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Perfect Squares, Infinite Integers and Integers
steven_zhang123   4
N 38 minutes ago by ohiorizzler1434
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
4 replies
steven_zhang123
Yesterday at 12:06 PM
ohiorizzler1434
38 minutes ago
another geometry problem with sharky-devil point
anyone__42   11
N 40 minutes ago by zhenghua
Source: The francophone mathematical olympiads P1
Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$.
Prove that $B,D,G$ and $X$ are concylic
11 replies
anyone__42
Jun 27, 2020
zhenghua
40 minutes ago
IMO Shortlist 2011, Algebra 5
orl   18
N an hour ago by mathfun07
Source: IMO Shortlist 2011, Algebra 5
Prove that for every positive integer $n,$ the set $\{2,3,4,\ldots,3n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.

Proposed by Canada
18 replies
orl
Jul 11, 2012
mathfun07
an hour ago
Line Perpendicular to Euler Line
tastymath75025   55
N an hour ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
an hour ago
Foot from vertex to Euler line
cjquines0   31
N an hour ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
1 viewing
cjquines0
Jul 19, 2017
pUssydestroyer777
an hour ago
Inequality => square
Rushil   12
N 2 hours ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
2 hours ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 2 hours ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
2 hours ago
H not needed
dchenmathcounts   44
N 3 hours ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
3 hours ago
IZHO 2017 Functional equations
user01   51
N 3 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
3 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 3 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
3 hours ago
Some properties of Hagge circle
TelvCohl   2
N Aug 4, 2015 by mhq
Some properties of Hagge circle :)

Theorem 1:

Let $ H $ be the orthocenter of $ \triangle ABC $ .
Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ .
Let $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB $, respectively .

Then $ A_2, B_2, C_2, H $ are concyclic .

Proof:

Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ .
Let $ \triangle A_1^*B_1^*C_1^* $ be the circumcevian triangle of $ P^* $ WRT $ \triangle ABC $ .
Let $ B_2^*, C_2^* $ be the reflection of $ B_2, C_2 $ in the midpoint $ D $ of $ BC $, respectively .
Let $ O $ be the circumcenter of $ \triangle ABC $ and $ A^* $ be the antipode of $ A $ in $ \odot (ABC) $ .

Easy to see $ A_2, B_2, C_2 $ is the reflection of $ A_1^*, B_1^*, C_1^* $ in the midpoint of $ BC, CA, AB $, respectively .

Since $ CB_2AB_1^*, CB_2BB_2^* $ are parallelogram ,
so $ ABB_2^*B_1^* $ is a parallelogram $ \Longrightarrow $ the midpoint of $ AB_2^* $ is the projection of $ O $ on $ BP^* $ .
Similarly we can get the midpoint of $ AC_2^* $ is the projection of $ O $ on $ CP^* $ ,
so the midpoint of $ AA_1^*, AB_2^*, AC_2^*, AA^* $ lie on a circle with diameter $ OP^* $ ,
hence after doing homothety $ \mathbf{H}(A,2) $ we get $ A_1^*, B_2^*, C_2^*, A^* $ are concyclic .

Notice that $ A_1^*, B_2^*, C_2^*, A^* $ is the reflection of $ A_2, B_2, C_2, H $ in $ D $ , respectively ,
so we get $ A_2, B_2, C_2, H $ are concyclic . $ \blacksquare $

The circle in Theorem 1 is called the Hagge circle $ \mathcal{H}_P $ of $ P $ WRT $ \triangle ABC $ .
____________________________________________________________
Theorem 2:

The center $ T $ of $ \mathcal{H}_P $ is the reflection of $ P^* $ in the 9-point center $ N $ of $ \triangle ABC $ .

Proof:

Let $ Q $ be the image of $ P^* $ under homothety $ \mathbf{H} (A, 2) $ .

From the proof of Theorem 1 we get the reflection of $ Q $ in $ D $ is the antipode of $ H $ in $ \mathcal{H}_P $ ,
so from $ HT=\tfrac{1}{2} QA^*=P^*O $ and $ HT \parallel QA^* \parallel P^*O \Longrightarrow P^*OTH $ is a parallelogram ,
hence we get $ T $ is the reflection of $ P^* $ in the midpoint $ N $ of $ OH $ . $ \blacksquare $
____________________________________________________________
Theorem 3:

Let $ A_3=\mathcal{H}_P \cap AH,  B_3=\mathcal{H}_P \cap BH,  C_3=\mathcal{H}_P \cap CH $ .

Then $ P \in A_2A_3, P \in B_2B_3, P \in C_2C_3 $ .

Proof:

Lemma:

Let $ S, S^* $ be the isogonal conjugate of $ \triangle ABC $ .
Let $ X=AS \cap \odot (ABC), X^*=AS^* \cap \odot (ABC), V=AS^* \cap BC $ .

Then $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $ .

Proof of the lemma:

Since $ \angle VCX^*=\angle BAX^*=\angle XAC $ ,
so we get $ \triangle VCX^* \sim \triangle CAX $ . ... $ (1) $
Since $ \angle X^*S^*C=\angle S^*AC+\angle S^*CA=\angle BCX+\angle SCB=\angle SCX $ ,
so we get $ \triangle CX^*S^* \sim \triangle SXC $ . ... $ (2) $

From $ (1), (2) \Longrightarrow X^*V \cdot XA=X^*C \cdot XC=X^*S^* \cdot XS $ . i.e. $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $

Back to the main proof:

Let $ H^* $ be the antipode of $ H $ in $ \mathcal{H}_P $ .

From symmetry it suffices to prove $ P \in A_2A_3 $ .

Since $ OP^* \parallel HH^* $ and $ HH^*=2OP^* $ ,
so $ H^* $ is the anti-complement of $ P^* $ WRT $ \triangle ABC $ ,
hence from $ H^*A_3 \parallel BC $ we get $ AA_3=2 \cdot \text{dist}(P^*, BC) $ ,

so combine with the lemma $ \Longrightarrow \frac{AP}{PA_1}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1^*, BC)}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1, BC)}=\frac{AA_3}{A_1A_2}  \Longrightarrow P \in A_2A_3 $ . $ \blacksquare $
____________________________________________________________
Theorem 4:

$ \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ .

Proof:

Let $ A_4, B_4, C_4 $ be the midpoint of $ AA_1^*, AB_2^*, AC_2^* $, respectively .

From the proof of Theorem 1 we get $ A_4, B_4, C_4 $ is the projection of $ O $ on $ AP^*, BP^*, CP^* $, respectively .

Since $ \angle A_4B_4C_4=\angle A_4P^*C_4=\angle (AP^*, CP^*)=\angle BAP+\angle BCP=\angle A_1B_1C_1 $ (Similarly $ \angle B_4C_4A_4=\angle B_1C_1A_1 $) ,
so we get $ \triangle A_1B_1C_1 \sim \triangle A_4B_4C_4 \Longrightarrow  \triangle A_1B_1C_1 \sim  \triangle A_1^*B_2^*C_2^* \sim \triangle A_2B_2C_2 $ .

Since $ \angle C_2A_2A_3=\angle C_2HA=\angle (OC_4, HA)=\angle P^*CB=\angle ACP=\angle C_1A_1A $ ,
so we get $ \triangle A_1B_1C_1 \cup A \sim \triangle A_2B_2C_2 \cup A_3 $ (Similar discussion for $ (B,B_3) $ and $ (C, C_3) $ ) ,
hence combine with Theorem 3 $\Longrightarrow \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ . $ \blacksquare $

____________________________________________________________
Some topics related to Hagge circle:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=444395
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=320075
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=318484
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=329272
2 replies
TelvCohl
Feb 11, 2015
mhq
Aug 4, 2015
Some properties of Hagge circle
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TelvCohl
2311 posts
#1 • 21 Y
Y by wanwan4343, buratinogigle, mineiraojose, mhq, Gryphos, buzzychaoz, Scorpion.k48, Vanescralet, baopbc, aopser123, JasperL, AlastorMoody, parola, RAMUGAUSS, AmirKhusrau, Functional_equation, PIartist, Multily, Adventure10, bin_sherlo, MS_asdfgzxcvb
Some properties of Hagge circle :)

Theorem 1:

Let $ H $ be the orthocenter of $ \triangle ABC $ .
Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ .
Let $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB $, respectively .

Then $ A_2, B_2, C_2, H $ are concyclic .

Proof:

Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ .
Let $ \triangle A_1^*B_1^*C_1^* $ be the circumcevian triangle of $ P^* $ WRT $ \triangle ABC $ .
Let $ B_2^*, C_2^* $ be the reflection of $ B_2, C_2 $ in the midpoint $ D $ of $ BC $, respectively .
Let $ O $ be the circumcenter of $ \triangle ABC $ and $ A^* $ be the antipode of $ A $ in $ \odot (ABC) $ .

Easy to see $ A_2, B_2, C_2 $ is the reflection of $ A_1^*, B_1^*, C_1^* $ in the midpoint of $ BC, CA, AB $, respectively .

Since $ CB_2AB_1^*, CB_2BB_2^* $ are parallelogram ,
so $ ABB_2^*B_1^* $ is a parallelogram $ \Longrightarrow $ the midpoint of $ AB_2^* $ is the projection of $ O $ on $ BP^* $ .
Similarly we can get the midpoint of $ AC_2^* $ is the projection of $ O $ on $ CP^* $ ,
so the midpoint of $ AA_1^*, AB_2^*, AC_2^*, AA^* $ lie on a circle with diameter $ OP^* $ ,
hence after doing homothety $ \mathbf{H}(A,2) $ we get $ A_1^*, B_2^*, C_2^*, A^* $ are concyclic .

Notice that $ A_1^*, B_2^*, C_2^*, A^* $ is the reflection of $ A_2, B_2, C_2, H $ in $ D $ , respectively ,
so we get $ A_2, B_2, C_2, H $ are concyclic . $ \blacksquare $

The circle in Theorem 1 is called the Hagge circle $ \mathcal{H}_P $ of $ P $ WRT $ \triangle ABC $ .
____________________________________________________________
Theorem 2:

The center $ T $ of $ \mathcal{H}_P $ is the reflection of $ P^* $ in the 9-point center $ N $ of $ \triangle ABC $ .

Proof:

Let $ Q $ be the image of $ P^* $ under homothety $ \mathbf{H} (A, 2) $ .

From the proof of Theorem 1 we get the reflection of $ Q $ in $ D $ is the antipode of $ H $ in $ \mathcal{H}_P $ ,
so from $ HT=\tfrac{1}{2} QA^*=P^*O $ and $ HT \parallel QA^* \parallel P^*O \Longrightarrow P^*OTH $ is a parallelogram ,
hence we get $ T $ is the reflection of $ P^* $ in the midpoint $ N $ of $ OH $ . $ \blacksquare $
____________________________________________________________
Theorem 3:

Let $ A_3=\mathcal{H}_P \cap AH,  B_3=\mathcal{H}_P \cap BH,  C_3=\mathcal{H}_P \cap CH $ .

Then $ P \in A_2A_3, P \in B_2B_3, P \in C_2C_3 $ .

Proof:

Lemma:

Let $ S, S^* $ be the isogonal conjugate of $ \triangle ABC $ .
Let $ X=AS \cap \odot (ABC), X^*=AS^* \cap \odot (ABC), V=AS^* \cap BC $ .

Then $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $ .

Proof of the lemma:

Since $ \angle VCX^*=\angle BAX^*=\angle XAC $ ,
so we get $ \triangle VCX^* \sim \triangle CAX $ . ... $ (1) $
Since $ \angle X^*S^*C=\angle S^*AC+\angle S^*CA=\angle BCX+\angle SCB=\angle SCX $ ,
so we get $ \triangle CX^*S^* \sim \triangle SXC $ . ... $ (2) $

From $ (1), (2) \Longrightarrow X^*V \cdot XA=X^*C \cdot XC=X^*S^* \cdot XS $ . i.e. $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $

Back to the main proof:

Let $ H^* $ be the antipode of $ H $ in $ \mathcal{H}_P $ .

From symmetry it suffices to prove $ P \in A_2A_3 $ .

Since $ OP^* \parallel HH^* $ and $ HH^*=2OP^* $ ,
so $ H^* $ is the anti-complement of $ P^* $ WRT $ \triangle ABC $ ,
hence from $ H^*A_3 \parallel BC $ we get $ AA_3=2 \cdot \text{dist}(P^*, BC) $ ,

so combine with the lemma $ \Longrightarrow \frac{AP}{PA_1}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1^*, BC)}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1, BC)}=\frac{AA_3}{A_1A_2}  \Longrightarrow P \in A_2A_3 $ . $ \blacksquare $
____________________________________________________________
Theorem 4:

$ \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ .

Proof:

Let $ A_4, B_4, C_4 $ be the midpoint of $ AA_1^*, AB_2^*, AC_2^* $, respectively .

From the proof of Theorem 1 we get $ A_4, B_4, C_4 $ is the projection of $ O $ on $ AP^*, BP^*, CP^* $, respectively .

Since $ \angle A_4B_4C_4=\angle A_4P^*C_4=\angle (AP^*, CP^*)=\angle BAP+\angle BCP=\angle A_1B_1C_1 $ (Similarly $ \angle B_4C_4A_4=\angle B_1C_1A_1 $) ,
so we get $ \triangle A_1B_1C_1 \sim \triangle A_4B_4C_4 \Longrightarrow  \triangle A_1B_1C_1 \sim  \triangle A_1^*B_2^*C_2^* \sim \triangle A_2B_2C_2 $ .

Since $ \angle C_2A_2A_3=\angle C_2HA=\angle (OC_4, HA)=\angle P^*CB=\angle ACP=\angle C_1A_1A $ ,
so we get $ \triangle A_1B_1C_1 \cup A \sim \triangle A_2B_2C_2 \cup A_3 $ (Similar discussion for $ (B,B_3) $ and $ (C, C_3) $ ) ,
hence combine with Theorem 3 $\Longrightarrow \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ . $ \blacksquare $

____________________________________________________________
Some topics related to Hagge circle:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=444395
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=320075
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=318484
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=329272
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jayme
9767 posts
#2 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

thank to remind to us the Hagge circle which appear as a generalization of the Fuhrmann's circle.
For me it appears also like a Mannheim's circle...

see: http://jl.ayme.pagesperso-orange.fr/Docs/Le%20cercle%20de%20Hagge.pdf
where you can find Theorem 1, 2, 3 and perhaps 4? with some examples and extension...

Sincerely
Jean-Louis
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mhq
16 posts
#3 • 2 Y
Y by Adventure10, Mango247
TelvCohl wrote:
Some properties of Hagge circle :)

Theorem 1:

Let $ H $ be the orthocenter of $ \triangle ABC $ .
Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ .
Let $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB $, respectively .

Then $ A_2, B_2, C_2, H $ are concyclic .

Proof:

Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ .
Let $ \triangle A_1^*B_1^*C_1^* $ be the circumcevian triangle of $ P^* $ WRT $ \triangle ABC $ .
Let $ B_2^*, C_2^* $ be the reflection of $ B_2, C_2 $ in the midpoint $ D $ of $ BC $, respectively .
Let $ O $ be the circumcenter of $ \triangle ABC $ and $ A^* $ be the antipode of $ A $ in $ \odot (ABC) $ .

Easy to see $ A_2, B_2, C_2 $ is the reflection of $ A_1^*, B_1^*, C_1^* $ in the midpoint of $ BC, CA, AB $, respectively .

Since $ CB_2AB_1^*, CB_2BB_2^* $ are parallelogram ,
so $ ABB_2^*B_1^* $ is a parallelogram $ \Longrightarrow $ the midpoint of $ AB_2^* $ is the projection of $ O $ on $ BP^* $ .
Similarly we can get the midpoint of $ AC_2^* $ is the projection of $ O $ on $ CP^* $ ,
so the midpoint of $ AA_1^*, AB_2^*, AC_2^*, AA^* $ lie on a circle with diameter $ OP^* $ ,
hence after doing homothety $ \mathbf{H}(A,2) $ we get $ A_1^*, B_2^*, C_2^*, A^* $ are concyclic .

Notice that $ A_1^*, B_2^*, C_2^*, A^* $ is the reflection of $ A_2, B_2, C_2, H $ in $ D $ , respectively ,
so we get $ A_2, B_2, C_2, H $ are concyclic . $ \blacksquare $

The circle in Theorem 1 is called the Hagge circle $ \mathcal{H}_P $ of $ P $ WRT $ \triangle ABC $ .
____________________________________________________________
Theorem 2:

The center $ T $ of $ \mathcal{H}_P $ is the reflection of $ P^* $ in the 9-point center $ N $ of $ \triangle ABC $ .

Proof:

Let $ Q $ be the image of $ P^* $ under homothety $ \mathbf{H} (A, 2) $ .

From the proof of Theorem 1 we get the reflection of $ Q $ in $ D $ is the antipode of $ H $ in $ \mathcal{H}_P $ ,
so from $ HT=\tfrac{1}{2} QA^*=P^*O $ and $ HT \parallel QA^* \parallel P^*O \Longrightarrow P^*OTH $ is a parallelogram ,
hence we get $ T $ is the reflection of $ P^* $ in the midpoint $ N $ of $ OH $ . $ \blacksquare $
____________________________________________________________
Theorem 3:

Let $ A_3=\mathcal{H}_P \cap AH,  B_3=\mathcal{H}_P \cap BH,  C_3=\mathcal{H}_P \cap CH $ .

Then $ P \in A_2A_3, P \in B_2B_3, P \in C_2C_3 $ .

Proof:

Lemma:

Let $ S, S^* $ be the isogonal conjugate of $ \triangle ABC $ .
Let $ X=AS \cap \odot (ABC), X^*=AS^* \cap \odot (ABC), V=AS^* \cap BC $ .

Then $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $ .

Proof of the lemma:

Since $ \angle VCX^*=\angle BAX^*=\angle XAC $ ,
so we get $ \triangle VCX^* \sim \triangle CAX $ . ... $ (1) $
Since $ \angle X^*S^*C=\angle S^*AC+\angle S^*CA=\angle BCX+\angle SCB=\angle SCX $ ,
so we get $ \triangle CX^*S^* \sim \triangle SXC $ . ... $ (2) $

From $ (1), (2) \Longrightarrow X^*V \cdot XA=X^*C \cdot XC=X^*S^* \cdot XS $ . i.e. $ \frac{AS}{SX}=\frac{S^*V}{VX^*} $

Back to the main proof:

Let $ H^* $ be the antipode of $ H $ in $ \mathcal{H}_P $ .

From symmetry it suffices to prove $ P \in A_2A_3 $ .

Since $ OP^* \parallel HH^* $ and $ HH^*=2OP^* $ ,
so $ H^* $ is the anti-complement of $ P^* $ WRT $ \triangle ABC $ ,
hence from $ H^*A_3 \parallel BC $ we get $ AA_3=2 \cdot \text{dist}(P^*, BC) $ ,

so combine with the lemma $ \Longrightarrow \frac{AP}{PA_1}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1^*, BC)}=\frac{\text{dist}(P^*, BC)}{\text{dist}(A_1, BC)}=\frac{AA_3}{A_1A_2}  \Longrightarrow P \in A_2A_3 $ . $ \blacksquare $
____________________________________________________________
Theorem 4:

$ \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ .

Proof:

Let $ A_4, B_4, C_4 $ be the midpoint of $ AA_1^*, AB_2^*, AC_2^* $, respectively .

From the proof of Theorem 1 we get $ A_4, B_4, C_4 $ is the projection of $ O $ on $ AP^*, BP^*, CP^* $, respectively .

Since $ \angle A_4B_4C_4=\angle A_4P^*C_4=\angle (AP^*, CP^*)=\angle BAP+\angle BCP=\angle A_1B_1C_1 $ (Similarly $ \angle B_4C_4A_4=\angle B_1C_1A_1 $) ,
so we get $ \triangle A_1B_1C_1 \sim \triangle A_4B_4C_4 \Longrightarrow  \triangle A_1B_1C_1 \sim  \triangle A_1^*B_2^*C_2^* \sim \triangle A_2B_2C_2 $ .

Since $ \angle C_2A_2A_3=\angle C_2HA=\angle (OC_4, HA)=\angle P^*CB=\angle ACP=\angle C_1A_1A $ ,
so we get $ \triangle A_1B_1C_1 \cup A \sim \triangle A_2B_2C_2 \cup A_3 $ (Similar discussion for $ (B,B_3) $ and $ (C, C_3) $ ) ,
hence combine with Theorem 3 $\Longrightarrow \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P $ . $ \blacksquare $

____________________________________________________________
Some topics related to Hagge circle:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=444395
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=320075
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=318484
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=329272

Could you please help me to find any information about anticompliment?
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