A convex hexagon $ABCDEF$ is given where $ \measuredangle FAB + \measuredangle BCD + \measuredangle DEF = 360^{\circ}$ and $ \measuredangle AEB = \measuredangle ADB$. Suppose the lines $AB$ and $DE$ are not parallel. Prove that the circumcenters of the triangles $ \triangle AFE, \triangle BCD$ and the intersection of the lines $AB$ and $DE$ are collinear.
Problem
Source: Polish Mathematical Olympiad finals 2021 P5
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Steve12345
23.04.2021 03:57
Let $O,O_1,O_2,O_k$ be the circumcenters of $ABD,BCD,AFE,SBD$ where $S$ is the intersection of $AB$ and $DE$. $SO_k || OO_2$ so it's enough to prove $\frac{SO_k}{O_kO_1}=\frac{OO_2}{OO_1}$ and this can be done with 3 law of sines.
Ru83n05
28.02.2022 21:02
Quadrilateral $(AEDB)$ is cylcic Let $P:=AB\cap DE$. Now consider the inversion $\Psi$ of radius $\sqrt{Pow_{(AEDB)}(P)}$. Claim: $\Psi: (BCD)\leftrightarrow (AFE)$ Proof: Let $F'=\Psi(F)$. Then $$\angle BF'D=\angle BF'F+\angle FF'D=180-\angle FAP+180-\angle PEF=\angle BCD$$so $(BF'CD)$ is cyclic. $\blacksquare$ Now, since the circles are inverses by an inversion at $P$, their circumcenters and $P$ must be colinear. $\blacksquare$
guptaamitu1
19.05.2022 00:46
Angle chase shows that this problem is equivalent to 2022 China TST, Test 1, Day 1 P1 (which also has another AoPS thread)