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Source: Serbia Additional IMO TST 2024, P3 (out of 4)

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677 posts

#1 h May 30, 2024, 9:27 AM • 1 Y

Y by GeoKing

Let $ABC$ be a triangle with circumcenter $O$ , angle bisector $AD$ with $D \in BC$ and altitude $AE$ with $E \in BC$ . The lines $AO$ and $BC$ meet at $I$ . The circumcircle of $\triangle ADE$ meets $AB, AC$ at $F, G$ and $FG$ meets $BC$ at $H$ . The circumcircles of triangles $AHI$ and $ABC$ meet at $J$ . Show that $AJ$ is a symmedian in $\triangle ABC$

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v4913

#2 h May 30, 2024, 1:59 PM • 1 Y

Y by GeoKing

(H, E; B, C) is a harmonic bundle since <HEA = 90 and EA bisects <FEG. This means H is collinear with the feet of the altitudes from B and C in ABC, and it is also the orthocenter of AH’M where H’ and M are the orthocenter of ABC and the midpoint of BC. If Ha is the reflection of H’ over BC then J is the intersection of HHa with (ABC), so the reflection of J over BC is the foot from H’ onto AM, which is the reflection of the intersection of AM with (ABC) over M, so AJ is a symmedian.

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716 posts

#3 h Jul 8, 2024, 10:06 AM

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Let $AK$ be $A-$ symmedian where $K\in (ABC)$ . Let $AD\cap (ABC)=A,M$ and $N$ be the midpoint of $BC$ . Let $MN\cap AC=X$ . We will show that $A,I,K,H$ are cyclic.
Take the inversion centered at $A$ with radius $\sqrt{AB.AC}$ and reflect inverted points according to $AD$ .
$D\leftrightarrow M,B\leftrightarrow B,C\leftrightarrow C,K\leftrightarrow N$ And $I^*$ is the intersection of $AE$ with $(ABC)$ . $F^*,M,G^*$ are collinear and $F^*G^*\perp AM$ . $H^*$ is the miquel point of $F^*G^*BC$ . Let's prove that $I^*,N,H^*$ are collinear.
Since $H^*$ is the center of spiral homothethy mapping $BC$ to $G^*F^*$ and $N,M$ are midpoints, we have $H^*BNC\sim H^*G^*MF^*$ .
$90-\angle XCH^*=90-\angle ACH^*=90-\angle AMH^*=\angle H^*MF^*=\angle H^*NC=90-\angle XNH^*\implies \angle XCH^*=\angle XNH^*$ Thus, $X,N,C,H^*$ are cyclic.
$\angle CH^*N=\angle CXN=90-\angle C=\angle CAI^*=\angle CH^*I^*$ Which gives that $H^*,N,I$ are collinear as desired. $\blacksquare$

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#4 h Jul 20, 2024, 5:27 AM • 1 Y

Y by ehuseyinyigit

Let $J'$ be point on $ABC$ such that $AJ'$ is symmedian. we will prove $J'$ is $J$ . Let $M$ be midpoint of $BC$ . Let $AE$ meet $ABC$ at $K$ .
Claim $1: H,K,J'$ are collinear.
Proof $:$ Let $KJ'$ meet $BC$ at $H'$ . Note that $\frac{H'B}{H'C}=\frac{BK}{CK}.\frac{BJ'}{CJ'} = \frac{\sin{90-B}}{\sin{90-C}}.\frac{\sin{C}}{\sin{B}}$ . Now note that $HE$ is the external angle bisector of $\angle FEG$ so $\frac{HF}{HG} = \frac{EF}{EG} = \frac{\sin{90-B}}{\sin{90-C}}$ . Also note that $HF = HB.\frac{\sin{B}}{\sin{AFG}}$ and $HG = HC.\frac{\sin{C}}{\sin{AGF}}$ so $\frac{HF}{HG} = \frac{HB}{HC}.\frac{\sin{B}}{\sin{C}}$ so $\frac{HB}{HC} = \frac{\sin{90-B}}{\sin{90-C}}.\frac{\sin{C}}{\sin{B}}$ which implies $H$ and $H'$ are same.
Claim $2: AHKM$ is cyclic.
Proof $:$ Note that $MEKJ'$ is cyclic so $\angle HKA = \angle EMJ' = \angle EMA = \angle HMA$ so $AHKM$ is cyclic.

Note that $\angle J'AI = \angle MAK = \angle MHK = \angle IHJ'$ so $HJ'IA$ is cyclic which implies $J'$ is $J$ as wated.

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#5 h Jul 20, 2024, 8:59 AM

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solution

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cursed_tangent1434

612 posts

cursed_tangent1434

#6 h Apr 25, 2025, 1:59 PM

Y by

Let $M$ denote the midpoint of segment $BC$ and $K$ the intersection of the $A-$ altitude with $(ABC)$ . The following is our key claim.

Denote by $Q$ the $A-$ Queue point of $\triangle ABC$ , and by $Q'$ the reflection of the $A-$ Queue Point across the perpendicular bisector of segment $BC$ . Let $A'$ denote the $A-$ antipode and $N$ the $BC$ minor arc midpoint in $(ABC)$ . Let $X$ and $Y$ be the intersections of line $NA'$ with $AC$ and $AB$ respectively.

Claim : Points $A$ , $Q'$ , $X$ and $Y$ are concyclic.

Proof : Note that $\measuredangle NYC = 90 + \measuredangle NAC = 90 + \measuredangle BAN = \measuredangle BXN$ . Hence,
$\frac{BX}{\sin(90-\angle C)} = \frac{BN}{\sin BXN)} \text{ and } \frac{CY}{\sin(90-\angle B)} = \frac{CN}{\sin \angle NYC}$ Obtaining the ratio of these two results we have,
$\frac{BX}{CY} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)}$ However, we further note that,
$\frac{Q'B}{Q'C} = \frac{QC}{QB} = \frac{P_bC}{P_cB} = \frac{\sin (90-\angle C) \cdot BC}{\sin (90 - \angle B) \cdot BC} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)}$ where $P_b$ and $P_c$ are the feet of the altitudes from $B$ and $C$ respectively. Thus,
$\frac{BX}{CY} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)} = \frac{Q'B}{Q'C}$ which in addition with
$\measuredangle Q'CY = \measuredangle Q'CA = \measuredangle Q'BA = \measuredangle Q'BX$ implies that $\triangle Q'CY \sim \triangle Q'BX$ which in turn implies the claim.

With this claim in hand, we are essentially done. We perform an inversion centered at $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection across the internal $A-$ angle bisector. Then, the problem rewrites to the following.

Inverted Problem wrote:

Let $\triangle ABC$ be an acute triangle with $BC$ minor arc midpoint $N$ and antipode $A'$ . Let the $A-$ altitude intersect $(ABC)$ at $K$ and let $X$ and $Y$ be the intersections of line $NA'$ with sides $AC$ and $AB$ respectively. If $Q'$ is the second intersection of circles $(AXY)$ and $(ABC)$ show that line $KQ'$ passes through the midpoint of segment $BC$ .

However, the previous claim shows that $Q'$ is simply the reflection of the $A-$ Queue Point across the perpendicular bisector of segment $BC$ , which since it is well known that $QA'$ passes through the midpoint of segment $BC$ implies that $KQ'$ also passes through the midpoint of segment $BC$ via reflection.

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225 posts

#7 h Apr 25, 2025, 11:26 PM

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Let $A'$ be the $A$ -antipode and let $H_A=\overline{AE} \cap \overline{BC}$ .

Claim: $(HE;BC)=-1$ .
Proof: By Ceva-Menelaus picture we just need to prove $\overline{BG}$ , $\overline{CF}$ , $\overline{AE}$ concur. Now see that $FB=BC \cos B=\frac{ac}{b+c} \cos B \text{ and similarly } CG=\frac{ab}{b+c} \cos C$ Hence see that $\begin{align*} \frac{BE}{CE} \cdot \frac{CG}{GA} \cdot \frac{AF}{FB} &=\frac{c \cos B}{b \cos C} \frac{ab \cos C}{b+c} \left(\frac{1}{b-\frac{ab \cos C}{b+c}} \right) \left(c-\frac{ac \cos B}{b+c} \right) \frac{b+c}{ac \cos B} = \frac{bc+c^2-ac \cos B}{b^2+bc-ab \cos C}=\frac{bc+c^2-\frac{a^2+c^2-b^2}2}{bc+b^2-\frac{a^2+b^2-c^2}2}=1 \end{align*}$ And by Ceva's theorem (bar directed lengths because bleh) we are done. $\square$

Now $\sqrt{bc}$ invert at $A$ in $\triangle ABC$ and $H_A$ replaces with $I$ , $A'$ replaces with $E$ , $J^*=\overline{H^*H_A} \cap \overline{BC}$ and we want to prove it is the midpoint of $\overline{BC}$ , but see that $-1=(H^*,A';B,C) \overset {H_A}= (J^*,\infty;B,C)$ And we are done.

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