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Source: India EGMO 2022 TST P5

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222 posts

#1 h Nov 28, 2021, 11:38 AM • 7 Y

Y by ike.chen, L567, tiendung2006, GuvercinciHoca, Euler365, sashamusta, cubres

Let $I$ and $I_A$ denote the incentre and excentre opposite to $A$ of scalene $\triangle ABC$ respectively. Let $A'$ be the antipode of $A$ in $\odot (ABC)$ and $L$ be the midpoint of arc $(BAC)$ . Let $LB$ and $LC$ intersect $AI$ at points $Y$ and $Z$ respectively. Prove that $\odot (LYZ)$ is tangent to $\odot (A'II_A)$ .

~Mahavir Gandhi

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222 posts

#2 h Nov 28, 2021, 11:42 AM • 1 Y

Y by L567

Solution 1(Rohan Goyal)

Let $I$ and $I_A$ denote the incentre and excentre opposite to $A$ of scalene $\triangle ABC$ respectively. Let $A'$ be the antipode of $A$ in $\odot (ABC)$ and $L$ be the midpoint of arc $(BAC)$ . Let $LB$ and $LC$ intersect $AI$ at points $Y$ and $Z$ respectively. Prove that $\odot (LYZ)$ is tangent to $\odot (A'II_A)$ .

Solution.
WLOG $AB<AC$ . Let $D=AI\cap BC, L'=AI\cap (ABC), P=DL'\cap (ABC), X=DL'\cap \ell$ where $\ell$ is tangent to $(ABC)$ at $L$ . $A'I\cap BC = R$ .

Claim 1. $BDPY$ is cyclic and similarly $CDPZ$ cyclic.
Proof. $\angle BYD=180-(C+\frac{A}{2})-(90-\frac{A}{2})=90-C=\angle BAA'=\angle BPA'=\angle BPD$ . $\square$

Claim 2. $PYLX$ is cyclic and thus similarly $PZLX$ is cyclic and thus $(LYZ)=(PYZLX)$ .
Proof. $\angle PYB=\angle PDB = \angle XDB=\angle LXD=\angle LXP$ . $\square$

Claim 3. $PIA'I_A$ is cyclic.
Proof. $DI\cdot DI_A= DB\cdot DC=DA'\cdot DI_A$ . $\square$

Claim 4. $A'I=A'I_A$ .
Proof. $A'L'\perp II_A$ but $L'I=L'I_A$ .

Claim 5. $\angle A'PL=\angle A'I_AP+\angle PXL$
Proof. $\angle PXL+\angle A'I_AP =\angle PDB+\angle II_AP+\angle A'I_AI=\angle A'DR+\angle RA'D+\angle I_AIA'=\angle IRD+\angle DIR=\angle IDB=\angle ADB=C+\frac{A}{2}=\angle A'PL$

Now, let $T$ be any point on the tangent at $P$ to $(A'II_A)$ inside $\angle A'PL$ . Now, $\angle A'PL=\angle A'I_AP+\angle PXL=\angle PXL+\angle A'PT=\angle PXL +\angle A'PL - \angle TPL\implies \angle PXL=\angle TPL\implies TP$ is tangent to $(PXL)$ i.e. $(LYZ)$ . Thus, $TP$ is tangent to $(A'II_A)$ and $(LYZ)$ at $P$ . Thus, they are both tangent to each other.

$\square$

Solution 2 (Mahavir Gandhi)
Let $L'$ be midpoint of arc $BC$ not containing $A$ in $\odot (ABC)$ and let $AD$ be altitude of $\triangle ABC$ with $D\in BC$ . Let $DL'$ intersect $\odot (ABC)$ again at point $P$ . Let $AI\cap BA'=K$ .

Claim 1: $P\in\odot (A'II_A)$ .
Proof: Let $AL'\cap BC=K$ . Then we have that $\triangle ADK\sim\triangle AL'A'$ . Hence there exists spiral similarity with centre $A$ taking $DK$ to $L'A'$ . Thus tehre exists spiral similarity with centre $A$ taking $DL'$ to $KA'$ . So we have that $DL'$ and $KA'$ intersect on $\odot (AL'A')$ ie $\odot (ABC)$ . Hence $P\in KA'$ .
Now consider cyclic quadrilaterals $BICI_A$ and $BPCA'$ . Since $K\in BC, II_A, PA'$ we have $KP.KA'=KB.KC=KI.KI_A$ . Hence $PIA'I_A$ is cyclic as desired.

Claim 2: $P\in\odot (LYZ)$
Proof: Let $\odot (LYZ)$ and $\odot (ABC)$ intersect at point $P'\neq L$ . Then we have that there exists spiral similarity with centre $P'$ taking $BY$ to $CZ$ . Hence $\triangle P'BY\sim\triangle P'CZ$ . Hence $\frac{}{}\frac{P'B}{P'C}=\frac{BY}{CZ}$ .
Now by applying sine rule in $\triangle ABY$ we have that $\frac{BY}{sin\left(\frac{A}{2}\right)}=\frac{AB}{sin(90^{\circ}+C)}$ , ie $BY=\frac{ABsin\left(\frac{A}{2}\right)}{cos(C)}$
By applying sine rule in $\triangle ACZ$ we have that $\frac{CZ}{sin\left(180^{\circ}-\frac{A}{2}\right)}=\frac{AC}{sin(90^{\circ}-B)}$ , ie $CZ=\frac{ACsin\left(\frac{A}{2}\right)}{cos(B)}$
Hence we have $\frac{P'B}{P'C}=\frac{BY}{CZ}=\frac{ABcos(B)}{ACcos(C)}=\frac{DB}{DC}$ .
Thus $P'D$ bisects $\angle BP'C$ or in other words $P',D,L'$ are collinear. Hence $P'\equiv P$ .
Thus we have $P\in\odot (LYZ)$ as desired.

Claim 3: $\odot (LYZ)$ and $\odot (A'II_A)$ are tangent at $P$ .
Proof: These circles are tangent iff $\angle PLY+\angle PA'I=\angle YPI$ . But $\angle PLY+\angle PA'I=\angle PLB+\angle PA'I=\angle PA'B+\angle PA'I=\angle BA'I=\angle KA'I$ .
$\angle YPI=\angle LPA'-\angle LPY-\angle IPA'=\left(C+\frac{A}{2}\right)-\angle LZY-\angle A'I_AI= \left(C+\frac{A}{2}\right)-(90^{\circ}-B)-\angle A'II_A=\left(90^{\circ}-\frac{A}{2}\right)-\angle A'IK=\angle AKB-\angle A'IK=\angle KA'I$ .
Thus $\angle PLY+\angle PA'I=\angle KA'I=\angle YPI$ as desired. This proves claim $3$ and hence the problem.
Q.E.D

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1184 posts

L567

#3 h Nov 28, 2021, 1:50 PM • 1 Y

Y by teomihai

Solved with bora_olmez, i3435

Let $AI \cap BC = K$ . Let $KA' \cap (ABC) = T$ . We claim $T$ is the desired point.

By PoP, we have $TK.KA' = BK.KC = KI.KI_A$ so $T \in (A'II_A)$ .

Note that since $\angle BTK = \angle BTA' = \angle BLA' = \angle LYA = \angle BYK$ so $BKYT$ cyclic and similarly $CKTZ$ cyclic.

So, $\angle YTL = \angle LTA' - \angle YTA' = \angle LBA' - \angle LBC = \angle CBA' = \angle CLA' = \angle CZA = \angle LZY$ so $T \in (LYZ)$ too.

For the tangency, we want $\angle LTA' = \angle LYT + \angle TI_AA'$ . But $\angle LYT = \angle TZC = \angle TKC$

Extend $I_AA'$ and $IA'$ to meet $(ABC)$ at $S,R$ . $R,S$ are symmetric wrt perpendicular bisector of $BC$ because they are the sharkydevil point and the excircle version of sharkydevil. So $\angle TI_AA' = \angle TI_AI + \angle AI_AS = \angle TA'R + \angle AI_AS = 90 - \angle RTA' + 90 - \angle MAS = 180 - \angle RTA' - \angle MAS = \angle RAA' - \angle MAS = \angle MAA'$

So $\angle LYT + \angle TI_AA' = \angle TKC + \angle MAA' = \angle LTA'$ , so the circles are indeed tangent, as desired. $\blacksquare$

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466 posts

#4 h Nov 28, 2021, 3:45 PM

Y by

L567 wrote:

Let $AI \cap BC = K$ . Let $KA' \cap (ABC) = T$ . We claim $T$ is the desired point.

Can smone tell me how to come up with this point?

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1079 posts

#8 h Nov 28, 2021, 6:53 PM

Y by

Let $E=\overline{AI}\cap\overline{BC}$ , let $M=\overline{AI}\cap\odot(ABC)$ and $D=\overline{A'E}\cap \odot (ABC)$ . I contend that $D$ is the desired tangency point.

Observe that $D$ lies on $\odot (A'II_A)$ by trivial PoP.
Claim: Let $F=\overline{LM}\cap\odot (LYZ)$ . Let $D'=\overline{FA}\cap\odot (LYZ)$ . Then, $D\equiv D'$ .
Proof. Firstly, I claim that $D'$ lies on $\odot (ABC)$ . Indeed, as $\triangle FLY\sim\triangle FYM$ by trivial angle chase, by Shooting Lemma and PoP, $\begin{align*} FA\cdot FD'=FY^2=FL\cdot FM, \end{align*}$ which means that $D'$ lies on $\odot (ABC)$ . Now, observe that $D'$ is the Miquel point of complete quadrilateral $CEBYZL$ and by trivial angle chase again, $\measuredangle ED'A=90^\circ$ , which means that $D'$ lies on $\overline{A'E}$ . Thus, $D\equiv D'$ . $\square$

Let $T$ be intersection of $\overline{AI}$ and tangent from $D$ to $\odot (LYZ)$ . Observe that $TD=TA$ , thus $T$ is the midpoint of $\overline{AE}$ . Now, as $(A,E;I,I_A)=-1$ , we get that $TE^2=TI\cdot TI_A$ , hence all in all, $TD^2=TI\cdot TI_A$ , thus $\overline{TD}$ is also tangent to $\odot(A'II_A)$ . $\blacksquare$

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1350 posts

i3435

#9 h Nov 28, 2021, 9:20 PM • 1 Y

Y by ike.chen

SPHS1234 wrote:

L567 wrote:

Let $AI \cap BC = K$ . Let $KA' \cap (ABC) = T$ . We claim $T$ is the desired point.

Can smone tell me how to come up with this point?

We thought that the circles might intersect on $(ABC)$ , and then Miquel/angle chase things would be of use. The radical axis of $(A'II_A)$ and $(ABC)$ goes through $K$ by radical axis with these circles and $(BIC)$ .

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1162 posts

#10 h Nov 29, 2021, 4:32 AM

Y by

Let $M_a$ be the midpoint of arc $BC$ , the reflection of $A'$ over $M_a$ be $A_2$ , $D = AA_2 \cap BC$ , and $E = AM_a \cap BC$ . Now, we consider an inversion about $(BICI_a)$ , which is centered at $M_a$ .

It's easy to see that $Y^*, Z^*$ are the projections of $B, C$ onto $AI$ respectively. We also know that $L^*$ is the midpoint of $BC$ and $A'^* = M_aA' \cap BC$ . Hence, it suffices to show $(L^*Y^*Z^*)$ and $(A'^*II_a)$ are tangent.

The Shooting Lemma and Fact $5$ implies $M_aA'^* \cdot M_aA_2 = M_aA'^* \cdot M_aA' = M_aB^2 = M_aI \cdot M_aI_a$ so $A_2$ lies on $(A'^*II_a)$ . In addition, we have $M_aI = M_aI_a$ and $\angle IM_aA'^* = \angle AM_aA' = 90^{\circ}$ so $A'^*A_2$ is actually a diameter of this circle.

It's easy to see that $ALA'M_a$ is a rectangle. Thus, we know $AL \parallel A'M_a \equiv A_2M_a$ and $AL = A'M_a = A_2M_a$ so $ALM_aA_2$ is a parallelogram. Now, we have $AA_2 \parallel LM_a \perp BC$ so $D$ is actually the projection of $A$ onto $BC$ . This yields $\angle A'^*DA_2 = 90^{\circ}$ , so $D$ also lies on $(A'^*IA_2I_a)$ .

Thales' implies $ACZ^*D$ is cyclic, and the Iran Lemma gives $L^*Y^* \parallel AC$ . Thus, $Y^*L^*Z^*D$ is cyclic by Reim's. Now, we will show that $D$ is the desired tangency point. Once again, Iran Lemma yields $L^*Z^* \parallel AB$ , and it's easy to see $A_2DEM_a$ is cyclic via Thales'. Hence, we have $\angle DZ^*L^* = \angle DZ^*A + \angle AZ^*L^* = \angle DCA + \angle Z^*AB$ $= \angle ECA + \angle EAC = \angle AED = \angle AA_2M_a = \angle DA_2A'^*$ which finishes since $D, L^*, A'^*$ are collinear. $\blacksquare$

This post has been edited 1 time. Last edited by ike.chen, Nov 29, 2021, 4:42 AM

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163 posts

Pindp

#11 h Nov 30, 2021, 10:44 AM • 2 Y

Y by Rg230403, phongzel34-vietnam

This is my generalization problem.
Generalization problem. Let $ABC$ be a triangle, $A'$ be the antipode of $A$ in $\odot (ABC)$ . $P$ , $Q$ are points such that $\angle BAP = \angle CAQ$ and $\angle PBQ = \angle PCQ = 90^\circ$ . Line passes through $A$ and perpendicular to $PQ$ cuts $\odot (ABC)$ at second point $L$ . $PQ$ cuts $LB, LC$ at $X,Y$ , respectively. Prove that $\odot (LXY)$ is tangent to $\odot (A'PQ)$ .

Attachments:

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#12 h Mar 1, 2022, 1:47 PM • 1 Y

Y by StarLex1

Really nice Problem
Let $I_aA'I$ meet $ABC$ at $S$ , $AI$ meet $BC$ at $K$ and $T$ be midpoint of arc $BC$ .
Claim1 : $A'I = A'I_a$ .
Proof : It's well-known that $T$ is midpoint of $II_a$ we also have $\angle A'TI = \angle A'TA = \angle 90$ so $A'$ lies on perpendicular bisector of $II_a$ so $A'I = A'I_a$ .

Claim2 : $A',K,S$ are collinear.
Proof : Let $A'K$ meet $I_aA'I$ at $S'$ . we have $S'K.KA' = IK.KI_a = BK.KC$ so $S'$ lies on $ABC$ so $S'$ is exact $S$ .

Claim3 : $SYKB$ is cyclic.
Proof : $\angle BYK = \angle A/2 + \angle B/2 - \angle C/2 = \angle 90 - \angle C = \angle BAA' = \angle BSA' = \angle BSK$ .

Claim4 : $ZLYS$ is cyclic.
Proof : $\angle YZL = \angle A/2 - \angle B/2 - \angle C/2 = (\angle 180 - \angle B/2 - \angle C/2) - (\angle B + \angle A/2) = \angle BSL - \angle BSY = \angle YSL$ .

Let $\ell$ be tangent to $I_aA'I$ at $S$ and $X$ an arbitrary point on it.
Claim5 : $\ell$ is tangent to both $I_aA'I$ and $LYZ$ .
Proof : $\angle YSX = \angle YSK - \angle XSK = \angle B/2 + \angle C/2 - \angle XSI - \angle ISA' = \angle B/2 + \angle C/2 - \angle IKS = \angle SLB = \angle SLY = \angle SZY$ so $\ell$ is tangent to $LYZ$ .

we're Done.

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1517 posts

#13 h Sep 3, 2022, 9:13 PM

Y by

I mean...its ok to get scared of this but an inversion will calm ur soul. (Wlog $AB \le AC$ )
Step 1: Identify the point of tangency $X$
Let $H$ be the projection of $A$ over $BC$ , $AI$ hits $BC,(ABC)$ in $K,N$ respectivily, now let $(IA'I_A)$ hit $(ABC)$ again at $X$ , by radax on $(ABC),(IA'I_A),(BICI_A)$ we get that $X,K,A'$ are colinear and that means $AXHK$ cyclic. Now by angle chase:
$\angle BYK=90-\angle ALB=\angle HAC=\angle BAA'=\angle BXK \implies BYXK \; \text{cyclic}$ Hence $X$ is the miquel point of $KYLC$ meaning that $CZXK$ and $ZXYL$ are cyclic.
Step 2: Invert, no more words lol
Make a $\sqrt{bc}$ inversion then $NA' \cap BC=X'$ and it holds that $(HIX'I_A)$ is cyclic and using the pervious cyclic quads we get that $Y',Z'$ are points in $AI$ such that $Z'ABL', Y'ACL', L'X'Y'Z'$ is cyclic, now by I-E lemma $IN=NI_A$ hence $NX'$ is the perpendicular bisector of $II_A$ and so, it passes through the center of $(HIX'I_A)$ , also notice that $\angle Z'L'X'=\angle BAI=\angle CAI=\angle X'L'Y'$ so $X'$ is midpoint of arc $Z'Y'$ in $(L'Z'X'Y')$ and that means $NX'$ is the perpendicular bisector of $Y'Z'$ meaning that $NX'$ passes through the center of $(L'Z'X'Y')$ hence the centers of $(L'X'Y'Z'),(HIX'I_A)$ and $X'$ are colinear hence those circles are tangent, inverting back we get $(IA'I_A),(LYZ)$ tangent as desired.
Thus we are done

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522 posts

#15 h Jul 11, 2023, 1:27 PM • 1 Y

Y by GeoKing

https://cdn.aops.com/images/2/2/c/22c2849e889d6f9901dd808852a2eba47f580d51.png

This post has been edited 2 times. Last edited by trinhquockhanh, Jul 11, 2023, 2:36 PM

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#16 h Oct 18, 2023, 1:07 PM

Y by

IMHO this is > IMOP5 level.
Will add soln later

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164 posts

Om245

#17 h Jan 10, 2024, 6:49 PM • 1 Y

Y by Deadline

Rename $Z$ as $X$ .Let $R= (LXY) \cap (ABC)$ , $D = AI \cap BC$ .

$\angle XYL = \angle ADB - \angle LCB = 90 - \angle C$ and $\angle YXL = 180 - \angle ADC - \angle XBC = 90 - \angle B$

Claim : $\triangle RXY \sim \triangle RBC$

$\angle RXY = \angle RLY = \angle RBC$ , $\angle RYX = \angle RLX = \angle RCB$ and $\angle XRY = \angle BAC$

Hence Spiral Similarty At $R$ give us $XY \rightarrow BC$

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Claim : This Spiral Similarity also give us $\overline{RL} \rightarrow {RD}$ (that mean image of $L$ will lie on $RD$ only)

Note some spiral similarity $B \rightarrow X$ and $C \rightarrow Y$ .Now
$\angle BRX = 180 - \frac{\angle A}{2} - \angle C$ $\angle LRD = \angle LRX + \angle XRD = \angle LYX + \angle XBL = 90 - \angle C + 90 - \frac{A}{2} = 180 - \frac{\angle A}{2} - \angle C$
Let $L' = RD \cap (ABC)$ .From our claims we get this similarity also send $L$ to $L'$

Note $\angle XRL = 90 - \angle C$ and $\angle LRY = 90 - \angle B$ .
so we get $\angle BRL' = 90 - \angle C$ and $\angle L'RC = 90 - \angle B \implies L' = A'$

Claim: $R$ lie on $(AI_AI)$

By power of point
$DR.DA'=DB.DC=DI.DI_A$
Which give us $(XLY) , (ABC)$ and $(AI_AI)$ intersect on $R$

Note if $(XLY)$ and $(AI_AI)$ are tangent to each other then $\angle LRA'$ is sum of angle made by respective circle tangents which is $\angle RYL + \angle RI_AA'$

$\angle RYL = \angle RXB = 180 - \angle RDB$
$\angle RI_AA' = \angle IA'R + \angle IRA' = \angle IA'R + \angle II_AA' = \angle IA'R + \angle I_AIA' = \angle A'DI_A = \angle ADR$
So $\angle RYL + \angle RI_AA' = 180 - \angle C - \frac{\angle A}{2}$
As we know $\angle LRA' = 180 - \angle C - \frac{\angle A}{2}$

We get $(XLY)$ and $(AI_AI)$ are tangent to each other.

This post has been edited 1 time. Last edited by Om245, Jan 11, 2024, 2:22 AM
Reason: rename variable

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#18 h Nov 4, 2024, 3:33 PM

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Let $(LYZ)\cap (ABC)=T,AI\cap BC=X$ and the perpendicular line from $A$ to $BC$ intersect $(ABC)$ at $H$ .

Lemma: $ABC$ is a triangle with incenter $I$ and $AI$ intersects $BC,(ABC)$ at $X,M$ respectively. $A'$ is the antipode of $A$ on $(ABC)$ . Also $A'X$ intersects $(ABC)$ at $E$ and $EI$ meets $(ABC)$ at $F$ . If $N$ is the midpoint of arc $BAC$ , then show that $NF\parallel A'I$ .
Proof: Let $MF\cap BC=R,A'I\cap (ABC)=S,AS\cap BC=P,PM\cap (ABC)=U$ . $K$ lies on $BC$ such that $AK\perp BC$ . Note that $S$ is $A-$ sharky devil point and $\measuredangle PIA=90$ . $U$ is $A-$ mixtilinear touch point hence $N,I,U$ are collinear.
Also we observe that under the inversion centered at $M$ with radius $MI$ , $EIF$ swaps with $(MKIR)$ .
$\measuredangle SMN=\measuredangle SUN=\measuredangle SUI=\measuredangle SPI=\measuredangle API=\measuredangle AKI=90-\measuredangle IKR=90-\measuredangle IMR=\measuredangle FMA'$ Thus, $NF\parallel SA'$ . $\square$

Claim: $T,X,A'$ are collinear.
Proof:
$\measuredangle LZY=\measuredangle CAI-\measuredangle ACL=\frac{\measuredangle A}{2}-\frac{\measuredangle B-\measuredangle C}{2}=90-\measuredangle B=\measuredangle HCB$ $\measuredangle ZYL=\frac{\measuredangle A}{2}+\frac{\measuredangle B-\measuredangle C}{2}=90-\measuredangle C=\measuredangle CBH$ Hence $HBC\sim LYZ$ . Also since $HD\perp BC$ and $LA\perp YZ$ , we get $HBDC\sim LYAZ$ . $T$ is the center of spiral homothety sending $BC$ to $YZ$ thus, $TBDCH\sim TYAZL$ .
$\measuredangle XAT=\measuredangle YAT=\measuredangle BDT=180-\measuredangle TDX$ Which implies $A,X,D,T$ are concyclic. Note that $AX$ is diameter on this circle subsequently, $\measuredangle ATX=90=\measuredangle ATA'$ . So $T,X,A'$ are collinear. $\square$
Claim: $A',I_A,I,T$ are concyclic.
Proof:
$XI_A.XI=XB.XC=XA'.XT$ Which gives the expected result. $\square$
Claim: $(LYZT)$ and $(I_AA'IT)$ are tangent to each other.
Proof: Let $AH\perp BC$ and $H\in (ABC),TI\cap (ABC)=W$ .
$\measuredangle LZT+\measuredangle TA'I=(90-\measuredangle B+\measuredangle YZT)+\measuredangle TA'S=(90-\measuredangle B+\measuredangle BLT)+\measuredangle TLS=\measuredangle HLT+\measuredangle TLS=\measuredangle HLS$ By Lemma, we have that $LW\parallel A'S$ . Apply Pascal on $LLWHA'S$ to get that $BC_{\infty},A'S_{\infty},WH\cap SL$ are collinear. Since this line must be at infinity, $SL\parallel HW$ . Hence $\measuredangle LZT=\measuredangle HLS=\measuredangle LTW$ which gives the tangency of $(LYZT)$ and $(I_AITA')$ as desired. $\blacksquare$

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L13832

#19 h Jan 17, 2025, 12:32 PM • 2 Y

Y by alexanderhamilton124, GeoGuy3264

200th post :flex:

:flex:

Solved with math_holmes 15
Really nice problem :10:

:10:

, though only angle chasing was involved, it was fun.
$\rule{35cm}{0.1pt}$
Let $AM\cap BC=D$ , $A'D\cap \odot(ABC)=R$ where $M$ is the $L-$ antipode.
Claim: $\odot(RYBD)$ is cyclic.
$\angle BRD=\angle BLA'=\angle BYD$ Claim: $\odot(RYLZ)$ is cyclic.
$\angle ZLR=\angle RBC=\angle ZYR$ Claim: $\odot(RIA'I_A)$ is cyclic.
$RD\cdot RA'=BD\cdot CD=DI\cdot DI_A$ Let $N$ be the midpoint of $AD$ , then
$\rule{35cm}{0.1pt}$

Claim: $NR$ is tangent to $\odot(RYLZ)$ and $\odot(RIA'I_A)$ and $NR=ND$
$\begin{align*} \angle A'II_A=\angle A'RI_A=\angle A'RI=\angle A'I_AD=x\\ \angle A'I_AR=\angle A'DI_A=\angle RDN=\angle IRD=y \end{align*}$ then we have $\angle RIN=x+y$ and $NRD=y$ so $\angle NRI=y-x$ and similarly we also have $y-x=\angle A'I_AI$ .
So $NR$ is tangent to $\odot(A'I_ARI)$ .

Now we show $NR$ is also tangent to $\odot(LYZR)$ .
$\begin{align*} \angle YRN &= \angle YRD - \angle NRD\\ &= \angle LBD - (\angle NRI + \angle IRA')\\ &=\left(90^{\circ}-\frac{\angle A}{2}\right)-(\angle IA'R + \angle IRA')\\&=\left(90^{\circ}-\frac{\angle A}{2}\right)-\angle RI_AA'\\&=\left(90^{\circ}-\frac{\angle A}{2}\right)-\angle IDR\\&=\angle RAB\\&=\angle RLY \end{align*}$ Thus $\odot (LYZ)$ is tangent to $\odot (A'II_A)$ at $R$ and we are done.
$\rule{35cm}{0.1pt}$
$[asy] import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.71762211856687, xmax = 11.335548360553895, ymin = -12.376226640019961, ymax = 13.078331916213367; /* image dimensions */ draw((-6,4)--(-8,-2)--(0,-2)--cycle, linewidth(0.6)); draw(arc((-2.9196369730490948,-9.048627177541055),1.0432196129603841,79.6764268057746,103.282525588539)--(-2.9196369730490948,-9.048627177541055)--cycle, linewidth(0.6)); draw(arc((-8.373545369999913,0.9338634249997817),1.0432196129603841,-37.74402707459535,-14.137928291830944)--(-8.373545369999913,0.9338634249997817)--cycle, linewidth(0.6)); draw(arc((-8.373545369999913,0.9338634249997817),1.0432196129603841,-61.35012585735976,-37.74402707459536)--(-8.373545369999913,0.9338634249997817)--cycle, linewidth(0.6)); draw(arc((-5.080363026950906,0.10435526754189517),1.0432196129603841,-76.71747441146101,-53.111375628696614)--(-5.080363026950906,0.10435526754189517)--cycle, linewidth(0.6)); draw(arc((-2.9196369730490948,-9.048627177541055),1.0432196129603841,79.6764268057746,118.64987414264026)--(-2.9196369730490948,-9.048627177541055)--cycle, linewidth(0.6) + blue); draw(arc((-4.583592135001262,-2),1.0432196129603841,103.282525588539,142.25597292540465)--(-4.583592135001262,-2)--cycle, linewidth(0.6) + blue); /* draw figures */ draw((-6,4)--(-8,-2), linewidth(0.6)); draw((-8,-2)--(0,-2), linewidth(0.6)); draw((0,-2)--(-6,4), linewidth(0.6)); draw(circle((-4,0), 4.47213595499958), linewidth(0.6)); draw((-6,4)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw((-4,4.472135954999577)--(-8,-2), linewidth(0.6)); draw((0,-2)--(-7.416407864998738,10), linewidth(0.6)); draw((-7.416407864998738,10)--(-6,4), linewidth(0.6)); draw(circle((-8.472135954999578,5.527864045000421), 4.595058410947224), linewidth(0.6)); draw(circle((-8.236067977499792,-5.4721359549995805), 6.407474392457677), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-2,-4), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-5.52786404500042,2), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-4,4.472135954999577), linewidth(0.6)); draw((-2,-4)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-5.080363026950906,0.10435526754189517), linewidth(0.6)); draw((-5.291796067500631,1)--(-8.373545369999913,0.9338634249997817), linewidth(0.6)); draw((-4,4.472135954999577)--(-4,-4.472135954999577), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw(circle((-5.291796067500631,1), 3.0824588902380468), linewidth(0.6)); draw((-5.080363026950906,0.10435526754189517)--(-2,-4), linewidth(0.6)); /* dots and labels */ dot((-6,4),linewidth(3pt) + dotstyle); label("$A$", (-6.712150943660748,4.2109652060501155), NE * labelscalefactor); dot((-8,-2),linewidth(3pt) + dotstyle); label("$B$", (-8.659494221186797,-2.5351882910936903), NE * labelscalefactor); dot((0,-2),linewidth(3pt) + dotstyle); label("$C$", (0.31219445027250453,-2.430866329797652), NE * labelscalefactor); dot((-5.080363026950906,0.10435526754189517),linewidth(3pt) + dotstyle); label("$I$", (-4.938677601628095,0.3162786509980215), NE * labelscalefactor); dot((-2.9196369730490948,-9.048627177541055),linewidth(3pt) + dotstyle); label("$I_A$", (-2.5392724918192116,-9.281341788237496), NE * labelscalefactor); dot((-2,-4),linewidth(3pt) + dotstyle); label("$A'$", (-1.739470788549584,-4.552079542817096), NE * labelscalefactor); dot((-4,4.472135954999577),linewidth(3pt) + dotstyle); label("$L$", (-3.8606840015690316,4.697801025431628), NE * labelscalefactor); dot((-4.583592135001262,-2),linewidth(3pt) + dotstyle); label("$D$", (-5.112547537121492,-2.848154174981805), NE * labelscalefactor); dot((-5.52786404500042,2),linewidth(3pt) + dotstyle); label("$Y$", (-5.21686949841753,1.6376901607478391), NE * labelscalefactor); dot((-7.416407864998738,10),linewidth(3pt) + dotstyle); label("$Z$", (-7.268534737239619,10.192090987022974), NE * labelscalefactor); dot((-8.373545369999913,0.9338634249997817),linewidth(3pt) + dotstyle); label("$R$", (-9.042008079272271,1.359498263958404), NE * labelscalefactor); dot((-5.291796067500631,1),linewidth(3pt) + dotstyle); label("$N$", (-4.9039036145294155,1.0117583929716096), NE * labelscalefactor); dot((-4,-4.472135954999577),linewidth(3pt) + dotstyle); label("$M$", (-4.764807666134698,-5.282333271889364), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 7 times. Last edited by L13832, Feb 9, 2025, 3:27 AM

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#20 h Mar 4, 2025, 6:07 PM • 1 Y

Y by cursed_tangent1434

Feels a bit on the harder side onf P2/P5.

We will define some new points as usual.

$\bullet$ $N=\overline{AI} \cap (ABC)$ .
$\bullet$ $D=\overline{AI} \cap \overline{BC}$ .
$\bullet$ $S$ is the point on $(ABC)$ such that $\measuredangle ASD=90 ^{\circ}$ .
$\bullet$ $T=\overline{AI} \cap \overline{LL}$ .

Claim: $S \in (II_AA')$ .
Proof: Just PoP (and I-E lemma) to get $DS \cdot DA'=DB \cdot DC=DI \cdot DI_A$ . $\square$

Claim: $(TLDS)$ is cyclic.
Proof: Just angle chase $\measuredangle TLS=\measuredangle LNS=\measuredangle LAS=90 ^{\circ}+\measuredangle NAS=90^{\circ}+\measuredangle DAS=\measuredangle TDS$ And done. $\square$

Claim: $(LSYZ)$ is cyclic.
Proof: See that $(D,N;Y,Z) \overset L= (\overline{LD} \cap (ABC),N;B,C) \overset D= (L.A;C,B)=(A,L;B,C) \overset L= (A,T;Y,Z)$ And so we get that $\begin{align*} \frac{YA \cdot YN}{YT \cdot YD}=\frac{ZA \cdot ZN}{ZT \cdot ZD} \iff \frac{\text{Pow}(Y,(ABC))}{\text{Pow}(Y,(DTLS))}=\frac{\text{Pow}(Z,(ABC))}{\text{Pow}(Z,(DTLS))} \end{align*}$ And we are done by the OG Coaxiality lemma. $\square$

Claim: $(SBDY)$ is cyclic.
Proof: See that $\triangle SAD \overset + \sim \triangle SLN$ by easy angle chasing and so just $\measuredangle SBY=\measuredangle SNL=\measuredangle SDA=\measuredangle SDY$ . $\square$

To finally prove that the two circles are tangent to $S$ , it is just one big aah angle chase. First see that $\overline{A'N}$ is $\perp$ bisector of $\overline{II_A}$ and so $\begin{align*} \measuredangle LYS+\measuredangle SA'I= & \measuredangle BDS+\measuredangle DA'I \\ =& \measuredangle CDA+\measuredangle ADS+\measuredangle AIA'+\measuredangle A'DA \\ =&\measuredangle CDA+\measuredangle ADA'+\measuredangle AIA'+\measuredangle A'DA \\ =& \measuredangle CDA+\measuredangle AIA'=\measuredangle LNA'+\measuredangle A'I_AA \\ =& \measuredangle LNA'+\measuredangle A'SI=\measuredangle LSA'+\measuredangle DSI=\measuredangle LSI \end{align*}$ And finally done bruh.

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cursed_tangent1434

#21 h Mar 5, 2025, 5:45 PM • 1 Y

Y by ihategeo_1969

Not a bad problem, still it's pretty straightforward after guessing the tangency point. Let $D$ denote the foot of the $A-$ angle bisector and $X$ the second intersection of $\overline{DA'}$ with $(ABC)$ . We first have the following preliminary claims.

Claim : Quadrilaterals $BXYD$ and $CDXZ$ are cyclic.

Proof : Simply observe that,
$\measuredangle DXB = \measuredangle A'XB = \measuredangle A'AB = \measuredangle AYL = \measuredangle DYB$ which implies that points $B$ , $D$ , $X$ and $Y$ are concyclic. The other part of the claim also follows similarly.

We now show that $X$ is an intersection of the two circles.

Claim : Point $X$ lies on both circles $(I_aIA')$ and $(LYZ)$ .

Proof : For the first half, simply note that considering the power at $D$ we have,
$DI_A \cdot DI = DB \cdot DC = DX \cdot DA'$ which implies that $XIA'I_a$ is indeed cyclic. Next note,
$\measuredangle YZX = \measuredangle DZX = \measuredangle DCX = \measuredangle BCX = \measuredangle BLX = \measuredangle YLX$ which indicates that $X$ also lies on circle $(LYZ)$ .

All that remains is to show that the two given circles are indeed tangent to each other at $X$ . For this, we define $O$ as the midpoint of segment $AD$ , or equivalently the center of $(XAD)$ . We note the following.

Claim : Line $OX$ is the common internal tangent to circles $(II_aA')$ and $(LYZ)$ at $X$ .

Proof : It is well known that $(AD;II_a)=-1$ . Further,
$(AD;YZ)\overset{L}{=}(A,LD \cap (ABC);B,C)\overset{D}{=}(AD \cap (ABC),L;CB)=-1$ which implies that
$OY\cdot OZ = OA \cdot OD = OX^2$ and similarly,
$OI \cdot OI_a=OA\cdot OD = OX^2$ from which the claim clearly follows.

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