The question: Five points define a conic, so let points $A,B,C,D,E$ and $F,G,H,I,J$ define two conics $c$ and $d$. Is there a geometric construction that identifies the common points of $c$ and $d$?

(It's easy enough to do it in Geogebra, or algebraically. What I'm asking for here is a ruler and compass construction.)

Background and context Given, say, an ellipse $c$ it is not too difficult to construct tangents from a point $P$ to $c$ using just a straightedge. But this assumes that the ellipse has been drawn as a given. If only the points $A,B,C,D,E$ on $c$ have been given, you can easily construct the polar of $P$ but then have to construct the intersection of the polar and the ellipse. This is less straightforward but the construction is described in several 19th century projective geometry texts (e.g. Cremona or Russell),and requires a reference conic (or compass) in addition to a straightedge.

What I haven't been able to find is a construction for the intersection of two ellipses/conics that are not pre-drawn but given only as two sets of five points each. Hence my question. It came up in following a certain construction for "imaginary" chords that involved finding the common points of two conical loci (Russell, XXVII 6.).

Update: Given two or three points of intersection it is possible to construct the remaining points of intersection.

Update 2: This problem is discussed in Veblen and Young, Projective Geometry, Vol I, $\S102$.

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    $\begingroup$ If I remember correctly (please correct me if I'm wrong) the intersection of two conics, however given, cannot be constructed (in general) with straightedge and compass, because finding common points to two conics involves quartic equations. $\endgroup$ Commented Aug 8, 2019 at 7:28

2 Answers 2


Consider the following two sets of points: $$\{(-2,4),(-1,1),(0,0),(1,1),(2,4)\}$$ $$\{(2,2),(2,-2),(4,0),(4/5,8/5),(4/5,-8/5)\}$$ All these points are rational, so there is nothing that cannot be constructed by compass and straightedge alone that could potentially make the unconstructible constructible. The first set defines the parabola $y=x^2$ and the second defines the circle $(x-2)^2+y^2=4$. They meet in two places; one is the origin $(0,0)$ and the other is $$\left(\alpha-\frac1{3\alpha},\alpha^2-\frac23+\frac1{9\alpha^2}\right)$$ where $\alpha=\sqrt[3]{2+\frac{\sqrt{327}}9}$. The minimal polynomials of the coordinates of the second point of intersection are cubic, which means this second intersection cannot be constructed with compass and straightedge alone.

Thus the desired construction does not exist in general. Finding the coefficients of the implicit equations of the conics is possible, however, since it only requires basic linear algebra.


As a simpler counterexample, consider the parabola $y=x^2$ and the circle $y^2=x(1-x)$. Both have infinitely many constructible points and they intersect at the origin. But the second point of intersection requires solving $x^4=x(1-x)$ with $x$ nonzero, leading to the irreducible equation $x^3+x-1=0$ whose real root cannot be (Euclidean) constructed.


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