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Source: 2020 ISL G8

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dolphinday

1310 posts

dolphinday

#57 h Feb 6, 2024, 7:04 PM

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Let $K$ be the intersection point of the tangent at $I$ and $\Gamma$ . Then $\angle BIC = \angle BIK + \angle CIK = \angle IMB + \angle CNI = \angle YCI + \angle YBI$ . Note that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ so $\angle BYC = 360^{\circ} - (180^{\circ} - \angle A) = 180^{\circ} - \angle A \implies ABYC$ is cyclic.
Notice that $MNC_1B_1$ is an isosceles trapezoid due to the fact that the perpendicular bisectors of $MB_1$ and $NC_1$ coincide. Note that $PBCQ$ is cyclic.
We can then prove that $\triangle XMN \sim \triangle XQP$ . $\angle XNM = 180^{\circ} - (\angle MNC_1 + \angle C_1NQ) = 180^{\circ} - (\angle MBB_1 + 180^{\circ} - \angle C_1CQ)$ $= \angle BPM + \angle C_1CQ - 180^{\circ} = \angle BPM - \angle BPQ = \angle XPQ$ .
By using similar triangle ratios, we find that $XM \cdot XP = XN \cdot XQ$ , so $AX$ is the radical axis of $\omega_B$ and $\omega_C$ . Similarly, we have $XN \cdot XQ = XM \cdot XP \implies AX$ is the radical axis of $(AMP)$ and $(ANP)$ . All that is left is to show that $AY$ is the radical axis of $(AMP)$ and $(ANP)$ which is equivalent to $AY$ tangent to both circle. Notice that $PAYB$ is cyclic. Then $\angle PMB = 180^{\circ} - PBY = \angle PAY$ , which proves the tangency, so we are done.

This post has been edited 1 time. Last edited by dolphinday, Feb 6, 2024, 7:04 PM

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jp62

53 posts

jp62

#58 h Feb 18, 2024, 7:27 AM

Y by

Angle chase :sleeping:

We use directed angles mod $180^\circ$ throughout this solution.

ABCY is cyclic
I is the incenter of BCNM
(AMP) and (ANQ) are tangent at A
PMNQ is cyclic
AXY collinear

:wallbash:

mfw g8 is just a massive chase

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awesomeming327.

1664 posts

awesomeming327.

#59 h May 20, 2024, 10:02 PM

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https://cdn.aops.com/images/a/6/b/a6b5883bce870c8c62780cb1b6b03ad25f549101.png

Take an inversion about I.

In the inverted image, we have $I$ is the orthocenter of $\triangle ABC$ , $B,P,M$ collinear and $C,Q,N$ collinear. Note that $\angle APM=\angle ACB=180^\circ-\angle AIB=\angle AMP$ so $AP=AM$ and similarly $AQ=AN$ . This implies that $PMNQ$ is an isosceles trapezoid and $(MAP)$ is tangent to $(NAQ)$ . In our uninverted image, note that since $PMNQ$ is cyclic, $X$ is on the radical axis of $\omega_B$ and $\omega_C$ . Furthermore, it is on the radical axis of $(MAP)$ and $(NAQ)$ . It is also on the common tangents from $I$ to $\omega_B$ and $\omega_C$ and from $A$ to $(MAP)$ and $(NAQ)$ .
Note that
$\angle BIC+\angle MIN=180^\circ-\angle IBC-\angle ICB+\angle MIX+\angle XIN=180^\circ-\angle IBC-\angle ICB+\angle IBM+\angle ICN=180^\circ$ so $\angle BIM+\angle CIN = 180^\circ\implies \angle BPM+\angle CQN=180^\circ\implies \angle ABY+\angle ACY=180^\circ$ , so $Y$ is on $(ABC)$ .
Furthermore, $\angle AQN=\angle AQC-\angle NQC=180^\circ-\angle AYC-\angle ACY=\angle YAC$ so $YA$ is tangent to $(NAQ)$ . Similarly, it is tangent to $(MAP)$ and so $Y$ is also on the common tangent from $A$ to $(MAP)$ and $(NAQ)$ . We are done.

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awesomehuman

496 posts

awesomehuman

#60 h Aug 6, 2024, 1:54 AM • 2 Y

Y by OronSH, dolphinday

Invert about $I$ and rename $I$ to $H$ to get the following equivalent statement:

Let $\triangle ABC$ have orthocenter $H$ .
Let $\ell_B$ , $\ell_C$ be paralell lines through $B$ and $C$ , respectively.
Let $P$ and $Q$ be the intersections of
$\ell_B$ and $\ell_C$ with $(ABC)$ , respectively.
Let $M$ and $N$ be the intersections of
$\ell_B$ and $\ell_C$ with $(AHB)$ and $(AHC)$ , respectively.
Let $X$ be the intersection of $(HPM)$ and $(HQN)$ .
Let $Y$ be the second intersection of the circle through $H$ tangent to $\ell_B$ at $B$ and the circle through $H$ tangent to $\ell_C$ at $C$ .
Then, $AHYX$ is cyclic.

We will prove this statement instead of the original.
Let $H'$ be the reflection of $H$ over the angle bisector of $BC$ and $PQ$ .
Then, $\triangle BHC \cong PH'Q$ .

Claim:
$X$ is the intersection of $HH'$ with $(H'PQ)$ .

Proof:
We have
$\angle PXQ = \angle PXH + \angle HXQ = \angle BMH + \angle HNC$ $= \angle BAH + \angle HAC = \angle BAC = \angle CHB = \angle PH'Q.$ Thus, $X$ lies on $(PH'Q)$ .
We have
$\angle PXH = \angle PMH = \angle BMH = \angle BAH = \angle HCB = \angle PQH' = \angle PXH'.$ Thus, $X$ is on $HH'$ .

Claim:
$A$ is the orthocenter of $\triangle PXQ$ .

Proof:
Let $D$ be the intersection of $PQ$ and $HH'$ .
We have
$\angle APQ = \angle ACQ = \angle BCQ + \angle ACB$ $\angle PQX = \angle DXQ + \angle QDX = \angle HCB + \angle QCB.$ Thus,
$\angle APQ + \angle PQX = \angle BCQ + \angle QCB + \angle ACB + \angle HCB = 90.$ Repeating the same logic on the other side, $A$ is the orthocenter of $\triangle PXQ$ .

Claim:
$AHYX$ is cyclic

Proof:
We have
$\angle AXH = \angle AXP + \angle PXH' = \angle PQA + \angle PQH'$ $= \angle PBA + \angle HCB = \angle PBC + \angle CBA + \angle HCB = \angle PBC + 90$ $= \angle PBC + \angle CBH + \angle ACB = \angle PBH + \angle ACB = \angle BYH + \angle AYB$ $= \angle AYH.$

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kiemsibongtoi

25 posts

kiemsibongtoi

#62 h Aug 27, 2024, 10:06 AM

Y by

Eyed wrote:

Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$ . Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$ . Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$ , and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$ . Rays $PM$ and $QN$ meet at $X$ . Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$ .

Show that $A,X,Y$ are collinear.

We'll prove this problem follows these claims :
$\textbf{1/}$ Let $B'$ be the intersection of line $BI$ with $\omega_c$ ( $B' \neq I$ ); $C'$ be the intersection of line $CI$ with $\omega_b$ ( $C' \neq I$ )
$\hspace{1cm}$ $M'$ be the intersection of lines $C'M$ and $AC$ ; $N'$ be the intersection of lines $B'N$ and $AB$
Then $N'MM'N$ is a isosceles trapezoid and $I$ lie on circumcircle of $N'MM'N$
$\textit{Prove :}$
$\, \,$ Cuz $M$ , $I$ , $B$ , $C$ are concyclic and $I$ is incenter of $\triangle ABC$ , we see that :
$\angle AM'M = \angle MC'C + \angle ACC' = \angle IBC + \angle ICB = \angle C'IB = \angle C'MB$ $\, \,$ So $\triangle AMM'$ is a isosceles triangle at $A$ . Similar, $\triangle ANN'$ is a isosceles triangle at $A$
$\, \,$ Therefore, $N'MM'N$ is a isosceles trapezoid
$\, \,$ Next, let $\ell$ be the line which tangent both $\omega_b$ , $\omega_c$ at $I$ , we see that :
$\angle MIN = \angle MI\ell + \angle NI\ell = \angle MBI + \angle NCI = \angle IBC + \angle ICB = \angle AM'M$ $\, \,$ Therefore, $M'$ lies on circle $(IMN)$ . Similar, $N'$ lies on circle $(IMN)$ .
$\textbf{2/}$ $M$ , $N$ , $P$ , $Q$ are concyclic
$\textit{Prove :}$
$\, \,$ Let $\omega_b$ , $\omega_c$ meet line $BC$ at $T$ , $L$ respectively ( $T \neq B$ , $L \neq C$ )
$\hspace{1cm}$ $O_b$ , $O_c$ be center of $\omega_b$ , $\omega_c$ respectively
$\, \,$ Cuz $CI$ is the internal bisector of $\angle NCL$ so $O_cI \perp NL$ . Similar, $O_bI \perp MT$
$\, \,$ But $O_b$ , $I$ , $O_c$ are colibnear, so $NL$ $\|$ $MT$
$\, \,$ Combine with claim $\textbf{1/}$ , we see that :
$\angle N'IN = 2\angle NIA = 2(\angle AIC - \angle NIC) = 180^\circ + \angle BAC - 2\angle NIC = \angle BAC + \angle NQC - \angle NLC = \angle NQC - \angle BMT$ $\, \,$ So
$\angle PMN = \angle PMA + \angle AMN = \angle PC'B + \angle N'IN = \angle PBT + \angle NQC = 180^\circ - (\angle PQC - \angle NQC) = 180^\circ - \angle PQN$ $\, \,$ Therefore, $M$ , $N$ , $P$ , $Q$ are concyclic
$\textbf{3/}$ Let $V$ be the midpoint of arc $BAC$ of $\Gamma$
Then $V$ is the intersection of lines $B'Q$ , $C'P$
$\textit{Prove :}$
$\, \,$ Cuz $Q$ is the second intersection of $\omega_c$ and $\Gamma$ so $\angle B'QC = \angle BIC = 90^\circ + \frac12 \angle BAC = \angle VQC$
$\, \,$ Lead to $V$ , $B'$ , $Q$ are colinear. Similar, $P$ , $C'$ , $P$ are colinear
$\textbf{4/}$ $\triangle AXI$ is a isosceles triangle at $X$
$\textit{Prove :}$
$\, \,$ Let $V_b$ , $V_c$ be the intersections of lines $BI$ , $CI$ with $\Gamma$ respectively ( $V_b \neq B$ , $V_c \neq C$ )
$\hspace{0.7cm}$ $U_b$ be the intersection of lines $V_bQ$ and $B'N$
$\, \,$ We ez to see that $V_bV_c$ is the perpendicular bisector of $AI$ , so we just need to prove that $V_b$ , $X$ , $V_c$ are colinear
$\, \,$ Cuz $M$ , $N$ , $P$ , $Q$ are concyclic from $\textbf{2/}$ , so $X$ lie on radical axis of $\omega_b$ and $\omega_c$
$\, \,$ Which means $XI$ tangent with both $\omega_b$ , $\omega_c$ at $I$
$\, \,$ Lead to $\triangle XNI \sim \triangle XIQ$ . As a result : $\dfrac{XN}{XQ} = \dfrac{XN}{XI} .\dfrac{XI}{XQ} = \left( \dfrac{IN}{IQ} \right)^2$
$\, \,$ In the order hand :
$\hspace{0.5cm}$ From claim $\textbf{3/}$ and $NB' \perp AI$ at claim $\textbf{1/}$ , we see that :
$\angle V_bQB' = \angle V_bQV = 90^\circ - \frac12(\angle BAC + \angle ABC) = 90^\circ - \angle AIB' = \angle NB'I = \angle V_bB'U_b$ $\hspace{0.5cm}$ Lead to $\triangle V_bU_bB' \sim \triangle V_bB'Q$ . As a result : $\dfrac{V_bU_b}{V_bQ} = \dfrac{V_bU_b}{V_bB'} .\dfrac{V_bB'}{V_bQ} = \left( \dfrac{B'U_b}{B'Q} \right)^2$
$\, \,$ Therefore
$\dfrac{XN}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 = \left( \dfrac{\sin \angle NQI}{\sin \angle INQ} \right)^2 = \left( \dfrac{\sin \angle V_bB'U_b}{\sin \angle V_bB'Q} \right)^2 = \left( \dfrac{V_bU_b}{B'V_b} .\dfrac{V_bB'}{V_bQ} \right)^2 = \dfrac{V_bU_b}{V_bQ}$ $\, \,$ Hence, follow Thales theorem, we see that $XV_b$ $\|$ $NB'$ . Similar, $XV_c$ $\|$ $MC'$
$\, \,$ But from claim $\textbf{1/}$ , we already have that $B'N$ $\|$ $C'M$ , so $X$ , $V_b$ , $V_c$ are colinear
$\textbf{Finally, from these claims, we have : }$
$\, \,$ Cuz $\triangle XMI \sim \triangle XIP$ and $AX = XI$ , so $\triangle XMA \sim \triangle XAP$
$\, \,$ Lead to
$\angle MAX = \angle APM = \angle APB - \angle MPB = 180^\circ - \angle ACB - \angle MTC = \angle LNC = \angle BCY = \angle BAY$ $\, \,$ Similar, $\angle XAN = \angle CAY$ . Therefore $A$ , $X$ , $Y$ are colinear, done

Attachments:

imosl g8 2020.pdf (74kb)

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bin_sherlo

661 posts

bin_sherlo

#63 h Aug 29, 2024, 3:54 PM • 2 Y

Y by ehuseyinyigit, swynca

Claim: $Y$ lies on $(ABC)$ .
Proof:
$\measuredangle BYC= 180-\measuredangle CBY-\measuredangle YCB=180-(\measuredangle IMB-\frac{\measuredangle B}{2})-(\measuredangle CNI-\frac{\measuredangle C}{2})=270-\frac{\measuredangle A}{2}-\measuredangle CIB=180-\measuredangle A$ Which gives that $Y\in (ABC)$ . $\square$

Claim: $I$ is the $A-$ excenter on $\triangle AMN$ .
Proof: Let $O$ be the circumcenter of $(IMN)$ . Since $\measuredangle MIN=\measuredangle MBI+\measuredangle ICN=90-\frac{\measuredangle A}{2},$ we have $\measuredangle MON=180-\measuredangle MAN$ hence $O$ lies on $(AMN)$ . If $AI$ intersects $(IMN)$ at $J$ for second time, then
$\measuredangle IMB=\measuredangle MIJ+\frac{\measuredangle A}{2}=\measuredangle MNJ+\measuredangle ONM=\measuredangle ONJ=\measuredangle NJO=\measuredangle NMI$ Similarily we get $\measuredangle CNI=\measuredangle INM$ so $I$ is the $A-$ excenter on $\triangle AMN$ . $\square$

Claim: $M,N,P,Q$ are concyclic.
Proof: Let $MN$ and incircle of $ABC$ be tangent to each other at $S$ . Take the inversion centered at $I$ with radius $IS$ . Let $D,E,F$ be the tangency points of the incircle with $BC,CA,AB$ respectively. $A^*,B^*,C^*,M^*,N^*$ are the midpoints of $EF,FD,DE,SF,SE$ . $P^*,Q^*$ are the intersections of $B^*M^*,C^*N^*$ with $(A^*B^*C^*)$ . Note that $M^*B^*\parallel SD\parallel N^*C^*$ and $M^*N^*\parallel EF\parallel B^*C ^*$ .
$\measuredangle M^*P^*Q^*=180-\measuredangle Q^*P^*B^*=\measuredangle B^*C^*N^*=\measuredangle N^*M^*P^*$ Thus, $N^*M^*P^*Q^*$ is an isosceles trapezoid which yields $M,N,P,Q$ lie on a circle. $\square$

Claim: $A,X,Y$ are collinear.
Proof: Let $PM$ and $QN$ intersect $(ABC)$ at $K,L$ for second time.
$\measuredangle YCA=\measuredangle YCN=\measuredangle CQN=\measuredangle CQL$ $\measuredangle ABY=\measuredangle MBY=\measuredangle MPB=\measuredangle KPB$ These give that $AY=CL$ and $AY=BK$ . Hence $YK\parallel AB$ and $YL\parallel AC$ . Also
$\measuredangle KLN=\measuredangle KLQ=\measuredangle KPQ=\measuredangle MPQ=\measuredangle MNL$ Thus, $MN\parallel KL$ . By Desargues on $\triangle AMN$ and $\triangle YKL$ , since $AM\cap YK=AB_{\infty},MN\cap KL=MN_{\infty},NA\cap LY=AC_{\infty}$ are collinear, they are perspective. $AY,MK,NL$ are concurrent. $MK$ and $NL$ intersect at $X$ so $A,X,Y$ are collinear. $\blacksquare$

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GeorgeRP

130 posts

GeorgeRP

#64 h Oct 11, 2024, 10:23 PM

Y by

Why did I solve a G8 synthetically???

Let's denote with $t$ the common tangent of $w_b$ and $w_c$ .
Also let $U,V$ be the intersections of $BC$ with $w_b$ and $w_c$

Claim 1: $Y$ lies on $(ABC)$
Proof. $\angle BIU=\angle BCY=\phi$ and $\angle CIV=\angle CBY=\psi, \angle UIt=\angle IBC=\frac{\beta}{2}, \angle VIt=\angle ICB=\frac{\gamma}{2}$
Now:
$90+\frac{\alpha}{2}=\angle BIC=\angle BIU+\angle UIV+\angle CIV= \phi+\psi+90-\frac{\alpha}{2}$ So $\angle BYC=180-\phi-\psi=180- \alpha=\angle BAC \text{. }\square$

Claim 2: $MU$ || $NV$ || $t$
Proof. $\angle MUI=\angle MBI= \angle UBI = \angle UIt \Rightarrow$ $MU$ || $t$ and similarly $NV$ || $t$ . $\square$

Claim 3: $MUVN$ is an isosceles trapezoid
Proof. As $I$ is the midpoint of arcs $MU$ and $NV$ in $w_b,w_c$ , this means that $MI=UI, IN=IV \Rightarrow \triangle MIN$ and $\triangle UIV$ are congruent $\Rightarrow MN=UV$ . $\square$

Let $Q'$ and $P'$ be the intersections of $QX$ and $PX$ with $(ABC)$ .

Claim 4: $BQ'$ || $QC$ || $t$
Proof. Reims theorem on $(ABC), (NQCV)$ and lines $QX,CV$ gives that $BQ'$ || $NV$ and similarly $CP'$ || $UM$ . $\square$

Let $K;G$ be the intersections of $AB;BC$ and $NV;MU$

Claim 5: $PMAG$ and $QNAK$ are cyclic
Proof. $\angle KNQ=\angle VCQ=180-\angle BQ'Q=180-\angle BAQ=\angle KAQ$ and similarly $\angle PMG=\angle PAG$ . $\square$

Claim 6: $(PMAG)$ and $(NAKQ)$ are tangent at $A$
Proof. As $GMNK$ is a trapezoid we have that the circumcenters of $GMA$ and $KAN$ lie on a line with $A$ . $\square$

Claim 7: $PMNQ$ is cyclic
Proof. $\angle MPQ= \angle BPQ-\angle BPM=(180-\angle BCQ)-\angle MUV=$
$(180-\angle KNQ)-\angle UMN=\angle XNK-\angle MNK=\angle MNX$ . $\square$

Claim 8: $A,X,Y$ are collinear
Proof. By radical axises on $(PMNQ), (PMAG), (NAKQ)$ we get that $AX$ is the tangent of $(PMAG)$ and $(NAKQ)$ .
$\angle YAC=\angle YBC= \angle BMU=\angle Q'BM=\angle Q'QA=\angle XAC$ . $\square$

Attachments:

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L13832

245 posts

L13832

#65 h Nov 12, 2024, 12:20 PM • 2 Y

Y by radian_51, S_14159

Amazing problem! also my first G8 :showoff:

We show $AY$ is tangent to $\odot(AMP), \odot(ANQ)$
$\angle YAM+\angle MAP = 180^{\circ} - \angle YBP=180^{\circ}-\angle PBI-\angle BPI = \angle BIP = 180^{\circ} - \angle PMA$ Note that $Y\in{\odot(ABC)}$ as
$\begin{align*} & \angle BIC=\angle IMB+\angle INC=\angle IBY+\angle ICY=90^{\circ}+\frac{\angle A}{2}\\ & \angle BYC=360-\angle YBI-\angle BIC-\angle ICY=360^{\circ}-2\left(90^{\circ}+\frac{\angle A}{2}\right)=180^{\circ}-\angle A \end{align*}$ By another angle chase we show $\odot(I)$ is the incircle of $MNCB$ . $\angle MIN = \angle MBI + \angle NCI = \angle IBC + \angle ICB = 180^{\circ}-\angle BIC=90-\frac{\angle A}{2}$ If we show $\odot(MPNQ)$ we get $X$ as the radical center of $(MNQP)$ , $(AMP)$ , and $(ANQ)$ , which will clearly lie on radax of $(AMP)$ , and $(ANQ)$ which is clearly $\overline{AY}$ implying $\overline{A-X-Y}$ is collinear.
So finally we go on to another angle chase to prove $\odot(MPNQ)$ which took me a lotta time to figure out :stretcher:

$\begin{align*} \angle PQN=\angle CQN-\angle CQP&= 180^{\circ}-\angle CIN-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle INC-\angle ICN)-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle ICY-\angle ICN)-\angle CQP\\&=\angle ICY+\angle ICN-\angle CQP\\&=\angle ACY-\angle CQP\\&=(180^{\circ}-\angle ABY)-(180^{\circ}-\angle CBP)\\&=\angle CBP-\angle ABY\\&=\angle CBA+\angle ABP-\angle CBA-\angle YBC\\&=\angle ABP-\angle YBC\\ \angle PMN=\angle PMB+\angle BMN&=\angle PIB+2\angle BMI\\&=180^{\circ}-(\angle IPB+\angle PBI)+2\angle BMI\\&=180^{\circ}-(\angle IBY+\angle PBI)+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle CBY+\angle CBA\\&=180^{\circ}-\angle PBA+\angle CBY \end{align*}$ and we are finally done! :stretcher:

This post has been edited 2 times. Last edited by L13832, Nov 21, 2024, 5:31 AM

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Saucepan_man02

1294 posts

Saucepan_man02

#66 h Nov 12, 2024, 3:20 PM

Y by

Here's my solution (angle-chase+invert+radical-axis+trig) :

Claim-1: $Y \in (ABC)$

Notice that: $\angle ABY+\angle ACY = 360^\circ - (\angle MIB + \angle NIC) = \angle MIC + \angle BIC = \angle MIX+\angle XIN + 90^\circ + \angle A/2$ $= \angle MBI + \angle NCI + 90^ + \angle A/2 = 180^\circ.$
Claim-2: $MNQP$ is cyclic + $(AMP), (ANQ)$ are tangent.

Invert around incircle. Let $X'$ denote the inverse of point $X$ in this inversion. Notice that:

- $I$ is the orthocenter of $\triangle A'B'C'$ .
- $M'A'B'I$ and $N'A'C'I$ are cyclic ( $A, M, B$ and $A, N, C$ are collinear).
- $M', P', B'$ and $N', Q', C'$ are collinear ( $MPIB$ and $NIAC$ are cyclic).
- $M'P'B' \parallel N'Q'C'$ .
- $A',P',B',C',Q'$ lie on same circle.

We will prove $M'N'Q'P'$ is cyclic. It suffice to show $M'N'=P'Q'$ . Note that: $\angle M'IB'+\angle N'IC' = 360^\circ - (\angle B'M'I + \angle C'N'I) - (\angle M'B'I + \angle N'C'I) = 180^\circ.$
Now, we have: $M'B' = \frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I}$ Note that: $\frac{B'I}{\sin \angle M'B'I} = 2R = \frac{C'I}{\sin \angle N'C'I}.$ Thus, we have: $\frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I} = \frac{C'I \sin \angle N'IC'}{\sin \angle N'C'I} = N'C'$ which implies $M'B'C'N'$ is a parallelogram.

Note that, from the following sub-claim, we will have: $P'Q' \parallel M'N' \parallel BC$ and $(A'M'P'), (A'N'Q')$ to be tangent at $A'$ which finishes the main-claim.

Sub-Claim: $\triangle A'M'P'$ and $\triangle A'N'Q'$ are isosceles.
Note that: $\angle A'M'B' = 180^\circ - \angle A'IB' = \angle A'C'B'$ and $\angle A'P'M' = \angle A'C'B'$ which shows that $A'M' = A'P'$ . Similarly, $A'N'=A'P'$ and we are done.
Hence, Claim-2 is true.

Thus, from Claim-2, $AX$ is the radical axis of $(APM), (ANQ)$ along with $AX$ being tangent to both circles $(APM), (ANQ)$ .

Note that, the following claim finishes the problem:

Claim-3: $AY$ is tangent to both circles $(APM), (ANQ)$ .

Let $R$ be a point of line $AP$ such that $A, R$ are on different sides with-respect to $P$ .
We will show it with an one-liner angle-chase:
$\angle BAY = 180^\circ - \angle AYB - \angle ABY = 180^\circ - \angle RPB - \angle BPM = \angle APM.$ Thus, $AY$ is tangent to $(APM)$ . Similarly, $AY$ is tangent to $(ANQ)$ and we are done.

Remarks

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Likeminded2017

391 posts

Likeminded2017

#67 h Dec 24, 2024, 10:50 AM

Y by

We prove that $Y$ lies on $(ABC)$ through angle chasing - $\angle ABY+\angle ACY=360-\angle MIB-\angle NIC=180^\circ.$ Next, $YA$ is tangent to both $(ANQ)$ and $(AMP)$ and is thus the radical axis, as $\angle APM=\angle BPA-\angle BPM=(180-\angle BYA)-\angle YBA=\angle YAB=\angle YAM$ and the same works for $(ANQ).$ Finally, we show $X$ lies on the radical axis by showing $MNQP$ are concyclic to finish. Let $\omega_b$ and $\omega_c$ intersect $BC$ at $R,S$ respectively. Observe by Fact 5 $\triangle IRS \cong \triangle IMN$ so $\angle INM=\angle ISR=\angle INC.$
$\begin{align*} \angle QPM &= \angle MPB-\angle QPB \\ &= \angle NIC+\angle QCB-180 \\ &= \angle NIC+\angle BCI-\angle QNI \\ &=\angle NIC+\angle NCI-\angle QNI \\ &=180-\angle INC-\angle QNI \\ &=180-\angle INM-\angle QNI \\ &=180-\angle QNM \end{align*}$ so we are done.

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cj13609517288

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cj13609517288

#68 h Jan 15, 2025, 9:15 PM

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Part 1 - The spiral similarity
First, note that
$\angle BIC=\angle BMI+\angle CNI=\angle YBI+\angle YCI.$ Therefore,
$\angle BYC=360^{\circ}-\angle BIC-\angle BIC=180^{\circ}-\angle A.$ Therefore, $Y\in (ABC)$ .

Now note that
$\angle PBY=180^{\circ}-\angle PMB=\angle PMA$ $\angle PYB=\angle PAB=\angle PAM$ so $\triangle PBY\sim\triangle PMA$ .

So there exists a spiral similarity centered at $P$ sending $AM$ to $YB$ .

Part 2 - Redefining $X$
Redefine $X$ to be the intersection of the perpendicular bisector of $AI$ with $AY$ . Then $\angle IXY=2\angle IAX$ , so $IX$ is parallel to the isogonal of $AY$ wrt angle $BAC$ . In particular,
$\angle(BA,IX)=\angle YAC=\angle YBC=\angle(YB,CB)\Longrightarrow \angle(BI,IX)=\angle(YB,IB)\Longrightarrow \angle BIX=180^{\circ}-\angle YBI=180^{\circ}-\angle BPI.$ Therefore, $XI$ is tangent to $(BPI)$ .

Part 3 - The finish
By our spiral similarity from earlier,
$\angle YAM=\angle YAB=\angle YPB=\angle APM.$ Therefore, $AY$ is tangent to $(APM)$ . Then the powers from $X$ to $(APM)$ and $(BPM)$ are $AX^2$ and $IX^2$ , respectively, which are equal, so $X$ lies on their radical axis $PM$ . Similarly, $X$ lies on $QN$ , as desired. $\blacksquare$

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swynca

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swynca

#69 h Feb 13, 2025, 8:23 PM • 1 Y

Y by bin_sherlo

A different approach,
After a simple angle chasing, it appears that $K=AC \cap YP$ and $L=AB \cap YQ$ is on $(MNPQ)$ . Using pascal with points $P,K,N,Q,L,M$ gives the desired result.

This post has been edited 1 time. Last edited by swynca, Feb 13, 2025, 8:23 PM
Reason: typo

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Iveela

115 posts

Iveela

#70 h Feb 28, 2025, 12:37 PM

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The ultimate angle chasing.

Let $O_B$ and $O_C$ be the circumcenters of $\omega_B$ and $\omega_C$ respectively. Recall that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ which implies $\angle O_BIB + \angle O_CIC = 90^{\circ} - \frac{\angle A}{2}$ and hence
$\angle IBY + \angle ICY = \angle IPB + \angle IQC = 180 - \angle O_BIB - \angle O_CIC = 90^{\circ} + \frac{\angle A}{2}.$ Consequently, $\angle ABY + \angle ACY = 180^{\circ}$ or $ABCY$ cyclic.

Let $T = BP \cap CQ$ . Since $TP \cdot TB = TQ \cdot TC$ , it must lie on the radical axis of $\omega_B$ and $\omega_C$ . In other words, $TI$ is tangent to both circles. We will now show that the quadrilateral $TPQX$ is cyclic. Indeed, we have
$\angle XQC = \angle NQC = \angle ACY = 180^{\circ} - \angle ABY = 180^{\circ} - \angle XPB = \angle TPX$ as desired. Now, observe that
$\angle XTQ = \angle QPX = \angle QPB - \angle ABY = \angle BCY - \angle ACQ = \angle ICY - \angle ICQ = \angle ACQ$ which implies that $T$ , $X$ , and $I$ are collinear. In particular, $XI$ is tangent to $\omega_B$ and $\omega_C$ . This implies $XM \cdot XP = XI^2 = XN \cdot XQ$ or $MNPQ$ cyclic.

Now, we are well equipped to prove that $XA = XI$ . We once again, angle chase. Recall that $XI$ is tangent to $\omega_B$ and $\omega_C$ . Indeed, we have
$\angle XIA = \angle AIB - \angle XIB = (90^{\circ} + \frac{\angle C}{2}) + \angle IBY - 180^{\circ} = \angle YAC - \frac{\angle A}{2} = \angle YAI.$ Recall that $XN \cdot XQ = XI^2 = XA^2$ . Therefore, $(ANQ)$ is tangent to $XA$ . Angle chasing one final time, we receive
$\angle XAC = \angle AQN = \angle AQC - \angle NQC = \angle BAC + \angle ACB - \angle ACY = \angle CAY$ which implies the result.

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EeEeRUT

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EeEeRUT

#71 h Mar 4, 2025, 8:57 AM

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Denote $\ell$ a line tangent to $\omega_B$ and $\omega_C$ at $I$ .

Claim : $\angle MIB + \angle NIC = 180^{\circ}$
Proof : Consider $\begin{align*} \angle MIB + \angle NIC &= 360^{\circ} - \angle MIN - \angle BIC\\ &= 360^{\circ} - (\angle IBM + \angle ICN) - (180^{\circ} - \angle IBC - \angle ICB)\\ &= 360^{\circ} - (\angle IBC + \angle ICB) - (180^{\circ} - \angle IBC - \angle ICB) = 180^{\circ} \end{align*}$ Let $BP$ intersect $CQ$ at $T$ . By radical axis, $T$ lies on $\ell$ . Note that $\angle TPX + \angle TQX = 360^{\circ} - (\angle MIB + \angle NIC) = 180^{\circ}$ Thus, $TPXQ$ is cyclic.

Claim : $X$ lies on $\ell$ .
Proof : Let $\ell$ intersect $BC$ at $S$ .
Consider $\angle BMX = 180^{\circ} - \angle PIB = 180^{\circ} - \angle ISB$ Hence, $PXSB = \Omega_B$ is cyclic. Similarly, $QXSC = \Omega_C$ is cyclic.
That is the radical axis of these $2$ circles is $XS$
By radical center of $\Omega_B, \Omega_C$ and $(BCPQ)$ is $T$ . Thus, $T,X,S$ are collinear, which is the line $\ell$ .

Claim : $XA = IA$ .
Proof : Consider $\begin{align*} \angle MPA + \angle NQA &= \angle APB - \angle BPM + \angle AQC - \angle CQN\\ &= -\angle ABC - \angle ACB + \angle MIB + \angle NIC\\ &= \angle CAB \end{align*}$ Thus, $(PMA)$ is tangent to $(QNA)$
By power of point, the segment $XA = XI$ is tangent to $(PMA), (QNA)$ .

Claim : $A,B,C,Y$ cyclic.
Proof : $\angle ABY + \angle YCA = 360^{\circ} - \angle MIB - \angle NIC = 180^{\circ}$ Thus, $A,B,C,Y$ are concyclic.

Let $\angle XAI = \gamma$
Now, we are left to consider the following angle chasing $\begin{align*} \angle XAC &= \angle CAB - \angle BAX\\ &= \angle CAB -(360^{\circ} - \angle AMI - \angle PIX - \angle IXA)\\ &= \angle CAB + 180^{\circ} - \angle IAM - \gamma - \angle IBC\\ &= \angle CAB - \angle IMB - \gamma\\ &= \angle CAB - \angle IBY\\ &= \angle CAB - \angle YAB\\ &= \angle YAC \end{align*}$
Hence, $A,X,Y$ are collinear as desired.

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This post has been edited 1 time. Last edited by EeEeRUT, Mar 4, 2025, 9:44 AM
Reason: Diagram

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joshualiu315

2513 posts

joshualiu315

#72 h Today at 12:51 AM • 1 Y

Y by OronSH

first g8 yippee! :coolspeak:

Denote the centers of $\omega_B$ and $\omega_C$ as $O_B$ and $O_C$ , respectively.

Claim: $Y$ lies on $\Gamma$ .

Proof: Notice that

$\begin{align*} \angle ABY + \angle ACY &= \angle BPM + \angle CQN \\ &= (180^\circ - \angle BIM) + (180^\circ - \angle CIN) \\ &= \angle BIC + \angle MIN \\ &= \angle BIC + \angle MBI + \angle NCI \\ &= \angle BIC + \angle IBC + \angle ICB = 180^\circ, \end{align*}$
so $ABYC$ is cyclic. $\square$

Let $D \neq B = \omega_B \cap \overline{BC}$ and $E \neq C = \omega_C \cap \overline{BC}$ .

Claim: $MDEN$ is an isosceles trapezoid.

Proof: Since $\overline{BI}$ and $\overline{CI}$ are angle bisectors, $I$ is the midpoint of $\widehat{MD}$ in $\omega_B$ and the midpoint of $\widehat{NE}$ in $\omega_C$ . Thus, $\overline{O_BO_C}$ is the perpendicular bisector of both $\overline{MD}$ and $\overline{NE}$ . $\square$

Claim: Points $M$ , $N$ , $P$ , and $Q$ are concyclic.

Proof: It suffices to show that $\triangle XMN \sim \triangle XQP$ , or that $\angle XMN = \angle XQP$ . We proceed with angle chasing:

$\begin{align*} \angle XMN &= 180^\circ - \angle PMD - \angle DMN \\ &= \angle PBD - \angle MDE \\ &= \angle PBD - \angle NEC \\ &= (180^\circ - \angle PQC) - (180^\circ - \angle NQC) \\ &= \angle NQC - \angle PQC = \angle NQP, \end{align*}$
as desired. $\square$

Then, consider the radical axis of $(APM)$ and $(AQN)$ , denoted as $\ell$ .

Claim: Points $A$ , $X$ , and $Y$ all lie on $\ell$

Proof: Since $A$ lies on both circles, it obviously lies on the radical axis. Also, we have $XM \cdot XP = XN \cdot XQ$ from Power of a Point with respect to $(MNQP)$ . This implies $X$ has equal power with respect to $(APM)$ and $(AQN)$ , meaning $X$ lies on $\ell$ .

Finally, it suffices to show that $Y$ lies on $\ell$ . We prove the stronger statment that $\overline{AY}$ is the common tangent of $(APM)$ and $(AQN)$ . Notice that

$\begin{align*} \angle YAP &= 180^\circ - \angle YBP \\ &= 180^\circ - (\angle PBD + \angle YBD) \\ &= 180^\circ - (180^\circ - \angle PMD + \angle BMD) \\ &= \angle PMD - \angle BMD \\ &= \angle PMB = 180^\circ - \angle AMP. \end{align*}$
Thus, $\overline{AY}$ is tangent to $(APM)$ , and similarly we can find that $\overline{AY}$ is tangent to $(AQN)$ . This means $Y$ lies on $\ell$ as well. $\square$

Since $A$ , $X$ , and $Y$ all lie on $\ell$ , they are obviously collinear. $\blacksquare$

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