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Source: TST IMO CHILE 2025

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136 posts

#1 h Mar 29, 2025, 2:35 AM

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Let $ABC$ be a triangle with $AB < AC$ . Let $M$ be the midpoint of $AC$ , and let $D$ be a point on segment $AC$ such that $DB = DC$ . Let $E$ be the point of intersection, different from $B$ , of the circumcircle of triangle $ABM$ and line $BD$ . Define $P$ and $Q$ as the points of intersection of line $BC$ with $EM$ and $AE$ , respectively. Prove that $P$ is the midpoint of $BQ$ .

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#2 h Mar 29, 2025, 4:02 AM

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I think we could do some side length bash to get $(A, D; F, M) = -1$ , where $F$ is a point on segment $AC$ so that $FE \parallel BC$ . Then, we take perspective at $E$ to get $(A, D; F, M) = (B, Q; P, \infty) = -1$ which yield the midpoint.

This post has been edited 1 time. Last edited by EeEeRUT, Mar 29, 2025, 4:04 AM

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#3 h Mar 29, 2025, 4:07 AM • 2 Y

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Let $N$ midpoint of $AB$ , also consider $AE \cap BM=R$ and also let $(ABM)$ hit $BC$ again at $G$ and consider $F$ such that $FEGB$ is an isosceles trapezoid.
Notice that from subtracting arc $EM$ from the angles we get $\triangle BRQ$ isosceles and thus from arcs $FG \parallel AC$ , now to finish just do this:
$-1=(A, C: M, \infty_{AC}) \overset{G}{=} (A, B; M, F) \overset{E}{=} (Q, B; P, \infty_{BC}) \implies BP=PQ$ Thus we are done :cool:

:cool:

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(P.D. I got half sniped wtf, (half cuz above did not elaborate on the first part))

This post has been edited 1 time. Last edited by MathLuis, Mar 29, 2025, 4:08 AM

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#4 h Mar 29, 2025, 11:31 AM

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Let $A'$ be the reflection of $A$ across perpendicular bisector of $BC$ , and let $P'$ be the midpoint of $AA'$ . Obviously $A, D, A'$ are collinear.

Claim: $P', E, M$ are collinear
Proof: Note that:
$\measuredangle AMP' = \measuredangle ACA' = \measuredangle ABA' = \measuredangle ABE = \measuredangle AME$ Hence $P', E, M$ are collinear.

It follows that $P', P, E$ are collinear because they all lie on line $EM$ . But then, homothety at $E$ sending $A$ to $Q$ and $A'$ to $B$ sends $P'$ to $P$ . As $P'$ is the midpoint of $AA'$ , $P$ is the midpoint $BQ$ , done.

This post has been edited 2 times. Last edited by Pekiban, Mar 29, 2025, 11:32 AM
Reason: typo 2

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#5 h Mar 29, 2025, 4:14 PM

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Too easy for TST?
Denote $L$ the reflection of $B$ in $M$ , then $ALCB$ is a parallelogram.
It's anglechasable to get $\angle AQB = \angle MBQ = \angle LBQ$ , i.e $ALQB$ is an isoceles trapezoid.
Also anglechasable to obtain $\angle EPB = \angle ABP = \angle LQB$ . Thus $MP \parallel LQ$ , implying that $P$ is midpoint of $BQ$ .

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