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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 $
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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 $
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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
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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 $ .
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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
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#2 • 2 Y
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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
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#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 $
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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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