OIMC 2021 P5: HAT's have tangencies too!

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OIMC 2021 P5: HAT's have tangencies too!

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Source: OIMC 2021 P5 (Mock Contest)

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L567

#1 h Aug 18, 2021, 12:08 PM • 9 Y

Y by oVlad, EpicNumberTheory, i3435, mathmax12, Orestis_Lignos, centslordm, megarnie, MathLuis, MatBoy-123

Let $H$ be the orthocenter of $\triangle ABC$ and let $AH$ intersect the circumcircle of $\triangle ABC$ at $H_A$ . Let $A'$ be the reflection of $A$ across $BC$ , let $H_B$ , $H_C$ be the feet of perpendiculars from $H$ onto $A'B$ and $A'C$ , respectively. Let $T$ be the intersection of lines $BH_C$ and $CH_B$ . Prove that $TH_A$ is tangent to the circumcircle of $\triangle ABC$ .

Proposed by Bora Olmez, Switzerland

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#2 h Aug 18, 2021, 12:22 PM • 5 Y

Y by oVlad, mathmax12, centslordm, betongblander, megarnie

Solution, during contest :blush:

:blush:

Define the following points:
Let $D,E,F$ the feet of altitudes from $A,B,C$ to $BC,CA,AB$ respectivily. Let $EF$ meet $BC$ at $G$ , $M$ is the midpoint of $BC$ , $I$ the miquel point of $BFEC$ , $L$ the miquel point of $BH_BCH_C$ , $K$ the reflection of $I$ over $BC$ , $H_BH_C$ meets $AT$ at $J$ and $N,O$ are centers of $(BH_BCH_C)$ and $(IHH_A)$ respectivily.
First than anything wlog $AB<AC$ (we will check the sad case when $AB=AC$ later xd), i will present 2 useful lemmas i used:
Lemma 1: Let $ABCD$ be a cyclic quadrilateral, $AB$ meets $CD$ at $E$ and $AD$ meets $BC$ at $F$ , let $G$ be the miquel point of $ABCD$ , then $E,F,G$ are colinear.
Proof: Let $EF$ meet $(EBC)$ at $G'$ , then by power of point $FG' \cdot FE=FB \cdot FC=FA \cdot FD$ thus we have that $G'EDA$ is also cyclic but this means that $(EBC)$ and $(EDA)$ meet at $G'$ thus we have that $G=G'$
Lemma 2: Let $ABCD$ be a cyclic quadrilateral with center $O$ , $AB$ meets $CD$ at $E$ , $AD$ meets $BC$ at $F$ and $AC$ meets $BD$ at $G$ , let $H$ be the miquel point of $ABCD$ , then $G$ is the inverse of $H$ w.r.t. $(ABCD)$
Proof: Denote $M,N$ as the mid points of $AB$ and $CD$ , then its known that $H$ is the center of sprial simlarity that sends $AB$ to $CD$ , then it also sends $AM$ to $DN$ becuase $\angle HAM=\angle HDN$ and that if you divide by $2$ and then cross multiply the ratio formed of the spiral similarity that sends $AB$ to $CD$ you get $HA \cdot DN=HD \cdot AM$ . Now using the new spiral similarity and knowing that $\angle OND=\angle OMA=90$ we have that $OMHEN$ is cyclic with diameter $OE$ , thus $\angle OHE=90$ , and since by Lemma 1 we have that $E,F,H$ are colinear and by brockard the polar w.r.t. $(ABCD)$ of $G$ is $EF$ we are done.
Now we proced to solve the main problem...
Claim 1: $H_B,D,H_C$ are colinear.
Proof: Since $\angle HH_BB=\angle HDB=\angle HDC=\angle HH_CC=90$ then we have that $HH_BBD$ and $HCH_CD$ are cyclic and by power of point $AH_B \cdot AB=AH \cdot AD=AC \cdot AH_C$ thus $H_BBH_CC$ is cyclic. Now it follows that the radical axis of $(H_BBH_CC)$ and $(HH_BA'H_C)$ its $H_BH_C$ .
Since its known that $H_A$ is the reflection of $H$ over $BC$ we have that $BHCA'$ is cyclic, thus by power of point $DH \cdot DA'=DB \cdot DC$ and this means $D$ lies on the radical axis of $(H_BBH_CC)$ and $(HH_BA'H_C)$ thus we are done.
Claim 2: $J,T,G,L,K,A'$ are colinear
Proof: Using Lemma 1 we have that $A,I,G$ are colinear and if you reflect this over $BC$ then $G,K,A'$ are colinear, we also have that $T,L,A'$ are colinear.Let $BC$ meet $A'J$ at $G'$ , now by a well known lemma on $\triangle TCA'$ we have that $-1=(G', B; D, C)$ thus $G'=G$ , thus the rest of the colinearities also follows.
Claim 3: $TH$ is tangent to $(AIFHE)$ at $H$ .
Proof: Taking $(H_BBH_CC)$ as circle of inversion and using Lemma 2 we have that $L,D$ are inverses, now by power of point $A'K \cdot A'T=A'H_B \cdot A'B=A'H \cdot A'D$ we have that $TLDH$ is cyclic and since $\angle TLD=90$ we have that $\angle THA=90$ thus we are done since $(AIFHE)$ has $AH$ as diameter.
Claim 4: $TIHH_AK$ is cyclic.
Proof: Taking inversion w.r.t. $(BFEC)$ and using Lemma 2 we have that $I,H$ are inverses, and reflecting this on $BC$ gives us that $H_A,K$ are inverses, since $M$ is the center of $(BFEC)$ then $MH \cdot MI=MH_A \cdot MK$ thus $K$ lies on $(IHH_A)$ , but since $\angle H_AKT=90=\angle THH_A$ we have that $T$ lies on $(KH_AHI)$ and we are done.
Final step: Using Claim 3 and Claim 4 we have that $\angle IAH=\angle THI=\angle TH_AH$ thus $TH_A$ is tangent to $(ABC)$ and we are done :blush:

:blush:

.
Wait, wait, wait!!, but what if $AB=AC$ ?
Well, We have that $D=M$ , since $H_A$ is the reflection of $H$ with respect of $BC$ and that $A$ , the center of $(ABC)$ and $M$ are colinear, then $AH_A$ is diameter on $(ABC)$ and if you reflect this with respect of $BC$ then $HA'$ is diameter on $(HBC)$ , this means that $H_B=B$ and $H_C=C$ then $T$ becomes a point at infinity on $BC$ , thus the problem is now proving that the parallel of $BC$ that passes $H_A$ its tangent to $(ABC)$ but this follows becuase $\angle AMC=90$ and by the parallel line $\angle TH_AA=90$ thus we are done since $AH_A$ is diameter of $(ABC)$ on this special case.
Now we are really done :blush:

:blush:

, i hate and love this problem lol.

PD: as you guys can see, i literaly spammed all i know to solve this lol, this was solved with hate and brute force...

This post has been edited 1 time. Last edited by MathLuis, Aug 18, 2021, 12:24 PM

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892 posts

#3 h Aug 18, 2021, 12:28 PM • 3 Y

Y by oVlad, mathmax12, centslordm

$AH$ intersects $BC$ at $D$ . $AH$ intersects $H_CB$ at $U$ . $HH_C$ intersects $BC$ at $V$ . $TH$ intersects $H_BH_C$ at $X$ .

$\textbf{Claim:}$ $X$ is the midpoint of $TH$ .

$\textbf{Proof:}$ We have $\angle HH_BA'=90, \angle HH_CA'=90, HD\perp BC \implies HH_BBD, HDH_CC, HH_BA'H_C$ are cyclic quadrilaterals. Thus $\angle BHD=\angle BH_BD=\angle A'HH_C=\angle DHV$ and $HD \perp BC$ together implies $|BD|=|DV|$ .
Because $A'U, BC$ and $H_BH_C$ are concurrent we have $-1=(T,U;B,H_C)\overset{H} =(UT\cap BC,D;B,V)$ . With $|BD|=|DV|$ , we see that $UT\cap BC$ is a point at infinity $\implies UT \parallel BC$ . $D$ is the midpoint of $BV \implies X$ is the midpoint of $TH$ . $\square$

It is well known that $D$ is the midpoint of $HH_A$ . Because $X$ is the midpoint of $TH$ we have $TH_A\parallel DX$ . $\angle TH_AA=\angle XDH=\angle A'DH_C=\angle A'CH=\angle ACH_A \implies TH_A$ is tangent to the circumcircle of $ABCH_A$ .

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#4 h Aug 18, 2021, 12:40 PM • 3 Y

Y by mathmax12, oVlad, centslordm

diagram
Let WLOG $AB>AC$ (if $AB=AC)$ everything meets at the point of infinity on line $BC$ ).
Let $\angle BAC=\alpha ; \angle ABC=\beta$ and $\angle ACB=\gamma$ .Let $X$ be the foot of the perpendicular from $A$ to $BC$ .

Now $\angle BA'C + \angle BHC =\alpha + 180 - \alpha = 180$ , thus $BA'HC$ is cyclic. Also from all the perpendiculars we get that $A'H_BH_CH; BH_BXH$ and $CH_CHX$ are cyclic.

Claim: $H_B; X$ and $H_C$ are collinear.
Proof: $\angle HH_BX=\angle HBX= \angle HA'H_C=\angle HH_BH_C$
Claim: $BH_BCH_C$ is cyclic.
Proof: $\angle H_BBC= \angle XHC=\angle H_BH_CC=\beta$
Now let $O$ be the circumcenter of $(BH_BH_CC)$ .
Claim: $OH$ is perpendicular to $A'H$ .
Proof: $\angle H_BOC= 2\angle H_BBC= 2\beta = \angle H_BHC$ so we get that $H_BCHO$ is cyclic. Now $90 - \beta = \angle OCH_B=\angle OHH_B$ , thus $\angle OHX= 90 - \beta +\beta = 90$ . $\square$

Now Brocard's theorem on the cyclic quadrilateral $BH_BCH_C$ gives us that $OX\perp A'T$ and $OT\perp A'X$ ,thus $H$ lies on $OT$ .

Now let $M$ be the midpoint of $BC$ and let $Y$ be the reflection of $H_A$ across point $M$ and $Z$ be the reflection of $H$ across $M$ . $O$ lies on the perpendicular bisector of $BC$ ,thus $OM\perp BC$ and so $OM||HH_A$ and $HH_A=2OM$ , thus $Y$ lies on $OH$ and $O$ is the midpoint of $HY$ . But we know that $X$ is the midpoint of $HH_A$ , thus $OX||YH_A$ and $YH_A\perp A'T$ .
Now because $A'H\perp YT$ and $YH_A\perp A'T$ we get that $H_A$ is the orthocenter of $A'TY$ .

Claim: $TH_A$ is tangent to $(ABC)$ .
Proof: It's enough to prove that $\angle TH_AA=\angle H_AZA$ .
Because $H_A$ is the orthocenter of $A'TY$ we get that $\angle TH_AH=\angle HYA'$ . We also know that $H_AZYH$ is a rectangle, thus $\triangle H_AZA \cong A'YH$ , i.e $\angle AZH_A=\angle A'YH= \angle AH_AT$ and we're done!

This post has been edited 1 time. Last edited by Pitagar, Aug 18, 2021, 12:45 PM

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hyay

#5 h Aug 18, 2021, 12:40 PM • 3 Y

Y by mathmax12, oVlad, centslordm

Solution

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PUjnk

#6 h Aug 18, 2021, 12:47 PM • 2 Y

Y by oVlad, centslordm

My solution during test:

Claim 1: $BH_BCH_C$ is cyclic.

proof: Firstly note that since A' is the reflection of A about BC, $H \in \circledcirc{A"BC}$ . So $H_B,D,H_C$ is the simson line of H with respect to $\triangle{A"BC}$ and are hence collinear.
Now note that $\triangle{A'H_AF} \sim \triangle{A'HH_B}$ and $\triangle{A'H_AE} \sim \triangle{A'HH_C} \Longrightarrow EF \Vert H_BH_C$ .
But now we have $BCEF$ is cyclic so by Reim's Theorem, $BH_BCH_C$ is also cyclic as desired.

Claim 2: $A',S,T$ are collinear.

proof: Firstly note that $H=\circledcirc{BH_BD} \cap \circledcirc{CH_CD}$ (since $\angle HH_BB=\angle HDC= \angle HH_CC=90$ .)
So H is the miquel point of the quadrilateral $BH_BH_CC$ .
But now since $BH_BCH_C$ is cyclic, we also have that the miquel point is the foot of perpendicular from $T$ onto $DA'$ .
So $\angle THD=90$ . - (1)
Now we show that $HD$ bisects angle $BHH_C$ .
For this note that $\angle BHD=\angle BH_BD=\angle DCH_C=\angle DHH_C \Longrightarrow \angle BHD=\angle DHH_C$ .
So $HD$ bisects angle $BHH_C$ .
Thus $(HT,HD;HB,HH_C)=-1$ .
Projecting on $BH_C$ , we obtain:
$-1= (T,K;B,H_C) \overset{A'}{=}(A'T \cap BC,D;B,C)$ . (since $A',H_A,H$ are collinear)
But $(S,D;B,C)=-1 \Longrightarrow S=TA' \cap BC$ . This proves the claim.

Claim 3: $TH_A \Vert SL$

proof: Note that since $\angle THA'=\angle SDA'=90$ , we have $TH \Vert SD$ . Also we have $T-S-A'$ by Claim 2. So it suffices to show :
$\frac{A'L}{A'D}=\frac{A'H_A}{A'H}$ .
Now note that $(A',H_A;L,D)=-1 \Longrightarrow \frac{A'L}{A'D}=\frac{H_AL}{H_AD}=\frac{LF}{DF}$ (since $H_A$ is the incenter) $=\frac{FE}{DF+DE}$ .
$\iff \frac{A'H}{A'H_A}=\frac{DF+DE}{FE} \iff \frac{HH_A}{A'H_A}= \frac{DF+DE-FE}{FE} \iff \frac{DH_A}{A'H_A}=\frac{\frac{DF+DE-FE}{2}}{FE}$ .
Now we first prove the following lemma :

Lemma: In a triangle ABC with side lengths a,b,c and semiperimeter s,if I is the incenter and $I_A$ the A-excenter, then $\frac{IA}{II_A}=\frac{b+c-a}{2a}$

proof:
Drop perpendiculars from I and $I_A$ onto AB at X and Y respectively. Now it is well known that $AX=s-a$ and $AY=s$ .
Now $IX \Vert I_AY$ and $A,I,I_A$ are collinear. So $\frac{AI_A}{AI}=\frac{s}{s-a}$ .
Subtracting one from both sides gives:
$\frac{II_A}{AI}=\frac{a}{s-a} \Longrightarrow \frac{AI}{II_A}=\frac{s-a}{a}=\frac{b+c-a}{2a}$ as desired.

Now observe that $H_A$ is the incenter of $\triangle{DEF}$ and $A'$ the D-excenter since $\triangle{DEF}$ is the orthic triangle of $\triangle{A'BC}$ .
This proves the claim.
\end{proof}

Back to the main problem, note that $\angle TH_AB=\angle SEB$ (since $TH_A \Vert SL$ by Claim 3) $=\angle FEB$ (since $S \in EF$ ) $=\angle FCB=\angle H_ACB$ since $BEFC$ is cyclic.
Thus $\angle TH_AB=H_ACB$ .
Thus by Tangent-secant theorem, $TH_A$ is tangent to $\circledcirc{ABC}$ .

This post has been edited 1 time. Last edited by PUjnk, Aug 21, 2021, 4:11 PM

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#7 h Aug 18, 2021, 12:48 PM • 4 Y

Y by oVlad, centslordm, CrazyMathMan, GeoKing

Let $D$ be the foot of perpendicular from $A$ onto $\overline{BC}$ and consider a reflection over $\overline{BC}$ . Then $H_B$ and $H_C$ become foots of perpendiculars from $H_A$ onto $\overline{AB}$ and $\overline{AC}$ , say $B'$ and $C'$ . $T'$ is the intersection of $\overline{BC'}$ and $\overline{CB'}$ . $(ABC)$ becomes $(BHC)$ , so it suffices to show that $\overline{T'H}$ is tangent to $(BHC)$ .

The Simson line of $H_A$ is $\overline{B'DC'}$ and since $H_A$ lies on both $(BDB')$ and $(CDC')$ , it follows that $H_A$ is the Miquel point of quadrilateral $BB'C'C$ . As $AB \cdot AB' = AD \cdot AH_A = AC \cdot AC'$ , it follows that $B$ , $B'$ , $C$ , $C'$ are concyclic as well. Let $\ell$ be the Aubert (or Steiner) line of this quadrilateral. This passes through $H$ by definition, and since this is the radical axis of circles with diameters $BC'$ and $CB'$ , it follows by radical axes that it also passes through $T'$ . Thus $\overline{T'H} \equiv \ell$ . Meanwhile, $\ell$ is the dilation of $\overline{B'DC'}$ with factor $2$ centered at $H_A$ , so $T'H \parallel B'C'$ . Therefore, finally,
$\measuredangle T'HB = \measuredangle DC'H_A = \measuredangle DCH_A = \measuredangle HAB,$ and we have the desired tangency. $\blacksquare$

This post has been edited 1 time. Last edited by KST2003, Aug 18, 2021, 12:52 PM

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MP8148

#8 h Aug 18, 2021, 1:50 PM • 2 Y

Y by centslordm, oVlad

https://artofproblemsolving.com/community/c6h2036412

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#9 h Aug 18, 2021, 3:14 PM • 2 Y

Y by oVlad, centslordm

Notice that $D$ and $H$ are the intersections of $(H_bBH)$ and $(H_cCH)$
By simple angle chase $HH_cC$ and $HH_bB$ are in a spiral similarity, so by simple angle chase we can conclude that $H_b, H_c$ and $D$ are collinear.
It is a known fact that $H_a \in (ABC)$ , and it is easy to prove that $AH_bHH_c$ is a cyclic quadrilateral. By power of point:
$DH_a.DA=DB.DC=DH.DA'=DH_b.DH_c$ so $H_bBH_cC$ is a cyclic quadrilateral, and let $I$ be his circumcenter. Let $ID\cap A'P=M$ and $PI\cap AD =H'$ , i will prove that $H'=H$ .

By brocard theorem, $D$ is the ortocenter of $IPA'$ , in fact, using inversion with center $I$ and ray $IB$ we can prove that the inverse of $D$ is the miquel point, and using also brocard theorem we can easily prove that $M$ is this inverse, because $PQ$ is the polar of $D$ . This is very know, so I don't need to go into more detail. Anyway, as $ID.IM=IB^2=IC^2$ , $\angle IMC= \angle ICD= \angle IBD$ , so $IBMC$ is cyclic. $IH'MA'$ is also cyclic, so by power of point we have:
$H'D.DA'=DI.DM=DC.DB=DA'.DH$ and then $H'=H$
and $PH//BC$ .

Now let $O$ be the circumcenter of $ABC$ , by pitagoras theorem $\angle OH_aP=90$ if and only if
$OI^2 + IP^2 =OH_a^2 +PH_a^2$ some replacements:

$PH_a^2=4HD^2 +HP^2$

$OH_a^2= IH^2 + \frac{AH_a^2}{4}$

$OI=2HD - \frac{AH_a}{2}$

$IP=IH+HP$

So we have:
$OH_a^2 +PH_a^2=IH^2 + \frac{AH_a^2}{4} +4HD^2 +HP^2 - 2AH_a.HD + 2AH_a.HD$ Notice that since $D$ is ortocenter of $IA'P$ , the reflection of $D$ by $IP$ lies on $(Ia'P)$ , and then, by power of point, $HD.HA'=HI.HP=AH_a.HD$ . SO:
$OH_a^2 +PH_a^2=IH^2 + \frac{AH_a^2}{4} +4HD^2 +HP^2 - 2AH_a.HD + 2AH_a.HD = (IH+HP)^2 + ( 2HD - \frac{AH_a}{2})^2= OI^2 + IP^2$ then
$\angle IH_aP=90$ and, finally, $PH_a$ is tangent to $(ABC)$ .

This post has been edited 1 time. Last edited by betongblander, Aug 18, 2021, 3:14 PM

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#10 h Aug 30, 2021, 8:41 PM • 2 Y

Y by centslordm, Mango247

Let $D$ be the foot from $A$ to $BC$ .
Observe that $A'BCH$ , $A'H_BHH_C$ , $HDBH_B$ , $HDH_CC$ are cyclic quadrilaterals. Also, by Simson Line, $H_B,H_C,D$ are collinear, and, by radical axis theorem, $H_BH_CCB$ is cyclic quadrilateral.

Let $M$ be the midpoint of $TH$ , let $P$ be the center of $(H_BH_CCB)$ .
Claim. $HP\perp HD$ .
Proof. $\begin{align*} \measuredangle H_BPC=2\measuredangle H_BH_CC=\measuredangle H_BH_D+\measuredangle DHC=\measuredangle H_BHC, \end{align*}$ thus $H_BHPC$ is cyclic quadrilateral. Similarly, $BHPH_C$ is cyclic quadrilateral.
Now, $\begin{align*} \measuredangle CHP=\measuredangle CH_BP=90^\circ+\measuredangle CH_CH_B=\measuredangle HH_C=\measuredangle HCD, \end{align*}$ hence $HP\parallel BC\perp HD$ . $\square$
Note that the center of $(H_BH_CCB)$ lies on $A'H$ and as $A'D$ is the polar of $T$ wrt $(H_BH_CCB)$ , we get that $H,P,T$ are collinear and $HT$ is tangent to $(HH_BH_C)$ .

Also, $\begin{align*} \measuredangle H_BTH=\measuredangle H_BCB=\measuredangle H_BH_CT, \end{align*}$ thus $HT$ is tangent to $(H_BH_CT)$ .

As $M$ is the midpoint of $HT$ , we get that $M$ lies on the radical axis of $(H_BH_CT)$ and $(HH_BH_C)$ , which is $H_BH_C$ .

Because $H_AT$ is parallel to $MD$ , it is sufficient to show that $\measuredangle HDH_B=\measuredangle ACH_A$ . Indeed,
$\begin{align*} \measuredangle HDH_B&= \measuredangle HBH_B=\measuredangle HBA'=\measuredangle HCA'\\&=\measuredangle HCB+\measuredangle BCA'=\measuredangle ACB+\measuredangle BCH_A=\measuredangle ACH_A, \end{align*}$ we are done. $\blacksquare$

$[asy]import olympiad; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,H,Ha,Hb,Hc,a,D,T,M,P; O=(0,0);A=dir(125);B=dir(195);C=dir(345);path w=circumcircle(A,B,C);D=foot(A,B,C);a=2D-A;H=orthocenter(A,B,C);Hb=foot(H,a,B); Hc=foot(H,a,C);Ha=2D-H;T=extension(Hb,C,Hc,B);M=midpoint(T--H);P=circumcenter(Hb,Hc,B); draw(A--B--C--cycle,red+1);draw(w,heavyblue+1); draw(Hb--a--C--H--cycle,orange);draw(circumcircle(B,C,Hb),lightgreen+1);draw(A--a,orange);draw(T--Hc,olive);draw(T--P,olive);draw(C--T,olive);draw(T--Ha,olive);draw(M--Hc,heavycyan+1); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$A'$",a,dir(a)); dot("$H$",H,dir(H)); dot("$H_A$",Ha,dir(Ha)); dot("$H_B$",Hb,dir(Hb)); dot("$H_C$",Hc,dir(Hc)); dot("$T$",T,dir(T)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); [/asy]$

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i3435

#11 h Aug 30, 2021, 10:18 PM • 4 Y

Y by Mango247, Mango247, Mango247, MS_asdfgzxcvb

$(P_1P_2P_3\dots)$ represents the circumcircle of $P_1,P_2,P_3,\dots$ and $\measuredangle$ represents directed angle $\mod 180$ .

$\measuredangle CA'B=\measuredangle BAC=\measuredangle CHB$ , and $\measuredangle A'H_BH=90=\measuredangle A'H_CH$ , so $B,H,C,A'$ and $H,H_B,A',H_C$ are cyclic. If $D$ is the foot from $A$ to $\overline{BC}$ , then $H_B-D-H_C$ since they lie on the Simson line from $H$ to $\triangle BCA'$ .

Let $B'$ be the reflection of $B$ over $D$ . $\measuredangle DHH_C=\measuredangle A'HH_C=90-\measuredangle CA'H=\measuredangle DCA'=\measuredangle ACB=\measuredangle BHD$ , so $H-B'-H_C$ . If $M$ is the intersection of the line through $H$ parallel to $\overline{BC}$ and $\overline{H_BH_C}$ , then by homothety at $H_C$ , $\overline{BH_C}\cap\overline{HM}$ is the reflection of $H$ over $M$ . By symmetry, $T$ is the reflection of $H$ over $M$ .

By a homothety at $H$ with factor $\frac{1}{2}$ , it suffices to show that $\overline{MD}$ is parallel to the tangent from $H_A$ to $(ABC)$ . $\measuredangle BH_BH=90=\measuredangle BDH$ , so $H_B,H,D,B$ are cyclic. $\measuredangle ADH_B=\measuredangle HDH_B=\measuredangle HBH_B=\measuredangle HBA'=\measuredangle ABH_A$ , as desired.

$[asy] unitsize(3cm); import olympiad; import geometry; pair A,B,C,D,H_A,H_B,H_C,H,M,T,B_A,A_A; A=dir(110); B=dir(220); C=dir(320); D=foot(A,B,C); H=A+B+C; H_A=2*D-H; A_A=2*D-A; H_B=foot(H,B,A_A); H_C=foot(H,C,A_A); B_A=2*D-B; M=extension(H_B,H_C,H,H+B-C); T=2*M-H; draw(unitcircle); draw(M--H_C); draw(H--T); draw(C--T--H_C,dotted); draw(circumcircle(H_B,H_C,A_A)); draw(A--A_A); draw(H_B--A_A--C); draw(A--B--C--cycle); draw(circumcircle(H_B,B,D)); draw(circumcircle(B,H,C)); draw(H--H_C,dotted); draw(T--H_A,dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,NE); dot("$H_A$",H_A,dir(H_A)); dot("$H_B$",H_B,N); dot("$H_C$",H_C,SE); dot("$H$",H,NE); dot("$B'$",B_A,SW); dot("$A'$",A_A,S); dot("$M$",M,N); dot("$T$",T,N); [/asy]$

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