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NT_G

#1 h Dec 31, 2023, 1:13 PM • 2 Y

Y by GeoKing, SBYT

1. I is incenter of triangle $ABC$ . $K$ is the midpoint of arc $BC$ not containing $A$ . $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$ . $D$ is projection of $I$ onto $BC$ . $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$ .

2. Circle $w$ with center $I$ is incircle of triangle $ABC$ . $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$ , $CI$ are following condition: $IE = IF$ . $M$ is the midpoint of $EF$ . $P$ is the second intersection of $(IMD)$ and $w$ . $Q$ is the second intersection of $AP$ and $(IMD)$ .
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.

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NT_G

#2 h Dec 31, 2023, 8:11 PM • 1 Y

Y by GeoKing

P1 has a generalisation:
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SBYT

#3 h Jan 2, 2024, 10:54 AM • 3 Y

Y by GeoKing, vuanhnshn, Om245

Proof of part 1

It is well know that $\angle ASI=\frac{\pi}{2}$ (Sharky Devil).Let $AS$ meets $BC$ at $E$ ,so $EI$ is the diameter of $\odot ISD$ .
By $KI^2=KB^2=KD\cdot KS$ , we can know that $KI$ is a tangent line of $\odot IDS$ ,so $AK\perp IE$ .
Let $\odot ISD$ meets $\odot ABC$ at $S,F$ , $\angle SFI=\angle SEI=\frac{\pi}{2}-\angle SAI=\frac{\pi}{2}-\angle SLK=\angle SKL=\angle SFL$ ,so $F,I,L$ are conlinear.
By $MF=MI=MK$ , we can get $MF$ is another tangent line of $\odot IDS$ .
It means that $FI$ is the polar of $M$ to $\odot ISD$ ,and $L$ lies on this line.
By $N$ is the midpoint of $ML$ , we know $N$ lies on the radical axis of $\odot ISD$ and $\odot M$ .
Let $NI$ meets $\odot ISD$ at $G$ again,then $NI\cdot NG=NM^2=NL^2=NA^2$ .
So $A,N,M,G$ are concyclic.
Let $GP$ is a tangent line of $\odot ISD$ ,then $\angle IGP=\angle GIK=\angle AIN=\angle NAG$ ,so $IG$ tangents $\odot AMN$ ,too. $\Box$ .

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Om245

#4 h Jan 2, 2024, 2:53 PM • 2 Y

Y by ATGY, GeoKing

Solution of frist part

It well known that $S$ is Sharky-Devil point (as if $S$ is Sharky-Devil point then $SD$ is angle bisector so $S - D - K$ )
$\angle SIA = \angle AES = \angle AEF - \angle SEF$
Spiral similarity at $S$ sends $\overline{EF}$ to $\overline{BC} \implies \angle SEF = \angle SCB$
so $\angle SIA = 90 - \frac{\angle A}{2} - \angle SCB$ . From $S - D - K \implies \angle KSC = \angle KAC$

So $\angle SIA = 90 - \angle SDA$ from $\overline{ID} \perp \overline{BC}$ we get $\angle SIA = \angle SDI$ hence we get $\overline{AK}$ tangent to $(SID)$ .

Observe $\angle LAM = 90$ and $N$ is midpoint of $LM$ we get $AN = NM$ .

Let $Y =\overline{LI} \cap (ABC)$ from $\angle LYK = 90$ . $Y$ also lie on circle $w$ with center $M$ with radius $MI$ .
$\overline{MI}$ is tangent to $(SID)$ and $MI = MY$ so $MY$ also tangent to $(SID)$ .

Let $X = \overline{LI} \cap (SID)$ and $P = \overline{MX} \cap (SID)$
$(Y,I;X,P)\stackrel{I}{=}(L,M;N,\infty{\overline{LM}}) = -1 \implies \overline{IP} \parallel \overline{LM}$
$\angle IXM = \angle PIM = \angle IMN \implies \angle IXM = \angle XMN$
Now as $AN=NM$ we get $\angle IXM = \angle XMN = \angle MAN$ so $X,A,N,M$ cyclic points.

Now as tangent to $I$ and $N$ are parallel, Homothety at $X$ sends $(SID)$ to $(ANM)$ hence $X$ is tangent point of $(SID)$ and $(ANM)$

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NT_G

#5 h Jan 11, 2024, 2:18 PM • 3 Y

Y by Grifon, SBYT, mathlove_13520

My first solution of P1 was similar to SBYT's. I found it accidently while Cartesian bashing the problem for the New Year stream on NeuroGeometry. Trick with proving tangency using radical axis was new for me...

My second solution:
Due to the fact that $ID \parallel KL$ , $\angle IDS = \angle LKS$ , therefore $LI$ passes through the second intersection of $(SID)$ and $(ABC)$ . It means that $(SID)$ contains the point of tangency of circumcircle and mixtilinear circle of $\triangle ABC$ .
Now we invert the problem with center in $A$ . Problem turns into:
In triangle $ABC$ $I_a$ is the $A$ - excenter. $N$ is the projection of $I_a$ onto $BC$ . $L, L_a$ are foots of bisectors of $\angle BAC$ . K is a point on $AI_a$ such that $(I,I_a; K, A) = -1$ . $A'$ is a point on $\Gamma = (AL_{a}K)$ suiting: $KA = KA'$ . $\gamma$ is a circle passing through $N, I_a$ that is tangent to $AI_a$ . We have to prove that $KA'$ is tangent to $\gamma$ .
Let's spot that due to the fact that $\angle L_{a}AK = \frac{\pi}{2}$ , $A'$ is the reflection of $A$ in $KL_a$ .Therefore, we have to prove that line $O_{\Gamma}K$ passes through $O_{\gamma}$ . $L_a, K, O_{\Gamma}$ are obviously collinear. Let $I_{a}'$ be the reflection of $I_{a}$ over $O_{\Gamma}$ . $I_{a}'$ lies on $BC$ . Then after spotting that $I_{a}I_{a}' \parallel AL_{a}$ and projecting $(I,I_a; K, A)$ from $L_a$ onto $I_{a}I_{a}'$ we get what we needed.

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#6 h Jan 11, 2024, 7:46 PM • 1 Y

Y by GeoKing

First problem. $G=AS \cap BC$ is on $(SID)$ . $LI \cap ABC = J$ . $A'$ is the antipode of $A$ and $S,I,A'$ are collinear. By radical axis theorem on $(ASFIE), (IJK), (ABC)$ , $G$ is on $JK$ . $H = IN \cap (SID)$ . Let $I'$ be the reflection of $I$ across $N$ .

$LIMI'$ is parallelogram so $LI' = MI = MK$ and $I'M \parallel JL \implies LJMI'$ is cyclic.

$\angle I'HJ + \angle JMI' = \angle ISJ + \angle JMI + \angle IMI' = \angle JAA' +\angle JMI + \angle JIM = \angle JAA' + 90 + \angle JKA = \angle JAA' + 90^{\circ} + \angle JA'A = 180^{\circ} \implies H \in (LJMI')$ .

$NA^2 = NM^2 = NM \cdot NL = NI' \cdot NH = NI \cdot NH \implies HMNA$ is cyclic. If tangents to $(SID)$ at $I$ and $H$ intersect at $Q$ . $\angle NHQ = \angle HIQ = \angle HAN \implies HQ$ is tangent to $(HMNA)$ .

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#7 h Jan 13, 2024, 7:59 PM • 1 Y

Y by GeoKing

Second problem. Let $G= EF \cap BC$ . $(IMDG) \cap (I) = P$ , $(IMDG) \cap (IEF) = Q$ , $(IMDG) \cap AI = L$ , $PM \cap GL = S$ , $N$ is the antipode of $I$ in $(IEF)$ and $AI \cap (I) = J$ . I will prove that $A \in PQ$ .

$\angle IQG = 90^{\circ} = \angle IQN \implies N \in QG$ .

$\angle EID = 90^{\circ} - \frac{B}{2} = \angle JIF \implies J$ is the reflection of $D$ across $AI$ . $\angle IJM = \angle IDM = \angle IPM = \angle IJP \implies J \in PM$ . $\angle LGD = \angle DIJ = 2 \cdot \angle DIM = 2 \cdot \angle DGM \implies ML = MD \implies M$ is the center of $(SLJD)$ .

$N,S,A$ are collinear $\iff$ (by Menelaus)
$\frac{NM}{NI} \cdot \frac{IA}{AJ} \cdot \frac{JS}{SM} = 1$ This is true because $\frac{NM}{NI} = \cos^2 (\angle IFE) = \frac{1+\cos (90^{\circ} - \angle A/2) }{2}= \frac{JA}{IA} \cdot \frac{JS}{SM}$ .

Applying Pascal on $PQG LIM$ shows that $PQ \cap LI \in SN \implies A=PQ \cap LI$ as desired.

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NT_G

#9 h Jan 15, 2024, 7:17 PM • 1 Y

Y by GeoKing

P2: (R.Prozorov's solution)
Let $R$ be the second intersection of $(IMD)$ , $BC$ . Obviously, $R$ lies on $EF$ , and $IR$ is diameter of $(IMD)$ .
$\angle IQD = \angle PQI$ . So, using DIT for $ABDC$ , we get that $\angle BQI = \angle CQI$ .
$K = IQ \cap BC$ . $IQ \perp QR$ , therefore $(R,K; B,C) = -1 \Rightarrow$ $CE$ , $BF$ , $IQ$ are concurent. Using isogonal theorem for $\angle BQC$ and points $E$ , $F$ , we prove that $\angle EQI = \angle IQF$ . Therefore, due to the equality: $IE = IF$ , we get what we needed.

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SBYT

#10 h Jan 16, 2024, 2:37 AM • 1 Y

Y by GeoKing

Another solution which is similar to NT_G's:
We get $(R,K;B,C)=-1$ the same way, $IK$ meets $EF$ at $N$ . $(R,K;B,C)=-1\implies(IR,IK;IB,IC)=-1\implies(R,N;E,F)=-1\implies(QR,QN;QE,QF)=-1$ .
Due to $IQ\perp RQ$ , $IQ$ is the angle bisector of $\angle EQF$ ,and $IE=IF$ ,so $I,E,Q,F$ are concyclic. $\Box$

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#11 h Apr 10, 2025, 11:05 AM • 1 Y

Y by MS_asdfgzxcvb

I 'll present a solution to the first problem.
Let $AS\cap BC=R,MR\cap (ABC)=T$ . Let $R,W'$ be the feet of the perpendicular from $P,N$ to $IN,PI$ respectively. Let $W,I'$ be the reflections of $I$ over $W',N$ . Note that $P,T,I,R,S,D$ are concyclic.
Claim: $A,M,N,R$ are concyclic.
Proof: $A,P,K,W$ are concyclic because $IP.IW=IL.IT=IA.IK$ so $A,P,M,W'$ are concyclic. $IA.IM=IP.IW'=IN.IR$ which gives the result.
Let $PR\cap (AMNR)=V$ which is the antipode of $N$ . Since $\measuredangle LAM=90$ , we have $NA=NM$ and $NV$ is diameter thus, $NV\perp AI\perp IP$ . Hence $(RPI)$ and $(RNV)$ are tangent to each other as desired. $\blacksquare$

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