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About a year ago, I discovered a property related to quatrains, but I haven't been able to prove it. The property is rather strange, and I don't know where to begin proving it. If anyone can help me, please do.

Let $ABCD$ be a quadrilateral whose vertices are not concyclic. For each vertex, consider the triangle formed by the other three vertices, and denote its circumcenter as follows:

$O_A$ is the circumcenter of $\triangle BCD$,

$O_B$ is the circumcenter of $\triangle ACD$,

$O_C$ is the circumcenter of $\triangle ABD$,

$O_D$ is the circumcenter of $\triangle ABC$.

Prove that the four points $O_A, O_B, O_C, O_D$ are not concyclic.

Prove that no circle passes through all four points $O_A, O_B, O_C, O_D$.

Also, if the property is known in advance, please add a source that mentions it.

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    $\begingroup$ One of my go-to coordinatizations for quadrilaterals is to let the coordinate axes bisect the angles made by the diagonals; eg, $$\begin{align} A&:= \phantom{-}a(\cos t,\sin t) & B &:= \phantom{-}b(\cos(-t),\sin(-t) \\ C&:= -c(\cos t,\sin t) & D &:= -d (\cos(-t),\sin(-t)) \end{align}$$ Then $\square ABCD$ is cyclic iff $ac=bd$. From this starting point for your question, I'd find the circumcenters of the your four triangles, then check that (barring other degeneracies) they're cyclic iff $ac=bd$. $\endgroup$ Commented Apr 24 at 17:09
  • $\begingroup$ I can't use this method, but it would be good to investigate it from multiple perspectives, for me I came to this conclusion experimentally, and I have a hunch that there is a relationship between the measurements of the angles of the original quadrant and the measurements of the angles of the new quadrant which prevents the new quadrant from being circular but I have not been able to reach that relationship yet @Blue $\endgroup$ Commented Apr 24 at 17:27
  • $\begingroup$ "I can't use this method" Why "can't"? $\endgroup$ Commented Apr 24 at 18:22
  • $\begingroup$ Since I haven't developed my skills in using parameters yet, I hope to learn to deal with it later @Blue $\endgroup$ Commented Apr 25 at 3:11
  • $\begingroup$ Inshaallah, I will learn a lot of techniques that will help me solve problems as soon as I finish some of the projects I am working on. @Blue $\endgroup$ Commented Apr 25 at 7:15

2 Answers 2

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The statement is in fact equivalent to the cyclicity of $ABCD$, with a degenerate case.

Since the property is invariant under similarities, we may assume $$ A=(a,am),\quad B=(b,-bm),\quad C=(-c,-cm),\quad D=(-d,dm), $$ where $a,b,c,d>0$ and $m\neq 0$.

Let $O_A,O_B,O_C,O_D$ be the circumcenters of triangles $BCD,ACD,ABD,ABC$, respectively. A straightforward computation using perpendicular bisectors gives $$ O_A=\left(\frac{(b-c)(c+d)(1+m^2)}{4c},\ \frac{(b+c)(d-c)(1+m^2)}{4cm}\right), $$ $$ O_B=\left(\frac{(a-d)(c+d)(1+m^2)}{4d},\ \frac{(d-c)(a+d)(1+m^2)}{4dm}\right), $$ $$ O_C=\left(\frac{(a+b)(a-d)(1+m^2)}{4a},\ \frac{(a-b)(a+d)(1+m^2)}{4am}\right), $$ $$ O_D=\left(\frac{(a+b)(b-c)(1+m^2)}{4b},\ \frac{(a-b)(b+c)(1+m^2)}{4bm}\right). $$

We use the standard determinant criterion: four points $(x_i,y_i)$ are concyclic iff $$ \det\begin{pmatrix} x_1 & y_1 & x_1^2+y_1^2 & 1\\ x_2 & y_2 & x_2^2+y_2^2 & 1\\ x_3 & y_3 & x_3^2+y_3^2 & 1\\ x_4 & y_4 & x_4^2+y_4^2 & 1 \end{pmatrix}=0. $$

For $A,B,C,D$, this determinant simplifies to $$ -2m(1+m^2)(a+c)(b+d)(ac-bd), $$ hence $$ ABCD \text{ is cyclic } \iff ac=bd. $$

For $O_A,O_B,O_C,O_D$, the same determinant becomes $$ -\frac{(1+m^2)^5}{128\,a^2b^2c^2d^2m^3}(a+c)(b+d)(ac-bd)^5. $$

Thus $$ O_A,O_B,O_C,O_D \text{ are concyclic } \iff ac=bd. $$

If $ac=bd$, then $ABCD$ is cyclic and all four circumcenters coincide (they are the common circumcenter), so the result is trivially true.

If $ac\ne bd$, then the determinant is nonzero, hence $O_A,O_B,O_C,O_D$ are not concyclic.

$\square$

Note: This proof was done by ChatGPT

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    $\begingroup$ Math.SE policy states: "Generative artificial intelligence (a.k.a. GPT, LLM, generative AI, genAI) tools may not be used to generate content [...]." ... That said, ChatGPT essentially ran with my suggestion, so I guess I'm happy that it saved me the trouble of typing-up a full solution. :) ... That said, you commented "I can't use [the suggestion ... s]ince I haven't developed my skills in using parameters yet". This suggests that you're relying solely on an LLM for the content of this answer, which is inappropriate. $\endgroup$ Commented Apr 25 at 7:08
  • $\begingroup$ BTW: It's not necessary that $a$, $b$, $c$, $d$ be positive. (That condition forces the quadrilateral to be convex, but the result —and the generated proof— holds more generally.) $\endgroup$ Commented Apr 25 at 7:11
  • $\begingroup$ Out of curiosity: What prompt(s) did you provide ChatGPT? $\endgroup$ Commented Apr 25 at 7:17
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    $\begingroup$ I gave him the link to the question and asked him to understand and explain it. Then I asked him to solve it in whatever way he saw fit and then write it in a format ready to paste into math.stackexchange.com, with easy-to-copy formatting and a dollar sign ($) before and after each equation. I did this in Arabic. @blue $\endgroup$ Commented Apr 25 at 7:22
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    $\begingroup$ This is the link to the conversation if you are interested chatgpt.com/share/69ec6d96-6fd0-83eb-a956-337163f901cb @Blue $\endgroup$ Commented Apr 25 at 7:31
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Now, by experimenting with Geogebra, I've come up with a relationship around the angles, which I think will bring us a lot closer to the solution.

$\angle{DAB}-\angle{BCD}=\angle{O_DO_AO_B}-\angle{O_BO_CO_D}$

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