IMO Shortlist 2014 G7

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209 posts

#1 h Jul 11, 2015, 8:59 AM • 12 Y

Y by shinichiman, Akatsuki1010, anantmudgal09, Tumon2001, e_plus_pi, magicarrow, A-Thought-Of-God, HWenslawski, Adventure10, Mango247, Rounak_iitr, Funcshun840

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$ . Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$ , and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$ , respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

Proposed by David B. Rush, USA

This post has been edited 3 times. Last edited by hajimbrak, Jul 15, 2016, 9:20 AM
Reason: Added proposer

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TelvCohl

2312 posts

TelvCohl

#2 h Jul 11, 2015, 12:02 PM • 29 Y

Y by shinichiman, NewAlbionAcademy, andria, Mediocrity, utkarshgupta, Abubakir, don2001, e_plus_pi, BobaFett101, Phie11, AlastorMoody, Greenleaf5002, math-lover, enhanced, Siddharth03, IFA, magicarrow, KST2003, rcorreaa, mijail, jeteagle, rayfish, Adventure10, Mango247, math_comb01, hectorleo123, aidan0626, Rounak_iitr, crazyeyemoody907

My solution :

Let $T \equiv XU \cap AB$ .

Since $\angle BIV=\angle BIC-90^{\circ}=\tfrac{1}{2} \angle BAC=\angle BAI=\angle BYV=\angle BTU$ ,
so we get $B, I, V, Y$ are concyclic ... (1) and $B, I, T, U$ are concyclic ... (2)

From $XT:IA=YT:YA=VU:VI=XU:IA \Longrightarrow XT=XU$ ,
so from (2) and notice $\angle UBI=\angle TBI \Longrightarrow XI$ is the perpendicular bisector of $TU$ ,
hence $\angle AIY=90^{\circ} \Longrightarrow$ combine (1) we get $V$ is the tangency point of A-mixtilinear circle with $\Omega$ ,
so $AV$ is the isogonal conjugate of the A-Nagel line of $\triangle ABC$ WRT $\angle BAC$ (well-known) . ... (3)

Since $IZ$ is parallel to the A-Nagel line of $\triangle ABC$ (well-known) ,
so combine (3) $\Longrightarrow \angle ZIA=180^{\circ}-\angle VAI=180^{\circ}-\angle AIW \Longrightarrow I, W, Z$ are collinear .

Q.E.D

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shinichiman

3212 posts

shinichiman

#3 h Jul 12, 2015, 10:07 PM • 3 Y

Y by NewAlbionAcademy, Adventure10, Mango247

Since $X,Y,I$ are collinear and $UX \parallel YV \parallel AI$ so $\tfrac{UI}{UV}= \tfrac{AI}{VI} \left( = \tfrac{XI}{XY} \right)$ .
If $IU \cap AC \equiv K$ then $IK=IU$ . We then have $\tfrac{IK}{UV}= \tfrac{AI}{VI}$ . Note that $\angle AIK= \angle YVU$ so $\triangle AIK \sim \triangle YVU$ . This follows $YU \parallel AC$ .

$YI \cap AC=L$ then since $YU \parallel AC$ and $IU=IK$ so $IY=IL$ . Hence, $AI \perp YL$ .

Since $VY \parallel AI$ so $\angle YVI= \angle AIK= \tfrac 12 \angle B= \angle IBY$ . This follows $Y,I,V,B$ are concyclic. Therefore $\angle YIV= \angle LCI (=180^{\circ}- \angle ABV)$ . This means $I,L,C,V$ are concyclic. Therefore $\angle ICL= \angle IVL$ . Note that $\angle IVL= \angle YVA= \angle VAI$ so $\angle VAI= \tfrac 12 \angle C$ . Since $XI \perp AI$ and $W$ is the midpoint of $AX$ so $\angle VAI= \angle WIA= \tfrac 12 \angle C$ . since $\angle YIB= \tfrac 12 \angle C$ so $\angle BIV= \angle ICV (= \angle ACV- \tfrac 12 \angle C)$ . Therefore $\angle BVI= \angle IVC (=90^{\circ}- \angle BIV)$ or $VI$ passes through the midpoint $R$ of arc $BAC$ .

We have $MB^2=MI^2=MZ \cdot MR$ so $\angle MIZ= \angle MRI= \angle MAV= \angle WIA$ or $\angle WIA= \angle MIZ$ . This follows $W,I$ and $Z$ are collinear.

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pi37

2079 posts

pi37

#4 h Jul 12, 2015, 11:57 PM • 1 Y

Y by Adventure10

A slightly different finish: you can get $BIVY$ cyclic as others have shown, and it is clear that $B(I,V;U,IV\cap AB)$ is harmonic, so $(BA,BC;BI,BV)$ is harmonic and $V$ is the midpoint of arc $ABC$ . So $\angle AIX=\angle VYI=\angle VBI=90$ . Then angle chasing yields it suffices to show that $\angle BIZ=90$ , or that $Z$ is the $B$ -mixtilinear touchpoint on $BC$ . This is equivalent to the midpoint $T$ of arc $BAC$ being the $B$ -mixtilinear touchpoint, but $VT\perp CI$ so $V,I,T$ are collinear. But the fact that $V$ is the midpoint of arc $ABC$ implies that the intersection of $VI$ and the circumcircle is indeed the $B$ -mixtilinear touchpoint, so we are done.

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livetolove212

859 posts

livetolove212

#5 h Jul 14, 2015, 5:02 AM • 4 Y

Y by shinichiman, nguyendangkhoa17112003, Adventure10, Mango247

Let $R$ be the midpoint of $AC$ . After proving the concyclic of 4 points $B,Y,I,V$ we get $\angle IBV=\angle IYV=90^\circ$ , then $V$ is the midpoint of arc $ABC$ . Then applying problem from here, we get $AC+BC=3AB$ . This follows that the $C$ -midline $ZR$ of triangle $ABC$ is tangent to $(I)$ . From trapezoid $ABZR$ we get angle bisectors of $\angle ABZ$ and $\angle BZR$ are perpendicular. Therefore $\angle BIZ=90^\circ$ . On the other side, since $VA=VC$ , $\angle VAC=90^\circ-\frac{1}{2}\angle ABC$ . Then $\angle WIA=\angle WAI=\angle VAC-\angle IAC=\frac{1}{2}\angle ACB=\angle XIB$ . Thus $\angle BIW=\angle XIA=90^\circ$ . This means $WI$ and $ZI$ are perpendicular to $BI$ or $Z,I,W$ are collinear.

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houssam9990

33 posts

houssam9990

#6 h Dec 24, 2015, 8:07 PM • 1 Y

Y by Adventure10

a different approach:
-Let $T$ the midpoint of arc $BAC$ , $D$ the midpoint of minor arc $BC$ , $N=XI\cap AC$
Lemma 1: $V,I,T$ are collinear
proof:
using the fact that $DI=DC$ and $AIB=90+B/2$ one gets by straightforward angle chase: $AIV=180-VID=90+DIC=90+DCI=90+C/2+A/2=180-B/2=180-(AIC-90)=180-AIT$ as desired.
Next suppose $I,X,Y$ are collinear
by lemma 1, $V$ is nothing but the intersection of $(ABC)$ and the $A$ _mixtilinear incircle.
also, it is clear that $IN=IY$ since $AI\perp NY$ , therefore $(VNY)$ is the $A$ _mixtilinear incircle.
lemma 2: $CIZ\sim CVD$
proof:
$ITZ=ITD=ATD-ATI=C+A/2-(90-B/2)=C+A/2+B/2-90=C/2=ICZ$ and therefore $T,I,Z,C$ are concyclic.
Henceforth and since $T,Z,D$ are collinear, the center of spiral similarity sending $D\rightarrow Z$ and $V\rightarrow I$ is $(TIZ)\cap (TVD)=C$ ,i.e $CIZ\sim CVD$ .
Now we angle chase the final result:
$DIZ=180-IZD-IDZ=IZT-ADT=ICT-ACT=ICA=VCD=VAD$ QED.

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Complex2Liu

83 posts

Complex2Liu

#7 h Dec 25, 2015, 10:32 AM • 1 Y

Y by Adventure10

houssam9990 wrote:

a different approach:
-Let $T$ the midpoint of arc $BAC$ , $D$ the midpoint of minor arc $BC$ , $N=XI\cap AC$
Lemma 1: $V,I,T$ are collinear
proof:
using the fact that $DI=DC$ and $AIB=90+B/2$ one gets by straightforward angle chase: $AIV=180-VID=90+DIC=90+DCI=90+C/2+A/2=180-B/2=180-(AIC-90)=180-AIT$ as desired.

Hey! It's wrong

, since $\angle CIT$ isn't invariant( $\equiv 90^{\circ}$ ). Your proof just regard $T$ as $VI\cap \Omega$ .

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nunoarala

106 posts

nunoarala

#9 h Jan 31, 2016, 3:08 AM • 2 Y

Y by Adventure10, AlephG_64

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We denote $\alpha=\angle{CAB}$ , $\beta=\angle{ABC}$ , $\gamma=\angle{BCA}$ . Let $T=IU\cap AC$ , $K=UX\cap AB$ and let $S$ be the midpoint of side $AC$ .

First observe that
$\angle{AIT}=\angle{AIB}-\frac{\pi}{2}=(\pi-\frac{\alpha}{2}-\frac{\gamma}{2})-\frac{\pi}{2}=\frac{\beta}{2}$ hence $\angle{KUI}=\frac{\beta}{2}$ (because $KU$ is parallel to $AI$ ); thus $\angle{KUI}=\angle{KBI}$ and quadrilateral $BUIK$ is cyclic. Since $I$ lies on the angle bissector of $\angle{KBU}$ , $I$ is the midpoint of arc $UK$ , hence $IK=IU$ . On the other hand, triangles $CIU$ and $CIT$ are congruent, yielding $IT=IU$ . Thus $IK=IU=IT$ , yielding that $K$ lies on the circle with diameter $TU$ ; it follows that $TK\perp KU$ . Since $KU$ is parallel to $AI$ , $TK\perp AI$ , hence due to symmetry $AT=AK$ .

Since $I, X, Y$ are collinear, triangles $AXI$ and $BXY$ are similar, as well as triangles $UXV$ and $IAV$ . It follows that
$\frac{IA}{YV}=\frac{AX}{XV}=\frac{IU}{UV}=\frac{IT}{UV}$ or, equivalently, $\frac{IA}{IT}=\frac{VY}{VU}$ . Since $AI$ and $YV$ are parallel, triangles $AIT$ and $YVU$ are similar; hence $UY$ is parallel to $AC$ .

Thus $\angle{UYB}=\alpha$ , and $\angle{KYU}=\pi-\alpha$ ; moreover, since $AK=AT$ , $\angle{AKT}=\frac{\pi}{2}-\frac{\alpha}{2}$ and $\angle{UKY}=\pi-\frac{\pi}{2}-(\frac{\pi}{2}-\frac{\alpha}{2})=\frac{\alpha}{2}$ . This implies $\angle{YUK}=\pi-\frac{\alpha}{2}-(\pi-\alpha)=\frac{\alpha}{2}$ , and $YK=YU$ . Since $IK=IU$ , the line containing $I$ , $X$ and $Y$ is actually the perpendicular bissector of $KU$ , and hence is perpendicular to $KU$ ; it follows that it is also perpendicular to $AI$ and $YV$ .

Now note that
$\angle{YVI}=\angle{KUI}=\frac{\beta}{2}=\angle{YBI}$ implying that quadrilateral $BVIY$ is cyclic. Since $\angle{IYV}=\frac{\pi}{2}$ , we also have $\angle{IBV}=\frac{\pi}{2}$ ; since $BI$ is the internal angle bissector of $\angle{ABC}$ , it follows that $BV$ is the external angle bissector of the same angle, and thus $V$ is the midpoint of the arc $AC$ (containing $B$ ) of $\Omega$ .

Since $VA=VC$ and $\angle{AVC}=\beta$ , $\angle{CAV}=\frac{\pi}{2}-\frac{\beta}{2}$ . Since $\angle{CAI}=\frac{\alpha}{2}$ , $\angle{IAV}=\frac{\pi}{2}-\frac{\beta}{2}-\frac{\alpha}{2}=\frac{\gamma}{2}=\angle{ICA}$ , hence $VA$ is tangent to the circumcircle of $AIC$ . Similarly, $VC$ is tangent to such circle as well. It follows from a well-known lemma that $VI$ is the $I$ -symmedian of $AIC$ . Since $VI\perp CI$ , the corresponding median $IS$ is perpendicular to $AI$ ; that is, $\angle{AIS}=\frac{\pi}{2}$ .

Consider the homothety centered at $C$ mapping the excenter $I_C$ to $I$ . Since $AI_C\perp AI$ , this homothety maps $A$ to $S$ . Thus it maps $B$ to the midpoint of $BC$ , that is, to $Z$ . Hence $IZ$ is parallel to $BI_C$ ; since $BI_C$ is perpendicular to $BI$ , it follows that angle $\angle{ZIB}$ is right.

Since $\angle{XIA}$ is right, point $W$ is actually the circumcenter of $XIA$ . We have previously noticed that $\angle{IAW}=\frac{\gamma}{2}$ , hence $\angle{WIA}=\frac{\gamma}{2}$ . On the other hand, $\angle{BIY}=\angle{BIA}-\frac{\pi}{2}=(\pi-\frac{\alpha}{2}-\frac{\beta}{2})-\frac{\pi}{2}=\frac{\gamma}{2}=\angle{WIA}$ , hence $\angle{BIW}=\angle{XIA}=\frac{\pi}{2}$ . It follows that $IW\perp BI$ . Since $IZ\perp BI$ , it follows that $I$ , $W$ and $Z$ are collinear, as desired.

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bojler

16 posts

bojler

#10 h Apr 11, 2016, 3:53 PM • 2 Y

Y by Adventure10, Mango247

Let $N,N_1$ be midpoint of arc $BC$ not containing $A$ and midpoint of arc $BAC$ . After proving that $BYIV$ is cyclic with $VI$ as its diameter (angle and length chase), we get $\angle NVI=\angle NVB-\angle BVI=180-\angle A/2-(90-\angle A/2)=90$ so $V,I,N_1$ are collinear. Since $N_1C$ and $N_1B$ are tangents to circumcircle of $BIC$ , we get that $IN_1$ is symmedian to triangle $BIC$ and thus $\angle BIZ=\angle CIU=90$ . On the other hand, easy angle chase yealds $\angle WIB=90$ , so $W,I,Z$ are collinear.

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tastymath75025

3223 posts

tastymath75025

#11 h Apr 24, 2016, 12:51 AM • 4 Y

Y by FlakeLCR, BobaFett101, Adventure10, Mango247

Astounding.

sol

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gighiuhui

64 posts

gighiuhui

#12 h May 13, 2016, 5:16 AM • 2 Y

Y by Systematicworker, Adventure10

official proposal

Attachments:

dbrimo2014.pdf (96kb)

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uraharakisuke_hsgs

365 posts

uraharakisuke_hsgs

#13 h May 14, 2016, 3:38 PM • 2 Y

Y by Adventure10, Mango247

My solution :
First we prove that quadrilateral $IYBV$ is cyclic
$IV$ cuts $(O)$ again at $S$ then : $\angle YVI = \angle AIS = \angle YBI \Rightarrow IYBV$ is cyclic
$XU$ cuts $AB$ at $T$ then : $\angle TUI = \angle YVI = \angle YBI \Rightarrow$ quadrilateral $TIUB$ is cyclic
Three points $I,X,Y$ are collinear , by Thales : $\frac{XT}{IA} = \frac{YX}{YI} = \frac{VX}{VA} = \frac{XU}{IA}$ $\Rightarrow$ $X$ is the midpoint of $UT$ $(1)$
Quadrilateral $TIUB$ is cyclic and $BI$ is the bisector of $\angle TBU$ then $IT = IU$ $(2)$
Combine $(1)$ with $(2)$ $\Rightarrow$ $YI$ is the perpendicular bisector of $UT$
$\Rightarrow$ $UT \perp YI$
Since $UT$ is parallel with $AI$ then we have $YI$ is perpendicular with $IA$ $\Rightarrow$ $\angle YIB = \angle ICB$
But we have $Y,I,V,B$ are concyclic , therefore $\angle YIB = \angle YVB$ then $\angle YVB = \angle ICB$
$\Rightarrow$ $YV$ passes through $R$ which is the midpoint of arc $AB$ not containing $C$
Let $S'$ be the midpoint of arc $BC$ containing $A$ so $\angle RVS' = \frac{1}{2} \angle B = \angle ABI = \angle RVS$ then we have $S \equiv S'$
$IU$ passes through $S$ . Therefore, $IU$ is the symmedian line of $\triangle IBC$ ; it follows that $\angle BIZ = \angle CIV = 90^{\circ}$ $\Rightarrow$ $IZ$ is perpendicular to $IB$ $(3)$
Note that $VR$ is the bisector of $\angle BVA$ then $\angle RVB = \angle RVA = \angle VAI = \angle WIA$
Hence , $BYIV$ is cyclic so $\angle BVY = \angle BIY$ and it follows that : $\angle BIY = \angle AIW$
$\Rightarrow$ $\angle WIB = \angle AIY = 90^{\circ}$ then $WI$ is also perpendicular with $BI$
Combine $(3)$ with $(4)$ , we come to a conclusion that $W,I,Z$ are collinear

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Saki

14 posts

Saki

#14 h Jun 20, 2016, 2:06 PM • 2 Y

Y by GuvercinciHoca, Adventure10

An interesting (probably not, yeah..) property about $IX$ :
Define $M_c,M_a$ by midpoint of $\overarc{AB}, \overarc{BC}$ .

\Claim: $IX,AB,M_cV$ are always concurrent.
Define $E=IX\cap M_aV$ . Applying Desargues' involution theorem on quadrilateral $EXUV$ and line $\ell = AI$ , we get an involution $\sigma:\ell\rightarrow \ell$ with fixed point $I$ and $\sigma(M_a)=\infty_\ell$ , which is harmonic conjugation wrt $(I,I_A)$ where $I_A$ is $A$ -excenter. Therefore $EU\cap\ell=\sigma(A)=BC\cap AI$ which tells $E\in BC$ . Now applying Pascal's theorem on $ABCM_cVM_a$ we get $IE,AB,M_cV$ are concurrent, hence \End Claim.

As a direct consequence, we get $Y\in VM_c$ in the problem. \return;

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mathtastic

3258 posts

mathtastic

#15 h Jun 21, 2016, 2:24 AM • 2 Y

Y by Adventure10, Mango247

Saki wrote:

Applying Desargues' involution theorem

Hi,

What is this!

Thanks

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Saki

14 posts

Saki

#16 h Jun 30, 2016, 2:00 PM • 1 Y

Y by Adventure10

\Above:
Desargues' involution theorem states:
"The eight points of intersection of a line $\ell$ with the three pairs of opposite sides of a complete quadrangle and a conic section circumscribed about the quadrangle form four pairs of an involution on $\ell$ ".
where involution means projectivity of period 2. For some proof, try this paper. \return;

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L567

1184 posts

L567

#39 h May 4, 2021, 9:08 AM

Y by

Wow, really nice problem!

Extend $UX$ to meet $AB$ at $P$ and let $AI$ meet $BC$ at $D$ .

Since $\angle PUI = \angle UID = 90 - \angle DIC = 90 - (\frac{A}{2} + \frac{C}{2}) = \frac{B}{2} = \angle PBI$ and so $PIUB$ is cyclic.

Also since $\angle YVI = \angle XUI = \angle PUI = \angle PBI = \angle YBI$ , $BVIY$ is also cyclic.

Observe that since $BUIP$ is cyclic, and $BI$ is angle bisector, $IP = IU$ by Fact 5.

Due to the parallel lines, $\frac{AI}{PX} = \frac{YI}{YX} = \frac{VI}{VU}$ and since $UX || AI$ in $\triangle AVI$ , we have $\frac{VI}{VU} = \frac{AI}{XU}$ and so this means $\frac{AI}{PX} = \frac{AI}{XU} \implies PX = XU$ .

Now, since $P,X,U$ are collinear and $PX = XU$ in isosceles $\triangle IPU$ , we have that $\angle IXP = 90^\circ$ , which means $\angle AIY = \angle IYV = \angle IXU$ are all $90^\circ$ .

Obviously, since $\angle YIA = 90^\circ$ , $Y$ is the point where the mixtilinear incircle touches $AB$ and so since $BYIV$ is cyclic, $V$ is the mixtilinear touchpoint. Let $Q$ be the ex-touch point. Its well known that $IZ || AQ$ and $AQ, AV$ are isogonal.

So, $\angle DIZ = \angle DAQ = \angle DAV = \angle IAW = \angle WIA$ and so $I,W,Z$ are collinear, as required

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lneis1

243 posts

lneis1

#40 h Jul 20, 2021, 5:01 AM

Y by

Nice problem.

First of all note that $\angle BAI=\angle BIV$

Let $U'=UX \cap AB$ ,

Then because of the given conditions $BUIU'$ and $BYIV$ are cyclic. Notice that due to similarity we have

$UX=\frac{XI}{YI} \times VY$ and $U'X=\frac{XY}{YI} \times AI$ .

Also $\frac{AI}{VY}=\frac{XI}{XY} \implies UX=U'X$

Clearly in $\triangle BUU'$ , $BI$ is the angle bisector of $\angle UBU' \implies XI$ is the perpendicular bisector of $UU'$
$\implies YX \perp VY \implies XY \perp AI \implies \angle BIY=\angle BVY=\frac{C}{2} \implies VY \cap CI \in (ABC)$ But it is well known (trivial angle chase) that $(BYI) \cap (ABC)$ is the mixtilinear intouch point.

Also note that because of concyclicity of $B,Y,I,V$ , we get $\angle YVI =\frac{B}{2} \implies \angle VBI=90$

By tangent secant theorem, since $VI \perp CI$ , $CI$ is tangent to $(BYVI)$

Let $D= VY \cap CI$ . Then we have $\angle DVA=\angle VAI=\angle WAI=\angle AIW \implies BI \perp WI$
Note that it suffice to show that $BI \perp IZ \iff \angle ZIC=\frac{A}{2} \iff IU$ is symmedian of $\triangle BIC$ .

Clearly $V$ is the centre of spiral similarity sending $BI$ to $CI$ , hence we can say that $IV$ is the symmedian.

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MathLuis

1475 posts

MathLuis

#41 h Mar 24, 2022, 6:40 PM

Y by

Proposed by USA=-10 mohs (jk dont kill me lmao)
The proposer is already a hint as u can think on spamming usa config, and more especificaly mixtilinear incircle config. These days i dont count with so much time so yeah, me typin fast.
Let $XU \cap AB=D$ , $M,N$ the midpoint of the arcs $BC, ABC$ on $(ABC)$ , respectivily also let $IC \cap (ABC)=M_C$ also let $VM \cap BC=G$ , so now we being with the most important claim
Claim 1: $V$ is the touch point of the $A$ -mixtilinear incircle with $(ABC)$
Proof: Its enough to show that $\overline{XYI} \perp AI$ , clearly $\angle BYV=\angle BAI=\angle BIV$ so $BYIV$ is cyclic and now $\angle DBI=\angle YVI=\angle DUI$ so $BDIU$ is cyclic, now since $BI$ bisects $\angle DBU$ we have $\triangle DIU$ isosceles and now by Thales spam on those parallels we can get $DX=XU$ so indeed $\overline{XYI} \perp AI$ which means that $Y$ is the touch point of the $A$ -mixtilinear incircle with $AB$ and by well known mixtilinear config that means $V$ is the touch point of the $A$ -mixtilinear incircle with $(ABC)$ as desired.
Finishing: Since $\angle AIX=90$ we have that $W$ is center of $(AIX)$ so $\triangle AWI$ is isosceles, now by well known usa mixtilinear config we have $V,U,I,N$ colinear, $G,Y,X,I$ colinear and $M_C,Y,V$ colinear. So using these colinearities by angle chase $\angle GVN=90=\angle GZN$ so $GVZN$ is cyclic and now since $\angle GIM=90=\angle GZM$ we have $GIZM$ cyclic so now by the final angle chasing:
$\angle WIA=\angle IAW=\angle VNM=\angle VGZ=\angle MIZ \implies W,I,Z \; \text{colinear}$ Thus we are done

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HoRI_DA_GRe8

593 posts

HoRI_DA_GRe8

#42 h Apr 16, 2022, 2:13 PM • 1 Y

Y by Rounak_iitr

Finally did something worthy in this holiday,so yayyy!(Even tho I might have glanced on Telv Cohls diagram,yes taking a small hint is ok right?).Also it seems like I have lost every skill except angle chasing

ISL 2014 G7 wrote:

Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$ . Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$ , and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$ , respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

Proposed by David B. Rush, USA

Say $UX \cap AB=D$ .Notice that by angle chasing we have $\{B,Y,I,V \},\{B,D,I,X\}$ as cyclic quadraples.Also by Fact 5 $IU=ID$ and as we have $VY || UX || AI$ by similarity ratios, $\frac{XU}{YV}=\frac{IU}{IV}=\frac{AX}{AV}=\frac{XD}{YV} \implies XU=XD$ (i.e. $X$ is the midpoint of $UD$ ).So we get $IX \perp UD \implies YI \perp AI$ .Now we present a claim which is the crux of the problem.

Claim : $V$ is the $\text{A-Mixtillinear touch point}$
Proof : By angle chasing $\angle IBV=\angle IYV=\angle YIA=90$ .So,
$\begin{align*} \angle BVI &=90-\angle BIV \\ &=90-(\angle BIC-\angle CIV) \\ &=90-(\frac{1}{2}\angle A+90-90)\\ &=90-\frac{1}{2}\angle A \end{align*}$ So $VI$ bisects $\angle BVC$ so if $M$ is the midpoint of $\widehat{BAC}$ we have $V-I-M$ which implies our claim $\blacksquare$

Now say $G$ is the midpoint of $\widehat{BC}$ ,note that we have $\angle MIC=90=\angle MZC \implies \text{ M,I,Z,C are concyclic}$
So now as $\triangle AIX$ is right angled,
$\begin{align*}\angle WIX &=\angle WXI \\ &=90-\angle XAI \\ &=90-\angle VAI \\ &=90-(\angle VAC-\angle CAI)\\ &=90+\frac{1}{2}\angle A-\angle VBC \\ &=\angle IBC+\frac{1}{2}\angle A \\ &=\frac{1}{2}\angle B+\frac{1}{2}\angle A \end{align*}$ Again by angle chasing,
$\angle XIU=180-(\angle ABV)=180-(\frac{1}{2}\angle B+90)=90-\frac{1}{2}\angle B \implies \angle WIU=90+\frac{1}{2}\angle A$ Again as $MIZC$ is cyclic we get,
$\angle CIZ=\angle CFZ=\angle CFG=\angle CAG=\frac{1}{2}\angle A \implies \angle ZIU=90-\frac{1}{2}\angle A$ Finally $\angle ZIU+\angle WIU=90+\frac{1}{2}\angle A+90-\frac{1}{2}\angle A =180$ Which means $I,W,Z$ are collinear.The End $\blacksquare$

This post has been edited 2 times. Last edited by HoRI_DA_GRe8, May 25, 2022, 6:08 PM

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NoctNight

108 posts

NoctNight

#43 h May 24, 2022, 3:22 AM • 2 Y

Y by PRMOisTheHardestExam, Rounak_iitr

Solution

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JAnatolGT_00

559 posts

JAnatolGT_00

#44 h Jul 5, 2022, 7:23 PM

Y by

Let $T=UX\cap AB.$

Step 1. By $\angle BTU=\angle BYV=\frac{1}{2}\angle BAC=\angle BIV=\angle BIU$ sets $(BIYV),(BITU)$ are concyclic.

Step 2. Since $BI$ bisects angle $TBU$ we conclude $|IT|=|IU|.$ Now observe that $|TX|:|AI|=|YT|:|YA|=|VU|:|VI|=|UX|:|AI|\implies TU\perp IY\implies AI\perp IY.$
Step 3. Since $\angle BVI=\angle IYA=90^\circ -\frac{1}{2}\angle BAC=\frac{1}{2}\angle BVC$ point $V$ is the $A-\text{Mixtilinear touch point}.$

Step 4. From two previous steps $IW$ is parallel to isogonal of $AV$ in $BAC,$ which is $A-\text{Nagelian} \text{ } \ell.$
But well-known that $IZ\parallel \ell,$ thus we are done.

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Inconsistent

1455 posts

Inconsistent

#45 h Sep 15, 2022, 7:26 PM

Y by

Missing configs... :noo:

Did I mention I HATE BARY?

Let $V = I + t(U-I)$

Then $(ABCV)$ gives $(a+b+c)c + (a+b+c)(a-b)t + (a^2-2ab+b^2-c^2)t^2 = 0$
And $\overline{IXY}$ gives $2c + (a-b-c)t = 0$

So taking a polynomial gcd gives $a + b = 3c$

Notice $\overline{IWZ}$ is equivalent to $(a+b+c)(c-b) + (a^2-ab-ac+2b^2-2c^2)t = 0$ which follows from substitution, so we are done. Thank goodness.

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MathsLover04

95 posts

MathsLover04

#46 h Dec 5, 2022, 7:29 PM • 2 Y

Y by AngleChasingXD, seker.brun

As I have seen so far, nobody had my solution, so here we go :-D

Let $I_a$ be the $A-$ excentre of $ABC$ .

Claim 1: $V\in BI_a$

Proof:
Consider the point $\{N\}=XU\cap AB$
$\angle{BYV}=\angle{BAI}=\angle{BIV}$ resulting in $BYIV$ being cyclic. On the other hand, $\angle{BNU}=\angle{BYV}=\angle{BIU}$ so $BUIN$ is cyclic as well.
Now a simple angle chasing leads to $XI\perp AI$ and $\angle{ABV}=\angle{ABI_a}\Rightarrow V\in BI_a$ .

Now, consider $YV\cap CI=\{T\}$

Claim 2: $T\in (ABC)$

Proof:
$\angle{BVY}=\angle{BI_aI}=\angle{BCI}=\angle{BCT}$

Claim 3: $WI\parallel VI_a$

Proof:
Angle chase again:
$\angle{TVC}=\angle{TBC}=\angle{ABC}+\frac{\angle{ACB}}{2}\Rightarrow \angle{YVA}=\frac{\angle{ACB}}{2}=\angle{VI_aA}$ Since $WI=WA=WX$ , we obtain $\angle{WIA}=\angle{VI_aA}\Rightarrow WI\parallel VI_a$ .

Finally, construct the following points: $S$ the midpoint of $\overline{CI_a}$ , $M$ the midpoint of $\overline{II_a}$ and $\{K\}=IV\cap (BIC)$

Claim 4: $V$ is the midpoint of $KI$

Proof:
$KI\perp IC$ and $I_aC\perp IC$ , thus $ICI_aK$ is a rectangle centred at $M$ . So it is enough to prove that $MV\perp IK$ and the claim would be proved. But just use another angle chase, THIRD TIME:
$\angle{MVI}=180-\angle{MIV}-\angle{VMI}=180-\angle{VCA}-\frac{\angle{ABC}}{2}=\angle{ABV}-\frac{\angle{ABC}}{2}=90$

So I have that $VI=SI_a=\frac{KI}{2}$ and $VI\parallel I_aS\Rightarrow VISI_a$ is a parallelogram. This means nothing else than $SI\parallel VI_a$ .

In conclusion, $ZS\parallel BI_a$ , $IS\parallel BI_a$ , and $WI\parallel BI_a\Rightarrow W, I$ and $Z$ are collinear.
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popop614

270 posts

popop614

#47 h Jun 28, 2023, 4:02 AM

Y by

Absolute beauty.

sol

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Reason: silly

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Leo.Euler

577 posts

Leo.Euler

#48 h Dec 28, 2023, 7:20 PM

Y by

oops I forgot to post this one.

Suppose that $I$ , $X$ , and $Y$ are collinear. We are required to show that $I$ , $W$ , and $Z$ are collinear. Let $P=\overline{UX} \cap \overline{AB}$ .

Claim: $BUIP$ is cyclic.
Proof. Angle chase: $\angle BPU = \angle BAI = \angle A/2 = \angle BIC - 90^{\circ} = \angle BIU,$ as desired.
:yoda:

Claim: $X$ is the midpoint of $UP$ .
Proof. Since $\overline{AI} \parallel \overline{PU} \parallel \overline{YV}$ , and noting that $X = \overline{AV} \cap \overline{YI}$ , we readily have $XU=XP$ , as desired.
:yoda:

By the Incenter-Excenter lemma on $(BUIP)$ , we have $\overline{IX} \perp \overline{XU} \parallel \overline{AI}$ , so $\angle AIY = 90^{\circ}$ . Let $M$ denote the midpoint of arc $BAC$ .

Claim: $M$ lies on $\overline{IUV}$ . In particular, $V$ is the $A$ -mixtillinear touchpoint of $\triangle ABC$ .
Proof. It suffices to show that $\overline{VI}$ is an angle bisector in $\triangle BVC$ . First, I contend that $BYIV$ is cyclic. This is easy to show by angle chase: $\angle YBI = \angle PBI = \angle PUI = \angle XUI = \angle YVI,$ which implies the claimed concyclicity. The proof of the claim is easy from here: write $\angle BVI = \angle AYI = 90^{\circ} - \angle A/2 = \frac{1}{2}\cdot (180^{\circ}-\angle A) = \frac{1}{2} \angle BVC,$ as desired.
:yoda:

Note that $V$ is the center of the spiral similarity that maps $\overline{BI} \mapsto \overline{IC}$ . Thus $\overline{IV}$ is the symmedian of $\triangle BIC$ . Angle chasing shows that $\angle WIB = 90^{\circ}$ so $WI$ and $IU$ are isogonal conjugates in $\triangle BIC$ . Thus $W$ , $I$ , and $Z$ are collinear, and we are done.

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Shreyasharma

668 posts

Shreyasharma

#49 h Feb 14, 2024, 5:00 AM • 1 Y

Y by GeoKing

Pretty sure I, like many others, just used angle-chasing.

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Fun. Define $T = \overline{UX} \cap \overline{AB}$ and $E = \overline{UV} \cap \overline{AC}$ .

Note that we easily find $(BVYI)$ and $(BUIT)$ as,
$\begin{align*} (\angle YVI = \angle XUI) &=\angle EIA\\ &= \angle AIC - 90\\ &= (180 - \angle A/2 - \angle C/2)\\ &= \angle B/2 \end{align*}$ from the parallel condition.

Claim: $(UTE)$ has center $I$ .
Proof. Indeed note that to show $IU = UT$ we have,
$\begin{align*} \angle IUT &= \angle B/2\\ &= \angle IBU\\ &= \angle ITU \end{align*}$ Then to show that $IU = IE$ note that $CU = CE$ as $\overline{EU} \perp \overline{IC}$ and $\overline{IC}$ bisects $\angle ACB$ . Thus our claim follows. $\square$

Claim: $X$ is the midpoint of $\overline{TU}$ and equivalently $\overline{IXY} \perp \overline{UXT}$
Proof. To show this let $D \neq B$ be the intersection of $(BVIY)$ with $\overline{BC}$ . Then note that $\triangle TIY \overset{+}{\sim} \triangle IUD$ and $\triangle TIU \overset{+}{\sim} \triangle YID$ . Now some standard angle chasing gives,
$\begin{align*} \angle IXT &= 180 - \angle XTI - \angle XIT\\ &= 180 - \angle DYI - \frac{1}{2}(\angle YIZ)\\ &= 180 - \angle DBI - \frac{1}{2}(180 - \angle YBD)\\ &= 180 - \angle B/2 - 90 + \angle B/2\\ &= 90 \end{align*}$ Note then that $(XIA)$ has center $W$ .

Moving forward, denote by $F$ the second intersection of $(TDU)$ with $\overline{AB}$ . Now we aim to show that $\overline{EF} \perp \overline{IW}$ so we can rewrite $\overline{WI}$ as the perpendicular bisector of $\overline{EF}$ and delete $W$ from the picture. To do this we will first show,

Claim: $E, F \in (XIA)$
Proof. Note that,
$\begin{align*} \angle EAV &= \angle CBV\\ &= \angle VID\\ &= \angle VIY\\ &= \angle VIX \end{align*}$ where the last step follows from noting that under the aforementioned spiral similarities we have $\overline{IV} \perp \overline{YD}$ in isosceles $\triangle YID$ . Thus $E \in (XIA)$ . Now we show the claim for $F$ :
$\begin{align*} \angle FDI &= \angle FDU\\ &= \angle FTU\\ &= \angle BIU\\ &= \angle BIC - 90\\ &= \angle A/2 \end{align*}$ as desired. Then we find that $\overline{WZ}$ is the perpendicular bisector of $\overline{EF}$ as claimed. $\square$

Now to show the collinearity we may focus on just $Z$ . We will show it lies on the perpendicular bisector of $\overline{EF}$ . To do this first note,

Claim: $\overline{EF} \perp \overline{BI}$ .
Proof. Simply note that,
$\begin{align*} \angle AFE &= \angle AIE\\ &= \angle AIC - 90\\ &= \angle B/2 \end{align*}$ as desired. $\square$

Then all that remains is to show that $\overline{IZ} \perp \overline{IB}$ . This is then equivalent to showing that $\overline{IZ}$ and $\overline{IU}$ are isogonal with respect to $\angle BIC$ . Then as $\overline{IZ}$ is a median, it suffices to show that $\overline{IU}$ is the symmedian of $\triangle IBC$ . Then noting that the tangents to $(IBC)$ at $B$ and $C$ meet at the midpoint of arc $\widehat{BAC}$ , say $K$ , it suffices to show that $\overline{IVK}$ collinear. This is easy as,

Claim: $\overline{IVK}$ collinear.
Proof. Simply note that,
$\begin{align*} \angle AIV &= 90 + \angle C/2 + \angle BIV\\ &= 90 + \angle C/2 + \angle BYV\\ &= 90 + \angle C/2 + BAI\\ &= 90 + \angle C/2 + \angle A/2\\ &= 180 - \angle B/2 \end{align*}$ but we also have,
$\begin{align*} \angle AIK &= \angle AIC - \angle CIK\\ &= 90 + \angle B/2 - 90\\ &= \angle B/2 \end{align*}$ so we are done. $\square$

This completes the proof.

This post has been edited 2 times. Last edited by Shreyasharma, Feb 14, 2024, 5:28 PM

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thdnder

194 posts

thdnder

#50 h Mar 19, 2024, 6:10 PM

Y by

We begin with the following claim:

Claim: The points $B, Y, I, V$ are concyclic.

Proof. Note that $\angle BYV = \angle BAI = \frac{\angle BAC}{2} = \angle BIV$ , so $BYIV$ is cyclic. $\blacksquare$

Now let $N = XU \cap AB$ . Then $XU \parallel AI \implies \angle BNU = \angle BAI = \angle BIU$ , hence $BNIU$ is cyclic. Since $\angle IUN = \angle INU = \frac{\angle ABC}{2}$ , so $IN = IU$ . Now since the lines $AI, NU, YV$ are parallel, thus $\frac{XU}{AI} = \frac{XV}{VA} = \frac{XY}{YI} = \frac{XN}{AI}$ , so we get $XU = XN$ . Combined with the fact that $IN = IU$ , we see that $XI \perp NU$ , i.e $\angle AIX = \angle AIY = 90^{\circ}$ . Therefore $A-$ mixtilinear circle touches $AB$ at $Y$ and since $BYIV$ are cyclic, so $A$ -mixtilinear circle touches $\Omega$ at $V$ . Let $M = AI \cap \Omega$ and let $I_A$ be the $A$ -excenter and assume $A$ -excircle touches $BC$ at $D$ .

Claim: $IZ \parallel AD$ .

Proof. Let $L = AI \cap BC$ . Then $\frac{LI}{LA} = \frac{\frac{BL \cdot LC}{LI_A}}{\frac{BL \cdot LC}{LM_A}} = \frac{LM_A}{LI_A}$ , so $\frac{IM_A}{AI_A} = \frac{ZM_A}{DI_A}$ and $\angle IM_AZ = \angle AI_AD \implies \triangle IM_A$ is similar to $\triangle AI_AD$ . Therefore $\angle LIZ = \angle LAD$ , so $IZ \parallel AD$ . $\blacksquare$

Now it's well-known that $AV$ and $AD$ are isogonal. Combining with the fact that $\angle AIX = 90^{\circ}$ and $\angle WAI = \angle WIA$ , we see that $I, Z, W$ are collinear, as desired. $\blacksquare$

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awesomeming327.

1692 posts

awesomeming327.

#51 h Apr 15, 2024, 3:10 AM • 2 Y

Y by DU4532, aidan0626

Diagram

Let $IV$ intersect $AB$ at $T$ and $M_B$ be the midpoint of arc $AC$ not containing $B$ . Clearly, we have
$\frac{TI}{TV}=\frac{AI}{YV}=\frac{AX}{XV}=\frac{UI}{UV}$ so $(T,U;I,V)=-1$ . Taking perspectivity through $B$ , $(A,C;M_B,V)=-1$ . This implies that $V$ is the midpoint of arc $ABC$ .

Now, observe that
$\angle BIV=\angle BIC-90^\circ=\frac12 \angle BAC=\angle BAI=\angle BYV$ so $BYIV$ is cyclic. Thus, $\angle VYI=\angle VBI=90^\circ$ . Since $AI\parallel YV$ , $\angle AIY=90^\circ$ .

Since $VA=VC$ and $\angle AIC=90^\circ+\frac12 \angle AVC$ , $VA$ and $VC$ are tangent to $(AIC)$ . Thus, $VI$ is symmedian of $\triangle AIC$ . Let $Y$ be the midpoint of $AC$ then $\angle AIY'=\angle CIT=90^\circ$ . Therefore, $Y,I,Y'$ are collinear. Furthermore, $Y'$ is the reflection of $Y$ over $I$ , and $Y'$ is on $VM_B$ .

We have
$\angle YVI=\angle YBI=\angle ABM_B=\angle AVM_B=\angle XVY'$ so $VX$ is symmedian in $\triangle YVY'$ . This implies $AY$ , $AY'$ tangent to $(YVY')$ . Thus, $(YVY')$ is the mixtillinear incircle of $\triangle ABC$ . Let $E$ be the point where the $A$ -excircle is tangent to $BC$ . Using a $\sqrt{bc}$ inversion, we can see that $AV$ and $AE$ are isogonal. Let $X'$ be the reflection of $X$ over $I$ . We know that $X'$ is on $AE$ , so $WI\parallel AE$ . We know that $AE\parallel IZ$ so we are done.

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ohhh

46 posts

ohhh

#52 h Jun 24, 2024, 2:48 AM

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Define $R = UI \cap AC$ , $T = UX \cap AB$ and suppose $I-X-Y$ .
$\frac{VY}{IA} = \frac{VX}{XA} = \frac{VU}{UI} = \frac {VU}{IR}$
$\implies \frac{VY}{VU} = \frac {IA}{IR}$
Since $VY \parallel IA$ , $\angle YVI = \angle AIR \implies\triangle YVU \sim \triangle AIR$ (because the angle + above equality).
From this we get $\angle VYU = \angle IAC \implies AC \parallel YU$ !
$\implies \angle AIY = 90$ (Ez to prove with trig).
Now, notice that $\angle BIV = \angle VYB \implies BYIV$ cyclic.
$\implies 90 = \angle AIY = \angle IYV = \angle IBV = \frac{\angle B}{2} + \angle VAC = \frac{\angle B}{2} + \frac{\angle A}{2} + \angle VAI$
$\implies 90 = \frac{\angle B}{2} + \frac{\angle A}{2} + \angle VAI \implies \angle VAI = \frac{\angle C}{2}$
With simple angle chasing we can get from this:
$\cdot$ Line $WI$ bissects the segment $VC$ ;
$\cdot$ And $WI \parallel BV$ ;
So $WI$ bissects $BC$ too, as desired.

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bin_sherlo

687 posts

bin_sherlo

#54 h Sep 4, 2024, 7:18 PM

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Let $M,N$ be the midpoint of the arcs $BC,BAC$ . $VM \cap BC=T$ .

Claim: $V$ is the midpoint of the arc $ABC$ .
Proof: Let $AB\cap VI=K$ .
$\frac{KV}{KI}=\frac{VY}{IA}=\frac{XV}{XA}=\frac{UV}{UI}\implies -1=(K,U;V,I)\overset{B}{=}(A,C;V,BI\cap (ABC))$ Since $BI\cap (ABC)$ is the midpoint of the arc $AC$ , we get that $VA=VC$ . $\square$

Claim: $V,I,N$ are collinear.
Proof:
$\measuredangle MIV=\measuredangle ADC-\measuredangle IUC=\frac{\measuredangle A}{2}+\measuredangle B-\frac{\measuredangle A+\measuredangle B}{2}=\frac{\measuredangle B}{2}$ $\measuredangle VMA=\measuredangle VCA=\frac{\measuredangle A+\measuredangle C}{2}$ Hence $\measuredangle IVM=180-\measuredangle MIV-\measuredangle VMA=90=\measuredangle NVM$ yields $V,I,N$ are collinear. $\square$

Claim: $\measuredangle XIA=90$ .
Proof: Since $\measuredangle UXV=\measuredangle IAV=\frac{\measuredangle C}{2}=\measuredangle CBM-\measuredangle VMB=\measuredangle UTV,$ we conclude that $T,X,U,V$ are concyclic. $\measuredangle TVU=90$ gives $\measuredangle UXT=90$ .
$\frac{TU}{TD}=\frac{MI}{MD}.\frac{VU}{VI}=\frac{AI}{ID}.\frac{XU}{AI}=\frac{XU}{ID}$ Thus, $T,X,I$ are collinear. $180-\measuredangle XIA=\measuredangle DIT=\measuredangle UXT=90$ , this gives us $\measuredangle XIA=90$ which is the desired angle condition. $\square$

$MI^2=MZ.MN$ yields $\measuredangle ZIM=\measuredangle MNI=\measuredangle MNV=\frac{\measuredangle C}{2}$ . Also since $\measuredangle XIA=90$ and $W$ is the midpoint of $AX,$ we have $WA=WI=WX$ .
$\measuredangle WAI=\measuredangle IAW=\frac{\measuredangle C}{2}=\measuredangle ZIM$ Thus, $I,Z,W$ are collinear as desired. $\blacksquare$

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