Problem 1

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5 posts

#1 h Jul 10, 2012, 5:24 PM • 17 Y

Y by buratinogigle, Mualpha7, Amir Hossein, iarnab_kundu, Davi-8191, math1181, AlastorMoody, A-Thought-Of-God, jhu08, megarnie, HWenslawski, ImSh95, Adventure10, Tastymooncake2, ItsBesi, Rounak_iitr, and 1 other user

Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$ , and to the lines $AB$ and $AC$ at $K$ and $L$ , respectively. The lines $LM$ and $BJ$ meet at $F$ , and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$ , and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$ , to the ray $AB$ beyond $B$ , and to the ray $AC$ beyond $C$ .)

Proposed by Evangelos Psychas, Greece

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FelixD

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FelixD

#2 h Jul 10, 2012, 5:43 PM • 20 Y

Y by UraharaKisuke, buratinogigle, AndrewTom, biomathematics, karitoshi, Mathasocean, A-Thought-Of-God, translate, jhu08, yshk, ImSh95, Adventure10, Mango247, shafikbara48593762, Tastymooncake2, and 5 other users

If I'm not mistaken, it's easy to see that $\angle LFJ=\alpha/2$ and so $AFJL$ is cyclic. But $\angle JLA=90$ and so $\angle AFJ=90$ . Thus, $AB=BS$ and hence $MS=AK$ . Similarly, $MT=AL$ , but $AK=AL$ , so we are done.

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AndreiAndronache

88 posts

AndreiAndronache

#3 h Jul 10, 2012, 5:54 PM • 7 Y

Y by jhu08, ImSh95, Adventure10, Tastymooncake2, and 3 other users

Otherwise: $m(\angle{JFL})=m(\angle{JAL})=\frac{A}{2}\Rightarrow A,F,J,L$ are concyclic $\Rightarrow S,J,M,F$ are concyclic $\Rightarrow m(\angle{JST})=m(\angle{JTS})=\frac{A}{2}\Rightarrow JS=JT\Rightarrow SM=MT$ .

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WakeUp

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WakeUp

#4 h Jul 10, 2012, 6:03 PM • 9 Y

Y by buratinogigle, jhu08, metricpaper, ImSh95, Adventure10, Tastymooncake2, and 3 other users

Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$ . Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ( $PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$ , which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $AJ$ is clearly the diameter of $(JKL)$ , so the points $A,F,K,J,L,G$ are concyclic.

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Concyclicboy

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Concyclicboy

#5 h Jul 10, 2012, 6:11 PM • 2 Y

Y by jhu08, Adventure10

Yes, we only need that <MFB= <A/2 and then AFJL cyclic then <AFJ=<ALJ=90 then <SFJ=<SMJ=90 then SFMJ cyclic with <MSJ=<A/2. Similar to <JTM=<A/2

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Zhero

2043 posts

Zhero

#6 h Jul 10, 2012, 6:14 PM • 7 Y

Y by buratinogigle, yugrey, jhu08, Adventure10, Tastymooncake2, and 2 other users

By a simple angle chase, $\angle FBM = \angle ABJ = 90 + \angle B/2$ and $\angle FMB = \angle AJB = \angle C / 2$ , so $\triangle BFM \sim \triangle BAJ$ . Thus, $\triangle BFA \sim \triangle BMJ$ (since $BF/BA = BM/BJ$ and $\angle FBA = \angle MBJ = 90 - \angle B / 2$ .) Since $BM \perp MJ$ , we have $BF \perp AF$ . Since $\angle ABF = \angle FBS = 90 - \angle B/2$ , we have $BS = BA$ . Similarly, $CA = CT$ , so $SM = (s-c) + c = s = (s-b) + b = MT$ .

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golem24

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golem24

#7 h Jul 10, 2012, 6:20 PM • 19 Y

Y by yugrey, amar_04, ETS1331, Aryan-23, jhu08, mistakesinsolutions, Adventure10, Mango247, MS_Kekas, Tastymooncake2, Funcshun840, and 8 other users

The solutions are smaller than the problem statement.

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#8 h Jul 10, 2012, 6:27 PM • 7 Y

Y by AceOfDiamonds, buratinogigle, jhu08, yshk, Adventure10, Mango247, Tastymooncake2

Here is the attached figure.

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Orin

22 posts

Orin

#9 h Jul 10, 2012, 6:30 PM • 3 Y

Y by buratinogigle, Adventure10, Mango247

It is well-known that $\angle BJC=90^{\circ}-\frac {1}{2}\angle A$
now $\angle LJC=\angle MJC=90^{\circ}-\angle MCJ=90^{\circ}-(90^{\circ}- \frac {1}{2}\angle C)=\frac {1}{2} \angle C$
so $\angle FJL=\angle BJC+\angle LJC=90^{\circ} -\frac {1}{2} \angle A+\frac {1}{2} \angle C$ ....(1)
now $\angle MLJ=90^{\circ}- \angle LJC=90^{\circ}-\frac {1}{2} \angle C$ ....(2)
so from (1),(2) we get in $\triangle LJF,\angle LFJ=\frac {1}{2} \angle A$ which inplies $AFJL$ is cyclic.
since $\angle JLA=90^{\circ}$ we get $AFJ=90^{\circ}$
now $\angle FAB=\angle FAL- \angle A=\angle \frac {1}{2} \angle B$
and $\angle FSB=90^{\circ}-\angle FBS=90^{\circ}-\angle CBJ=\angle \frac {1}{2} \angle B$
hence $\angle FAB=\angle FSB$
so $\triangle FBS \cong \triangle FBA$ [ASA] implying $AB=BS$
now $MS=MB+BS=BK+AB=AK$
simimlarly $MT=AL$
but $AK=AL$ [tangent to the excircle from $A$ ]
hence $MS=MT$ which means $M$ is the midpoint of $ST$
P.S.:Edited only to correct a typo.

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math_genie

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math_genie

#10 h Jul 10, 2012, 6:42 PM • 2 Y

Y by Adventure10, Mango247

Its quite easy to see that
$A,K,J,G$ is cyclic. (1)
$A,F,J,L$ is cyclic. (2)

Since $\angle AKJ = 90^{\circ}$ , this implies that $\angle AGJ = 90^{\circ}$ from (1)
Similarly $\angle AFJ = 90^{\circ}$ .
Hence we get
$A,F,K,J$ is cyclic (3)
$A,G,L,J$ is cyclic (4)

Thus from (1),(2),(3),(4) we get that all six points $A,G,F,J,K,L$ to be cyclic.
Now, note that from triangle ABS, $AB = BS = c$ .
Similarly, $AC = CT = b$ .
Thus $SM = SB + BM = c + (s-c) = s$
Similarly $TM = TC + CM = b + (s-b) = s$ .
Hence proved.

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sjaelee

485 posts

sjaelee

#11 h Jul 10, 2012, 7:05 PM • 2 Y

Y by Adventure10 and 1 other user

Solution

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Neerjhor

12 posts

Neerjhor

#12 h Jul 10, 2012, 7:10 PM • 1 Y

Y by Adventure10

It was really easy. Mine is almost similar as everyone. I proved that $AFJL$ and $AGJK$ are cyclic, which concludes that $AB=BS$ and $AC=CT$ . Then $BM=s-c$ and $CM=s-b$ . Now, we add these and $MS=MT$ .

This was quite easy for IMO 1 I guess.

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pi37

2079 posts

pi37

#13 h Jul 10, 2012, 7:19 PM • 1 Y

Y by Adventure10

Slightly different approach

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guysfromKIROVOGRAD

3 posts

guysfromKIROVOGRAD

#14 h Jul 10, 2012, 8:20 PM • 2 Y

Y by Adventure10, Mango247

WakeUp wrote:

Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$ . Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ( $PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$ , which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $A$ is clearly the diameter of $(JKL)$ , so the points $A,F,K,J,L,G$ are concyclic.

Cool solution, bro. But ... too many letters as for me. Ты ещё не задолбался точечки отмечать? ( Use russian translator, bro)

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AV-2

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AV-2

#15 h Jul 10, 2012, 8:55 PM • 2 Y

Y by Adventure10, Mango247

Notice that $MK$ and $ML$ are parallel to the internal bisectors of $B$ and $C$ respectively. Then $MK \perp BJ$ , $ML \perp CJ$ , so $M$ is the orthocenter of $JFG$ . Since $JM \perp BC$ from tangency, and $JM \perp FG$ from the orthocenter, $BC \parallel FG$ . Now we can prove that $MF \parallel AG$ and $MG \parallel AF$ with any of the methods above, or by considering the excenters $I_b, I_c$ , the fact tht $I_bI_c$ is antiparallel to $BC$ and a short angle chase. With this, $FMG$ is the medial triangle of $AST$ and we are done.

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surpidism.

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surpidism.

#141 h Jun 3, 2024, 7:15 PM

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Claim 1: $FG \parallel BC$
Proof: Clearly $BJ \perp MK$ and $CJ \perp ML$ , meaning $M$ is the orthocentre of $\triangle FGJ$ . So $MJ \perp BC$ and $MJ \perp FG$ implies that $FG \parallel BC$ .

Claim 2: $A$ , $G$ , $L$ , $J$ , $K$ , $F$ are concylic
Proof: $\angle GFJ = \angle MBJ = \angle MKJ = \angle GKJ \implies JKFG$ is cyclic. Similarly, $JLGF$ is cyclic. Thus with $AKJL$ cyclic as well, $A$ , $G$ , $L$ , $J$ , $K$ , $F$ are concylic.

Claim 3: $AT \parallel FM$
Proof: $\angle GFM = \angle CML = \angle CLM = \angle ALF = \angle AGF$ .

Claim 4: $F$ is the midpoint of $\overline{AS}$
Proof: $\angle AKJ = \angle AFJ = \angle AFB = 90^{\circ} = \angle SFB$ and $\angle FBS = \angle MBJ = \angle KBJ = \angle FBA$ . So, $\triangle FBS \cong \triangle FBA$ , which implies $SF = FA$ .

Hence, claim 3 and 4 follows that $M$ is the midpoint of $ST$ . $\square$

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PEKKA

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PEKKA

#142 h Jun 27, 2024, 9:54 PM

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By right angles, notice that $AKJL$ is a cyclic quad. Let the circumcircle of this quadrilateral be $\omega.$
Then we can angle chase to get $\angle JFM= \frac{A}{2}$ and $\angle LKJ=\frac{A}{2}.$ This implies that $F$ lies on $\omega.$
By symmetry $G$ lies on $\omega$ too.
Then by cyclic quads, $\frac{B}{2}=\angle BJK=\angle BAF.$
By definition then vertical angles, $\angle KBJ=\angle FBA=90-\frac{B}{2}.$
Therefore $ABF$ is a right triangle and therefore $FJ \perp AS.$
By definition then vertical angles again, $\angle SBF=90-\frac{B}{2}.$
Therefore by ASA, triangles $SBF$ and $ABF$ are congruent.
By symmetry, triangles $ACG$ and $TCG$ are congruent too.
By tangents excircle properties, $SM=SB+SM=AB+BK=s$ and $AL=AC+CL=CM+CT=MT=s$ where $s$ is the semiperimeter of ABC.
Therefore $SM=MT$ and we are done.

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G0d_0f_D34th_h3r3

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G0d_0f_D34th_h3r3

#143 h Jul 10, 2024, 10:10 AM

Y by

Let's solve this easy angle chase problem.

First we notice that $\angle SMJ = \angle SFJ$ . So, quadrilateral $FSJM$ is cyclic.

Now, Since $\angle CML = 180^{\circ} - \frac{180^{\circ}-\angle C}{2}$ and
$\angle FSJ = 90^{\circ}- \angle SJF = 90^{\circ}-\angle SMF = 90^{\circ}-\angle CML = 90^{\circ} - \frac{\angle C}{2}$
Also,
$\angle ACJ = \angle C + \frac{180^{\circ}-\angle C}{2}= 90^{\circ} +\frac{\angle C}{2}$
Therefore, quadrilateral $ACJS$ is cyclic. Since $A$ , $I$ and $J$ are collinear, we get,
$\angle JSM = \angle JSC = \angle JAC = \angle IAC = \frac{\angle C}{2}$
Similarly, $\angle JTM = \frac{\angle C}{2}$ . Hence by congruency, we get that $M$ is the midpoint of $ST$ .

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ezpotd

1286 posts

ezpotd

#144 h Sep 6, 2024, 2:09 AM

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bad formatting because i copied my solution from math dash

let JC meet ML at X, then JXM is right, so JFL = 90 - BJX = 90 - BJC = a/2. since JAL is also a/2, we see that (AFJL) is cyclic, and since (AKJL) is also cyclic so is (AFJKL). now AFJ is right, then so is SFJ, so (SFJM) is cyclic. since JFM = JFL = JAL = a/2, we have JSM = JFM = a/2. repeating this argument gives JTM = a/2. since JMS = JMT = 90, by AAS congruency we have JSM and JTM are congruent, so SM = TM.

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AshAuktober

1009 posts

AshAuktober

#146 h Sep 18, 2024, 8:33 AM

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We present three claims, all of which may be proven by angle chasing with respect to the angles of the triangle $ABC$ .
Claim 1: $A, F, G, J, K, L$ are concylic.
Claim 2: $AFMG$ is a parallelogram.
Claim 3: $FG \parallel ST$ .

Note that from Claim 2, $AM$ bisects $FG$ . But now taking the homothety at $A$ sending $FG$ to $ST$ (which exists by Claim 3), $AM$ bisects $ST$ , so $M$ is the midpoint of $ST$ . $\square$

Remark(due to Rijul Saini): The above claim 2 and 3 may essentially be summarised as Iran Lemma on the $A-$ excircle.

This post has been edited 1 time. Last edited by AshAuktober, Sep 18, 2024, 8:34 AM

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shendrew7

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shendrew7

#147 h Oct 22, 2024, 1:34 PM

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Using angle chase, notice that
$\angle LFJ = \angle XBJ - \angle CXL = 90 - \frac B2 - \frac C2 = \frac A2 = \angle LAJ,$
so $AFJL$ is cyclic. Notice that $\angle ALJ = \angle AKJ = 90$ implies $AFKJL$ is also cyclic. Instead exploiting symmetry like a normal person, we can instead notice that the converse of Pascal's on hexagon $AKGJFL$ implies these six points lie on a conic, which must be a circle as we know five of the points define a circle.

Hence we have $\angle AFB = \angle AGC = 90$ , giving us isosceles trapezoids $AXKS$ and $AXLT$ , so
$XS = AK = AL = XT. \quad \blacksquare$
$[asy] size(330); pair A, B, C, J, K, L, F, G, S, T, X; A = dir(120); B = dir(210); C = dir(330); J = extension(A, incenter(A, B, C), B, rotate(90, B)*incenter(A, B, C)); K = foot(J, A, B); L = foot(J, A, C); X = foot(J, B, C); F = extension(B, J, X, L); G = extension(C, J, X, K); S = extension(A, F, B, C); T = extension(A, G, B, C); draw(A--K--J--L--A--J--X^^K--G--J--F--L^^A--S--T--cycle); draw(circumcircle(A, K, L), dashed); draw(circumcircle(X, K, L), dotted); draw(A--K^^S--X, red+linewidth(1.5)); draw(A--L^^T--X, blue+linewidth(1.5)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$K$", K, W); dot("$L$", L, E); dot("$J$", J, SE); dot("$X$", X, N); dot("$F$", F, NW); dot("$G$", G, NE); dot("$S$", S, W); dot("$T$", T, E); [/asy]$

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Bonime

37 posts

Bonime

#148 h Nov 12, 2024, 1:14 AM • 1 Y

Y by Jorge_144p

Claim 1: $AF\bot BF$ and $AG\bot CG$
Prove: Let $P=AC\cap KM$ and $Q=AB\cap LM$ . Clearly, $BJ$ is a perpendicular bissector of $KM$ e $CJ$ of $LM$ , so $\angle KFB=\angle BFQ$ and $\angle LGC=\angle CGP$ . Also, by ceva and then menelaus with the points $K$ , $M$ and $L$ and $Q$ , $M$ and $L$ likewise $P$ , $M$ and $P$ , we get that $(A, B; K, Q)=(A, C; P, L)=-1$ , so we´re done $\blacksquare$

Claim 2: $FG//BC$
It´s easy to see that $GK\bot FJ$ and $FL\bot GJ$ , therefore $M$ is the orthocenter of $\triangle FJG$ so $JM\bot FG$ , but $JM\bot BC$ . $\blacksquare$ .

Hence, from the Claim 1, we get that $AGMF$ is a paralelogram. We want to prove that $(S,T;M, P_{\infty, BC})=-1$ but $(S, T; M, P_{\infty, BC})\stackrel{A}{=}(F, G; AM\cap FG, P_{\infty, FG})=-1$ so we´re done. $\blacksquare$

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lnzhonglp

120 posts

lnzhonglp

#149 h Nov 26, 2024, 5:48 AM

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Note that $\angle JFL = \angle JAL = \angle KAJ = \angle KGJ,$ so $AKJG$ and $AFJL$ are cyclic. Then $\angle AGM = \angle AJK = \angle AJL = \angle AFL = 90^\circ - \frac{\angle A}{2}$ and $\angle FAG = \angle FAJ + \angle GAJ = \angle FLJ + \angle GKJ = \angle FMG = 90^\circ + \frac{\angle{A}}{2},$ so $AFMG$ is a parallelogram. We also have $AS \perp FJ$ and $AT \perp JG$ , so since $\angle ABF = \angle FBS$ and $\angle ACG = \angle GCT$ , we get $SF = AF = MG$ and $FM = AG = GT$ . Then $\triangle SFM \cong \triangle MGT,$ so $SM = MT$ , as desired.

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cj13609517288

1922 posts

cj13609517288

#150 h Dec 19, 2024, 2:40 PM

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This is really a problem about triangle $MKL$ . We claim that $BA=BS$ and $CA=CT$ , which will clearly finish.

Note that if we show that $\triangle BKM\sim\triangle BSA$ , we will win. Thus it suffices to show that $KM\parallel AF$ , or $\angle AFJ=90^{\circ}$ , or $F\in(AKJL)$ .

Now let's invert around the excircle. Then $F'\in (JLM)$ , so
$\angle JF'L=\angle JML=90^{\circ}-\angle K,$ so since $F'$ lies on $BJ$ , it is exactly $BJ\cap KL$ , as desired. $\blacksquare$

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Maximilian113

575 posts

Maximilian113

#152 h Jan 10, 2025, 11:01 PM

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Notice that $BJ$ is the perpendicular bisector of $MK$ so $FK=FM.$ Therefore $\angle MFK = 2 \angle JFM = 2(90^\circ - \angle BJC) = \angle A.$ Thus $KFAL$ is cyclic. Similarly $LGAK$ is cyclic and it follows that pentagon $FAGLK$ is cyclic.

Thus, $\angle FAG = 180^\circ - \angle FLG = 90^\circ +\frac12 \angle JGL = 90^\circ + \frac{\angle A}{2}.$ Also, $\angle AGK = \angle ALK = 90^\circ - \frac{\angle A}{2},$ so $\angle FAG + \angle AGK = 180^\circ \implies AS \parallel KG.$ But $BK = BM$ so by homothety $BS=BA.$ Therefore $MS = AK = s$ where $s$ denotes the semiperimeter.

Similarly $MT=s$ too, and we are done. QED

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KevinYang2.71

428 posts

KevinYang2.71

#153 h Mar 10, 2025, 4:09 PM

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We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$ . Then
$\begin{align*} J&=(-a:b:c)\\ M&=(0:s-b:s-c)\\ K&=(c-s:s:0)\\ L&=(b-s:0:s). \end{align*}$ $F$ is given by $(-a:t:c)$ for some $t$ where
$\begin{vmatrix} -a & t & c\\ 0 & s-b & s-c\\ b-s & 0 & s \end{vmatrix}=0.$ It follows that $-as(s-b)+t(b-s)(s-c)-c(b-s)(s-b)=0$ so $t=\frac{cs-bc-as}{s-c}=-a-c$ . Hence
$S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)$ and similarly
$T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).$ Thus $\frac{1}{2}(S+T)=\left(0,\,\frac{s-b}{a},\,\frac{s-c}{a}\right)=M$ , as desired. $\square$

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imagien_bad

58 posts

imagien_bad

#154 h Mar 10, 2025, 5:03 PM

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KevinYang2.71 wrote:

We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$ . Then
$\begin{align*} J&=(-a:b:c)\\ M&=(0:s-b:s-c)\\ K&=(c-s:s:0)\\ L&=(b-s:0:s). \end{align*}$ $F$ is given by $(-a:t:c)$ for some $t$ where
$\begin{vmatrix} -a & t & c\\ 0 & s-b & s-c\\ b-s & 0 & s \end{vmatrix}=0.$ It follows that $-as(s-b)+t(b-s)(s-c)-c(b-s)(s-b)=0$ so $t=\frac{cs-bc-as}{s-c}=-a-c$ . Hence
$S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)$ and similarly
$T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).$ Thus $\frac{1}{2}(S+T)=\left(0,\,\frac{s-b}{a},\,\frac{s-c}{a}\right)=M$ , as desired. $\square$

bh bary is banned

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Sadigly

227 posts

Sadigly

#155 h Mar 25, 2025, 12:22 PM

Y by

Honestly it is easy with syntethic but I'm trying to improve my complex bash

Let $(KLM)\in\mathbb{S}^1\hspace{5mm}K=k\hspace{2mm}L=l\hspace{2mm}M=m\hspace{2mm}J=0$

$A=\frac{2kl}{k+l}\hspace{2mm}B=\frac{2lm}{l+m}\hspace{2mm}C=\frac{2mk}{m+k}$

$F=LM\cap BJ=\frac{k(m+l)}{k+l}\hspace{5mm}G=KM\cap CJ=\frac{l(m+k)}{k+l}$

$S=AF\cap BC=\frac{2km}{k+l}\hspace{5mm}T=AG\cap BC=\frac{2lm}{k+l}$

$\frac{\frac{2km}{k+l}+\frac{2lm}{k+l}}{2}=m$ , so midpoint of $ST$ is $M$

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IndexLibrorumProhibitorum

10 posts

IndexLibrorumProhibitorum

#156 h Apr 23, 2025, 2:42 PM

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Claim I. $A , F , J , L$ are concyclic.
This is because $\angle AJF = \pi-\angle BAI-\angle ABI-\angle IBJ=\pi-\frac{A}{2}-\frac{B}{2}-\frac{\pi}{2}=\frac{C}{2}=\angle FLA$
$\implies A , F , J , L$ are concyclic. $\quad \square$

Claim II. $FA \parallel BI \parallel KM$ .
At first, we have $\measuredangle JBI =\measuredangle (JB,KM)=\frac{\pi}{2} \implies BI \parallel KM.$
Then by Claim I, we have $\angle JLM = \angle JFB= \frac{\pi}{2}=\angle JBI \implies BI \parallel FA.\quad \square$

Claim III. Quadrilateral $FGMS$ is a parallelogram
By Claim II, we have $-1=( K , M ; BJ \cap KM , \infty )= ( A , S ; F , \infty ) \implies AF=SF$ .
The same is true of $AG=GT$ , so $FG \parallel ST$ .
Hence, Quadrilateral $FGMS$ is a parallelogram. $\quad \square$

Thus now it's clear to know $FG = SM$ by Claim III. The same is true of $FG = TM$ , therefore, $SM=TM$ .

Q.E.D.

$[asy] import olympiad; import cse5; size(11cm); pair A = dir(75); pair B = dir(210); pair C = dir(-30); pair X= dir((B+C)/2); pair I = incenter(A,B,C); pair J = (X-I)*2+I; pair L= foot(J,A,C); pair K= foot(J,A,B); pair M= foot(J,B,C); pair F =IP(L(L,M,5,5),L(J,B,5,5)); pair G =IP(L(K,M,5,5),L(J,C,5,5)); pair S =IP(L(A,F,5,5),L(B,C,5,5)); pair T =IP(L(A,G,5,5),L(B,C,5,5)); draw(A -- C--B--cycle); draw(unitcircle); draw(circumcircle(K,L,M),blue); draw(circumcircle(A,J,F),red + dotted); draw(L(A,C,0,1.5)); draw(L(A,B,0,1.5)); draw(L--F); draw(J--F); draw(J--G); draw(K--G); draw(S--T); draw(A--T); draw(S--A); dot("$A$", A, dir(A)); dot("$B$", B,dir(145)*2); dot("$C$", C, dir(33)*2); dot("$I$",I,dir(45)); dot("$J$",J,dir(270)); dot("$K$", K, dir(180)); dot("$L$",L,dir(45)); dot("$M$",M,N); dot("$F$",F,dir(135)); dot("$G$", G, dir(45)); dot("$S$",S,dir(180)); dot("$T$",T,E); [/asy]$

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YaoAOPS

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YaoAOPS

#157 h Apr 23, 2025, 5:38 PM

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We claim that $JG \perp AG$ . To prove this, note that if $K'$ is the reflection of $K$ over $JG$ , then by Brokard's, $G' \coloneqq KL \cap MK' \cap GJ$ is the polar of $AG$ since $A$ is the polar of $KL$ . Similarly, $AF \perp FM$ . Since $AL$ is the reflection of $BC$ over $GJ$ , $G$ is the midpoint of $AT$ , and similarly $F$ is the midpoint of $AS$ . To show that $M$ is the midpoint of $ST$ , it thus remains to show that $AFMG$ is a parallelogram which follows by the perpendicularities.

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