<BAK = <CAL wanted, 3 midpoints, centroid, 2 circumcircles,

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<BAK = <CAL wanted, 3 midpoints, centroid, 2 circumcircles,

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Source: 2023 Grand Duchy of Lithuania, MC p3 (Baltic Way TST)

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#1 h Mar 23, 2024, 10:04 PM • 1 Y

Y by Roh_Hoi_Chan

The midpoints of the sides $BC$ , $CA$ and $AB$ of triangle $ABC$ are $M$ , $N$ and $P$ respectively . $G$ is the intersection point of the medians. The circumscribed circle around $BGP$ intersects the line $MP$ at the point $K$ (different than $P$ ).The circle circumscribed around $CGN$ intersects the line $MN$ at point $L$ (different than $N$ ). Prove that $\angle BAK = \angle CAL$ .

This post has been edited 1 time. Last edited by parmenides51, Mar 24, 2024, 8:01 AM
Reason: typo, incenter -> centroid

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#2 h Mar 24, 2024, 6:55 AM • 2 Y

Y by parmenides51, Roh_Hoi_Chan

I think there is some typo.

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#3 h Mar 24, 2024, 7:22 AM • 1 Y

Y by Roh_Hoi_Chan

I think $G$ should be the centroid. Then the problem is fine and here is the solution.

Firstly, note that $\angle BPK = \angle LNC = \alpha$ . Also, we have that $\angle PKB = \angle PGB = \angle NGC = \angle NLC.$
From here it follows that $\triangle NCL \sim \triangle PBK$ . In particular, we have that $\angle PBK = \angle NCL$ .

Claim: $\triangle ACL \sim \triangle ABK$ .
Proof: Since $\angle ACL = \angle ABK$ , it is enough to show that $\frac{AC}{CL}=\frac{AB}{BK}\iff \frac{AC}{AB}=\frac{CL}{BK}$ .
However, we know that $\frac{NC}{CL}=\frac{PB}{BK}$ . Hence, $\frac{CL}{BK}=\frac{NC}{PB}=\frac{2NC}{2PB}=\frac{AC}{AB},$ so we are done.

From the claim it follows that $\angle CAL=\angle BAK$ and we are done.

This post has been edited 1 time. Last edited by rstenetbg, Mar 24, 2024, 7:23 AM

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#4 h Mar 24, 2024, 8:02 AM • 1 Y

Y by Roh_Hoi_Chan

$G$ is centroid and not incenter as I had written at the start. It has been corrected. Thanks for noticing.

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#5 h Mar 24, 2024, 11:43 AM • 1 Y

Y by Roh_Hoi_Chan

Let $Q=(BPG)\cap (CGN)$ ,
it s clear that $Q$ is the Miquel point of $CNBP$ then $QCN\sim QPB$ so the ratio of the respective
$Q$ -heights is equal to $\frac{CN}{PB} =\frac{CA}{AB}$ which means that $Q$ is on the $A$ -symmedian of $ABC$
more since $Q$ is the Miquel point of $CNBP$ we deduces that $APQC,BQNA$ are cyclic thus $\angle KQG=\angle KNG=\angle MNG=\angle PBG=\angle PQG$ so $P,K,Q$ are collinear similarly we get $N,L,Q$ are collinear
hence :
since $(AN,AP)$ and $(AM,AQ)$ are 2 pairs of isogonals wrt $\angle BAC$ and $PQ\cap MN=L, NQ\cap MP=K$
then by the isogonal property we deduce $AK,AL$ are also isogonal.

Best regards.
RH HAS

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