# cyclical points on a conic section

$$P_1, P_2, P_3, P_4$$ 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 $$xy=1$$?

• 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.
– Blue
Dec 22 '19 at 5:45
• @Blue that is indeed interesting. I wonder if it makes some complex number bashing easier. Dec 22 '19 at 14:06
• 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. Dec 22 '19 at 17:18
• @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.. Dec 22 '19 at 19:49

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.

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).

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.)