# A generalization of the Tucker circle theorem and the Thomsen theorem associated with a conic

I gave a generalization of the Tucker circle theorem and the Thomsen theorem at here. Now, I give a more generalization of these theorems as following:

Problem: Let $A_1A_2A_3A_4A_5A_6$ be a hexagon, $L$ be a line on the plane. Let $L$ meets $A_1A_2$, $A_2A_3$, $A_3A_4$, $A_4A_5$, $A_5A_6$, $A_6A_1$ at $B_2$, $B_3$, $B_4$, $B_5$, $B_6$, $B_1$ respectively.

Let $C_1$ be a point on the line $A_1A_4$. Let $C_1B_2$ meets $A_2A_5$ at $C_2$. Let $C_2B_3$ meets $A_3A_6$ at $C_3$. Let $C_3B_4$ meets $A_1A_4$ at $C_4$. Let $C_4B_5$ meets $B_2B_5$ at $C_5$. Let $C_5B_6$ meets $A_3A_6$ at $C_6$. Let $C_6B_1$ meets $A_1A_4$ at $C_7$. Then:

1. Six points $C_1$, $C_2$, $C_3$, $C_4$, $C_5$, $C_6$ lie on a conic if only if six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic.

2. $C_7 \equiv C_1$ if only if six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic.

3. if $C_7 \equiv C_1$ then the Pascal line of hexagon $A_1A_2A_3A_4A_5A_6$ and $C_1C_2C_3C_4C_5C_6$ and $L$ are concurrent.

Remark: When the conic through $A_1, A_2, A_3, A_4, A_5, A_6$ is the circumcircle and $L$ at infinity, the item 1 is the Tucker circle theorem. When the conic through $A_1, A_2, A_3, A_4, A_5, A_6$ is the Steiner inellipse and $L$ at infinity, item 2 is the Thomsen theorem.

My question: can you give a proof for the problem? Before I start, just want to comment that the three claims of the statement above are not quite true. Claim 2 is true only in the direction "if the hexagon $A_1...A_6$ is inscribed then $C_7=C_1$". The converse is not true. The proof goes as follows: first I prove Claim 2 in the proper direction, disproving the converse, and then I prove Claim 1 and Claim 3.

Projectivities and central projectivities: By projectivity I mean a projective transformation of the projective plane. They map points to points and lines to lines. By central projectivity with center $B$ (a point) and axis $b$ (a line), or in short just central projectiviy, I mean a projective transformation that fixes $B$ and fixes $b$ point-wise. Its existence is equivalent to Desargue's theorem. Basically, given the point $B$, the axis $b$ and two points $A$ and $A'$ not equal to $B$ and not on $b$, then the central projectivity $\sigma$ with center $B$ and axis $b$ is uniquely defined by the condition $\sigma(A)=A'$. Central projectivities are involutions, projective generalizations of Euclidean reflections in a point). Observe that because $\sigma^2 = id$ then $\sigma(A') = A$.

Some constructions and notations: Now, let us have an arbitrary hexagon $A_1A_2A_3A_4A_5A_6$, a line $L$ disjoint from the vertices of the hexagon, and a construction of the points $B_1,..., B_6 \in L$ and $C_1,..., C6, C_7$ as described in the problem. Denote by $E_{36}$ the intersection point of lines $A_1A_4$ and $A_2A_5$ (from now on, I denote the point of intersection of any pair of lines by $E_{36} = A_1A_4 \cap A_2A_5$). Analogously, let $E_{14} = A_2A_5\cap A_3A_6$ and $E_{25} = A_3A_6\cap A_1A_4$. Also, let $$D_{12} = A_6A_1 \cap A_2A_3, \,\, D_{23} = A_1A_2 \cap A_3A_4, \,\, D_{34} = A_2A_3 \cap A_4A_5,$$ $$D_{45} = A_3A_4 \cap A_5A_6, \,\, D_{56} = A_4A_5 \cap A_6A_1, \,\, D_{61} = A_1A_2 \cap A_5A_6.$$ Define the lines $$b_2 = D_{12}E_{36}, \,\, b_3 = D_{23}E_{14}, \,\, b_4 = D_{34}E_{25},$$ $$b_5 = D_{45}E_{36}, \,\, b_6 = D_{56}E_{14}, \,\, b_1 = D_{61}E_{25}.$$

Denote by $\sigma_i$ the central projectivity with center $B_i$ and axis $b_i$ such that $\sigma_i(A_{i}) = A_{i-1}$ for $i=1,..,6$ where in the case of $\sigma_1$ we have $\sigma_1(A_{1}) = A_{6}$. By construction $\sigma_i(L)=L$ (not point-wise) for all $i=1,..,6$. Moreover, since $\sigma_2(A_2) = A_1$ and $\sigma_2(E_{36})=E_36$, the line $A_1A_4$ gets mapped to the line $A_2A_5$ and $A_2A_5$ gets mapped to $A_1A_4$ by $\sigma_2$. Absolutely analogously, the same is true for the rest of the central projectivites $\sigma_3,..., \sigma_6, \sigma_1$ and the appropriate pairs of lines formed from the set of three lines $A_2A_5, A_3A_6$ and $A_1A_4$.

Define the projectivity $f = \sigma_1 \circ \sigma_6 \circ \sigma_5 \circ \sigma_4 \circ \sigma_3 \circ \sigma_2$. Let $M_{14} = L \cap A_{1}A_{4}, M_{25} = L \cap A_{2}A_{5}$ and $M_{36} = L \cap A_{3}A_{6}$. Then $$\sigma_2(M_{14}) = M_{25}, \sigma_3(M_{25}) = M_{36}, \sigma_4(M_{36}) = M_{14},$$ $$\sigma_5(M_{14}) = M_{25}, \sigma_6(M_{25}) = M_{36}, \sigma_2(M_{36}) = M_{14}.$$ Therefore $f(M_{14}) = M_{14}.$ In addition to that, if we look at the images of the points $B_1$ and $B_2$ under the consecutive application of the projectivites $\sigma_2,..., \sigma_6, \sigma_1$ in this order, we arrive at the chain of images $$B_1 \mapsto B_3 \mapsto B_3 \mapsto B_5 \mapsto B_5 \mapsto B_1 \mapsto B_1,$$ $$B_2 \mapsto B_2 \mapsto B_4 \mapsto B_4 \mapsto B_6 \mapsto B_6 \mapsto B_2.$$ Therefore, $f(B_1) = B_1$ and $f(B_2) = B_2$. But a projective transformation, in this case $f$, which fixes three points on a line, fixes the whole line point-wise. Hence, $f$ fixes $L$ point-wise. Moreover, by construction, $f(A_1) = A_1$. Therefore, $f$ is a central projectivity with a center $A_1$ and axis $L$. Also by construction, $f(C_1) = C_7$.

Lemma 1: If the six lines (the axes of the central projectivites $\sigma_1,...,\sigma_6$) $b_1,...,b_6$ intersect at a common point $O$, then $f$ is the identity and so $C_7 = f(C_1) = C_1.$

Proof of Lemma 1: Indeed, by construction $f(O)=O$ and since $O$ is different from $A_1$ and does not lie on $L$, the transformation $f$ is a central projectivity which fixes one extra point, which is possible only when $f$ is the identity.

Corollary 1: If the hexagon $A_1...A_6$ is superscribed around a conic, then $f$ is an identity and $C_7 \equiv C_1$ which means the polygonal chain $C_1...C_6$ closes up to a hexagon.

Proof of corollary 1: By Brianchon's theorem the diagonal lines $A_1A_4, A_2A_5$ and $A_3A_6$ intersect in a common point, call it $O$. Then $O\equiv E_{36} \equiv E_{14} \equiv E_{25}$ and therefore by construction the axes $b_1,..., b_6$ intersect in $O$.

As you can see by the latter corollary, $C_7 \equiv C_1$ occurs in the case of any superscribed around a conic hexagon and since there are definitely superscribed hexagons that are not inscribed in conics, the statement that "if $C_7 \equiv C_1$ then $A_1...A_6$ is inscribed in a conic" cannot be true.

Proof of Claim 2 from the main statement.

Let $A_1...A_6$ be a hexagon inscribed in a conic.

Lemma 2: The three lines $D_{12}D_{45}, D_{23}D_{56}$ and $D_{34}D_{61}$ intersect in a common point called $O$.

Proof of Lemma 2: Look at triangles $D_{12}D_{34}D_{56}$ and $D_{23}D_{45}D_{61}$. By Pascal's theorem for $A_1...A_6$ the intersection points $D_{12}D_{56} \cap D_{23}D_{45}, \,\, D_{12}D_{34} \cap D_{45}D_{61}$ and $D_{34}D_{56} \cap D_{23}D_{61}$ lie on a common line. By Desargue's theorem, applied to the triangles $D_{12}D_{34}D_{56}$ and $D_{23}D_{45}D_{61},$ the three lines $D_{12}D_{45}, \, D_{23}D_{56}$ and $D_{34}D_{61}$ intersect at a common point, denoted by $O$.

Lemma 3: I claim that $E_{25} \in D_{61}D_{34}, \, E_{36} \in D_{12}D_{45}$ and $E_{14} \in D_{23}D_{56}$. Consequently $D_{61}D_{34} = b_1=b_4$ and $D_{12}D_{45} = b_2=b_5$ and $D_{23}D_{56} = b_3=b_6$.

Proof of Lemma 3: The (self intersecting) hexagon $A_1A_2A_3A_6A_5A_4$ (observe the order of the vertices!) is inscribed in a conic, because $A_1A_2A_3A_4A_5A_6$ is inscribed by assumption. By Pascal's theorem for $A_1A_2A_3A_6A_5A_4$, the points $E_{25}, D_{61}$ and $D_{34}$ are collinear, that is $E_{25} \in D_{61}D_{34}$ and thus $D_{61}D_{34} = b_1=b_4$. The rest of the claims follow analogously.

By combining Lemma 2 with Lemma 3, one concludes that the axes of the central projectivities $\sigma_1,...,\sigma_6$ pass through a common point $O$ and therefore by Lemma 1 it follows that the projectivity $f$ is the identity and $C_7 = f(C_1) = C_1$.

Proof of Claim 1 from the main statement.

Let $A_1A_2A_3A_4A_5A_6$ be inscribed in a conic. By Claim 2, $C_7 \equiv C_1$ and so $B_1$ lies on the line $C_1C_6$. By Desargue's theorem applied to the triangles $A_1B_1C_1$ and $A_2B_3C_2$ it follows that the three intersection points $E_{36}=A_1C_1 \cap A_2C_2, \,\, D_{12}=A_1B_1 \cap A_2B_3$ and $F_{12}=B_1C_1 \cap B_3C_2$ are collinear and since $D_{12}E_{36}=b_2,$ clearly $F_{12} \in b_2.$ Next, apply Desargue's theorem to the triangles $A_4B_4C_4$ and $A_5B_6C_5$ in order to conclude that $E_{36}=A_4C_4 \cap A_5C_5, \,\, D_{45}=A_4B_4 \cap A_5B_6$ and $F_{45}=B_4C_4 \cap B_6C_5$ are collinear. Since $D_{45}E_{36}=b_4=b_2$ by Lemma 3, as before $F_{45} \in b_2.$ Therefore the three points $F_{12}, \, F_{45}$ and $E_{36}$ lie on the line $b_2$. Consequently, if one looks at the hexagon $C_2C_3C_4C_1C_6C_5$ (observe the order of the points!) one can conclude that since $F_{12}, \, F_{45}, \, E_{36} \in b_2$ by Pascal's theorem $C_2C_3C_4C_1C_6C_5$ is inscribed in a conic. Hence, the hexagon $C_1C_2C_3C_4C_5C_6$ is inscribed in the same conic.

Conversely, let $A_1A_2A_3A_4A_5A_6$ be an arbitrary hexagon and $C_1C_2C_3C_4C_5C_6$ be constructed by following the procedure given in the main statement but stopping after constructing the point $C_6$ and then simply connecting $C_6$ to $C_1$, assuming that $C_7$ and $C_1$ may not coincide. Furthermore, assume that $C_1C_2C_3C_4C_5C_6$ is inscribed in a conic. Denote by $B'_1$ the intersection point of $C_6C_1$ with $L$. By interchanging the two hexagons in the construction given in the main statement, we can use $C_1C_2C_3C_4C_5C_6$ as a starting hexagon, the line $L$ and then starting from point $A_1$ follow the same method of constructing $A_2, A_3, A_4, A_5, A_6$ and then using $B'_1$ construct $A_7$. But by Claim 1, $A_7\equiv A_1$ because $C_1C_2C_3C_4C_5C_6$ is inscribed in a conic. By the already proven direction of Claim 2, the hexagon $A_1A_2A_3A_4A_5A_6$ is also inscribed in a conic. Finally, that guarantees that $B_1, C_6$ and $C_1$ are collinear and hence $B'_1 \equiv B_1$.

Proof of Claim 3 from the main statement.

Assume $A_1A_2A_3A_4A_5A_6$is inscribed in a conic. By Claim 2, $B_1 \in C_6C_1$ and by Claim 1 $C_1C_2C_3C_4C_5C_6$ is inscribed in a conic.

Some notations: Let $P_{14} = A_2A_3 \cap A_5A_6, \, P_{25} = A_1A_2 \cap A_4A_5$ and $P_{36} = A_3A_4 \cap A_6A_1.$ By Pascal's theorem the three intersection points $P_{14}, P_{25}$ and $P_{36}$ lie on a common line $p$. Furthermore, let $Q_{14} = C_2C_3 \cap C_5C_6, \, Q_{25} = C_1C_2 \cap C_4C_5$ and $Q_{36} = C_3C_4 \cap C_6C_1.$ By Pascal's theorem the three intersection points $Q_{14}, Q_{25}$ and $Q_{36}$ lie on a common line $q$.

Claim 3 from the main statement: The three lines $p, q$ and $L$ are concurrent.

Let $H_{14} = A_1A_4 \cap D_{23}D_{56} = A_1A_4 \cap b_3, \, H_{25} = A_2A_5 \cap D_{34}D_{61} = A_2A_5 \cap b_1$ and $H_{36} = A_3A_6 \cap D_{45}D_{12} = A_3A_6 \cap b_2$.

Lemma 4: If $A_1A_2A_3A_4A_5A_6$ is inscribed in a conic then

- the three points $H_{14}, H_{25}$ and $P_{36} = A_1A_2 \cap A_4A_5$ are collinear;

- the three points $H_{36}, H_{14}$ and $P_{25} = A_3A_4 \cap A_6A_1$ are collinear;

- the three points $H_{25}, H_{36}$ and $P_{14} = A_2A_3 \cap A_5A_6$ are collinear;

Proof of Lemma 4: Look at triangles $A_4D_{23}H_{14}$ and $A_5D_{61}H_{25}$. Then $A_4H_{14} \cap A_5H_{25} = A_4A_1 \cap A_5A_2 = E_{36}$, as well as $D_{23}H_{14} \cap D_{61}H_{25} = b_2 \cap b_1 = O$, and finally $D_{23}A_4 \cap D_{61}A_5 = A_3A_4 \cap A_5A_6 = D_{45}$. By Lemma 2, $O \in b_3 = D_{45}E_{36}$, therefore by Desargue's theorem the three intersection points $D_{61}D_{23} \cap A_4A_5 = A_1A_2 \cap A_4A_5 = P_{36}$ are collinear. The rest of the statements are proven analogously.

Next, apply Lemma 4 to the inscribed in a conic hexagon $C_1C_2C_3C_4C_5C_6$ in place of $A_1...A_6$ and observe that by construction $A_1A_4=C_1C_4, \, A_2A_5 = C_2C_5, \, A_3A_6 = C_3C_6$. Then one concludes that $Q_{36} \in H_{14}H_{25}, \, Q_{25} \in H_{36}H_{14},$ and $Q_{14} \in H_{25}H_{36}$ because the points $F_{12}, D_{12}, E_{36}, O, D_{45}, F_{45} \in b_2$, as well as $F_{23}, D_{23}, E_{14},$ $O, D_{56}, F_{56} \in b_3$ and $F_{34}, D_{34}, E_{25}, O, D_{61}, F_{61} \in b_1$ according to Lemma 2, Lemma 3 and the facts that the points $F_{ij}$ lie on the appropriate axes $b_j$ established in the proof of Claim 1 from the statement. Therefore $H_{25} \in P_{36}Q_{36}$ and $H_{25} \in P_{14}Q_{14}$. Since $H_{25} = A_2A_5 \cap b_1$ and $A_2A_5 = C2C_5$, one concludes that $H_{25} \in C_2C_5$ and thus $H_{25} = P_{36}Q_{36} \cap P_{14}Q_{14} \in C_2C_5$. Look at triangles $B_2P_{36}Q_{36}$ and $B_3P_{14}Q_{14}.$ Then $B_2Q_{36} \cap B_3Q_{14} = C_2 \in C_2C_5$, furthermore $B_2P_{36} \cap B_3P_{14} = A_2 \in C_2C_5 = A_2A_5$ as well as $P_{36}Q_{36} \cap P_{14}Q_{14} = H_{25} \in C_2C_5$. By (the converse) Desargue's theorem, the three lines $B_2B_3 = L, \, P_{36}P_{14} = p$ and $Q_{36}Q_{14} = q$ are concurrent.

• Dear @Futurologist Do you agree which I publish your proof with your name on a journal? – Đào Thanh Oai Sep 13 at 7:43
• @ĐàoThanhOai Hi, yes I agree. Let me know what information you would need from me. – Futurologist Sep 18 at 7:46