Some properties of Hagge circle

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#1 h Feb 11, 2015, 1:21 PM • 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 $\triangle ABC$ .
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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 $OH$ . $\blacksquare$
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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 $\Longrightarrow \triangle ABC \cup \triangle A_1B_1C_1 \cup P \sim \triangle A_3B_3C_3 \cup \triangle A_2B_2C_2 \cup P$ . $\blacksquare$

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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

#2 h Feb 11, 2015, 2:35 PM • 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

#3 h Aug 4, 2015, 10:37 AM • 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 $\triangle ABC$ .
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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 $OH$ . $\blacksquare$
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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

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

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