This worksheet explores how transformations can simplify geometric proofs. We prove Ceva’s theorem by affine normalization and prove that five points determine a conic by projective normalization.

1. Transformations as a proof strategy

A recurring idea in this course is that transformations allow us to replace a complicated configuration by a simpler one.

For example, in the proof of Pappus’ theorem, we may send two projective lines to convenient coordinate axes.

The general strategy is:

(a)
Normalize the configuration.
(b)
Prove the theorem in the normalized case.
(c)
Transform back to the original configuration.

2. Ceva’s theorem by affine normalization

We begin with a classical theorem about concurrence in a triangle.

Theorem (Ceva’s theorem). Let \(ABC\) be a triangle, and let

\[ D\in BC,\qquad E\in CA,\qquad F\in AB \]

be points on the three sides, none of them equal to a vertex. Then the three lines

\[ AD,\qquad BE,\qquad CF \]

are concurrent if and only if

\[ \frac {BD}{DC}\cdot \frac {CE}{EA}\cdot \frac {AF}{FB}=1. \]

In this section, we prove Ceva’s theorem using transformations. The key idea is that affine transformations preserve:

  • collinearity;
  • concurrency of lines;
  • ratios along a line.

Therefore, instead of proving Ceva’s theorem in an arbitrary triangle, we may use an affine transformation to send the triangle to the standard triangle

\[ (0,0),\qquad (1,0),\qquad (0,1). \]

The proof then becomes a coordinate calculation.

Affine normalization of a triangle.

Let \(A,B,C\in \RR ^2\) be three non-collinear points.

(a)
Explain why the vectors \(B-A\) and \(C-A\) form a basis of \(\RR ^2\).
(b)
Define
\[ \Phi (u,v)=A+u(B-A)+v(C-A). \]
Show that \(\Phi \) is an affine transformation.
(c)
Compute
\[ \Phi (0,0),\qquad \Phi (1,0),\qquad \Phi (0,1). \]
Conclude that any nondegenerate triangle can be sent by an affine transformation to the standard triangle
\[ (0,0),\qquad (1,0),\qquad (0,1). \]
Ceva’s theorem in the standard triangle.

We now prove Ceva’s theorem for the standard triangle. Let

\[ A=(0,0),\qquad B=(1,0),\qquad C=(0,1). \]
Let \(D\in BC\), \(E\in CA\), and \(F\in AB\), with none of \(D,E,F\) equal to a vertex. Write
\[ D=(1-t,t),\qquad E=(0,u),\qquad F=(v,0), \]
where \(0<t,u,v<1\).
(a)
Show that
\[ \frac {BD}{DC}=\frac {t}{1-t}, \qquad \frac {CE}{EA}=\frac {1-u}{u}, \qquad \frac {AF}{FB}=\frac {v}{1-v}. \]
(b)
Show that the equation of the line \(AD\) is
\[ tx-(1-t)y=0. \]
(c)
Show that the equations of the lines \(BE\) and \(CF\) are
\[ ux+y-u=0 \qquad \text {and}\qquad x+vy-v=0. \]
(d)
Explain why the three lines \(AD\), \(BE\), and \(CF\) are concurrent exactly when the system
\[ \begin{cases} tx-(1-t)y=0,\\ ux+y-u=0,\\ x+vy-v=0 \end{cases} \]
has a common solution \((x,y)\).
(e)
Rewrite this system as
\[ \begin{pmatrix} t & -(1-t) & 0\\ u & 1 & -u\\ 1 & v & -v \end{pmatrix} \begin{pmatrix} x\\ y\\ 1 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}. \]
Deduce that concurrency implies
\[ \det \begin{pmatrix} t & -(1-t) & 0\\ u & 1 & -u\\ 1 & v & -v \end{pmatrix} =0. \]
(f)
Compute this determinant and show that the condition above is equivalent to
\[ u(1-t)(1-v)=tv(1-u). \]
(g)
Conclude that \(AD\), \(BE\), and \(CF\) are concurrent if and only if
\[ \frac {BD}{DC}\cdot \frac {CE}{EA}\cdot \frac {AF}{FB}=1. \]
Ceva’s theorem in an arbitrary triangle.

Explain why the result for the standard triangle implies Ceva’s theorem for any nondegenerate triangle.

Your answer should mention:

  • affine transformations;
  • preservation of collinearity;
  • preservation of concurrency;
  • preservation of ratios along a line.

3. Five points determine a conic

We now move from affine normalization to projective normalization.

Recall that a conic in \(\PP ^2(\FF )\) is the zero set of a nonzero homogeneous quadratic equation

\[ aX^2+bY^2+cZ^2+dXY+eXZ+fYZ=0. \]

Two equations which differ by a nonzero scalar define the same conic.

Theorem. Let

\[ P_1,P_2,P_3,P_4,P_5\in \PP ^2(\FF ) \]

be five points such that no three are collinear. Then there is a unique conic passing through them.

We will prove this theorem by using a projective transformation to simplify four of the points.

Normalizing four points.

Let \(P_1,P_2,P_3,P_4\in \PP ^2(\FF )\) be four points such that no three of them are collinear.

(a)
Choose vector representatives
\[ P_i=[v_i], \qquad v_i\in \FF ^3\setminus \{0\}. \]
Explain why \(v_1,v_2,v_3\) form a basis of \(\FF ^3\).
(b)
Write
\[ v_4=\alpha v_1+\beta v_2+\gamma v_3. \]
Explain why
\[ \alpha ,\beta ,\gamma \neq 0. \]
(c)
Define a linear map \(A:\FF ^3\to \FF ^3\) by
\[ A(\alpha v_1)= \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \qquad A(\beta v_2)= \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \qquad A(\gamma v_3)= \begin{pmatrix} 0\\0\\1 \end{pmatrix}. \]
Explain why \(A\) is invertible.
(d)
Let \([A]\) be the induced projective transformation of \(\PP ^2(\FF )\). Show that
\[ [A](P_1)=[1:0:0], \qquad [A](P_2)=[0:1:0], \qquad [A](P_3)=[0:0:1], \]
and
\[ [A](P_4)=[1:1:1]. \]
Conics through four normalized points.

Consider the four points

\[ [1:0:0],\qquad [0:1:0],\qquad [0:0:1],\qquad [1:1:1] \]
in \(\PP ^2(\FF )\).
(a)
Let
\[ aX^2+bY^2+cZ^2+dXY+eXZ+fYZ=0 \]
be a conic passing through
\[ [1:0:0],\qquad [0:1:0],\qquad [0:0:1]. \]
Show that
\[ a=b=c=0. \]
(b)
Deduce that the equation has the form
\[ dXY+eXZ+fYZ=0. \]
(c)
Now impose that the conic passes through \([1:1:1]\). Show that
\[ d+e+f=0. \]
Adding the fifth point.

Let

\[ P=[u:v:w]\in \PP ^2(\FF ) \]
be a fifth point, and suppose that the five points
\[ [1:0:0],\qquad [0:1:0],\qquad [0:0:1],\qquad [1:1:1],\qquad [u:v:w] \]
are distinct and no three are collinear.
(a)
Explain why
\[ u,v,w\neq 0. \]
(b)
Show that a conic passing through the five points must have equation
\[ dXY+eXZ+fYZ=0, \]
where
\[ d+e+f=0 \]
and
\[ duv+euw+fvw=0. \]
(c)
Explain why the two equations
\[ d+e+f=0 \qquad \text {and}\qquad duv+euw+fvw=0 \]
are independent.

Hint: If they were dependent, then \((uv,uw,vw)\) would be proportional to \((1,1,1)\).

(d)
Conclude that the space of triples \((d,e,f)\) satisfying these equations is one-dimensional.
(e)
Explain why this means that there is a unique conic passing through the five points.
The five-point theorem for conics.

Prove the following theorem.

Theorem. Let

\[ P_1,P_2,P_3,P_4,P_5\in \PP ^2(\FF ) \]
be five points such that no three are collinear. Then there is a unique conic passing through them.

Your proof should use:

  • the projective normalization of four points;
  • the normalized computation above;
  • the fact that projective transformations send conics to conics.