6. PASCAL’S THEOREM


Blaise Pascal (1623 – 1662) discovered this beautiful theorem when he was just 16. Although the result seems very simple, it is hard to see where to start with a proof.

In the figure at right, a hexagon AB'CA'BC' is inscribed in a circle. (The figure shows a rather zig-zag hexagon. This is not necessary, but it makes the figure neater.) Opposite edges of the hexagon meet in points L, M and N. What appears to be true about these points?

Try some hexagons of your own. Can you make a conjecture?

Check ...

Now check out this applet ...

Amazingly, this result appears to remain true if we replace the circle by an ellipse, or even by a pair of straight lines. In this last case, the result is usually known as Pappus’s theorem. Pascal’s theorem belongs to projective geometry – it really has nothing to do with lengths or angles, and this is why it is initially hard to see how we might prove it. It will take us several easy initial steps, but then the actual proof is very easy.

                                                                                                                                                            





























Cross Ratio on a Line

We begin be defining what appears to be a completely unrelated and unmotivated concept – cross ratio.
Let A, B; C, D be two pairs of points on a straight line with coordinates a, b, c and d respectively.
Then we define the cross ratio of the two pairs to be

               (A, B; C, D) = .

Alternatively, we can define the cross ratio in terms of the lengths of the segments, but obviously these need to be signed lengths, since for example, a – c is negative and d – b is positive. We might write this as

              (A, B; C, D) =      – a sort of ratio of ratios.

As an example, if A(1), C(2), B(3), and D(4), what is (A, B; C, D) ?   Check ...

Of course the cross ratio can be defined for any placing of the points on the line, not necessarily in the separated positions shown.

Why the interest in cross ratio? We shall see that this curious quantity has some important invariant properties: under certain simple transformations of the points the value of the cross ratio remains the same.

                                 
(A, B; C, D) =

  (1, 3; 2, 4)   = 
                
































Cross Ratio of a Pencil

Let a, b, c, d be four lines passing through point O, and draw an arbitrary line (transversal) not through O meeting them in points A, B, C, D as in the diagram. We prove:

Theorem 6.1 The value of the cross ratio (A, B; C, D) is independent of the position of the transversal. This value is then a characteristic of the four lines, and we can speak of the cross ratio of the pencil, (a, b; c, d).

Proof Draw through B a line parallel to a meeting c and d in points P, Q respectively, as in the diagram. We assume that lines a and PQ are directed in the same sense.
Triangles ACO, BCP are similar, so .
Similarly, triangles ADO and BDQ are similar, giving . Hence . It follows that the cross ratios of points on all transversals through the given point B are equal. In fact, this value doesn't depend on the actual position of B since the value on the line parallel to a remains constant for all positions of B: both numerator and denominator of the fraction are musltiplied by the same constant which then cancels out.

                                   










































A possible converse?

Using the adjacent diagram we have hown that the value of the cross ratio (A, B; C, D) is independent of the position of the transversal. We can prove an important partial converse result. Use this diagram.

Theorem 6.2 If A, B, C, D and A, B', C', D' are two sets of collinear points on lines meeting in A, and (A, B; C, D) = (A, B'; C', D'), then BB', CC' and DD' are concurrent in some point O. That is, we have two transversals of a pencil of lines.

Proof Let BB' and CC' meet in point O, and let OD meet line AB'C' in D".

Now (A, B'; C', D") = (A, B; C, D) (property of the pencil)
                                      = (A, B'; C', D') (given).

We conclude that D" = D'   , so O, D and D' are collinear as required. This completes the proof of the theorem.

                                   

The easiest way to see this is probably to use coordinates. With the obvious notaion we have



whence we obtain  d' = d".
































Proof of Pascal’s Theorem

We first observe that in the adjacent diagram, the green pencil and the yellow pencil are actually congruent – the angles between corresponding lines are equal, using the subtending property of chords in a circle. Obviously then theses two pencils have the same cross ratio.

Let’s now prove Pascal’s theorem. It’s easy! We use this diagram.

From our above argument, the two pencils A'(A, B, C, B') and C'(A, B, C, B') are congruent and so have the same cross ratio.

The value of this common cross ratio is the same as the cross ratio of the points on any transversals. So we get

           (A, N; U, B') = (V, L; C, B').

Now these two sets of points have common point B'. From Theorem 6.2 it follows that lines AV, NL and UC meet in a common point. This point is clearly M, showing that points L, M, N are collinear.
What a wonderful proof!                                                                                                                   






























P

Extensions

1. The cross ratio of four points on a line is worth investigating in its own right. For example, if the cross ratio is negative, what can you say about the two pairs of points? You should be able to show that the value of the cross ratio of four points remains unchanged when two pairs of points are interchanged.

2. Again, there are 24 possible cross ratios associated with four distinct points on a line. How many of these values can be distinct? Can you show that the 24 possible cross ratios generally yield just six (different) values: m, 1 – m, 1/m, 1/(1 – m), (m–1)/m, m/(m – 1).

3. Show that we need at least four points on a line to obtain a quantity which remains the same under projection (that is, changing the position of the transversal on the given pencil). Investigate properties of two points, three points.

4. Investigate Pascal’s theorem for special positions of the six given points. What happens when two points coincide (think of introducing a tangent)?

5. Show from Pascal’s theorem that the tangents to the circumcircle at the vertices of a given triangle meet the opposite sides in collinear points.

                                                                        Hints and Solutions ...

For looking up ...

http://en.wikipedia.org/wiki/Pascal's_theorem

http://en.wikipedia.org/wiki/Brianchon's_theorem     

Geometry for Advanced Pupils, Maxwell, E. A. (Oxford 1963)
                                                                                                                                                                           
                                                                                                                                                     


































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Hints and Solutions

1. The cross ratio of four points on a line is worth investigating in its own right. For example, if the cross ratio is negative, what can you say about the two pairs of points? You should be able to show that the value of the cross ratio of four points remains unchanged when two pairs of points are interchanged.

•  The two pairs separate one another. For example, show (A, B; C, D) = (B, A; D, C) by rearranging the formula for the second one.

2. Again, there are 24 possible cross ratios associated with four distinct points on a line. How many of these values can be distinct? Can you show that the 24 possible cross ratios generally yield just six (different) values: m, 1 – m, 1/m, 1/(1 – m), (m–1)/m, m/(m – 1).

•  24 = 4! For example, if (A, B; C, D) =  = m, then (A, B; D, C) = = 1/m.

3. Show that we need at least four points on a line to obtain a quantity which remains the same under projection (that is, changing the position of the transversal on the given pencil). Investigate properties of two points, three points.

•  The distance between 2 points on a transversal varies by changing the position of the transversal. With 3 points A, B, C you might hope for B lying between the A and C to be invariant, but it is easy to find an example where this fails.

4. Investigate Pascal’s theorem for special positions of the six given points. What happens when two points coincide (think of introducing a tangent)?

•  The theorem states that L = BC'.B'C, M = CA'.A'C, N = AB'.B'A are collinear. If we take B' = A, and interpret AA as the tangent to the circle at A we find that L = BC'.AC, M = CA'.A'C, N = AA.B'A are still collinear.

5. Show from Pascal’s theorem that the tangents to the circumcircle at the vertices of a given triangle meet the opposite sides in collinear points.

•  Take ABC with the given circumcircle, and take the Pascal triangle AABBCC(A). The result follows immediately.