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Source: Mongolian TST P3

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B1t

#1 h Apr 26, 2025, 7:20 AM

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Let $ABC$ be an acute triangle with $AB \neq AC$ and orthocenter $H$ . Let $B'$ and $C'$ be the feet of the altitudes from $B$ and $C$ onto sides $AC$ and $AB$ , respectively. Let $M$ be the midpoint of $BC$ , and $M'$ be the midpoint of $B'C'$ . Let the perpendicular line through $H$ to $AM$ meet $AM$ at $S$ and $BC$ at $T$ . The line $MM'$ meets $AC$ at $U$ and $AB$ at $V$ . Let $P$ be the second intersection point (different from $M$ ) of the circumcircles of triangles $BMV$ and $CMU$ . Prove that the points $T$ , $P$ , $M'$ , $S$ , and $M$ lie on the same circle.

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#2 h Apr 26, 2025, 7:28 AM

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it suffices to show $\angle TPM = 90$ by well-known configuration properties
redefine $P$ to be the $A$ -queue point in $ABC.$
let $D$ be the foot from $A$ to $BC.$
consider reference triangle $ATM,$ and observe that $DPS$ is the orthic triangle
thus, $HP\cdot HM = AH\cdot HD = HB\cdot HB',$ implying that $PB'BM$ is cyclic.
similarly, $PCC'M$ is cyclic

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B1t

#3 h Apr 26, 2025, 7:31 AM

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Ilikeminecraft wrote:

it suffices to show $\angle TPM = 90$ by well-known configuration properties
redefine $P$ to be the $A$ -queue point in $ABC.$
let $D$ be the foot from $A$ to $BC.$
consider reference triangle $ATM,$ and observe that $DPS$ is the orthic triangle
thus, $HP\cdot HM = AH\cdot HD = HB\cdot HB',$ implying that $PB'BM$ is cyclic.
similarly, $PCC'M$ is cyclic

you make it more easily using miquel point . but this was easy too.

This post has been edited 1 time. Last edited by B1t, Apr 26, 2025, 7:32 AM

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#5 h Apr 27, 2025, 3:38 AM

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Notice $S$ is the A-humpty point of $\triangle ABC$ so by Radax it is known that $T,B',C'$ are colinear, redefine $P$ as the A-queuepoint, we prove it is also the one on the problem statement. To start it is known that $A,P,T$ are colinear by radax again, but also $\measuredangle B'MV=\measuredangle B'BV$ and from invert $-\sqrt{BH \cdot HB'}$ we have that $PVB'MB$ is cyclic and similarily $UPC'MC$ is cyclic so indeed $P$ is the one from the problem statement which basically finished as the desired cyclic is just on a circle with diameter $TM$ thus we are done :cool:

:cool:

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