In triangle $ABC$ let $N, M, P$ be the midpoints of the sides $BC, CA, AB$ and $G$ be the centroid of this triangle. Let the circle circumscribed to $BGP$ intersect the line $MP$ in point $K$, $P \neq K$, and the circle circumscribed to $CGN$ intersect the line $MN$ in point $L$, $N \neq L$. Prove that $ \angle BAK = \angle CAL $.
Problem
Source: czechoslovak national mo round
Tags: geometry
ricarlos
09.04.2023 05:45
This problem is not true, but if $AC=BC$ then $\angle BAK=\angle CAL$ holds. prvocislo wrote: In triangle $ABC$ let $N, M, P$ be the midpoints of the sides $BC, CA, AB$ and $G$ be the centroid of this triangle. Let the circle circumscribed to $BGP$ intersect the line $MP$ in point $K$, $P \neq K$, and the circle circumscribed to $CGN$ intersect the line $MN$ in point $L$, $N \neq L$. Prove that $ \angle BAK = \angle CAL $. _______________________________________________________________________________________________________________________________ Let $K'=PN\cap (BGP)$ and $P'=MK'\cap (BGP)$, since $\angle KPG=\angle K'PG$ and $BG$ is diameter of $(BGP)$ then $MK=MK'$ and $MP=MP'$, thus $PK=P'K'$. $PKK'P'$ is an isosceles trapezoid and its diagonals meet at $Q$. Let $D=KP'\cap BC$, $\angle DP'M=\angle K'PM=\angle MCD$, also $\angle MP'G=\angle MPG=\angle MCG$, then $M,G,D,P',C$ are concyclic points. We observe that $\angle BQN=\angle MQP'$, $\angle QBN=\angle QMP'$ and $QB=QM$ then $\Delta BQN\cong\Delta MQP' \rightarrow QN=QP', QK'=QD$ and $DN=P'K'$. Let $E=AC\cap (CGN)$, since $\Delta CGN\cong\Delta CGM$ it is easy to see that $(CGN)$ is the reflection of $(CGM)$ w.r.t. $CG$ therefore as $CM=CN$ also $EM=DN$, that is, $EM=PK$ then $\frac{CM}{ML}=\frac{MN}{EM}=\frac{BP}{PK}=\frac{AP}{PK}=\frac{AM}{ML}$, and also $\angle APK=\angle AML$ so $\Delta APK\sim\Delta AML\rightarrow \angle BAK=\angle CAL$
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Ritwin
29.03.2024 03:09
Rename $MNP$ to $DEF$ and $(K,L)$ to $(P,Q)$. Let $\triangle DEF$ be the medial triangle of $\triangle ABC$ with centroid $G$. Define $P = DF \cap (BFG)$ and $Q = DE \cap (CEG)$ to be second intersections. Show that $AP$ and $AQ$ are isogonal with respect to $\angle BAC$.
Let $X = (BFG) \cap (CEG)$ be a second intersection.
Claim: $P$ lies on $EX$ and $Q$ lies on $FX$.
Proof. Define $Q' = DE \cap FX$ and compute \[ \measuredangle XQ'E = \measuredangle FQ'D = \measuredangle Q'FA + \measuredangle AFD + \measuredangle FDQ' = \measuredangle XFB = \measuredangle XGB = \measuredangle XGE, \]so $Q'$ lies on $(CEGX)$, meaning $Q' = Q$. The proof of $P \in EX$ is analogous. $\square$
Claim: $AFBP \sim AECQ$ are (inversely) similar
Proof. First, $\triangle FBP \sim \triangle ECQ$ because \begin{align*} \measuredangle FPB = \measuredangle FGB = \measuredangle CGE = \measuredangle CQE = -\measuredangle EQC, \\ \measuredangle BFP = \measuredangle BFD = \measuredangle DEC = \measuredangle QEC = -\measuredangle CEQ. \end{align*}Then the affine transformation sending $\triangle FBP$ to $\triangle ECQ$ sends $\overline{AFB}$ to $\overline{AEC}$, so $AFBP \sim AECQ$. $\square$
Hence $AP$ and $AQ$ are isogonal, as desired. $\blacksquare$
extra observation 1Since $FP \parallel AE$ we have $ABXE$ cyclic, and similarly $ACXF$ is also cyclic. Also, $P$ and $Q$ lie on $(DEG)$ and $(DFG)$ respectively, since \[ \measuredangle DPG = \measuredangle EPG = \measuredangle ABG = \measuredangle DEG \]and symmetric.
extra observation 2Since $DEGP$ and $FXGP$ are cyclic and $P = EX \cap DF$, there is a spiral similarity sending $\triangle GFX \sim \triangle GDE$, which implies $GD$ and $GX$ are isogonal because $\angle FGB = \angle EGC$.
Since $GEDP$ and $GEQX$ are cyclic and $E = DQ \cap PX$, there is a spiral similarity sending $\triangle GPD \sim \triangle GXQ$, which implies $GP$ and $GQ$ are isogonal because $GD$ and $GX$ are already isogonal.
extra observation 3 (hinted by Celestia)Actually, $AG$ and $AX$ are isogonal. We can prove this by a force-overlay inversion, which sends $(ABXE)$ and $(ACXF)$ to $\overline{CB'}$ and $\overline{BC'}$ respectively, where $B' = E^*$ and $C' = F^*$ are the reflections of $A$ over $B$ and $C$ respectively, so $X^* = G'$ is the reflection of $A$ over $G$.
old_csk_mo
02.07.2024 20:51
ricarlos wrote: This problem is not true, but if $AC=BC$ then $\angle BAK=\angle CAL$ holds. The official solution does not require the triangle to be isosceles. https://www.matematickaolympiada.cz/media/3549099/a72e.pdf#page=28 I guess the problem is in the statement -- M is midpoint of BC and N is midpoint of CA, not vice versa.