\color{blue} OBAC$is cyclic. The rest is straightforward angle chasing. \begin{align*}\measuredangle OBA &= \measuredangle O_1DO + 90^{\circ} \\ &= \measuredangle O_2DO + 90^{\circ} \\ &= \measuredangle CDO + \measuredangle O_2DC + 90^{\circ} \\ &= \measuredangle OCD + \measuredangle DCO_2 + 90^{\circ} \\ &= \measuredangle OCO_2 + 90^{\circ} \\ &= \measuredangle OCA,\end{align*}and we're done. $\square$

Let $I$ be the incenter of $ABC$, segment $AC$ meet $S_1$ again at $Y$ and $Z$ so that $AY < AZ$, the common tangent to $S_1$ and $S_2$ meet $AC$ at $M$, segment $BC$ meet $S_1$ again at $X$, the midpoint of arc $YZ$ not containing $D$ be $N$, the reflection of $C$ over $M$ be $E$, and $B_1$ be the point on $S_1$ such that $BB_1 \parallel YZ$. The parallel condition implies $N$ is also the midpoint of arc $BB_1$. Because $ME = MC = MD$ follows from equal tangents, we have $\angle CDE = 90^{\circ}.$ Now, observe $$MC^2 = |Pow_{S_2}(M)| = MD^2 = |Pow_{S_1}(M)| = MY \cdot MZ$$so $-1 = (C, E; Y, Z)$. Thus, the Right Angle and Bisectors Lemma implies $DE$ bisects $\angle YDZ$, which means $D, E, N$ are collinear. Now, we know $$\angle BDB_1 = 180^{\circ} - \angle BNB_1 = 180^{\circ} - \angle BNX - \angle XNB_1$$$$= 180^{\circ} - \angle ABX - \angle XBB_1 = 180^{\circ} - \angle ABC - \angle ACB = \angle BAC.$$This gives $$\angle BDC = \angle BDN + \angle NDC = \frac{\angle BDB_1}{2} + \angle EDC$$$$= \frac{\angle A}{2} + 90^{\circ} = \angle BIC$$which finishes via the Incenter-Excenter Lemma. $\blacksquare$ Remark: This solution is a bit overkill.

Let $O_1$ be the center of $S_1$, $O_2$ be the center of $S_2$, and $O$ be the circumcenter of $BCD$. We have \begin{align*} \measuredangle BOC &= 2 \measuredangle BDC \\ &=( \measuredangle BCD + \measuredangle DBC ) \\ &=2( \measuredangle BCA + \measuredangle ACD + \measuredangle DBA + \measuredangle ABC ) \\ &= 2(\measuredangle BAC + \measuredangle DBA + \measuredangle ACD) \\ &= 2 \measuredangle BAC + 2 \measuredangle DB O_1 + \measuredangle CO_2D \\ &= 2 \measuredangle BAC + \measuredangle DO_1B + \measuredangle CO_2D \\ &= 2\measuredangle BAC + \measuredangle CBO_1 + \measuredangle O_2CB \\ &= 2 \measuredangle BAC + \measuredangle CBA - 90 + \measuredangle ACB +90 \\ &= \measuredangle BAC \\ \end{align*}This means $B,A,O,C$ are concyclic as desired so we are done.

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Let $O$ be the circuncenter of $BCD$. It would be sufficient to prove that $360^{\circ}-2\angle CDB=\angle DOC=180^{\circ}- \angle BAC$ $\iff 180^{\circ}=2\angle CDB - \angle BAC$ Now invert at $C:$ We need $180^{\circ}=2\angle C'B'D' - \angle A'B'C' \iff 90^{\circ} = \angle C'B'D' - \frac{\angle A'B'C'}{2}$ Let the line parallel to the tangent at $D'$ through $B'$ meet the circles $(B'D'), (B'A'C')$ at $E, F$. Let M be the midpoint of the arc $A'C'$ not containing $B'$ $A'C' \parallel EF \Rightarrow \angle EB'D' + \angle C'B'M = \angle D'B'M = 90 $ But $90^{\circ}=\angle D'B'M = \angle C'B'D' - \frac{\angle A'B'C'}{2}$

Nice angle chasing. We have <BFC=360-2<BDC=2(<BCD+<CBD)=2(<ACB+<ACD+<ABC-<ABD)=2(180-<BAC+<ACD-<ABD)=360-2<BAC+<CED-<BOD (tangent angle rules)=... I'm not sure how to continue, I ask someone to post a solution like this. Edit: I saw lrjr's but i didnt get exactly how he got the third to last line from the fourth Anyways, second attempt: First, we have <FBD=<FDB, <ODB=<OBD. Then we have <FBO=<FDO. We have <FBA=90+<FBO=90+180-<EDF=180+90-<ECF=180-<FCA, so ABCF is cyclic, as desired. $\blacksquare$ Edit: After comparing with others, it seems like directed angles are necessary, but I don't want to do them right now.

only angle chasing!

I HAD POSTED SOLUTION IN PHOTO CHECK IT

Draw in the circle centers $O_1, O_2$, let $O_1DO_2$ meet $S_1$ again at $D'$, we then see that $\angle CAB = \angle DO_2C - \angle D'O_1B$ (the condition on $C$ guarantees this is positive) by comparing arguments of lines. Now $\angle CDB = 180 - \frac 12 (180 - (180 - \angle D'O_1 B)) - \frac 12 (180 - \angle DO_2C) = 90 + \frac 12 (\angle DO_2C - \angle D'O_1B)$. Since this is guarantee to be obtuse, we can conclude that the desired circumcenter $O_3$ is on the opposite side of $BC$ as $D$, and $D$ is on the same side of $BC$ as $A$, so we it remains to check $\angle CO_3B + \angle CAB = 180$, which is trivial since $CO_3B = 2(180 - \angle CDB) = 180 - (\angle DO_2C - \angle D'O_1B)$ as desired.