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I know how to construct an ellipse inscribed into a triangle, given its perspector or its center. Suppose that on each sideline of a triangle we are given a point which is not a triangle's vertex, and that the three given points satify the Ceva's condition (this is the same as if we were given the common point of the three cevians from each of the vertices to the given point on the opposite sideline). There exists a (unique) nondegenerate inconic of the triangle which touches its sides at the given points. Assuming that this inconic is a hyperbola, how we construct it? Equivalently, how we find the foci of the hyperbola, or its asymptotes (once we have one of these, it is easy to finish the construction).

The following is an uninspired "construction" which is not what I am looking for. Apply to the given situation a projective transformation that maps the vertices of the triangle to the vertices of an equilateral triangle, and maps the perspector (the common point of the cevians) to the center of the equilateral triangle. Now the inconic is the inscribed circle of the equilateral triangle; apply to it the inverse projective transformation.

Added on May 4, 2018

Below you'll find the proof of correctness of the construction of asymptotes which I described in the third comment to the answer by Aretino.

Referring to the figure in the Aretino's answer, we choose the Cartesian coordinate system with
the origin at the center $O$ of the hyperbola, and with the first axis the line connecting the origin
to $N$ (which is different from $O$, see below), and directed from $O$ towards $N$.

Let $n:=|NO|$, $d:=|NC|$, and $(u,v):=Q-N=N-R$, so that $O=(0,0)$, $N=(n,0)$, $C=(n-d,\,0)$, $Q=(n+u,\,v)$, and $R=(n-u,\,-v)$. We have $n>0$ since the tangents $CQ$ and $CR$ of the hyperbola intersect; $n=0$ is possible only if $Q$ and $R$ are diametrically opposite points of the hyperbola, and then the hyperbola's tangents at $Q$ and $R$ are parallel. Either $d<n$ when the two lines $CQ$ and $CR$ are tangents of the same branch of the hyperbola, or $d>n$
when the two lines are tangents of different branches, but never $d=n$ since the hyperbola is nondegenerate. Also note that $v>0$ $($or $v<0$ if the points $Q$ and $R$ are swapped$)$.

We seek the equation $f((x,y))=1$ of the hyperbola, where $f((x,y))=\alpha\, x^2+2\,\beta\, x\, y+\gamma\, y^2$,
and hence $df((x,y),(dx,dy))=\alpha\, x\, dx+\beta\,(x\, dy+y\, dx)+\gamma\, y\, dy$ $($actually this $df(X,\,dX)$
is half the differential of $f\,$$)$. Solving the system of equations \begin{equation*} f(Q)=1~,\quad f(R)=1~,\quad df(Q,\,Q-C)=0 \end{equation*} for $\alpha$, $\beta$, $\gamma$, we get \begin{equation*} \alpha=\frac{1}{n\,(n-d)}~,\quad \beta=-\,\frac{u}{n\,(n-d)\,v}~,\quad \gamma=\frac{u^2-d\,n}{n\,(n-d)\,v^2}~. \end{equation*} These values of $\alpha$, $\beta$, and $\gamma$ also satisfy the equation $df(R,\,R-C)=0$.

Now, with the values for the coefficients of $f$ in hand we solve $f((x,y))=0$ and get for $x:y$
two solutions $(u-\sqrt{d\,n}\,):v$ and $(u+\sqrt{d\,n}\,):v$. And that's it.

Once we have the asymptotes of the hyperbola, it is easy to construct its foci which lie on the bisector of the two vertically opposite angles between the asymptotes which contain the points
$Q$, $R$, and $S$. Choose any of the three given tangents, say the tangent $CQ$, and let this tangent intersect the asymptotes at the points $X$ and $Y$. The linear eccentricity of the hyperbola is then
the geometric mean of $|OX|$ and $|OY|$.

Added on May 5, 2018

I unearthed this ancient result. Suppose that $d<n$, so that both lines $CQ$ and $CR$ are tangents of the same branch of the hyperbola. Let the line segment $CN$ intersect the hyperbola in the point $W$; we know that the tangent of the hyperbola at $W$ is parallel to the line $QR$. Moreover, the positive solution of the equation $f((x,0))=1$ is $x=\sqrt{n(n-d)}$, which means that \begin{equation*} |OC|\cdot|ON|=|OW|^2~; \end{equation*} this is half of Proposition 37 in Book I of Conics by Apollonius of Perga, the other half being the same identity for an ellipse (which is very easy to prove).

Is it possible that the construction of asymptotes which I 'discovered' is equally ancient?

Added on May 8, 2018

I see now - everything is much simpler if we choose, instead of a Cartesian coordinate system, an affine coordinate system whose axes are aligned with conjugate diameters. So we let $O$ be the origin, then lay the first axis from $O$ through $C$ and $N$ and the second axis through $O$ parallel to the line $QR$. On each axis we choose a unit (any unit) for measuring coordinates independently of the unit on the other axis. The featured points have the coordinates $O=(0,0)$, $C=(c,0)$, $N=(n,0)$, $Q=(n,v)$, and $R=(n,-v)$, where $n$ and $v$ are positive, and $c$ is either negative or it is positive but different from $n$. In the spirit of Apollonius of Perga we also consider the case when the conic is an ellipse. All in all there are three cases:

  1. $c<0\,$: hyperbola with $Q$ an $R$ on different branches;
  2. $0<c<n\,$: hyperbola with $Q$ and $R$ on the same branch;
  3. $n<c\,$: ellipse.

This time we seek the conic's equation $\alpha\, x^2+\beta\, y^2=1$, with the equation $\alpha\, x_0\,x+\beta\, y_0\,y=1$ of the tangent at a point $(x_0,y_0)$ on the conic. From the equations \begin{equation*} \alpha\,n^2+\beta\,v^2=1~,\quad \alpha\,n\,c=1 \end{equation*} (where the first equation says that $Q$ and $R$ lie on the conic and the second equation says that the tangents to the conic at $Q$ and $R$ pass through $C\,$) we obtain, in each of the three cases:

  1. $c<0\,$: hyperbola $-x^2/g^2+y^2/h^2=1$ with $g=\sqrt{(-c)n}$, $h=v\sqrt{(-c)/(n-c)}$, and $h : g = v : \sqrt{n(n-c)}\,$;

  2. $0<c<n\,$: hyperbola $x^2/g^2-y^2/h^2=1$ with $g=\sqrt{c\,n}$, $h=v\sqrt{c/(n-c)}$, and still $h : g = v : \sqrt{n(n-c)}\,$;

  3. $n<c\,$: ellipse $x^2/g^2+y^2/h^2=1$ with $g=\sqrt{c\,n}$, $h=v\sqrt{c/(c-n)}$, and $h : g = v : \sqrt{n(c-n)}\,$.

Note that $g$ is homogeneous of degree $1$ in $n$ and $c$, and is homogeneous of degree $0$ in $v$, while $h$ is homogeneous of degree $0$ in $n$ and $c$, and is homogeneous of degree $1$ in $v$.

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  • $\begingroup$ @Aretino I added to my question the proof of the correctness of the construction of asymptotes which I described in a comment to your answer. The proof is so simple that somebody - perhaps quite many somebodys - have surely already stumbled over the construction and described it. $\endgroup$ – chizhek May 4 '18 at 10:32
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I would construct the hyperbola the same way as the ellipse. Let $P$, $Q$, $R$ be the tangency points, on sides $AB$, $BC$ and $CA$ respectively. If $M$ is the midpoint of $PQ$ and $N$ is the midpoint of $QR$, it is well known that lines $BM$ and $CN$ meet at the center $O$ of the conic (both for an ellipse and a hyperbola).

Once you know the center it is easy to find the rest of the conic. For instance, you can construct the symmetric points of $P$, $Q$, $R$ with respect to $O$ to have three more points on the conic.

In the case of a hyperbola, to find the asymptotes you can construct point $L$ on line $OM$ such that $RL\parallel PQ$ and compute $$ \alpha={RL^2OM^2-OL^2QM^2\over OM^2-OL^2}, \quad \beta={RL^2OM^2-OL^2QM^2\over QM^2-RL^2}. $$ Draw then on line $OM$ two points $T$ and $T'$ such that $OT=OT'=\sqrt{\beta}$, and on the line through $O$ parallel to $PQ$ two points $S$ and $S'$ such that $OS=OS'=\sqrt{\alpha}$. The parallels to the conjugate axes $TT'$ and $SS'$ through their endpoints form a parallelogram, whose diagonals are the asymptotes of the hyperbola.

enter image description here

EDIT.

As pointed out by chizhek a much simpler construction of the asymptotes is possible. It relies on two properties of any hyperbola: $$ {QM^2\over OS^2}-{OM^2\over OT^2}=1 \quad\hbox{and}\quad OM\cdot BM=OM^2+OT^2. $$ From these it follows that if we construct a point $U$ on line $OM$ such that $UM^2=OM\cdot BM$, then lines $PU$ and $QU$ are parallel to the asymptotes, because $UM/QM=OT/OS$.

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  • $\begingroup$ Not sure where the triangle is. Can't see point A. $\endgroup$ – Oscar Lanzi Apr 29 '18 at 21:00
  • $\begingroup$ Point $A$ is outside the diagram, at the intersection between $CR$ and $BP$. But it is not needed in the construction. $\endgroup$ – Aretino Apr 29 '18 at 21:07
  • $\begingroup$ Discovered a simple construction of asymptotes... Referring to your figure, let $d$ be the geometric mean of $NO$ and $NC$, and let $U$ be a point on the line $ON$ such that $NU=d$ (it does not matter on which side of the point $N$ the point $U$ lies). Then the asymptotes are parallel to the lines $UQ$ and $UR$. I suspect that this construction must be already known. If anybody, you will know whether it is part of the conics folklore or not. Let me know. Cheers. $\endgroup$ – chizhek May 2 '18 at 12:29
  • $\begingroup$ @Aretino I added a comment with a simple construction of asymptotes. I would appreciate some answer - any answer - from you about a possible provenance of this construction. I somehow cannot believe that it is not known, since it is so obvious (once you see it). $\endgroup$ – chizhek May 3 '18 at 13:27
  • $\begingroup$ @chizhek Sorry for the delay - I'm quite busy at the moment. As a matter of fact I don't remember such a construction for the asymptotes, but of course that doesn't imply it wasn't known in the literature: I'll check a couple of books when I have some time left. $\endgroup$ – Aretino May 3 '18 at 20:05

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