The points $P$ and $Q$ are placed in the interior of the triangle $\Delta ABC$ such that $m(\angle PAB)=m(\angle QAC)<\frac{1}{2}m(\angle BAC)$ and similarly for the other $2$ vertices($P$ and $Q$ are isogonal conjugates). Let $P_{A}$ and $Q_{A}$ be the intersection points of $AP$ and $AQ$ with the circumcircle of $CPB$, respectively $CQB$. Similarly the pairs of points $(P_{B},Q_{B})$ and $(P_{C},Q_{C})$ are defined. Let $PQ_{A}\cap QP_{A}=\{M_{A}\}$, $PQ_{B}\cap QP_{B}=\{M_{B}\}$, $PQ_{C}\cap QP_{C}=\{M_{C}\}$. Prove the following statements: $1.$ Lines $AM_{A}$, $BM_{B}$, $CM_{C}$ concur. $2. $ $M_{A}\in BC$, $M_{B}\in CA$, $M_{C}\in AB$
Problem
Source: MDA TST for EGMO 2017, problem 4
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RagvaloD
26.04.2017 15:42
Can someone give image? Because, I can`t even understand, how to build it.
GGPiku
26.04.2017 17:58
Is the statement correct? I don't think the initial condition is enough, I think we need $P,Q$ the be isogonal conjugates. I made a Geogebra figure and $M_C$ doesn't seem to be on AB$, or did I make a mistake in the construction?
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Snakes
26.04.2017 20:54
@above, yes you are right. I checked it one more time but as I remember the condition for $P$ and $Q$ must go cyclically for all the 3 vertices. With other words, they are isogonal conjugates.
rmtf1111
27.04.2017 00:00
$b)$ Obviously $P_A$ and $Q_A$ are isogonal conjugates. Now applying Isogonality Lemma on the pair of isogonal conjugates $(P,Q)$ and $(P_A,Q_A)$ wrt we get that $PQ_A \cap QP_A = M_A$ and $PP_A \cap QQ_A = A$ are isogonal conjugates, so the result follows immediately. $a)$ I'm not too sure about that , but I think the following argument works: Since $M_A$,$M_B$ and $M_C$ are defined in a cyclic way, we deduce that the trilinear coordinates of the intersection of $AM_A$ and $BM_B$ obeys homogeneity, cyclicity and bisymmetry, so it must be a triangle center, that is also incident with $CM_C$.
utkarshgupta
01.06.2017 08:24
Lemma : $PQ || P_AQ_A$
Since $P,Q$ are isogonal,
$$\angle BAP_A = \angle CAQ$$$$\angle BP_AP = \angle BCP = \angle ACQ$$$$\boxed {\triangle ABP_A \sim \triangle AQC}$$$$\implies \frac{\overline{AP}}{\overline{AP_A}} = \frac{\overline{AP} \times \overline{AQ}}{\overline {AB} \times \overline{AC}}$$But similarly
$$\frac{\overline{AQ}}{\overline{AQ_A}} = \frac{\overline{AP} \times \overline{AQ}}{\overline {AB} \times \overline{AC}}$$$$\implies \frac{\overline{AP}}{\overline{AP_A}} = \frac{\overline{AQ}}{\overline{AQ_A}}$$$$\boxed {PQ || P_AQ_A}$$
Since $\overline{PP_A} \cap \overline {QQ_A} = A$ and $\overline {PQ_A} \cap \overline {QP_A} = M_A$ $\implies AM_A$ passes through the midpoint of $PQ$. That is $AM_A, BM_B, CM_C$ concur on the midpoint of $PQ$.