posted for the image links

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This is a very interesting problem. The concurrence in the second quesiton is the generalized Napolean point for $\theta = \frac{\pi}{4} $ or Vectan point . Lets look at line $BB_2$ in the original diagram. It must be the angular bisector of right angle $A B_2 C$. Therefore $BB_2$ meet the midpoint of semicircle with diameter AC and excluding $B2$. See attached. Same thing applies to $AA_2$ and $CC_2$. A simplified version is attached. Notice in the attachment triangle AHC, CGB and ABI are isoceles right triangle. Thus concurrence is Vectan point in the original problem.

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parmenides51 wrote: Let $ABC$ be an acute triangle. Squares $AA_1A_2A_3$, $BB_1B_2B_3$ and $CC_1C_2C_3$ are located such that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ pass through the points $B$, $C$ and $A$ respectively and the lines $A_2A_3$, $B_2B_3$, $C_2C_3$ pass through the points $C$, $A$ and $B$ respectively. Prove that (a) the lines $AA_2$, $B_1B_2$ and $C_1C_3$ intersect at one point. (b) the lines $AA_2$, $BB_2$ and $CC_2$ intersect at one point. (Mykhailo Plotnikov) Use the attached diagram we can also show the concurrence in the first quesiton. Simliar to the argument above, X,Y,Z are the midpoint of the semicircle and $AA_2$ is the angular bisector of right angle $BA_2C$ and $B_1$ $A_2$ $C_3$ are on the same circle centered at $D$ with diameter $BC$. Also $C_3 C_1$ bisects right angle $B C_3 C$, thus $C_3 C_1 X$ are colinear. Simlilarly, $B_3 B_1 X$ are colinear. Therefore the concurrence in the first question is at point $X$

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Please note that "B1B2" in a) shoud obviously read B1B3