经典的射影几何定理

Desargues Theorem
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If the lines joining corresponding vertices of two triangles are concurrent, then the intersections of corresponding sides (P, Q and R) are collinear.
The dual of Desargues Theorem states that:
If the intersections of corresponding sides of two triangles are collinear, then the lines joining corresponding vertices are concurrent.
This is the converse of Desargues Theorem!
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La Hire's Theorem

If P lies on the polar of Q, then Q lies on the polar of P.
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This the key result for duality - it shows that incidence is preserved.

Pappus's Theoren

A, B and C are collinear, as are A',B' and C'. AB' and A'B meet in P, AC' and A'C in Q, and BC' and B'C in R. Then P, Q and R are collinear.
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To allow us to apply duality, Pappus's Theorem may be restated as
If alternate vertices of a hexagon are collinear, then the intersections of opposite sides are collinear.
The dual is then
If alternate sides of a hexagon are concurrent, then the lines joining opposite vertices are concurrent.
which is Brianchon's First Theorem.
Tell me about Pappus

Brianchon's First Theorem

If AB, CD and EF are concurrent at X, and BC, DE and FA are concurrent at Y, then AD, BE and CF are concurrent.
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This is the dual of Pappus's Theorem.

Pascal's Theorem

A,B,C,A',B',C' lie on a conic, AB',A'B meet in P, AC',A'C meet in Q, and BC',B'C meet in R. Then P, Q and R are collinear.
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Note that, if we take the points A,B',C,A',B,C' (in that order) as the vertices of a hexagon, then the line pairs (AB', A'B, etc) are pairs of opposite sides of the hexagon.
Thus the theorem can be stated as If the vertices of a hexagon lie on a conic, then the intersections of opposite sides are collinear.
See Brianchon's Theorem.
The points do not have to be distinct. For example, if we have B'=A, then the result holds with the "chord" AB' being the tangent at A.
See The 3 Tangents and 3 Chords Theorem.
Tell me about Pascal.

Brianchon's Theorem

If a each side of a hexagon touches a conic, then its diagonals are concurrent.
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This is the dual of Pascal's Theorem.
To see this easily, we restate Pascal's Theorem as If the vertices of a hexagon lie on a conic, then the intersections of opposite sides are collinear.
Tell me about Brianchon

The 3 Tangents Theorem

If the triangle ABC touches the conic at P, Q, and R as shown, then AP, BQ and CR are concurrent.
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This can be taken as a special case of Brianchon's Theorem.
In detail, take the hexagon ARBPCQ. Then the diagonals are AP, BQ and CR.

The 3 Tangents and 3 Chords Theorem

Triangle ABC is inscribed in a conic. If the tangent at A meets the chord BC at P, the tangent at B meets CA at Q, and the tangent at C meets AB at R, then P, Q and R are collinear.
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This is the dual of The 3 Tangents Theorem.
It can also be seen as a degenerate case of Pascal's Theorem.
In detail, apply Pascal's Theorem to the "hexagon" AABBCC. Then the "side" AA is the tangent at A, and the opposite side is BC, and similarly for BB and CC.

The Common Tangents Theorem
Suppose that C and D are distinct conics. Then the common tangents (if any) to D and C occur at the points where C intersects the dual of D with respect to C.

Proof
Throughout the proof duality means duality with respect to C.
Let D' be the dual of D. Then, of course, D is the dual of D'.

Suppose first that D' meets C at T.
As T is on C, the dual of T is the tangent to C at T.
As D is the dual of D', the dual of T is a tangent to D,
so is a common tangent to C and D.

Suppose now that there is a common tangent L to C and D, meeting C at T, say.
As L is the tangent to C at T, T is the dual of L.
As L is also a tangent to D, T, the dual of L, must lie on D', the dual of D.
Thus T lies on the intersection of C and the dual of D.

Poles, polars and duality - the geometrical approach
Key Theorem
Suppose that C is a conic and that P is a fixed point. If the chord AB of C passes through P, then the tangents at A and B meet on a fixed line.

Theorem 2 amounts to a proof of this.

Note that, if P is on C, then the line is the tangent at P(since, for any point A on C, the tangents at A and P must meet on the tangent at P).

To allow us to state this theorem without the need for special cases, we must allow ideal points (also called points at infinity).
With each direction on the plane (i.e. each class of parallel lines), we associate an ideal point.
The set of ideal points is the ideal line or line at infinity.
This allows us to unify classical theorems. For example,
The Parallel Chords Theorem
Suppose that C is a central conic (an ellipse or hyperbola) with centre O. If the chord AB of C passes through O, then the tangents at A and B are parallel.

In the language of ideal points, this is just a special case of the Key Theorem(the tangents meet on the ideal line).

Ideal points also allow us to give a definition of tangent purely in terms of incidence(see Tangents and ideal points).

Poles and Polars

Provided we allow ideal points and lines, every point has a line associated with it by the Key Theorem.

Definition
Suppose that C is a conic, and that P is any point.
The polar of P with respect to C is the line specified in the Key Theorem.

By the remark after the Key Theorem, if P lies on C, then the polar of P is the tangent at P.

The main result on polars is La Hire's Theorem
Suppose that C is a conic, and that P and Q are points. Then P lies on the polar of Q if and only if Q lies on the polar of P.
Suppose that we can draw two tangents from the point P to the conic C, and that these meet C in Q and R. (Such a point P is said to lie outside C.)
Since the polars at Q and R are the tangents at these points, P lies on both polars.
Then, by La Hire's Theorem, Q and R lie on the polar of P, so if P lies outside C, then the polar of P is the chord of contact QR.

Suppose that C is a conic and that L is a line. Choose points A and B on L.
Let P be the point of intersection of the polars of A and B.
Then, by La Hire's Theorem, A and B must lie on the polar of P, so the polar of P is the line L.
Thus, every line is the polar of some point.

Definition
Suppose that C is a conic, and that L is a line. The pole of L with respect to C is the point with polar L.

Duality with respect to a conic C.
Suppose that C is a fixed conic.
Then we can define a map from the set of points (including ideal points) and lines on the plane to itself as follows
if P is a point, then the line P' is the polar of P with respect to C,
if L is a line, then the point L' is the pole of L' with respect to C.
If X belongs to the set, then X' is the dual of X with respect to C.
Of course, (X')' = X.

We say that a point and line are incident if the point lies on the line.
It is a consequence of La Hire's Theorem that duality preserves incidence.
Indeed, La Hire's Theorem yields the following central result.

The Principle of Duality
If a theorem involves only concepts defined in terms of incidence, then there is a dual theorem obtained by replacing each object and each concept by its dual.

Suppose that F is a geometrical figure composed of points and lines. Then the dual of F with respect to C is the figure F' consisting of of the polars of the points of F, together with the poles of the lines of F.

For convenience, we define a polygon as either an ordered list of points (a point-polygon), or an ordered list of lines (a line-polygon).
For example, if the points A,B,C,D define the quadrilateral Q, then Q may also de defined by the lines AB, BC, CD,DA.
If P is a polygon with n verices (and hence n sides), then so is P'.
Note that, since the objects are ordered, we can talk about, for example, opposite sides of a polygon with an even number of sides.

In using the Principle of Duality, it is useful to have a table of duals of various concepts.

concept	dual concept
point	line
line joining points	intersection of lines
collinear points	concurrent lines
polygon	polygon
vertex	side

Now it is routine to write down the dual of any theorem. For example,
Pappus's Theorem
If alternate vertices of a hexagon are collinear, then the intersections of opposite sides are collinear.
This has as its dual
Brianchon's First Theorem
If alternate sides of a hexagon are concurrent, then the lines joining opposite vertices are concurrent.

Using the traditional focus-directrix definition, we obtain a conic as a set of points (a point-conic), but we can equally well define a conic as the envelope of its tangents (a line-conic).
The figure on the right shows a conic with a few of its tangents.
This indicates that the envelope of tangents to a conic is the conic itself.
In the language of duality, for any conic C, the dual of C with respect to C is C itself.

It is now natural to ask what happens when we look at the dual of a conic with respect to a different conic.
Suppose that C and D are conics.
Consider D as a line-conic. Then D', the dual with respect to C, will be a set of points.
We can actually identify this dual as another conic, as the next theorem shows.

The Dual Conic Theorem
Suppose that C and D are conics. Then the dual of D with respect to C is a conic.

Proof of the Dual Conic Theorem
Let C be a fixed conic. Suppose that D is any conic, and that L is the polar of P with respect to D.
Choose A and B on D, so that P lies on AB. Now apply duality with respect to C.
By the definition of polar, the tangents T_A and T_B to D at A and B intersect on L.
Hence the line joining the points T_A' and T_B' must pass through L'. Also, A' and B' are tangents to D' at T_A' and T_B'.
The chord (line joining) AB dualizes to the intersection of the tangents A' and B', and so lies on the line P' (as P is on AB).
Thus, L' is the pole of P' with respect to D'.

We can now extend our table of dual concepts.

concept	dual concept
point-conic	line-conic
point on conic	tangent to conic
pole	polar

Poles, polars and duality -the algebraic version
In this section, we give an algebraic treatment of these topics.
A plane conic has an equation of the form $ax^2+bxy+cy^2+fx+gy+h=0$.
In terms of homogeneous coordinates, this becomes $ax^2+bxy+cy^2+fxz+gyz+hz^2=0$.
which can be written as $\vec{x}^{\top}M\vec x=0$ where $\vec x=\begin{pmatrix}x\\y\\1\end{pmatrix}$, and $M=\left[\begin{array}{ccc}a & \frac{1}{2} b & \frac{1}{2} f \\ \frac{1}{2}b & c & \frac{1}{2}g \\ \frac{1}{2}f & \frac{1}{2} g & h\end{array}\right]$ is a symmetric 3×3 matrix.

For a non-degenerate conic, M must be non-singular and have eigenvalues of different sign.

Note that, if a conic contains three (distinct) collinear points, then it must be degenerate.

Definition
If $C: \vec x^{\top}M\vec x=0$ is a non-degenerate conic and $U=[\vec u]$ is any point, then the algebraic polar of U with respect to C is the line $\vec u^{\top}M\vec x=0$.

Note that, as M is non-singular, we cannot have $\vec u^{\top}M=0$, so that the line always exists.

A line L has an equation $\vec a^{\top}\vec x=0$. Now, $\vec u^{\top}M\vec x=0$ and $\vec a^{\top}\vec x=0$ give the same line if and only if $[\vec u]=[M^{-1}\vec a]$.
Thus L is the polar of a unique point $U=[\vec u]$.

Definition
If $C:\vec x^{\top}M\vec x=0$ is a non-degenerate conic and L is any line, then the algebraic pole of L with respect to C is the point $U=[\vec u]$ such that L has equation $\vec u^{\top}M\vec x=0$.

Remark
If L has equation $\vec a^{\top}\vec x=0$, then, as we have seen, the pole of L is $U=[M^{-1}\vec a]$.

Theorem 1
If $C: \vec x^{\top}M\vec x=0$ is a non-degenerate conic and U is any point on C, then the algebraic polar of U with respect to C is the tangent to C at U .

Proof of Theorem 1: Let $U=[\vec u]$, so that the algebraic polar is L: $\vec u^{\top}M\vec x=0$.
Suppose that L cuts C again at $V=[\vec v]$.
As U lies on C, $\vec u^{\top}M\vec u=0$.
As V lies on C, $\vec v^{\top}M\vec v=0$.
As V lies on L, $\vec u^{\top}M\vec v=0$.
Transposing this, and using the fact that M is symmetric, $\vec v^{\top}M\vec u=0$.
Then, by expanding the left hand side, we see that
for any real a and b, $(a\vec u+b\vec v)^{\top}M(a\vec u+b\vec v)=0$.
Thus $W=[a\vec u+b\vec v]$ lies on C.
If U and V are distinct, then C contains the collinear points U, V and W,
but then C would be degenerate.
Thus U=V, so that L meets C just once, and hence is a tangent.

We now show that the algebraic polar coincides with the geometrical polar.

Theorem 2

Suppose that $C:\vec x^{\top}M\vec x=0$ is a non-degenerate conic and that VW is a chord of C passing through a fixed point U, then the tangents at V and W meet on the algebraic polar of U with respect to C.

Proof of Theorem 2: By Theorem 1, the tangents at $V=[\vec v]$ and $W=[\vec w]$ have equations $\vec v^{\top}M\vec x=0$ and $\vec w^{\top}M\vec x=0$, respectively.
Suppose that these meet in $T=[\vec t]$. By definition, T is on the geometrical polar of U.
As VW passes through $U=[\vec u],\vec u = a\,\vec v + b\,\vec w$, for some $a, b$.
Then $\vec u^{\top}M\vec t=a\vec v^{\top}M\vec t+b\vec w^{\top}M\vec t=0$ since T lies on each tangent.
Thus T lies on $\vec u^{\top}M\vec x=0$,the algebraic polar of U.

It follows readily that the algebraic and geometric poles of a line coincide.

We now have the idea of duality defined in algebraic terms.
The fundamental result is that duality preserves incidence.
Theorem 3 La Hire's Theorem
Suppose that $C:\vec x^{\top}M\vec x=0$ is a non-degenerate conic and that U and V are any points.
Then U lies on the polar of V if and only if V lies on the polar of U.

Proof of Theorem 3: Suppose that $U=[\vec u]$ and that $V=[\vec v]$.
Then each condition is clearly equivalent to $\vec u^{\top}M\vec v=0$.

From certain points in the plane, it is possible to draw tangents to the conic C.
Provided the point does not lie on the conic, there will be exactly two tangents, as the following theorem shows.
Theorem 4
Suppose that $C:\vec x^{\top}M\vec x=0$ is a non-degenerate conic and that U is any point.
Then the pair of tangents to C from U has equation $\vec u^{\top}M\vec u\,\vec x^{\top}M\vec x=\vec u^{\top}M\vec x\,\vec u^{\top}M\vec x$ .

Proof of Theorem 4: Suppose that the tangent at $V=[\vec v]$ on C passes through $U=[\vec u]$.
Then $\vec v^{\top}M\vec u=0$, and, transposing, $\vec u^{\top}M\vec v=0$.
Also, as V is on C, $\vec v^{\top}M\vec v=0$.
Any point T on the line UV may be written as $T=[\vec t]$, where $\vec t=a\vec u+b\vec v$, for some $a, b$.
Expanding each side, and using the equalities above,
we see that $\vec t$ satisfies the equation $\vec u^{\top}M\vec u\,\vec x^{\top}M\vec x =\vec u^{\top}M\vec x\,\vec u^{\top}M\vec x$.
Similarly, if the tangent at W passes through U, then each point of UW satisfies the equation.
Finally, the given equation is quadratic in $\vec x$, so can represent no more than two lines.

Of course, by La Hire's Theorem, the chord VW in the above theorem is the polar of U.

Together, Theorems 1, 2 and 4 establish
Joachimsthal's Formulae
Suppose that $C:\vec x^{\top}M\vec x=0$ is a non-degenerate conic. Then

The polar of $U=[\vec u]$ has equation $\vec u^{\top}M\vec x=0$.
If $U=[\vec u]$ is on C, then the tangent to C at U has equation $\vec u^{\top}M\vec x=0.$
The pair of tangents to C from $U=[\vec u]$ has equation $\vec u^{\top}M\vec u\,\vec x^{\top}M\vec x =\vec u^{\top}M\vec x\,\vec u^{\top}M\vec x $.

Tell me about Joachimsthal
Of course, the second is really a special case of the first (but we proved it as Theorem 1 to help with the proof of Theorem 2).

We now show that the dual of a conic with respect to a fixed conic is actually another conic.
Theorem 5
Suppose that $C:\vec x^{\top}M\vec x=0$ and $D:\vec x^{\top}N\vec x=0$ are non-degenerate conics.
Then the dual of D with respect to C is the non-degenerate conic with equation $\vec x^{\top}MN^{-1}M\vec x=0$.

Proof of Theorem 5: We think of D as the line-conic (the envelope of its tangents), $\{\vec u^{\top}N\vec x=0 :\vec u^{\top}N\vec u=0\}$.
The dual of $\vec u^{\top}N\vec x=0$ with respect to C is the point $V=[\vec v]$, where $\vec v= M^{-1}(N\vec u)$.
Then $\vec u= N^{-1}(M\vec v)$.
Since $[\vec u]$ is on D, $\vec u^{\top}N\vec u=0$, and so $\vec v^{\top}MN^{-1}NN^{-1}M\vec v=0$, i.e. $\vec v^{\top}MN^{-1}M\vec v=0$.
Thus the dual of D with respect to C has equation $\vec x^{\top}MN^{-1}M\vec x=0$, as required.

Tangents - a geometrical approach

The usual definition of a tangent to a curve is as the limiting position of a chord.
This assumes that the limit of the gradient exists.
Since a conic has an equation quadratic in the coordinates, the limit does exist.
Here we give a treatment without limits, showing that tangents can be defined purely in terms of incidence.
It also highlights the importance of ideal points.
An obvious first attempt would be the following
Temporary Definition
A line is a tangent to a conic if it meets the conic in only one point.
This works for an ellipse, but fails for parabolas and hyperbolas.

The Parabola	The Hyperbola

Here, any line parallel to the axis cuts the curve once, but we would not describe it as a tangent.	Here, any line parallel to an asymptote cuts the curve once, but we would not describe it as a tangent.

Ideal points on a conic
We say that an ellipse contains no ideal points, a parabola contains one ideal point, corresponding to the direction of the axis.
Now each line parallel to the axis meets the parabola twice,
once as shown above, and again at the ideal point for this direction.
Note that the ideal line meets the parabola once so is a tangent at this ideal point.
a hyperbola contains two ideal points, corresponding to the directions of the asymptotes.
Now each asymptote meets the hyperbola once, so is a tangent at the corresponding ideal point.
Other lines parallel to the asymptotes meet the hyperbola twice, once as above, and again at the ideal point.
The ideal line meets the hyperbola twice (at these ideal points), so is not a tangent.
Definition
A line is a tangent to a conic if it meets the conic in only one point, real or ideal.
This can be justified either on the grounds of geometric intuition, or via the algebra.
In the latter approach, we note that ideal ponts correspond to points with homogeneous coordinates [x,y,0].

Some results on quadrilaterals

The Inscribed Quadrilateral Theorem
If the vertices of a quadrilateral lie on a conic, then the line joining the intersections of opposite sides is the polar of the intersection of the lines joining opposite vertices.

Proof of the Inscribed Quadrilateral Theorem
Suppose that the points A,B,C and D lie on the conic E,
that AC,BD meet at P, AB,CD at Q, and AD,BC at R.
We have to show that QR is the polar of P.

Let the tangents at A,C meet at U and those at B,D at V.

Apply Pascal's Theorem to the degenerate hexagon AABCCD.
The "sides" AA and CC are the tangents at A and C respectively.
These meet at U, so that U,Q and R are collinear.
Similarly, using the hexagon ABBCDD, V,Q and R are collinear.
Thus, U and V lie on QR.

The tangents from U meet the conic at A and C, so AC is the polar of U.
Similarly, BD is the polar at V.
As AC and BD meet at P, P lies on both polars.
Then, by La Hire's Theorem, U and V lie on the polar of P.
Since U and V lie on QR, the result follows.

This suggests a construction for the polar of a point P:
Take two lines through P, meeting the conic at A,B,C and D. Then the polar of P is the line joining the intersections of AB,CD and of AD,BC.
Unfortunately, this will fail if P lies on the conic (when it should return the tangent).

The dual of this is

The Circumscribed Quadrilateral Theorem
If the sides of a quadrilateral touch a conic,
then the intersection of the lines joining opposite vertices
is the pole of the line joining the intersections of opposite sides.

Here, we may observe that to say that a point P is the pole of a line QR
is exactly the same as saying that QR is the polar of P.
Then we see that the two theorems assert the same property
for the quadrilateral, viz

If a quadrilateral is inscribed in, or circumscribed round a conic,
then the line joining the intersections of opposite sides is
the polar of the intersection of the diagonals.

Indeed, corresponding inscribed and circumscribed quadrilaterals have the following remarkable property.
I have chosen the name since it involves two quadrilaterals, each of which is the dual of the other.

The Dual Quadrilaterals Theorem
If the quadrilateral ABCD is inscribed in a conic, and the
tangents at A,B,C,D (in order) define the circumscribed quadrilateral A'B'C'D', then the diagonals of the quadrilaterals are concurrent.

Proof of the Dual Quadrilaterals Theorem
Let AB and CD meet in Q, AD and BC in R, and B'C' and A'D' in P'.
Observe that, since the tangents at B and C meet in B',
BC is the polar of B'.

Similarly, AD is the polar of D'.
By La Hire's Theorem, BC and AD meet in the pole of B'D'.
Thus R is the pole of B'D'.
Similarly, Q is the pole of A'C'.
Then, again by La Hire's Theorem, QR is the polar of P'.
But, by the Inscribed Quadrilateral Theorem,
the pole of QR is the intersection of AC and BD.
Thus the diagonals of the quadrilaterals are concurrent.