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Projection of a D4
Above is the general projection of the 12 points of the D4 configuration (also known as the Reye configuration) into a plane. You can control the five blue points and you can slide the green point along a line. This set of 12 points is a complete intersection because it is the intersection of a cubic curve (purple) and a quartic curve (orange, the union of four lines). This makes the D4 configuration a (3,4) geproci half grid.
The Geiser Involution Above is a demonstration of the Geiser involution on the plane, which uses the Cayley-Bacharach Theorem. Under the involution, the points P and Q map to each other.
You can control the orange line by dragging the orange point. The Unexpected Quartic Above is a demonstration of the unique (up to isomorphism) unexpected quartic in characteristic 0. Points 1, 2, 3, 4, and 10 are click-and-draggable, and point 10 will always be a triple point no matter where you drag it! (Although it may not always appear that way because non-real slopes don't render in Desmos.) You can read about its construction and uniqueness in the 2019 paper by Farnik, Galuppi, Sodomaco, and Trok.
Linked is an illustration of Pascal's theorem, which can be proven using the Cayley-Bacharach theorem. The points A1, A2, A3, B1, B2, and B3 are six points on a conic. The two orange lines connect A1 with B2 and A2 with B1. The intersection of those orange lines is X. The two purple lines connect A1 with B3 and A3 with B1. The intersection of those purple lines is Y. The two green lines connect A2 with B3 and A3 with B2. The intersection of those green lines is Z. No matter where you move the six points A1, A2, A3, B1, B2, and B3, the points X, Y, and Z will always be collinear! Centers of a Triangle A triangle's medicenter (green), circumcenter (blue), and orthocenter (purple) are always collinear, lying on the orange line. The center of the inscribed circle (dotted in red) is in general not collinear with the other three points. Monge's Theorem Each pair of circles (with neither properly contained in the other) share two external bitangents (a line that is tangent to both but does not cut between the two). For each pair of circles, the intersection of their two external bitangents is their external homothetic center. Given any three circles, the three external homothetic centers given by each pair are always collinear! You can control the circles by clicking and dragging the points to change their centers and radii. |
The Bertini Involution
The above is an example of the Bertini involution of the plane, which uses the group law of elliptic curves. Under the involution, the point P maps to the point Q. It begins by taking an elliptic fibration with nine base points and choosing one of the base points as the identity element under the group law for all of the curves in the fibration. This point is marked as O.
Then when P is not one of the nine base points, there is a unique elliptic curve that contains all nine base points and P. Then P is mapped to -P under a group homomorphism on the curve (which is why we have an involution: if P goes to -P, then -P goes to -(-P)=P). To show P+Q=O, we can connect P and Q with the green line, and label the third point where the green line meets the curve as R. Then to find P+Q, we connect R to the identity O with the orange line. This new line is tangent to the curve at O, and so the "third" point where the orange line meets the purple curve is O itself. This shows that P+Q=O, and so we truly have Q=-P. When P is O, the image of P is P. When P is one of the eight base points that are not O, the image of P is not defined. The Bertini involution is a rational transformation of the plane, and is just defined on a dense open subset of the plane. To have it be defined everywhere, you would have to blow up the eight points where it not defined. This one was really hard to make. I followed the construction outlined in these papers. Five Points Make a Conic The above is a demonstration of five points on the plane uniquely determining a conic. Three Points and Two Tangents Make Four Conics The above is a demonstration of choosing three points in the plane and two lines to become tangent lines. This yields exactly four conics. Note that sometimes the conics might disappear because their polynomials need complex coefficients. You can read about its construction here. You can control the orange and purple lines with the orange and purple points. Desargues's Theorem The above is a demonstration of Desargues's theorem. Observe the two triangles ABC and abc. Desargues's theorem says the two triangles are in perspective centrally (meaning that the lines Aa, Bb, and Cc meet at the same point, here called O) if and only if they are in perspective axially: the points X (where BC meets bc), Y (where AC meets ac), and Z (where AB meets ab) are collinear. A space where Desargues's Theorem holds is called Desarguesian. Any plane that embeds in a three-dimensional space is Desarguesian. An example of a non-Desarguesian space is the Moulton plane. Nine General Points Make a Cubic The above is a demonstration of nine general points uniquely determining a cubic curve. I specify that the points are general because the Cayley-Bacharach Theorem says that given eight points in the plane, there is a single ninth point such that all cubics that contain the first eight will also contain the ninth. In that case, the nine points do not uniquely determine a cubic, but rather a pencil (a one-dimensional family) of cubics. Rolling Parabola The path traced by the focus of a parabola as it rolls across the x-axis is a catenary. Specifically, the parabola y=x2 yields the catenary y=cosh(4x)/4. So that's pretty neat I would say. |