Everything you need to do is construct a parallelogram .

My solution: Lemma: $AT \parallel RK$. Proof: We have $\angle{JAT} = 180^{\circ} - \angle{JST} = \angle{JSR} = \angle{JKR}$, which implies the desired conclusion. Now suppose that the circumcircle of triangle $SKT$ meets $RK$ again at $X$. I claim that $X,S,A$ are collinear. To prove this, we will instead show that $RXTA$ is a parallelogram; we already know that $RX \parallel TA$, so now it suffices to show that $RA \parallel XT$. But this follows from $\angle{STX} = \angle{SKX} = \angle{ART}$, with the proof in other configurations following in a similar manner. Thus $\angle{KTR} = \angle{RXA} = \angle{SAT}$, implying that $KT$ is tangent to $\Omega$, so we are done.

Pretty easy for a 4, though despite geo being my strongest subject I think that this is harder than 1. Note that since $AJST$ is cyclic we have $\angle ATS = \angle SJK = \angle SRK$, so $RK\parallel AT$. Let $P$ be the point such that $RATP$ is a parallelogram, so that $A-S-T$ and $R-K-P$ are collinear. Now since $\angle SKR = \angle TRA$ we have $\triangle SKR\sim\triangle ART$, and so \[\dfrac{RK}{RS} = \dfrac{2RS}{AT}\quad\implies\quad RS\cdot RT = RK\cdot AT = RK\cdot RP.\]Thus $\odot(PKST)$ is a cyclic quadrilateral, which means that \[\angle KTS = \angle KPS = \angle SAT,\]implying the tangency.

Let $ M $ be the midpoint of $ KR. $ By Reim's theorem $ \Longrightarrow $ $ AT $ $ \parallel $ $ KR $ $ \Longrightarrow $ $ \measuredangle ATR $ $ = $ $ \measuredangle KRS, $ so combining $ \measuredangle TRA $ $ = $ $ \measuredangle SKR $ we get $ \triangle ART \cup S $ $ \stackrel{-}{\sim} $ $ \triangle SKR \cup M, $ hence $ \measuredangle STK $ $ = $ $ \measuredangle RSM $ $ = $ $ \measuredangle SAT. $ i.e. $ KT $ is tangent to $ \Gamma. $

EDIT: Michael Kural points out that the same parallelogram configuration appears in ELMO 2015. First Solution (elementary): First, note \[ \measuredangle RKA = \measuredangle RKJ = \measuredangle RSJ = \measuredangle TSJ = \measuredangle TAJ = \measuredangle TAK \]so $\overline{RK} \parallel \overline{AT}$. Now, $\overline{RA}$ is tangent at $R$ iff $\triangle KRS \sim \triangle RTA$ (oppositely), because both equate to $-\measuredangle RKS = \measuredangle SKR = \measuredangle SRA = \measuredangle TRA$. Similarly, $\overline{TK}$ is tangent at $T$ iff $\triangle KTS \sim \triangle ART$. The two similarities are equivalent because $RS = ST$ the SAS gives $KR \cdot TA = RS \cdot RT = TS \cdot TR$. [asy][asy] size(10cm); pair T = dir(100); pair S = dir(165); pair R = 2*S-T; pair E = dir(0); pair A = IP(unitcircle, R--(R+100*E)); pair B = OP(unitcircle, R--(R+100*E)); pair K = extension(T, T+dir(90)*T, R, R+T-A); draw(R--B, blue); draw(unitcircle, blue); draw(circumcircle(R, S, K), heavycyan); pair J = -A+2*foot(origin, A, K); draw(K--R--A--T--cycle, red); draw(A--K, orange); draw(R--T, orange); // invert pair Js = extension(T, T+A-B, R, J); pair Ks = extension(T, T+A-B, R, K); draw(K--Ks, heavygreen); draw(Ks--Js, heavygreen); draw(Js--B, dotted+blue); draw(R--Js, dotted+blue); dot("$T$", T, dir(T)); dot("$S$", S, dir(355)); dot("$R$", R, dir(270)); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$K$", K, dir(200)); dot("$J$", J, dir(180)); dot("$J^\ast$", Js, dir(Js)); dot("$K^\ast$", Ks, dir(Ks)); /* TSQ Source: !size(10cm); T = dir 100 S = dir 165 R355 R = 2*S-T R270 E := dir 0 A = IP unitcircle R--(R+100*E) R225 B = OP unitcircle R--(R+100*E) R315 K = extension T T+dir(90)*T R R+T-A R200 T--K blue R--B blue unitcircle 0.05 lightcyan / blue circumcircle R S K 0.05 paleblue / heavycyan J = -A+2*foot origin A K R180 K--R--A--T--cycle 0.1 lightred / red A--K orange R--T orange // invert J* = extension T T+A-B R J K* = extension T T+A-B R K K--Ks heavygreen Ks--Js heavygreen Js--B dotted blue R--Js dotted blue */ [/asy][/asy] Remark: The problem is actually symmetric with respect to two circles; $\overline{RA}$ is tangent at $R$ if and only if $\overline{TK}$ at $T$. Second Solution (inversion): Consider an inversion at $R$ fixing the circumcircle $\Gamma$ of $TSJA$. Then: $T$ and $S$ swap, $A$ and $B$ swap, where $B$ is the second intersection of $\ell$ with $\Gamma$. Circle $\Omega$ inverts to the line through $T$ parallel to $\overline{RAB}$, call it $\ell$. $J^\ast$ is the second intersection of $\ell$ with $\Gamma$. $K^\ast$ is the intersection of $\ell$ with the circumcircle of $RBJ^\ast$; this implies $RK^\ast J^\ast B$ is an isosceles trapezoid. In particular, one reads $\overline{RK^\ast} \parallel \overline{AT}$ from this, hence $RK^\ast TA$ is a parallelogram. Thus we wish to show the circumcircle of $RSK^\ast$ is tangent to $\Gamma$. But that follows from the final parallelogram observed: $S$ is the center of the parallelogram since it is the midpoint of the diagonal. Remark: This also implies $RKTB$ is cyclic, from $\overline{K^\ast SA}$ collinear.

My solution for this problem.

$\angle KRS=\angle KJS=\angle RTA$ so $KR$ is parallel to $AT$. So $AT$ cuts $KS$ at $D$ then $KRDT$ is a parallelogram. $\angle ARS=\angle RKS=\angle KDT$ so $R,A,D,S$ are cyclic. $\angle SAT=\angle SRD=\angle KTS$, we have $KT$ is tangent to $\Gamma$

My solution: Let $U=RK\cap (SKT).$ We know $\angle ATS=\angle SJK=\angle SRK\to RK\parallel AT.$ Also we know $\angle RAT=\angle RAJ+\angle JAT=\angle RAJ+\angle RSJ=\angle RAJ+\angle JRA=\angle RJK=\angle RSK=\angle KUT\to DUTA$ is parallelogram. Then $A,S,U$ collinear since $S$ is midpoint of $RT.$ Then $\angle KTS=\angle RUS\angle SAT.$ As desired.

From angle chasing we get $\triangle ART \sim \triangle SKR$ . Define point $X$ ,such that $A$ is the midpoint of segment $XT$. From the similar triangles we get that $\measuredangle RXT= \measuredangle KTS $ , but $AS$ is midline in $\triangle RXT \Rightarrow AS||XR \Rightarrow \measuredangle SAT= \measuredangle RXT=\measuredangle KTS $ .Therefore $KT$ is tangent to $\Gamma$.

A very easy question If you are lucky enough to read it correct (unlike me who read TK tangent \(\Omega\) )then three minutes.

It is probably the easiest problem in IMO history.My solution is just constructing parallelogram. The configuration is too similar to ELMO 2015 P3. I think there are about 500 14 points in this year´s IMO.

Murad.Aghazade wrote: It is probably the easiest problem in IMO history.My solution is just constructing parallelogram. The configuration is too similar to ELMO 2015 P3. I think there are about 500 14 points in this year´s IMO.

An easy question! But it truly cost me 15 mins to solve it.

Wow Im impressed people constructed an extra point for this. $\angle ART=\angle RKS$ and $\angle ATR=\angle KTS=\angle KRS$ thus $\triangle KRS\sim \triangle RTA$ so $\angle KRT=\angle ATR=\angle ATS$ and $$\frac{RT}{AT}=\frac{KR}{RS}=\frac{KR}{ST}$$Now these give $\triangle RKT\sim \triangle TSA$ so $\angle KTR=\angle SAT$ thus $KT$ is tangent to $\Gamma$.

dungnguyentien wrote: My solution for this problem. Wow.My solution is exactly the same.

Very nice, but easy question.

As before we have RK║AT Let RJ intersect circumcircle Γ in B. We have m(TBR) =m(TAK) =m(AKR)=m(ARC), so AK║BT. Let BT intersect RK in P and easily we notice that RATP parallelogram. m(STP)=m(ARS)=m(SKA) which implies that KSTP cyclic quadrilater. The final is the same as in the posts before.

Let $F$ be a point on $(ART)$ such that $AF$ is a symmedian. Let $TF$ meet $(RFS)$ again at $L$. Note that $R, S, F, K$ are concyclic. Now, we see that $$\measuredangle AKR=\measuredangle JSR=\measuredangle TAK,$$hence $RK \parallel TA$. Also, $$\measuredangle TRL=\measuredangle TFS=\measuredangle RTA,$$so $RL \parallel TA$ and $\boxed{K=L}$. Finally, we see that $$\measuredangle KTS=\measuredangle FAR=\measuredangle SAT,$$so $KT$ is tangent to $\Gamma$. $\blacksquare$

Let $X$and $Y$ be the midpoints of $AT$ and $RK$ respectively.$\angle TAK=\angle KJS=\angle KRS$,so $AT \parallel RK$. Let $AR\cap TK=P$,and $AK\cap RT=Z$.Its a well known fact that $X,Z,Y,P$ are collinear. From angle chasing,we get $ \triangle{XST}\sim\triangle{SKR}$,so $RS*TS=XT*KR$,which implies $RY*TA=RS*TS$,and since $AT\parallel KR$ we get $\triangle{RYS}\sim \triangle{TSA}$ which gives us the desired result.

After noticing that $RK\parallel AT$, we can complex bash. Let $\Omega$ be the unit circle and $k=1$. By Power of Point, we want $KT^2=KJ\times KA$, that is, \begin{align}((t-1)(\bar{t}-1))^2=(1-j)(1-\bar j)(a-1)(\bar a-1)\end{align} Now, since $A,J,K$ are collinear, we have $$\frac{j-1}{1/j-1}=\frac{a-1}{\bar a-1}\Leftrightarrow -j= \frac{a-1}{\bar a-1}\Leftrightarrow 1-j= \frac{a+\bar a-2}{\bar a-1}.$$ With this, $(1)$ becomes $$((t-1)(\bar{t}-1))^2= (a+\bar a-2)^2,$$so we just have to calculate $(t-1)(\bar{t}-1)$ and $a+\bar a-2$, and check that they are either equal or symmetric. Since $S$ is the midpoint of $RT$, we have $t=2s-r$, so \begin{align*} (t-1)(\bar{t}-1)&=(2s-r-1)(\frac{2}{s}-\frac{1}{r}-1) \\ &=6-\frac{2s}{r}-2s-\frac{2r}{s}+r-\frac{2}{s}+\frac{1}{r}. \end{align*} We now want to calculate $a+\bar a-2$. Since, by the tangency, $RA\perp RO$, we have $$\frac{a-r}{\bar a-1/r}=-\frac{r}{1/r}\Leftrightarrow a+r^2\bar a =2r.$$ On the other hand, since $AT\parallel RK$, we have $$\frac{a-t}{\bar a-\bar t}=\frac{r-1}{1/r-1}=-r\Leftrightarrow a+r\bar a=t+r\bar t.$$ With an appropriate linear combination of our last two equalities, we get \begin{align*} &\textcolor{white}{….}(1+r)(a+r\bar a)-(a+r^2\bar a)=(1+r)(t+r\bar t)-2r\\ &\Leftrightarrow r(a+\bar a)=t+rt+r\bar t+r^2\bar t -2r\\ &\Leftrightarrow a+\bar a =\frac{t}{r}+t+\bar t+r\bar t-2. \end{align*} Finally, \begin{align*} a+\bar a -2&=\frac{t}{r}+t+\bar t+r\bar t-4\\ &=\frac{2s-r}{r}+2s-r+\frac{2}{s}-\frac{1}{r}+r(\frac{2}{s}-\frac{1}{r})-4\\ &= -6+\frac{2s}{r}+2s+\frac{2r}{s}-r+\frac{2}{s}-\frac{1}{r}, \end{align*}which is indeed symmetric to $(t-1)(\bar t-1)$.

Using Reim on $\Omega$ and $\Gamma$ yields $KR \parallel AT$. Let $B = KR \cap AS$. Then since $RS = ST$ and $KB \parallel AT$ implies $ATBR$ is a parallelogram. Claim: $BKST$ is cyclic. Proof. Since $\angle BTS = \angle SRA = \angle SRJ + \angle JRA = \angle SKJ + \angle JKR = \angle SKR$ implies $BKST$ is cyclic. $\blacksquare$ Thus we have $\angle KTJ = \angle KTS + \angle STJ = \angle KBA + \angle SAJ = \angle KBA + \angle BAK = \angle AKR = \angle KAT$. Thus we get $KJ \cdot KA = KT^2$, so we're done. $\blacksquare$

Claim: $AT\parallel RK$ Proof: \[\angle JAT=180-\angle JST=\angle RSJ=\angle RKJ\]$\square$ Let the reflection of $K$ about $S$ be $K'$. Then $K'TKR$ is a parallelogram. Therefore, $K'$ lies on $AT$. Claim: $AK'SR$ is cyclic Proof: \[\angle ARS=\angle RKS=180-\angle SK'A\]$\square$ We can then say: \[\angle RTK=\angle K'RS=\angle TAS,\]as desired $\blacksquare$

Let $KS$ intersects $AT$ at $L$ Then from $\triangle LST$ similar to $\triangle KSR$ so $KS$=$SL$ From easy angle chase $ARSL$ is cyclic and it is very easy to finish

feels a lot like 2021 G1 this is the exact same as #113 wtf Reflect $K$ over $S$ to $K'$, so that $RKTK'$ is a parallelogram. By Reim's, $\overline{RK} \parallel \overline{AT}$, so $K'$ lies on $AT$. However, $$\measuredangle ARS = \measuredangle RKS = \measuredangle TK'S = \measuredangle AK'S,$$so $ARSK'$ is cyclic. To finish, note that $$\measuredangle KTS = \measuredangle K'RS = \measuredangle K'AS = \measuredangle TAS.$$

Let $\angle ARJ=\alpha , \angle DRS=\beta , \angle SRK=\gamma, \angle KJR=\delta , \angle ASJ=x$ From $\ell$ tangent to $\Omega$ we have: $\angle ARJ=\angle RAJ=\angle RSJ=\alpha$ $\angle JRS \stackrel{\Omega}{=} \angle JKS=\beta$ $\angle SRJ \stackrel{\Omega}{=} \angle SJK=\gamma$ $\angle ASR \stackrel{\Omega}{=} \angle AJS=\delta$ $\Omega$: $\angle RKS + \angle RJS=180 \implies \alpha+\beta+\gamma+\delta=180$ $\angle ASJ \stackrel{\Gamma}{=} \angle ATJ=x$ $180=\angle RST=\angle RSJ+\angle ASJ + \angle AST \implies \angle AST=\beta+\gamma+\delta-x$ $\angle AJT \stackrel{\Gamma}{=} \angle AST=\beta+\gamma+\delta-x$ From triangle $\triangle AJT$ we have: $\angle AJT+\angle ATJ + \angle TAJ=180 \implies \angle TAJ=\alpha$ $180=\angle AJK=\angle AJT + \angle TJS + \angle SJK \implies \angle TJS=\alpha+x-\gamma$ $\angle TAS \stackrel{\Gamma}{=} \angle TJS=\alpha+x-\gamma$ $\angle TAJ=\angle TAS + \angle SAJ \implies \angle SAJ=\gamma-x$ $\angle SAJ \stackrel{\Gamma}{=} \angle STJ=\gamma-x$ $\angle ATS=\angle ATJ+ \angle JTS=x+\gamma-x=\gamma \implies \angle ATS=\gamma$ Since $\angle ATS=\angle ATR=\angle SRK$ and $\angle ART=\angle ARJ + \angle JRT=\alpha+ \beta=\angle RKJ + \angle JKS=\angle RKS$ we get by $AA$ chriter that triangles $\triangle ATR$ and $\triangle SRT$ are simmilar $<=> \triangle ATR \sim \triangle SRK <=> \dfrac{AT}{SR}= \dfrac{TR}{RK}.$Combining with $SR=ST$ we get: $\dfrac{AT}{ST}= \dfrac{TR}{RK}$ Combining with $\angle ATS=\angle TRK$ we get by $SAS$ chriter that triangles $\triangle AST$ and $\triangle TKR$ are simmilar $<=> \triangle AST \sim \triangle TKR <=> \angle RTK=\angle TAK <=> \angle STK=\angle TAS <=> TK$- is tangent to $\Gamma$

Reflect $K$ over $S$ to form parallelogram $RK'TK$. Notice that \[\angle TAK = 180 - \angle JST = \angle JSR = \angle JKR\]so $\overline{AT} \parallel \overline{RK}$. Thus $K' \in \overline{AT}$. Thus \[\angle RAK' = 180 - \angle ARK = 180-\angle ARS - \angle SRK\]Notice that $\angle ARS = \angle RKS \implies \angle ARS+\angle SRK = \angle RSK'$. Combining with above gives $\angle RAK'+\angle RSK' = 180$, so $AK'SR$ cyclic. To finish we know \[\angle RTK = \angle TRK' = \angle K'AS = \angle TAS = \angle TJS\]so $KT$ is tangent to $\Gamma$.

This problem somehow took forever to solve.... By Reim's theorem, $\overline{AT} \parallel \overline{RK}$. We have $\triangle SKR \sim \triangle RTA$ since $\angle SKR = \angle ART$ (from the tangency) and $\angle KRS = \angle TAR$ (from the paralle lines). So, \[ \frac{AT}{TR} = \frac{SR}{RK} = \frac{TS}{RK} \implies \frac{AT}{TS} = \frac{TR}{RK}.\]Combined with the fact that $\angle ATS = \angle TRK$, we have $\triangle ATS \sim \triangle TRK$ from SAS similarity. So, it follows that $\angle KTR = \angle SAT$, which implies the tangency.

First of all, $$\measuredangle ATR = \measuredangle ATS = \measuredangle AJS = \measuredangle KJS = \measuredangle KRS = \measuredangle KRT,$$so $AT \parallel KR.$ Furthermore, since $RS = ST,$ the reflection $K'$ of $K$ across point $S$ lies on line $AT$ since then $K,S,K'$ are collinear and $TK' \parallel RK.$ In particular, we have that $RK' \parallel KT.$ The main claim is that $ARSK'$ is cyclic. Indeed, $$\measuredangle ARS = \measuredangle RKS = \measuredangle RKK' = \measuredangle AK'K = \measuredangle AK'S$$since $AR$ is tangent to $\Omega$ and $K'T \parallel RK.$ Finally, since $K'R \parallel KT$, we get $$\measuredangle KTA = \measuredangle RK'A = \measuredangle RSA = \measuredangle TSA,$$and we conclude.

This feels like someone covered up a triangle configuration by randomly generating point names and shuffling the order of definitions. Reformulated problem wrote: In $\triangle PRT$, let $S$ be the midpoint of $RT$, and let the circle through $S$ and $T$ tangent to $PT$ intersect line $PR$ at $A$ and $A'$ with $A$ closer to $P$. Define $K$ and $K'$ similarly on $PT$. Prove that $AK'$ passes through the other intersection $J$ of $(TSAA')$ and $(RSKK')$. It is well known that $J$ is the intersection of the $P-$ symmedian with $(PRT)$, and that $(RSJ)$ and $(TSJ)$ are $\sqrt{rt}$ inverses. It follows that $A$ and $K'$ are $\sqrt{rt}$ inverses, so $TA\parallel RK'$ and we're done by converse Reim.

Let $K'$ be the symmetric point of $K$ with respect to $S$. Then $RK'TK$ is a parallelogram because diagonals bisect each other, hence $TK' \parallel RK$. From the cyclic quadrilaterals we get that: $\angle ATR = \angle ATS = \angle SJK = \angle SRK = \angle TRK=> TA \parallel RK$ and hence $T,K',A$ are collinear. From the tagnency and the parallel lines we get that: $\angle AK'S = \angle AK'K = 180 - \angle RKK' = 180 - \angle RKS = 180 - \angle ARS => RAK'S$ is cyclic. From the cyclic quadrilaterals and the parallel lines we get that : $\angle TAS = \angle K'AS = \angle K'RS = \angle K'RT = \angle RTK = \angle SKT$ and, hence, $KT$ is tangent to $\Gamma$, as needed.

Why does the recurring theme for Problem 1/4 IMO geometry seem to be : All you need to do is construct a parallelogram?? We start off with some basic observations. Note that, \[\measuredangle AKR = \measuredangle JKR = \measuredangle JSR = \measuredangle JAT = \measuredangle KAT\]from which it follows that $RK \parallel AT$. Let $M$ denote the reflection of $K$ across $S$. Then, $M$ must lie on $\overline{AT}$, since $RMTK$ is a parallelogram. We can further note that, \[\measuredangle SRA = \measuredangle SKR = \measuredangle MKR = \measuredangle KMT\]so quadrilateral $RAMS$ must be cyclic. Finally note that, $\measuredangle RAS = \measuredangle RMS = \measuredangle RMK$ and $\measuredangle SRA = \measuredangle SKR = \measuredangle MKR$, so $\triangle RAS \sim \triangle RMK$, it thus follows that, \[\measuredangle ATK = \measuredangle MTK = \measuredangle KRM = \measuredangle ASR\]which implies that $KT$ is indeed tangent to $\Gamma$, as desired.

Heres an inversion: Invert around point $R$ with $r = \sqrt{RS \cdot RT}$. Therefore, it maps $\Gamma$ to $\Gamma$ itself. Note that $S, T$ map to each other in this inversion. Note that: $K',S',J'$ are collinear along with $K'S'J' \parallel RA$. Thus: $AA'J'S'$ is a cyclic trapezium. Also, note that $K, J, A$ were collinear which implies $RA'JK'$ is a cyclic trapezium. Therefore $RAS'K'$ is a parallelogram. It suffice to show that $(RSK'), (AST)$ are tangent to each other. Note that $RKS'A'$ is a cyclic quadrilateral which implies $A, S, K'$ to be collinear. Therefore $S$ is the center of parallelogram $RATK'$ which implies $(RSK'), (AST)$ are tangent to each other.