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$P_1, P_2, P_3, P_4$ are four arbitrary points on $xy = 1$. $P_1P_4$ intersects $P_2P_3$ at $D_1$, and similarly define $D_2, D_3$. Prove that $O, D_1, D_2, D_3$ are cyclical, where $O$ is the origin.

I have never seen a geometry problem asking to prove cyclical points on a conic section. So frankly not sure where to start. I suspect bashing coordinates would work, but that's quite a bit of efforts (and I am not sure how to go from coordinates to proving points are cyclical).

Is there some shortcut I could use by some geometric properties of $xy=1$?

enter image description here

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  • $\begingroup$ With the help of Mathematica, I verified the result via coordinate bashing; there must be a better way. That said, if the points are $A=(a,a')$, $B=(b,b')$, $C=(c,c')$, $D=(d,d')$, with $aa'=bb'=cc'=dd'=1$, then $$\overleftrightarrow{AB}\cap \overleftrightarrow{CD}=\left(\frac{(a'+b')-(c'+d')}{a'b'-c'd'},\frac{(a+b)-(c+d)}{ab-cd}\right)$$ Similarly for $\overleftrightarrow{AC}\cap\overleftrightarrow{BD}$ and $\overleftrightarrow{AD}\cap\overleftrightarrow{BC}$. So, there's some interesting coordinate structure, but it doesn't seem to make proving cyclicity particularly easy. $\endgroup$
    – Blue
    Dec 22 '19 at 5:45
  • $\begingroup$ @Blue that is indeed interesting. I wonder if it makes some complex number bashing easier. $\endgroup$
    – Chen Chen
    Dec 22 '19 at 14:06
  • $\begingroup$ Note that, by construction, $D_1,D_2,D_3$ form a self-polar triangle with respect to the given conic section, i.e. the pole $D_1$ has polar $D_2D_3$ etc. In a triangle coordinate system (e.g. barycentric or trilinear) based on the self-polar $D_1D_2D_3$, the symmetric $3\times3$ coefficient matrix of that conic section is diagonal. $\endgroup$
    – ccorn
    Dec 22 '19 at 17:18
  • $\begingroup$ @ccorn Mind to elaborate a bit or point me to more materials? sorry not even sure what "$D1$ has polar $D_2 D_3$" means.. $\endgroup$
    – Chen Chen
    Dec 22 '19 at 19:49
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Here's a proof without coordinates. I will relabel $D_1=P_1P_4\cap P_2P_3$, $D_2=P_2P_4\cap P_1P_3$ and $D_3=P_3P_4\cap P_1P_2$.

Lemma 1. The centre of any rectangular hyperbola $\mathcal H$ through points $A$, $B$, $C$ lies on the nine-point circle of $\triangle ABC$.

Proof. Let $(ABC)$ meet $\mathcal H$ again at $D$, and let $H_A$, $H_B$, $H_C$, $H_D$ be the orthocentres of $\triangle BCD$, $\triangle CDA$, $\triangle DAB$, $\triangle ABC$, which all lie on $\mathcal H$. There is a $180^{\circ}$ rotation mapping $ABCD$ to $H_AH_BH_CH_D$, and the centre of this rotation is both the centre of $\mathcal H$ and the midpoint of $\overline{DH_D}$, so it lies on the nine-point circle of $\triangle ABC$. $\square$

Lemma 2. The incentre $I$ and excentres $I_1, I_2, I_3$ (opposite $D_1$, $D_2$, $D_3$) of $\triangle D_1D_2D_3$ lie on the hyperbola.

Proof. Take a projective transformation sending $P_1P_2P_3P_4$ to a square. Then $D_1$ and $D_3$ go to infinity, so $I_1I_2I_3I$ becomes a rectangle with sides parallel to $P_1P_2P_3P_4$. Then $D_2$ maps to the common centre of rectangle $I_1I_2I_3I$ and square $P_1P_2P_3P_4$, so these eight points lie on a common conic. $\square$

Now $(D_1D_2D_3)$ is the nine-point circle of $\triangle I_1I_2I_3$, so our two lemmas imply the result.

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The following is definitely not a shortcut, but gives a self-contained reasoning (with the help of computer algebra system). Note that rectangular hyperbolas (the one with axes perpendicular to each other, or of eccentricity $\sqrt{2}$) are well studied due to its applications in other areas such as the study of centers of a given triangle. Examples include Feuerbach, Jerabek, and Kiepert hyperbolas, among others. The argument below is based on coordinates and all the computation is straightforward. A byproduct is that this also proves the Feuerbach's Theorem as a corollary. Note that since all rectangular hyperbolas are similar (being of the same eccentricity), we can use the one with equation $xy=1$ as our model without loss of generality.

Theorem. Let $P_i,i=1,\cdots,4$, be four points on a rectangular hyperbola. Let $A_i=\{P_1,P_{1+i}\},B_i=\{P_1,\cdots,P_4\}\setminus A_i,i=1,2,3$. Denote by $\ell(A_i)$ (resp. $\ell(B_i)$) the line joining the two points in $A_i$ (resp. $B_i$). Let $D_i:=\ell(A_i)\cap\ell(B_i)$. The the four points $D_1,D_2,D_3$ and the center $O$ of the rectangular hyperbola are concyclic.

Proof. As mentioned above, we use $xy=1$ as our model. Let $P_i=(t_i,1/t_i), i=1,\cdots 4.$ Then solving linear equations yields the following coordinates for $D_i=(x_i,y_i),i=1,2,3,$ where $$x_1=\frac{t_1t_2t_3+t_1t_2t_4-t_1t_3t_4-t_2t_3t_4}{t_1t_2-t_3t_4},\qquad y_1=\frac{t_1+t_2-t_3-t_4}{t_1t_2-t_3t_4}$$ $$x_2=\frac{t_1t_2t_3+t_1t_3t_4-t_1t_2t_4-t_2t_3t_4}{t_1t_3-t_2t_4},\qquad y_1=\frac{t_1+t_3-t_2-t_4}{t_1t_3-t_2t_4}$$ and $$x_3=\frac{t_1t_2t_4+t_1t_3t_4-t_1t_2t_3-t_2t_3t_4}{t_1t_4-t_2t_3},\qquad y_1=\frac{t_1+t_4-t_2-t_3}{t_1t_4-t_2t_3}.$$ Now to prove that four points ($D_1,D_2,D_3,D_0$ with $D_0=(x_0,y_0)$ ) are concyclic, it suffices to check the vanishing of the following determinant (this can be seen by working with an equation for a circle), namely $$\left|\begin{array}{cccc}x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ x_0^2+y_0^2&x_0&y_0&1\end{array}\right|=0,$$ where since $D_0=O=(0,0)$ (the center of the hyperbola), the result is reduced to showing $$\left|\begin{array}{cc}x_1^2+y_1^2&x_1&y_1\\ x_2^2+y_2^2&x_2&y_2\\ x_3^2+y_3^2&x_3&y_3\end{array}\right|=0,$$ which is true as can be checked directly by a computer algebra system (for example, in SAGE, use the factor command: it returns that a zero cannot be factored). QED

Lemma. For any triangle $P_1P_2P_3$ inscribed in a rectangular hyperbola $\mathcal{C}$, the orthocenter $H$ of $P_1P_2P_3$ lies on $\mathcal{C}$.

Proof. Without loss of generality, use the model $xy=1$ and the same parametrizations of $P_i, (i=1,2,3)$ as above. Then by direct computation, the orthocenter is given by $$H=(-1/(t_1t_2t_3),-t_1t_2t_3),$$ which indeed lies on $\mathcal{C}.$ QED

Corollary. (Feuerbach's Theorem) For a triangle $ABC$ inscribed in a rectangular hyperbola $\mathcal{C}$, its nine-point circle passes through the center of the hyperbola.

Proof. Given an inscribed triangle $ABC$ in $\mathcal{C}$, by the above lemma, the orthocenter $H$ of $ABC$ lies on $\mathcal{C}$. Now let $A,B,C,H$ be the four points $P_i$'s as in the Theorem. It is clear that the three points $D_i$'s correspond to the three feet of altitude from $C,B$ and $A$, therefore they lie on the nine-point circle of $ABC$. Since by the Theorem, this circle passes through $O$, the result is clear. QED

Remarks. The following gives some references without details (please look up undefined terms).

  1. In "Projective Geometry" by H.S.M. Coxeter, the author says "If $4$ points in a plane are joined in pairs by $6$ distinct lines, they are called the vertices of a complete quadrangle, and the lines are its $6$ sides." And a result on polarity induced by a conic is the following (See 8.21 of the same reference): If a quadrangle is inscribed in a conic, its diagonal triangle is self-polar.

  2. In "Introduction to Plane Geometry" by H.F. Baker (1943), it was mentioned in Ex. 12 on page 158 that "..., if $PQR$ be a self-polar triangle in regard to the rectangular hyperbola, the circumcircle of $PQR$ contains the center $C$" (of the hyperbola). The proof was given (but not mentioned here because it depends on other results).

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Here is an outline of a proof that still uses coordinates, but it is high-level enough to make the core of the proof an easy rearrangement without lengthy calculation.

For brevity, I rename the triangle $D_1D_2D_3$ to $ABC$. Its side lengths are denoted $a,b,c$. The rectangular hyperbola is named $K$ here.

I use barycentric coordinates with reference triangle $ABC$. I also presume a projective geometry context; that is, I will refer to points at infinity and the line at infinity. For the latter, I use the symbol $L_\infty$ and remark that in barycentric coordinates,

  • $L_\infty$ has coefficient tuple $[1:1:1]$;
  • the incenter of $ABC$ is at $(a:b:c)$; flipping the sign of one of these coordinates yields an excenter;
  • the points $(u:v:w)$ on the circumcircle of $ABC$ are those fulfilling $a^2 v w + b^2 w u + c^2 u v = 0$.

Notation: I use round parentheses for point coordinates and square brackets for coefficients of lines and curves. Both kinds are occasionally represented in matrix form without the $:$ separator; in such cases, I use $\cong$ to indicate linear dependence between two such matrices.

Some of the reasoning steps below may be regarded as reminders or observations left as exercises.

  1. Given $P_1=(u_1:v_1:w_1)$, then (a permutation of) $P_2,P_3,P_4$ can be given as $P_2=(-u_1:v_1:w_1)$, $P_3=(u_1:-v_1:w_1)$, $P_4=(u_1:v_1:-w_1)$. This has nothing to do with $K$; it follows from the construction of $ABC$.

  2. In barycentric coordinates, $K$ has its homogeneous equation in diagonal form $K(u,v,w) = Ru^2 + Sv^2 + Tw^2 = 0$ with nonzero coefficients $R,S,T$. (Nonzero because $K$ is not degenerate.) This diagonality follows from item (1) and $K(P_i) = 0$ or alternatively from the fact that the reference triangle $ABC$ is, by construction, self-polar with respect to $K$; thus the pole $A=(1:0:0)$ has associated polar $\overline{BC}=[1:0:0]$ etc.

  3. The center $O=(u_0:v_0:w_0)$ of $K$ is the pole of the line at infinity, thus:

    $$\begin{pmatrix}u_0\\v_0\\w_0\end{pmatrix} \cong \begin{bmatrix} R & 0 & 0 \\ 0 & S & 0 \\ 0 & 0 & T \end{bmatrix}^{-1} \begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix}$$

  4. As a hyperbola, $K$ has two distinct points at infinity, call them $E_1=(\hat{u}_1:\hat{v}_1:\hat{w}_1)$ and $E_2=(\hat{u}_2:\hat{v}_2:\hat{w}_2)$. Thus $\overline{E_1E_2}=L_\infty$, which in barycentric coordinates means

    $$\begin{pmatrix}\hat{u}_1\\\hat{v}_1\\\hat{w}_1\end{pmatrix} \times\begin{pmatrix}\hat{u}_2\\\hat{v}_2\\\hat{w}_2\end{pmatrix} = \begin{bmatrix}\hat{v}_1\hat{w}_2-\hat{w}_1\hat{v}_2 \\\hat{w}_1\hat{u}_2-\hat{u}_1\hat{w}_2 \\\hat{u}_1\hat{v}_2-\hat{v}_1\hat{u}_2\end{bmatrix} \cong \begin{bmatrix}1\\1\\1\end{bmatrix}$$

  5. Since both $E_i$ lie on $K$, we must have

    $$\begin{aligned} \begin{bmatrix}R\\S\\T\end{bmatrix} &\cong \begin{pmatrix}\hat{u}_1^2\\\hat{v}_1^2\\\hat{w}_1^2\end{pmatrix} \times\begin{pmatrix}\hat{u}_2^2\\\hat{v}_2^2\\\hat{w}_2^2\end{pmatrix} = \begin{bmatrix}\hat{v}_1^2\hat{w}_2^2-\hat{w}_1^2\hat{v}_2^2 \\\hat{w}_1^2\hat{u}_2^2-\hat{u}_1^2\hat{w}_2^2 \\\hat{u}_1^2\hat{v}_2^2-\hat{v}_1^2\hat{u}_2^2\end{bmatrix} \\&= \begin{bmatrix}(\hat{v}_1\hat{w}_2-\hat{w}_1\hat{v}_2)(\hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2) \\(\hat{w}_1\hat{u}_2-\hat{u}_1\hat{w}_2)(\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2) \\(\hat{u}_1\hat{v}_2-\hat{v}_1\hat{u}_2)(\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2)\end{bmatrix} \\&\cong \begin{bmatrix}\hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2 \\\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2 \\\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2\end{bmatrix} \text{ using item (4)} \end{aligned}$$

  6. Since points at infinity represent slopes of lines, it makes sense to ask whether two given points at infinity are orthogonal. This can be answered if the equation of a circle is given: $E_1,E_2$ (at infinity) are orthogonal if and only if $E_1$ lies on the polar of $E_2$ with respect to the circle. Concretely, given the symmetric bilinear form $U$ of any circle, then $E_1,E_2$ are orthogonal if and only if $U(E_1,E_2) = 0$.

  7. $K$ is a rectangular hyperbola, therefore $E_1,E_2$ must be orthogonal. Using item (6) with the symmetric bilinear form of the circumcircle $U$ of the reference triangle, we have

    $$\begin{align} 0 &= U(E_1,E_2) = \begin{pmatrix}\hat{u}_1\\\hat{v}_1\\\hat{w}_1\end{pmatrix}^\top \begin{bmatrix} 0 & c^2 & b^2 \\ c^2 & 0 & a^2 \\ b^2 & a^2 & 0 \end{bmatrix} \begin{pmatrix}\hat{u}_2\\\hat{v}_2\\\hat{w}_2\end{pmatrix} \\ &= \begin{pmatrix}a\\b\\c\end{pmatrix}^\top \begin{bmatrix} \hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2\hspace{-1em} & 0 & 0\quad \\ \quad 0 & \hspace{-1em}\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2\hspace{-1em} & 0\quad \\ \quad 0 & 0 & \hspace{-1em}\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2 \end{bmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix} \\ &\cong \begin{pmatrix}a\\b\\c\end{pmatrix}^\top \begin{bmatrix} R & 0 & 0 \\ 0 & S & 0 \\ 0 & 0 & T \end{bmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix} \tag{I} \\ &= \frac{RST}{2} \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix}^\top \begin{bmatrix} 0 & c^2 & b^2 \\ c^2 & 0 & a^2 \\ b^2 & a^2 & 0 \end{bmatrix} \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix} \tag{II} \end{align}$$

Now (II) tells us that the center $O$ of $K$ lies on the circumcircle of $ABC$. Furthermore, (I) tells us that $K$ contains the incenter $(a:b:c)$ of $ABC$, and we can generalize this to $K$ containing $(\pm a:\pm b:\pm c)$ which are the in- and excenters of $ABC$. (This is fitting: The in- and excenters form an orthocentric system.)

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