Pascal's theorem states that if six arbitrary points are chosen on a conic and joined by line segments in any order (here we choose A→B→C→D→E→F) to form a hexagon, then the three pairs of opposite sides (AB DE, BC EF and CD AF) of the hexagon meet at three points (G, H and I) which lie on a straight line.
Unlike the analytic geometry proof of butterfly theorem where equations can be simplified by Vieta's formulas, most theorems in projective geometry are too complicated to prove by analytic geometry by hand. Instead, we use SymPy to do most calculations.
Here are some simple cases:
- harmonic conjugate
- harmonic conjugate of pole-polar and converse for a circle
- Menelaus's theorem (and its converse) and Ceva's theorem (and its converse)
- Desargues's theorem (and its dual) and Pappus's theorem (and its dual)
- Pascal's theorem and Brianchon's theorem for a circle, and their quadrilateral form
- Brokard's theorem
- butterfly theorem, an analytic geometry proof without Vieta's formulas
However, it is more complicated to prove Pascal's theorem for a conic in a direct way, which means we don't reduce the conic to a circle by a projective transformation.
Let's put point I onto the origin, rotate the hexagon to make BE parallel to x-axis, denote the conic ADBFCE as:
and denote AF, CD and BE as:
We get two roots by solving the combination of a conic and a straight line. However, SymPy won't tell us whether the first root denotes the point in the left or right side. So we need to guess then verify by numerical evaluation.
Assume the conic is and the line AF is
, we should get A in quadrant I and F in quadrant III. So we should use
F, A = solve(...) but not A, F = solve(...). Then we get 6 points:
Without further simplification, SymPy can hardly solve the intersections G and H. (This may be due to too many fraction calculations.[1] I don't know if Mathematica or other alternatives can do this.) So we need to replace all square roots with:
Then the 6 points are simplified as:
Then we get AB, DE, BC and EF, and their intersections G and H, and the expression to check if G, H and I are collinear.
The numerator of this expression contains 422 terms, where so many P, Q and R appear. We replace them back with Eq. 2, then get the final result 0, which means G, H and I are collinear.
Here is the proof process.
Braikenridge-Maclaurin theorem is the converse to Pascal's theorem.
We use the diagram of Pascal's theorem and put I onto the origin again. But here we rotate the hexagon to make GH onto y-axis, and denote 6 lines as:
Let's assume the conic doesn't go through origin I, then we need to prove the 6 points:
lie on the a conic.
According to this rule, we just need to prove:
Here is the proof process.[2]
Because homogeneous coordinates have many advantages mentioned here, we can use SymPy to prove it in a very simple process.
Here we change all Cartesian to homogeneous from Eq. 1 to avoid fraction calculations.
Brianchon's theorem states that when a hexagon (marked as ad-db-bf-fc-ce-ea) is circumscribed around a conic section (where the 6 tangent points are ADBFCE), its principal diagonals (ad-fc, db-ce and bf-ea) meet in a single point.
We still use the 6 points in Eq. 1 but this time we draw their 6 tangent lines to form a hexagon.
We apply derivative of implicit function on curve , then the tangent line passing through point
is:
When the conic is denoted as , we have
and
, so the tangent line is:
(Note that .)
In this proof, each point is denoted as homogeneous coordinate , then the tangent line is:
We start from these 6 tangent lines, then get the 6 vertices of the hexagon, then prove the concurrency of its principal diagonals.
A conic can be represented as an equation in homogeneous coordinates:
If a point P(x,y,z) is on this conic, it should follow:
So we can rewrite the point as:
Either of the two roots can be used in the proof. For example, we can apply positive roots on points ABCD, and apply negative roots on points EF.
The proof process is very similar to Pappus's theorem, and proves both Pascal's theorem and Brianchon's theorem, because the conic equation can represent a point conic (where ABCDEF are 6 points) or a line conic (where ABCDEF are 6 straight lines).
We choose 3 arbitrary points A, C and E as 3 of 6 vertices on a hexagon, and set 3 collinear points GHJ. Put E onto x-axis and GHJ onto y-axis to reduce variables, then we have:
Then we can calculate the other 3 of 6 vertices:
After some calculations here, we get:
which implies all 6 vertices lying on a conic.
This process also proves its dual theorem (which is also the converse of Brianchon's theorem).