IMO 2017 Problem 4

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5452 posts

#1 h Jul 19, 2017, 4:30 PM • 26 Y

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Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$ . Point $T$ is such that $S$ is the midpoint of the line segment $RT$ . Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$ . Line $AJ$ meets $\Omega$ again at $K$ . Prove that the line $KT$ is tangent to $\Gamma$ .

Proposed by Charles Leytem, Luxembourg

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quangminhltv99

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quangminhltv99

#2 h Jul 19, 2017, 4:31 PM • 16 Y

Y by Snakes, claserken, Kunihiko_Chikaya, elnoga71, tenplusten, Tawan, TheImaginaryOne, Richangles, samrocksnature, tigerzhang, Quidditch, sabkx, Lamboreghini, Adventure10, Mango247, Funcshun840

Everything you need to do is construct a parallelogram

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wu2481632

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wu2481632

#3 h Jul 19, 2017, 4:32 PM • 8 Y

Y by temp73628, Tawan, Amir Hossein, samrocksnature, Teacher-TJB, Adventure10, Mango247, Rounak_iitr

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.

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djmathman

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djmathman

#4 h Jul 19, 2017, 4:33 PM • 5 Y

Y by Amir Hossein, samrocksnature, Adventure10, Mango247, sabkx

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.

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TelvCohl

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TelvCohl

#5 h Jul 19, 2017, 4:33 PM • 17 Y

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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 $\Gamma.$

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v_Enhance

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v_Enhance

#6 h Jul 19, 2017, 4:40 PM • 16 Y

Y by biomathematics, Kunihiko_Chikaya, samuel, e_plus_pi, Amir Hossein, Golub_Srecko, v4913, AllanTian, samrocksnature, HamstPan38825, Kontrol, Lamboreghini, Adventure10, Mango247, Rounak_iitr, zaidova

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 $KR \cdot TA = RS \cdot RT = TS \cdot TR$ .

$[asy] size(10cm); pair T = dir(100); pair S = dir(165); pair R = 2*S-T; pair E = dir(0); pair A = IP(unitcircle, R--(R+100*E)); pair B = OP(unitcircle, R--(R+100*E)); pair K = extension(T, T+dir(90)*T, R, R+T-A); draw(R--B, blue); draw(unitcircle, blue); draw(circumcircle(R, S, K), heavycyan); pair J = -A+2*foot(origin, A, K); draw(K--R--A--T--cycle, red); draw(A--K, orange); draw(R--T, orange); // invert pair Js = extension(T, T+A-B, R, J); pair Ks = extension(T, T+A-B, R, K); draw(K--Ks, heavygreen); draw(Ks--Js, heavygreen); draw(Js--B, dotted+blue); draw(R--Js, dotted+blue); dot("$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$", Ks, dir(Ks)); /* TSQ Source: !size(10cm); T = dir 100 S = dir 165 R355 R = 2*S-T R270 E := dir 0 A = IP unitcircle R--(R+100*E) R225 B = OP unitcircle R--(R+100*E) R315 K = extension T T+dir(90)*T R R+T-A R200 T--K blue R--B blue unitcircle 0.05 lightcyan / blue circumcircle R S K 0.05 paleblue / heavycyan J = -A+2*foot origin A K R180 K--R--A--T--cycle 0.1 lightred / red A--K orange R--T orange // invert J* = extension T T+A-B R J K* = extension T T+A-B R K K--Ks heavygreen Ks--Js heavygreen Js--B dotted blue R--Js dotted blue */ [/asy]$

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.

This post has been edited 2 times. Last edited by v_Enhance, Jul 28, 2017, 9:53 PM
Reason: ELMO 2015 reference

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dungnguyentien

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dungnguyentien

#7 h Jul 19, 2017, 4:42 PM • 12 Y

Y by biomathematics, guoh064, buratinogigle, mathematiculperson, BobaFett101, TomMarvoloRiddle, Peggy, samrocksnature, megarnie, Lamboreghini, Adventure10, Mango247

My solution for this problem.

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ntpt

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ntpt

#9 h Jul 19, 2017, 4:51 PM • 4 Y

Y by samrocksnature, Hopeooooo, Adventure10, Funcshun840

$\angle KRS=\angle KJS=\angle RTA$ 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 $\Gamma$

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Ferid.---.

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Ferid.---.

#10 h Jul 19, 2017, 5:18 PM • 2 Y

Y by samrocksnature, Adventure10

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.

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k.vasilev

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k.vasilev

#11 h Jul 19, 2017, 5:21 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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 $\Gamma$ .

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RC.

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RC.

#15 h Jul 19, 2017, 5:51 PM • 2 Y

Y by samrocksnature, Adventure10

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

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tenplusten

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tenplusten

#16 h Jul 19, 2017, 5:54 PM • 4 Y

Y by vjdjmathaddict, samrocksnature, Adventure10, Mango247

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.

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MathMath112233

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MathMath112233

#18 h Jul 19, 2017, 6:28 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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.

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liekkas

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#19 h Jul 19, 2017, 6:38 PM • 4 Y

Y by samrocksnature, keglesnit, Adventure10, Mango247

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

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leader

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leader

#20 h Jul 19, 2017, 6:57 PM • 6 Y

Y by PeppaBear, yugrey, samrocksnature, MrOreoJuice, hydrated, Adventure10

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 $\Gamma$ .

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eibc

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eibc

#120 h Feb 8, 2024, 3:02 AM

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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
$\measuredangle KTS = \measuredangle K'RS = \measuredangle K'AS = \measuredangle TAS.$

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ItsBesi

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ItsBesi

#125 h Feb 11, 2024, 12:16 PM

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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 $\Gamma$

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naonaoaz

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naonaoaz

#126 h Feb 25, 2024, 7:27 PM

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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 $\Gamma$ .

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ihatemath123

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ihatemath123

#127 h Apr 29, 2024, 4:10 PM

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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.

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EpicBird08

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EpicBird08

#128 h May 8, 2024, 2:08 AM

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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.

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CT17

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CT17

#129 h May 26, 2024, 8:40 PM

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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.

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giannis2006

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giannis2006

#130 h Jul 5, 2024, 6:41 PM

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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.

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cursed_tangent1434

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cursed_tangent1434

#131 h Aug 20, 2024, 4:44 AM

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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.

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Saucepan_man02

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Saucepan_man02

#132 h Nov 13, 2024, 6:32 AM

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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.

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cj13609517288

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cj13609517288

#133 h May 25, 2025, 9:19 PM

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what a weird problem

By Reim's we get $AT\parallel RK$ . Now complex with unit circle $(RSK)$ . Then $\frac{a-t}{r-k}$ is real and $\frac{a-r}{r}$ is purely imaginary. Set $a=2s-r+u(r-k)$ , after some algebra I did on paper we get
$u=\frac{2(r-s)^2 k}{s(r-k)^2}\Longrightarrow a-s=\frac{(r-s)(2rk-sk-sr)}{s(r-k)}.$ Therefore, $\frac{a-s}{a-t}$ has the same argument (modulo $180$ degrees) as
$\frac{(r-s)(2rk-sk-sr)}{s(r-k)^2}$ and $\frac{t-r}{t-k}$ has the same argument (modulo $180$ degrees) as
$\frac{r-s}{2s-r-k}.$ The quotient of these is indeed equal to its own conjugate. $\blacksquare$

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ezpotd

1344 posts

ezpotd

#134 h May 28, 2025, 7:10 AM

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Define $K', L'$ such that $K'$ is the closer intersection of $\Omega$ and the tangent to $\Gamma$ at $T$ to $T$ . Define $B$ to be the other intersection of $\Gamma, \ell$ . Extend $K'L', BA$ to meet at $Z$ . We show $K', J, A$ are collinear, which is sufficient.

Claim: $ZRK'L' \sim ZTBA$
Proof: We first show that the ratio of the circumradii of these triangles is the ratio $\frac{ZR}{ZT}$ , so taking a homothety on $\Gamma$ with this ratio and reflecting will map $\Gamma$ to $\Omega$ , giving the desired similarity. Of course, we can compute $R_{\Omega} = \frac{RS}{2 \sin RS } = \frac{RS}{\sin ZRT} = \frac{ST}{\sin TRZ} = \frac{ST}{2 \sin ST} = R_{\Gamma}$ , where the middle equality comes from Law of Sines on $ZRT$ .

Claim: $RL'TA$ is cyclic.
Proof: Trivial by the claimed similarity.

By radical axis, $JS$ , $RL'$ , $TA$ concur. By Pascal on $RRSJK'L'$ , $RR '\cap JK', RS \cap K'L' = T, JS \cap RL'$ concur. The unique line going through $JS \cap RL', T$ is $TA$ , since $RR' \cap K'J$ lies on $TA$ , we have $RR' \cap K'J = RR' \cap TA = A$ , so $A$ lies on $JK'$ , as desired.

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Sakura-junlin

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Sakura-junlin

#135 h Jun 17, 2025, 9:55 AM

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$[asy] import olympiad; import cse5; size(11cm); defaultpen(fontsize(10pt)); pair S = dir(-30); pair R = dir(210); pair T = 2*S-R; pair J = dir(-75); pair A = IP(circumcircle(J,S,T),L(R*dir(90)+R,R,5,5)); pair K = IP(L(J,A,5,5),unitcircle); pair K1 = 2*S-K; markscalefactor=0.01; draw(unitcircle); draw(R--T); draw(circumcircle(J,S,T)); draw(R--A); draw(K--T); draw(K--A); draw(A--T); draw(circumcircle(R,A,K1),blue+dashed); draw(K1--K,dotted); draw(R--K,red); draw(T--K1,red); draw(K--T,red); draw(K1--R,red); dot("$R$", R, W); dot("$S$", S, dir(-35)*2); dot("$T$", T, dir(60)); dot("$J$", J, dir(-25)*2); dot("$A$", A, dir(-75)); dot("$K$", K, dir(60)); dot("$K'$", K1, S); [/asy]$

Proof: Define $K' :=$ Reflection of point $K$ respecting to point $S$

Since $RS=ST , KS= SK'$ so $R,K,T,A$ is Parallelogram ,then we get $K'T \parallel RK$ , $KT \parallel RK'$

Claim1: $A,K',T$ is collinear.

Proof: it suffice to show $AT \parallel RK$ , and we have $\angle{JAT} = 180^{\circ} - \angle{JST} = \angle{JSR} = \angle{JKR}$

So $AT \parallel RK$ $\Rightarrow$ $A,K',T$ is collinear. $\blacksquare$

Claim2: $R,S,K',A$ is concyclic.

Proof: it suffices to show $\angle SRA = \angle SK'T$

Since $AT \parallel RK$ ,so we have $\angle SK'T = \angle KK'T = \angle RKS = \angle SRA$ (last Equals sign was by Alternate Segment Theorem). $\blacksquare$

Finally, we need to Proof KT is the tangent line of $\odot(JST)$ ,

it suffices to show $\angle KTS = \angle SAT$

Since $R,S,K',A$ is concyclic, so $\angle SAT = \angle SAK' = \angle SRK' = \angle KTR = \angle KTS$

So $KT$ is the tangent line of $\odot (JST)$ . $\square$

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Math_456

8 posts

Math_456

#136 h Jun 22, 2025, 12:28 PM

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It's easier to solve this problem than to draw its diagram

Constructing a parallelogram and angle-chasing solves the problem

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reni_wee

93 posts

reni_wee

#137 h Jul 6, 2025, 8:36 AM

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Claim: $\triangle ATS \sim \triangle TRK$
Proof:

We have that,
$\begin{align*} \angle KRS = \angle &KJS = \angle ATS \\ \angle ART &= \angle RKS \end{align*}$ Hence, $\triangle ART \sim \triangle SKR$ . From which we get that
$\begin{align*} \frac{AT}{RS} &= \frac{TR}{RK} \\ \\ \implies \frac{AT}{TS} &= \frac{TR}{RK} \end{align*}$ We also have that $\angle ATS = \angle KRT$ . Therefore $\triangle ATS \sim \triangle TRK$ .

From which we can get that $\angle AST = \angle KTR$ , directly implying that line $KT$ is tangent to $\Gamma$ .

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OronSH

1783 posts

OronSH

#138 h Today at 7:29 AM

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Let $P$ be the reflection of $A$ over $S$ .

Claim: $R,K,P$ collinear.
Proof. From $\measuredangle RKJ=\measuredangle RSJ=\measuredangle TSJ=\measuredangle TAJ$ we get $TA\parallel RK$ which implies the claim

Claim: $TSKP$ cyclic.
Proof. Follows from $\measuredangle PKS=\measuredangle RKS=\measuredangle ARS=\measuredangle PTS$

Claim: $KT$ tangent to $\Gamma$
Proof. Follows from $\measuredangle KTS=\measuredangle KPS=\measuredangle TAS$

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