I found a generalization of Thomsen theorem and Tucker circle as follows. I'm an electrical engineer, not a mathematician. I don't know how to prove these results.

Let $ABC$ be a triangle, let two points $A_1, A_4$ lie on BC, $A_2, A_5$ lie on $CA$, and two points $A_3, A_6$ lie on $AB$. Let $B_1$ on its edge $BC$ and $B_1 \neq A_1$. A sequence of points and parallel lines is constructed as follows. The parallel line to $A_1A_2$ through $B_1$ intersects $AB$ in $B_2$ and the parallel line to $A_2A_3$ through $B_2$ intersects $AB$ in $B_3$. Continuing in this fashion the parallel line to $A_3A_4$ through $B_3$ intersects $BC$ in $B_4$ and the parallel line to $A_4A_5$ through $B_4$ intersects $BC$ in $B_5$. Finally the parallel line to $A_5A_6$ through $B_5$ intersects AB in $B_6$ and the parallel line to $A_6A_1$ through $B_6$ intersects $BC$ in $B_7$. This problem states that: Six points $A_1, A_2, A_3, A_4, A_5, A_6$ lie on a conic iff $B_7=B_1$

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  • When the conic through $A_1, A_2, A_3, A_4, A_5, A_6$ is the Steiner inellipse this problem is Thomsen theorem
  • When the conic through $A_1, A_2, A_3, A_4, A_5, A_6$ is the circumcircle this problem is Tucker circle

For notational convenience, we'll take the vertices of the triangle to be $P_0$, $P_1$, $P_2$, and define further $P_{i+3} := P_i$, so that we can say that each $A_i$ and $B_i$ lie on $\overleftrightarrow{P_i P_{i+1}}$, for $i =0, 1, 2, 3, 4, 5$.

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The configuration of points gives rise to values $a_i$ and $b_i$ ($\neq -1$) defined by $$a_i := \frac{|P_i A_i|}{|A_i P_{i+1}|} \qquad\qquad b_i = \frac{|P_i B_i|}{|B_i P_{i+1}|}$$ (with subscript arithmetic done modulo $6$), where, in the tradition of Ceva's Theorem, we consider these signed ratios of oriented segments: the value is positive when the segments point in the same direction, and negative when they point in opposite directions. Note that we can write, for instance, $$\frac{|P_i A_i|}{|P_i P_{i+1}|} = \frac{|P_i A_i|}{|P_i A_i| + |A_i P_{i+1}|} = \frac{a_i}{a_i+1} \qquad\quad \frac{|A_iP_{i+1}|}{|P_i P_{i+1}|} = \frac{1}{a_i+1}$$

Now, to the problem at hand: The parallelism conditions $$\overrightarrow{A_i A_{i+1}} \parallel \overrightarrow{B_i B_{i+1}}$$ imply that $\triangle A_i P_{i+1} A_{i+1} \sim \triangle B_i P_{i+1} B_{i+1}$ for each $i$. Therefore, $$\frac{|A_i P_{i+1}|}{|P_{i+1} A_{i+1}|} = \frac{|B_i P_{i+1}|}{|P_{i+1} B_{i+1}|} \quad\to\quad \frac{|P_i P_{i+1}|/(a_i+1)}{|P_{i+1}P_{i+2}|\;a_{i+1}/(a_{i+1}+1)} = \frac{|P_i P_{i+1}|/(b_i+1)}{|P_{i+1}P_{i+2}|\;b_{i+1}/(b_{i+1}+1)}$$ so that $$(1+a_{i+1}) b_{i+1}(1+b_i) = (1+b_{i+1})a_{i+1}(1+a_i) \quad\to\quad b_{i+1} = \frac{(1 + a_i) a_{i+1}}{1 - a_i a_{i+1} + b_i + a_{i+1} b_i}$$

That is, we can write $b_1$ in terms of $b_0$; and we can write $b_2$ in terms of $b_1$ (which, in turn, is in terms of $b_0$), and so on, until we arrive at this expression for $b_6$ in terms of $b_0$: $$b_6 = a_0\;\frac{1 + b_0 - ( a_0 - b_0 ) a_1 a_2 a_3 a_4 a_5 \phantom{a_0}}{ 1 + b_0 + (a_0-b_0) a_0 a_1 a_2 a_3 a_4 a_5}$$

Now, point $B_6$ coincides with $B_0$ when (and only when) $b_6 = b_0$. This says $$( 1 + b_0 )( a_0 - b_0 )( 1 - a_0 a_1 a_2 a_3 a_4 a_5 ) = 0$$ Since $b_i\neq -1$, and we assume $A_0 \neq B_0$ so that $a_0 \neq b_0$, the condition reduces to

$$a_0\,a_1\,a_2\,a_3\,a_4\,a_5 \;=\; 1 \tag{$\star$}$$

By Carnot's Theorem for Conics (which I proved in another answer), relation $(\star)$ is satisfied if and only if the $A_i$ lie on a conic. This proves the result. $\square$

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  • $\begingroup$ I thank to Mister @Blue very much, You are a good solution geometer. I have some another problems sine two years ago. The problems are generalization of some classical theorem geometry. Can I post the problems in Math.Stackexchange.com? $\endgroup$ – Oai Thanh Đào Sep 17 '15 at 17:50
  • $\begingroup$ @OaiThanhĐào: Thank you for the compliment. (If I have solved your problem to your satisfaction, be sure to click the "√".) I think you should post all of your other problems to this site. (Do not post them all at once, though. One a week might be a good pace.) I find your problems quite interesting; they make me aware of ideas I hadn't seen (like Thebault circles) and remind me of ideas I'd nearly forgotten (like Steiner inellipses). $\endgroup$ – Blue Sep 18 '15 at 9:21
  • $\begingroup$ Mister @Blue , Can I using your solution to publish in Cut the Knot $\endgroup$ – Oai Thanh Đào Sep 18 '15 at 15:42
  • $\begingroup$ @OaiThanhĐào: So long as I get credit (a link to my profile here on M.SE is sufficient for a Cut-the-Knot post), you may publish my solution with your problem. Thanks for asking! $\endgroup$ – Blue Sep 18 '15 at 16:26
  • $\begingroup$ I have about one hundred my problems appear in Cut The Knot. But I need your really name? could you tell me? @Blue ? $\endgroup$ – Oai Thanh Đào Sep 18 '15 at 17:16

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