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Difference between revisions of "Pascal's Theorem"

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'''Pascal's Theorem''' is a result in [[projective geometry]].  It states that if a [[hexagon]] inscribed in a [[conic section]], then the points of intersection of the pairs of its opposite sides are collinear:
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'''Pascal's Theorem''' is a result in [[projective geometry]].  It states that if a [[hexagon]] is inscribed in a [[conic section]], then the points of intersection of the pairs of its opposite sides are collinear:
  
 
Since it is a result in the projective plane, it has a dual, [[Brianchon's Theorem]], which states that the diagonals of a hexagon circumscribed about a conic concur.
 
Since it is a result in the projective plane, it has a dual, [[Brianchon's Theorem]], which states that the diagonals of a hexagon circumscribed about a conic concur.
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''Proof.''  Since <math>ABNM, CDMN </math> are two sets of concyclic points and <math>A,M,C </math> and <math>B,N,D </math> are two sets of collinear points,
 
''Proof.''  Since <math>ABNM, CDMN </math> are two sets of concyclic points and <math>A,M,C </math> and <math>B,N,D </math> are two sets of collinear points,
 
<center>
 
<center>
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<math> \angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv \angle MCD \equiv \angle ACD </math>.
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<math> \angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv 180^\circ - \angle MCD \equiv 180^\circ - \angle ACD </math>.
 
</center>
 
</center>
 
Because alternate interior angles <math>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}}
 
Because alternate interior angles <math>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}}
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'''Theorem.''' Let <math>A_1A_2A_3A_4A_5A_6 </math> be a cyclic hexagon, and let <math> P_1 = A_1A_2 \cap A_4A_5 </math>, <math> P_2 = A_2A_3 \cap A_5A_6 </math>, <math> P_3 = A_3A_4 \cap A_6A_1 </math>.  Then <math>P_1, P_2, P_3 </math> are collinear.
 
'''Theorem.''' Let <math>A_1A_2A_3A_4A_5A_6 </math> be a cyclic hexagon, and let <math> P_1 = A_1A_2 \cap A_4A_5 </math>, <math> P_2 = A_2A_3 \cap A_5A_6 </math>, <math> P_3 = A_3A_4 \cap A_6A_1 </math>.  Then <math>P_1, P_2, P_3 </math> are collinear.
  
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''Proof.'' Let <math>\omega_1 </math> be the circumcircle of <math>A_1A_2A_3A_4A_5A_6 </math>, and let <math>\omega_2 </math> be the circumcircle of triangle <math>A_2A_5P_2 </math>. Let <math>B_1 </math> be the second intersection of <math>\omega_2 </math> with <math>A_1A_2 </math>, and let <math>B_2 </math> be the second intersection of <math>A_4A_5 </math> with <math>\omega_2 </math>. By lemma, <math>A_1P_3 = A_1A_6 </math> is parallel to <math>B_1P_2 </math>, and <math>A_1A_4 </math> is parallel to <math>B_1B_2 </math>, and <math>P_3A_4 = A_4A_3 </math> is parallel to <math>P_2B_2 </math>. It follows that triangles <math>P_3A_1A_4 </math> and <math>P_2B_1B_2 </math> are homothetic, so the line <math>P_3P_2 </math> passes through the intersection of lines <math>A_1B_1 </math> (which is the same as line <math>A_1A_2 </math>) and <math>A_4B_2 </math> (which is the same as line <math>A_4A_5 </math>), which interesect at <math>P_1 </math>.  {{halmos}}
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''Proof.'' Let <math>\omega_1 </math> be the circumcircle of <math>A_1A_2A_3A_4A_5A_6 </math>, and let <math>\omega_2 </math> be the circumcircle of <math>\triangle{A_2A_5P_2}</math>. Let <math>B_1</math> be the second intersection of <math>\omega_2 </math> with <math>A_1A_2 </math>, and let <math>B_2 </math> be the second intersection of <math>A_4A_5 </math> with <math>\omega_2 </math>. By lemma, <math>A_1P_3 \parallel B_1P_2 </math>, and <math>A_1A_4 \parallel B_1B_2 </math>, and <math>P_3A_4 \parallel P_2B_2</math>, see figure. It follows that triangles <math>\triangle{P_3A_1A_4}</math> and <math>\triangle{P_2B_1B_2}</math> are homothetic with a center <math>P_1</math>, because <math>A_1B_1 \cap A_4B_2 = P_1</math>. Therefore the line <math>P_3P_2</math> passes through the center of homothety, i.e. <math>P_1 </math>.  {{halmos}}
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As a side note: we can also see that triangles <math>P_2A_1A_4</math> and <math>P_2B_1B_1</math> are homothetic about center <math>P_1</math>
 
  
 
== Notes ==
 
== Notes ==
  
 
In our proof, we never assumed anything about configuration.  Thus the hexagon need not even be convex for the theorem to hold.  In fact, many useful applications of the theorem occur with degenerate hexagons, i.e., hexagons in which not all of the points are distinct.  In the case that two points are the same, we consider the line through them to be the tangent to the conic through that point.  For instance, when we let a triangle <math>ABC </math> be a "hexagon" <math>AABBCC </math>, Pascal's Theorem tells us that if <math> \ell_A, \ell_B, \ell_C </math> are the tangents to the circumcircle of <math>ABC </math> that pass through <math>A,B,C </math>, respectively, then <math> \ell_A \cap BC </math>, <math> \ell_B \cap CA </math>, <math> \ell_C \cap AB </math> are collinear; the line they determine is called the [[Lemoine Axis]].  In fact, Pascal's Theorem tells us that <math> \ell_A, \ell_B, \ell_C </math> can be the tangent lines to any conic circumscribed about triangle <math>ABC </math> and the result still holds.
 
In our proof, we never assumed anything about configuration.  Thus the hexagon need not even be convex for the theorem to hold.  In fact, many useful applications of the theorem occur with degenerate hexagons, i.e., hexagons in which not all of the points are distinct.  In the case that two points are the same, we consider the line through them to be the tangent to the conic through that point.  For instance, when we let a triangle <math>ABC </math> be a "hexagon" <math>AABBCC </math>, Pascal's Theorem tells us that if <math> \ell_A, \ell_B, \ell_C </math> are the tangents to the circumcircle of <math>ABC </math> that pass through <math>A,B,C </math>, respectively, then <math> \ell_A \cap BC </math>, <math> \ell_B \cap CA </math>, <math> \ell_C \cap AB </math> are collinear; the line they determine is called the [[Lemoine Axis]].  In fact, Pascal's Theorem tells us that <math> \ell_A, \ell_B, \ell_C </math> can be the tangent lines to any conic circumscribed about triangle <math>ABC </math> and the result still holds.
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==Small Pascal's theorem==
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Let <math>\triangle ABC</math> and point <math>P</math> be given. Let <math>\Omega</math> be the circumcircle of <math>\triangle ABC, A' = AP \cap \Omega, B' = BP \cap \Omega, C' = CP \cap \Omega.</math>
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Let the tangent line to <math>\Omega</math> at point <math>A</math> cross line <math> B'C'</math> at point <math>D.</math> Similarly denote points <math>E</math> and <math>F.</math>
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Prove that the points <math>D, E</math> and <math>F</math> are collinear.
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<i><b>Proof</b></i>
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<i><b>1. Solution in Barycentric coordinates.</b></i>
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[[Barycentric coordinates | Small Pascal's theorem]]
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<i><b>2. Simplest case -  Lemoine point.</b></i>
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Let <math>P</math> be the Lemoine point.

Latest revision as of 14:23, 6 July 2024

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Enlarge.png
A diagram of the theorem

Pascal's Theorem is a result in projective geometry. It states that if a hexagon is inscribed in a conic section, then the points of intersection of the pairs of its opposite sides are collinear:

Since it is a result in the projective plane, it has a dual, Brianchon's Theorem, which states that the diagonals of a hexagon circumscribed about a conic concur.

Proof

It is sufficient to prove the result for a hexagon inscribed in a circle, for affine transformations map this circle to any ellipse while preserving collinearity and concurrence in the projective plane, and projective transformations can map an ellipse to any conic while similarly preserving collinearity and concurrence in the projective sense. Thus we will prove the theorem for a cyclic hexagon, using directed angles mod $\pi$.

Lemma. Let $\omega_1, \omega_2$ be two circles which intersect at $M, N$, let $AB$ be a chord of $\omega_1$, and let $C, D$ be the second intersections of lines $AM, BN$ with $\omega_2$. Then $AB$ and $CD$ are parallel.

Proof. Since $ABNM, CDMN$ are two sets of concyclic points and $A,M,C$ and $B,N,D$ are two sets of collinear points,

$\angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv 180^\circ - \angle MCD \equiv 180^\circ - \angle ACD$.

Because alternate interior angles $CAB$ and $ACD$ are congruent, $AB || CD.$∎

Theorem. Let $A_1A_2A_3A_4A_5A_6$ be a cyclic hexagon, and let $P_1 = A_1A_2 \cap A_4A_5$, $P_2 = A_2A_3 \cap A_5A_6$, $P_3 = A_3A_4 \cap A_6A_1$. Then $P_1, P_2, P_3$ are collinear.

Proof. Let $\omega_1$ be the circumcircle of $A_1A_2A_3A_4A_5A_6$, and let $\omega_2$ be the circumcircle of $\triangle{A_2A_5P_2}$. Let $B_1$ be the second intersection of $\omega_2$ with $A_1A_2$, and let $B_2$ be the second intersection of $A_4A_5$ with $\omega_2$. By lemma, $A_1P_3 \parallel B_1P_2$, and $A_1A_4 \parallel B_1B_2$, and $P_3A_4 \parallel P_2B_2$, see figure. It follows that triangles $\triangle{P_3A_1A_4}$ and $\triangle{P_2B_1B_2}$ are homothetic with a center $P_1$, because $A_1B_1 \cap A_4B_2 = P_1$. Therefore the line $P_3P_2$ passes through the center of homothety, i.e. $P_1$. ∎

Notes

In our proof, we never assumed anything about configuration. Thus the hexagon need not even be convex for the theorem to hold. In fact, many useful applications of the theorem occur with degenerate hexagons, i.e., hexagons in which not all of the points are distinct. In the case that two points are the same, we consider the line through them to be the tangent to the conic through that point. For instance, when we let a triangle $ABC$ be a "hexagon" $AABBCC$, Pascal's Theorem tells us that if $\ell_A, \ell_B, \ell_C$ are the tangents to the circumcircle of $ABC$ that pass through $A,B,C$, respectively, then $\ell_A \cap BC$, $\ell_B \cap CA$, $\ell_C \cap AB$ are collinear; the line they determine is called the Lemoine Axis. In fact, Pascal's Theorem tells us that $\ell_A, \ell_B, \ell_C$ can be the tangent lines to any conic circumscribed about triangle $ABC$ and the result still holds.

Small Pascal's theorem

Let $\triangle ABC$ and point $P$ be given. Let $\Omega$ be the circumcircle of $\triangle ABC, A' = AP \cap \Omega, B' = BP \cap \Omega, C' = CP \cap \Omega.$ Let the tangent line to $\Omega$ at point $A$ cross line $B'C'$ at point $D.$ Similarly denote points $E$ and $F.$ Prove that the points $D, E$ and $F$ are collinear.

Proof

1. Solution in Barycentric coordinates.

Small Pascal's theorem

2. Simplest case - Lemoine point.

Let $P$ be the Lemoine point.

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