"The thing must be seen all at once, at a glance, and not as a result of progressive reasoning, at least up to a point." -Blaise Pascal in Pensées (1966, p. 211)
In 1639, at the age of 16, Pascal proposed his "mystic hexagon theorem" (Burton, 2007). The theorem posits that, given a hexagon inscribed in a circle, the three intersections of its three pairs of opposite sides will lie on the same line. Because the theorem is a projective theorem, it generalizes to all conic sections, such as ellipses and hyperbolas.
The illustration above shows the case of a convex hexagon inscribed in a circle, where P1, P2, and P3 are the three points of intersection from the three pairs of opposite sides of the hexagon inscribed in the circle (in blue). This figure was adapted from a relatively simple proof, given by Yzeren (1993), linked below. The proof involves the construction of a second (black dashed) circle whose introduction is not so intuitive.
Here, we'll prove Pascal's theorem transformationally, as inspired by Felix Klein's Erlangen Program (see Norton, 2019). Our proof will rely on two two-chords theorems, which we will also prove transformationally.
Two Two-Chords Theorems Suppose that two chords intersect within a circle, as illustrated in the app below. What can be said about the angles and segments formed by their intersection. As is turns out, a lot! The first two chords theorem states that the pairs of vertical angles formed by the chords are measured by the average of the two arcs that subtend them. For example, in the app below, one pair of vertical angles at P is measured by the average of the violet and brown acs, and the other pair is measured by the average of the blue and green arcs.
To prove this, we can translate the point P--along with the two chords that define it and the two pairs of vertical angles--to the center of the circle, at O. Check the "Show Chords" and "Show Values" boxes, and try dragging P to O. There, we know that each of the angles is measured by the arc that subtends it. In performing the translation, the angles don't change, but the arcs do. By the symmetry of the circle, we can see that what is taken from one arc is added to the opposite arc, until they each reach their average, when P reaches O.
The two chords theorem implies the inscribed angles theorem as a special case. The inscribed angles theorem is just the case where one of the arc lengths goes to zero, in which case the angle measure is half of the remaining arc (in turn, Thales' theorem, about the right angle formed by subtending the diameter of a circle, is just a special case of the inscribed angles theorem). Moreover, we can use the inscribed angles theorem to prove another two chords theorem.
The second two-chords theorem (the intersecting chords theorem) states that the chords form pairs of proportional segments. For example, click on the "Show Segments" box and consider the four segments, a, b, c, and d. The theorem states that a*d=b*c. To see this, we can draw two new dashed segments (as shown). By vertical angles and the inscribed angles theorem, we have two similar triangles, which have proportional sides a:d = c:b, or a*b=c*d.
Note that the two chords theorem appears as Proposition 35 in Book III of Euclid's Elements, as illustrated by David Joyce's excellent website. You can see another illustration of its proof at cut-the-knot. Here, we will use the two two-chords theorems to support an intuitive and transformational proof of Pascal's theorem.
The figure below illustrates the (self-intersecting) hexagon ABCDEF inscribed in a circle. Note that the three pairs of opposite sides intersect at P1, P2, and P3, which lie on the same (orange) line. We will prove the theorem by starting with the sides CD and FA removed, then showing that they intersect on the line defined by P1 and P2.
Two Pairs of Chords Note that the figure above includes two pairs of intersecting chords: AB & DE, and BC & EF. By the intersecting chords theorem, each pair of intersecting chords creates pairs of proportional sides. As labeled in the figure below, we have af=be and ch=dg. Thus, (d/a)(g/f)=(h/e)(c/b). Next, we'll consider what this equality implies for a transformation of triangles through the mystic hexagon.
Two Transformations In the two apps below, a (dashed) segment has been added to the left side of the hexagon by connecting opposite vertices, A and B, crossing the orange line at O. Point Q1 is the intersection of a ray through O--rotated clockwise from the orange line, through the blue angle (DEF)--with hexagon side DE. Point R1 is the intersection of a ray through P1--rotated clockwise from the orange line, through the blue angle (DEF)--with segment AD.
The first transformation rotates triangle DR1P1 through angle DEF, with a corresponding translation so that point D maps to point F, and then a dilation from point F by a factor of h/e. You can carry out this first transformation by using the blue slider (rotation and translation) followed by the orange slider (dilation). This pair of actions transforms segment R1P1 to a segment along the orange line, extending left from P2 to the point where hexagon side FA (pink segment) crosses the orange line.
For the second transformation, we refer to corresponding points Q2 and R2--constructed like Q1 and R1, but with respect to angle ABC rather than angle DEF. This transformation rotates triangle AR2P1 through angle ABC, with a corresponding translation, and then a dilation from point C by a factor of d/a. Like the first transformation, this second transformation will map segment R2P1 to a segment along the orange line, extending left from P2 to the point where hexagon side CD (purple segment) crosses the orange line.
Our claim is that the two transformed segments have the same length, and therefore, hexagon sides CD and FA intersect on the orange line defined by P1 and P2.
Proof of Claim Triangle P1OQ1 is similar to triangle P1EP2, and the scale factor between them is given by respective side lengths P1Q1 and P1P2. If we choose OP1 to have unit length, 1, and if P1Q1 has length j, then OQ1 will have length c/b and P1P2 will have length bj.
Now R1P1 is OQ scaled by a factor of e/(e-j), so it has length [e/(e-j)](c/b). The corresponding length for the other transformation will be [a/(a-k)](g/f), where k corresponds to j. The claim is that, when these two lengths are transformed (by the respective transformations described above), they will be equal.
The first transformation maps R1P1 to the orange line, with length [e/(e-j)](c/b)(h/e). The second transformation maps the corresponding segment, R2P1, to the orange line, with length [a/(a-k)](g/f)(d/a). Recalling that ch=dg, be=af, and bj=fk, a little algebra shows that the two transformed lengths are the same:
Closing Transformational proofs like this one illustrate the dynamic nature of mathematics, hidden within more formal arguments. They elucidate mathematics as a product of our own mental actions (Piaget, 1970).
References
Burton, D. M. (2007). The history of mathematics: An introduction. New York: McGraw-Hill.
Norton, A. (2019). An Erlangen Program That Empowers Students' Mathematics. For the Learning of Mathematics, 39(3), 22-27.
Pascal, B. (1963). Essai pour les coniques. Éditions du Seuil.
Pascal, B. (1966). Pensées (Trans. A. J. Krailsheimer). London: Penguin.
Piaget, J. (1970). Structuralism (C. Maschler, Trans.). New York: Basic Books (Original work published 1968).
Yzeren, J. (1993). A simple proof of Pascal's Hexagon Theorem. American Mathematical Monthly, 100(10), 930-931.