Let $ X', X'' \in XY$ be arbitrary. Let $ B_1', B_1'' \in CA$ and $ C_1', C_1'' \in AB$ be pedals of $ X', X''.$ Let $ P_2, P_2', P_2'' \in C_2A_2$ be pedals of $ B_1, B_1', B_1''$ and $ Q_2, Q_2', Q_2'' \in A_2B_2$ pedals of $ C_1, C_1', C_1''.$ Let $ D \equiv P_1P_2 \cap Q_1Q_2,$ $ D' \equiv P_1'P_2' \cap Q_1'Q_2',$ $ D'' \equiv P_1''P_2'' \cap Q_1''Q_2''.$ We have $ \frac {\overline{P_2P_2'} \cdot \overline{P_2A_2}}{\overline{P_2P_2''} \cdot \overline{P_2A_2}} = \frac {\overline{P_2P_2'}}{\overline{P_2P_2''}} = \frac {\overline{P_1P_1'}}{\overline{P_1P_1''}} = \frac {\overline{XX'}}{\overline{XX''}} = \frac {\overline{Q_1Q_1'}}{\overline{Q_1Q_1''}} = \frac {\overline{Q_2Q_2'}}{\overline{Q_2Q_2''}} = \frac {\overline{Q_2Q_2'} \cdot \overline{Q_2A_2}}{\overline{Q_2Q_2''} \cdot \overline{Q_2A_2}}$ It follows that circumcircles of $ (A_2P_2DQ_2), (A_2P_2'D'Q_2'), (A_2P_2''D''Q_2'')$ are coaxal, their centers are collinear. But their centers are midpoints of $ A_2D, A_2D', A_2D''$ $ \Longrightarrow$ $ D, D', D''$ are also collinear; locus of $ D$ is a line $ h_D.$ Midpoints of $ AY, BY, CY$ are circumcenters of quadrilaterals $ (AC_2YB_2), (BA_2YC_2), (CB_2YA_2)$ $ \Longrightarrow$ perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from $ A, B, C$ concur at the isogonal conjugate $ Z$ of $ Y$ WRT $ \triangle ABC$ $ \Longrightarrow$ perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from midpoints $ A', B', C'$ of $ BC, CA, AB$ concur at the complement $ Z'$ of $ Z.$ Let $ n_A, n_B, n_C$ be perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from $ A_1, B_1, C_1,$ respectively. Let $ D \equiv n_B \cap n_C,$ $ E \equiv n_C \cap n_A,$ $ F \equiv n_A \cap n_B.$ Then $ D, E, F$ are on 3 fixed lines $ h_D, h_E, h_F,$ which are obviously concurrent at the orthocenter $ H_2$ of the pedal $ \triangle A_2B_2C_2.$ Medial $ \triangle A'B'C'$ is pedal triangle WRT circumcenter $ O$ of $ \triangle ABC.$ $ O \in XY$ $ \Longrightarrow$ $ h_D, h_E, h_F$ are also concurrent at the complement $ Z'$ of the isogonal conjugate $ Z$ of $ Y.$ Consequently, $ h_D \equiv h_E \equiv h_F$ are identical and $ n_A, n_B, n_C$ concurrent for all $ X \in YO.$

Quote: $ \frac {\overline{P_2P_2'} \cdot \overline{P_2A_2}}{\overline{P_2P_2''} \cdot \overline{P_2A_2}} = \frac {\overline{P_2P_2'}}{\overline{P_2P_2''}} = \frac {\overline{P_1P_1'}}{\overline{P_1P_1''}} = \frac {\overline{XX'}}{\overline{XX''}} = \frac {\overline{Q_1Q_1'}}{\overline{Q_1Q_1''}} = \frac {\overline{Q_2Q_2'}}{\overline{Q_2Q_2''}} = \frac {\overline{Q_2Q_2'} \cdot \overline{Q_2A_2}}{\overline{Q_2Q_2''} \cdot \overline{Q_2A_2}}$ It follows that circumcircles of $ (A_2P_2DQ_2), (A_2P_2'D'Q_2'), (A_2P_2''D''Q_2'')$ are coaxal, their centers are collinear. But their centers are midpoints of $ A_2D, A_2D', A_2D''$ $ \Longrightarrow$ $ D, D', D''$ are also collinear; locus of $ D$ is a line $ h_D.$ Dear Yetti, sorry for my silly question but can you tell me clearer about this property? I notice it very nice and useful way to prove circles are coaxal. Thank you very much for your explanation.

Dear mathVNpro, it is not a silly question. When powers of 2 points to 2 given circles are in the same ratio, these 2 points are on a circle coaxal with the 2 given circles. See http://www.mathlinks.ro/viewtopic.php?t=121062 for a proof. See http://www.mathlinks.ro/viewtopic.php?t=121088, http://www.mathlinks.ro/viewtopic.php?t=190473 for nice applications.

Dear Mr. Yetti, thank you very much for your sincere help that you gave me. I am truly thank you for your careful explanation that you gave. By the way, do you have any file or ebook or any document realted to the above formula. Thank you in advance.

Actually, I do. But even in MS Word format (rather than PDF), it is larger than allowd attachment size - I cannot post it or send it by PM, only by e-mail.

Dear Yetti, it will be such great if you send me that file. I will send you my e-mail address by private massage. Thank you very very much in advance.

April wrote: Given triangle $ ABC$ and two points $ X,$ $ Y$ not lying on its circumcircle. Let $ A_1,$ $ B_1,$ $ C_1$ be the projections of $ X$ to $ BC,$ $ CA,$ $ AB,$ and $ A_2,$ $ B_2,$ $ C_2$ be the projections of $ Y.$ Prove that the perpendiculars from $ A_1,$ $ B_1,$ $ C_1$ to $ B_2C_2,$ $ C_2A_2,$ $ A_2B_2,$ respectively, concur if and only if line $ XY$ passes through the circumcenter of $ ABC.$ This configuration calls for the 2nd Fontene theorem. What do you think Vladimir?

This was the first thing I tried but I did not see a way. BTW, I did not prove "if and only if". I almost got it by contradiction, but could not complete the last step in a satisfactory way. Any idea ?

OK, I got a simple contradiction for "if and only if." yetti wrote: Let $ X', X'' \in XY$ be arbitrary. Let $ B_1', B_1'' \in CA$ and $ C_1', C_1'' \in AB$ be pedals of $ X', X''.$ Let $ P_2, P_2', P_2'' \in C_2A_2$ be pedals of $ B_1, B_1', B_1''$ and $ Q_2, Q_2', Q_2'' \in A_2B_2$ pedals of $ C_1, C_1', C_1''.$ Let $ D \equiv P_1P_2 \cap Q_1Q_2,$ $ D' \equiv P_1'P_2' \cap Q_1'Q_2',$ $ D'' \equiv P_1''P_2'' \cap Q_1''Q_2''.$ We have $ \frac {\overline{P_2P_2'} \cdot \overline{P_2A_2}}{\overline{P_2P_2''} \cdot \overline{P_2A_2}} = \frac {\overline{P_2P_2'}}{\overline{P_2P_2''}} = \frac {\overline{P_1P_1'}}{\overline{P_1P_1''}} = \frac {\overline{XX'}}{\overline{XX''}} = \frac {\overline{Q_1Q_1'}}{\overline{Q_1Q_1''}} = \frac {\overline{Q_2Q_2'}}{\overline{Q_2Q_2''}} = \frac {\overline{Q_2Q_2'} \cdot \overline{Q_2A_2}}{\overline{Q_2Q_2''} \cdot \overline{Q_2A_2}}$ It follows that circumcircles of $ (A_2P_2DQ_2), (A_2P_2'D'Q_2'), (A_2P_2''D''Q_2'')$ are coaxal, their centers are collinear. But their centers are midpoints of $ A_2D, A_2D', A_2D''$ $ \Longrightarrow$ $ D, D', D''$ are also collinear; locus of $ D$ is a line $ h_D.$ Midpoints of $ AY, BY, CY$ are circumcenters of quadrilaterals $ (AC_2YB_2), (BA_2YC_2), (CB_2YA_2)$ $ \Longrightarrow$ perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from $ A, B, C$ concur at the isogonal conjugate $ Z$ of $ Y$ WRT $ \triangle ABC$ $ \Longrightarrow$ perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from midpoints $ A', B', C'$ of $ BC, CA, AB$ concur at the complement $ Z'$ of $ Z.$ Let $ n_A, n_B, n_C$ be perpendiculars to $ B_2C_2, C_2A_2, A_2B_2$ from $ A_1, B_1, C_1,$ respectively. Let $ D \equiv n_B \cap n_C,$ $ E \equiv n_C \cap n_A,$ $ F \equiv n_A \cap n_B.$ Then $ D, E, F$ are on 3 fixed lines $ h_D, h_E, h_F,$ which are obviously concurrent at the orthocenter $ H_2$ of the pedal $ \triangle A_2B_2C_2.$ Medial $ \triangle A'B'C'$ is pedal triangle WRT circumcenter $ O$ of $ \triangle ABC.$ $ O \in XY$ $ \Longrightarrow$ $ h_D, h_E, h_F$ are also concurrent at the complement $ Z'$ of the isogonal conjugate $ Z$ of $ Y.$ Consequently, $ h_D \equiv h_E \equiv h_F$ are identical and $ n_A, n_B, n_C$ concurrent for all $ X \in YO.$ Assume $ O \not\in XY$ and perpendiculars $ n_A, n_B, n_C$ to $ B_2C_2, C_2A_2, A_2B_2$ from $ A_1, B_1, C_1$ concur at $ D \equiv E \equiv F.$ This means that the 3 fixed lines $ h_D, h_E, h_F$ through $ H_2, D \equiv E \equiv F$ coincide. Let $ X' \in XY$ be arbitrary and let $ A_1' \in BC, B_1' \in CA, C_1' \in AB$ be its pedals. Since $ h_D \equiv h_E \equiv h_F,$ it follows that perpendiulars $ n_A', n_B', n_C'$ to $ B_2C_2, C_2A_2, A_2B_2$ from $ A_1', B_1', C_1'$ also concur at $ D' \in H_2D.$ Consider now line $ X'O.$ Midpoints $ A', B', C'$ of $ BC, CA, AB$ are pedals of $ O$ and perpendiculars $ m_A, m_B, m_C$ to $ B_2C_2, C_2A_2, A_2B_2$ from $ A', B', C'$ concur at $ Z'.$ It follows that all points $ X'' \in X'O$ also have the same property: perpendiculars $ n_A'', n_B'', n_C''$ to $ B_2C_2, C_2A_2, A_2B_2$ from the pedals of $ X''$ concur at $ D'' \in D'Z'.$ Consequently, all points $ X$ in the plane have the same property. This includes the triangle vertices $ A, B, C$ in place of $ X.$ Pedal triangles of $ A, B, C$ WRT $ \triangle ABC$ degenerate to its altitudes $ AA_1, BB_1, CC_1.$ Let $ X \equiv A.$ Perpendiculars to $ C_2A_2, A_2B_2$ from $ B_1 \equiv C_1 \equiv A$ meet at $ A.$ Perpendicular to $ B_2C_2$ from $ A_1$ going through their intersection $ A$ is identical with the A-altitude of $ \triangle ABC$ $ \Longrightarrow$ $ (B_2C_2 \parallel BC) \perp AA_1.$ Similarly, $ C_2A_2 \parallel CA, A_2B_2 \parallel AB$ $ \Longrightarrow$ $ Y \equiv O,$ which is a contradiction.

Undoubtedly, it is one of the finest and elegant problems I have yet seen. The following solution is aided by the official solution as I was unable to solve it completely. WLOG, let $O \notin \{X, Y\}.$ Fix $Y$ and call a point $X$ in the plane as $Y$-good if and only if the pedal triangles of $X, Y$ w.r.t $\triangle ABC$ are orthologic. Lemma 1: Point $O$ is $Y$-good. (Proof) Note that the perpendicular from $A_2$ to the $A$-midline in $\triangle ABC$ passes through $Y$. It follows that $\triangle A_2B_2C_2$ and the medial triangle of $\triangle ABC$ are orthologic, as desired. $\square$ Lemma 2: Let $P$ be a $Y$-good point. For any $X \in \overline{YP};$ $X$ is also $Y$-good. (Proof) Vary $X$ with constant velocity along the line $\overline{YP}$. By Carnot's Theorem, $$X \, \text{is} \, Y \text{good} \iff \sum_{\text{cyc}} \left(A_1B_2^2-A_1C_2^2\right)=0.$$Since $(A_1B_2^2-A_1C_2^2)$ varies linearly as $X$ moves, and for $X \in \{Y, P\}$ the expression vanishes, it is identically zero. $\square$ Previous discussion shows that $X$ is $Y$-good if $X, O, Y$ are collinear. For the other direction, suppose there exists a point $X'$ not on line $\overline{OY}$ which works. By lemma 2, it follows that all points in the plane are $Y$-good. Evidently, this fails to hold for at least one element of the set $\{A, B, C\}$ as long as $Y \ne O,$ so we may conclude.

This solution is geotechspam.bin. What a problem. First suppose that $XY$ lies on a line with $O$. Note that the conclusion is equivalent to the two pedal triangles being orthologic. Animate $Y$ along $XO$, fixing $\triangle ABC$. It's equivalent to consider the perpendiculars from $A_2$ to $B_1C_1$ and so forth. Then the perpendiculars have degree $1$ so the conclusion has degree $3$. It remains to show four cases. Taking $Y \in (ABC)$ works because $\triangle A_2B_2C_2$ becomes a simson line. Taking $Y = O$ works because then they concur at $X$. Taking $Y = X$ works by the fact orthocenters exist. Conversely, suppose that $X$ and $Y$ are orthologic. It then follows by the same idea that the result also holds for line $XY$. Now, the locus of the orthology center mentioned in the problem statement has degree $2$ and can be checked to be $(A_1B_1C_1H\infty_1\infty_2)$ where $\infty_1, \infty_2$ are the directions of the Simson lines and $H$ is the orthocenter of $A_1B_1C_1$. However, this hyperbola must then be circumrectangular, so it follows that the Simson lines are perpendicular. Since the Simson line spins at twice the rate of the movement of the point, it must follow that $XY$ is a diameter.