Orthocenter is the midpoint of the altitude

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Orthocenter is the midpoint of the altitude

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Source: Mexican Quarantine Mathematical Olympiad P4

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#1 h Apr 26, 2020, 2:00 PM • 4 Y

Y by AlastorMoody, Mango247, Mango247, Mango247

Let $ABC$ be an acute triangle with orthocenter $H$ . Let $A_1$ , $B_1$ and $C_1$ be the feet of the altitudes of triangle $ABC$ opposite to vertices $A$ , $B$ , and $C$ respectively. Let $B_2$ and $C_2$ be the midpoints of $BB_1$ and $CC_1$ , respectively. Let $O$ be the intersection of lines $BC_2$ and $CB_2$ . Prove that $O$ is the circumcenter of triangle $ABC$ if and only if $H$ is the midpoint of $AA_1$ .

Proposed by Dorlir Ahmeti

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#2 h Apr 26, 2020, 2:49 PM • 1 Y

Y by amar_04

Easy problem . Here goes my solution.

We will use the following lemma thoughout the proof.

Lemma: In general $\overline{AB_2},\overline{AC_2}$ are isogonal w.r.t $\angle BAC$

Proof: This is just similiarity. Notice that $\triangle{ABB_1} \sim \triangle{ACC_1}$ and the result follows $\qquad \square$

Part 1: Assume that $H$ is the midpoint of $\overline{AA_1}$ then prove that $\overline{BC_2} \cap \overline{CB_2}$ is the circumcenter

Proof: Notice that by our lemma $\overline{BH},\overline{BC_2}$ are isogonal w.r.t $\angle ABC$ and $\overline{CH},\overline{CB_2}$ are isogonal w.r.t $\angle ACB$ . Hence $\overline{BC_2} \cap \overline{CB_2}$ is the isogonal conjugate of $H$ which is nothing but $O$ $\qquad \blacksquare$

Part 2: Assume that $\overline{BC_2} \cap \overline{CB_2}$ is the circumcenter then prove that $H$ is the midpoint of $\overline{AA_1}$

Proof: Notice that if $H'$ is the midpoint of $\overline{AA_1}$ then by our lemma we have that $\overline{BH'},\overline{BC_2}$ are isogonal w.r.t $\angle ABC$ and $\overline{CH'},\overline{CB_2}$ are isogonal w.r.t $\angle ACB$ . Hence $H'$ is the isogonal conjugate of $O$ . But its well known that $H$ is the isogonal conjugate of $O$ which yields that $H'=H$ as desired $\qquad \blacksquare$

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#3 h Apr 26, 2020, 3:21 PM • 2 Y

Y by Mango247, Mango247

Ez trigon

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#4 h Apr 26, 2020, 6:59 PM

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First suppose that $H$ is the midpoint of $AA_1$ let the parallel through $C$ to $AA_1$ intresect $(ABC)$ on $E$ and $AB$ on $D$ , note that $C(A_1, A; H, E)=-1 \implies (B, A; C_1, D)=-1$ , now it's easy to see that $AHCE$ is a parallelogram so $E(C, C_1; C_2, A)=-1 \implies (D, C_1; A, EC_2\cap AB)=-1$ , but we now that $(D, C_1; A, B)=-1$ , so $EC_2\cap AB=B$ , then $B-C_2-E$ is diameter of $(ABC)$ . Similarly if the parallel to $AA_1$ through $B$ intersects $(ABC)$ on $F$ we get $C-B_2-F$ is diameter $(ABC)$ , and the conclusion follows.

Now let $O$ be the circumcenter of $\triangle ABC$ ,let $E=BC_2\cap (ABC)$ , let $D=AB\cap CE$ , it's easy to see that $AHCE$ is a parallelogram so $E(C, C_1; C_2, A)=-1 \implies (D, C_1; A, B)=-1$ . Now $C(B, A; C_1, D)=-1 \implies (A_1, A; H, \infty)=-1$ and the result follows.

Attachments:

This post has been edited 1 time. Last edited by Ricochet, Apr 26, 2020, 7:13 PM

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#5 h Apr 26, 2020, 8:37 PM

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If $BO$ meets $AH$ at $X$ , we have $\frac{AX}{A_1X}=\frac{CH}{C_1H}$ since $AH$ , $AO$ isogonal and $\triangle{BA_1A} \sim \triangle{BC_1C}$ . Thus, projecting $(A,A_1;H,X)$ through $B$ onto $CH$ gives $B,O,C_2$ collinear iff $AH=A_1H$ , implying the result.

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#6 h May 7, 2020, 6:16 PM • 3 Y

Y by Moustafa, Muaaz.SY, Mango247

a one-line solution :blush:

:blush:

if $H$ is the midpoint of $AA_1$
$CB_1$ is median in $\triangle CBB_1$ but we have $(A,A_1;H,P_{\infty})=-1=(B,C'B_1C_1)$ so $CC_1$ is symmedian in $\triangle CBB_1$ so $CH,CO$ are isogonal
$\blacksquare$
if $O$ is the circumcenter
$CB_1$ is median in $\triangle CBB_1$ but we have $(CO,CH)$ are isogonal $\implies CH$ is symmedian so $-1=(B,C'B_1C_1)=(A,A_1;H,P_{\infty})$
and we win

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#7 h Apr 4, 2025, 6:49 AM

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First Part: Suppose that $O$ is the circuncenter of $\triangle ABC$ . Since $H$ and $O$ are isogonals. Then $\angle ABH=\angle OBC$ , so $\angle C_1BC_2=\angle HBC=\angle HBA_1$ . Now
$\frac{AH}{HA_1}=\frac{AB}{BA_1}\cdot\frac{\sin \angle ABH}{\sin \angle HBA_1},$ but we know that
$\frac{\sin \angle ABH}{\sin \angle HBA_1}=\frac{\sin \angle CBC_2}{\sin \angle C_2BC_1}=\frac{C_1B}{BC}.$ Since $\triangle ABA_1 \sim \triangle CBC_1 \Longrightarrow \frac{AB}{BA_1}=\frac{BC}{BC_1}$ , so
$\frac{AH}{HA_1}=\frac{AB}{BA_1}\cdot\frac{C_1B}{BC}=1$ and the result follows.

Second part: Now suppose that $H$ in the midpoint of $AA_1$ . Since $\triangle ABA_1 \sim \triangle CBC_1 \Longrightarrow \triangle ABH \sim \triangle CBC_2$ , then $\angle ABH = \angle CBC_2=\angle CBO$ . Anagously, $\angle ACH=\angle BCO$ . Then $H$ and $O$ are isogonals, and the conclusion follows. It's done.

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