isogonal geometry

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#1 h Apr 7, 2025, 4:27 AM

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Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$ . Let $\triangle P_1P_2P_3$ and $\triangle Q_1Q_2Q_3$ be their respective pedal triangles. Let $X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1$ Prove that the points $X_1$ , $X_2$ and $X_3$ lie on the line $PQ$ .

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#2 h Apr 7, 2025, 7:43 AM

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Let $M$ be the midpoint of $PQ$ . It is well known that $P_1, P_2, P_3, Q_1, Q_2, Q_3$ are cyclic, and call this circle $\Gamma$ . Note that $M$ is the center of $\Gamma$ . Let $R_i$ be the Miquel point of quadrilateral $P_{i+1}Q_{i+1}Q_{i+2}P_{i+2}$ . ( $P_{i+3}=P_{i}$ ) And rename $A,B,C$ as $A_1, A_2, A_3$ . Note that $\angle PR_{i}A_{i}= \angle QR_{i}A_{i}=90°$ by cyclic, $R_{i}$ lies on $PQ$ . And by Miquel’s theorem, $X_{i}$ lies on $R_{i}M$ , and we are done.
note

This post has been edited 5 times. Last edited by whwlqkd, Apr 7, 2025, 1:25 PM
Reason: Misspelling

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#3 h Apr 7, 2025, 9:47 AM

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Note that $\angle PR_{i}A_{i}= \angle QR{i}A{i}=90°$ by cyclic, $R_{i}$ lies on $PQ$ . And by Miquel’s theorem, $X_{i}$ lies on $R_{i}M$ , and we are done.
explain this part

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#4 h Apr 7, 2025, 10:26 AM

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Miquel’s theorem:
Let $ABCD$ be cyclic quadrilateral inscribed on circle $O$ . And let $X$ be the Miquel point of quadrilateral $ABCD$ . Then $AC$ , $BD$ , $OX$ are concurrent

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#5 h Apr 7, 2025, 1:11 PM

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Shouldnt $R_{i}$ exist in the line $X_{i}A_{i}$

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#6 h Apr 7, 2025, 1:15 PM

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whwlqkd wrote:

Miquel’s theorem:
Let $ABCD$ be cyclic quadrilateral inscribed on circle $O$ . And let $X$ be the Miquel point of quadrilateral $ABCD$ . Then $AC$ , $BD$ , $OX$ are concurrent

In here, $R_{i}=X$ , $A_{i}=AB\cap CD$ , $X_{i}=AC\cap BD$

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#8 h Apr 7, 2025, 1:57 PM • 1 Y

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I have nice solution with cool construction.

Obviously, $P_1Q_1P_2Q_2P_3Q_3$ is cyclic, let its center be $O$
So let $S$ be the antipode of $Q_{2}$ , and $R$ be the antipode of $Q_{3}$

Pascal on $Q_{2}P_{3}RQ_{3}P_{2}S$ gives that $X_{1} = Q_{2}P_{3} \cap Q_{3}P_{2}$ lies on $PO = PQ$ , so problem solved

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#9 h Apr 7, 2025, 2:32 PM

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aidenkim119 wrote:

I have nice solution with cool construction.

Obviously, $P_1Q_1P_2Q_2P_3Q_3$ is cyclic, let its center be $O$
So let $S$ be the antipode of $Q_{2}$ , and $R$ be the antipode of $Q_{3}$

Pascal on $Q_{2}P_{3}RQ_{3}P_{2}S$ gives that $X_{1} = Q_{2}P_{3} \cap Q_{3}P_{2}$ lies on $PO = PQ$ , so problem solved

Very nice solution with Pascal!!!!!!!!!

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