Difference between revisions of "Pascal's Theorem"

Latest revision as of 14:23, 6 July 2024

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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.

@@ Line 1: / Line 1: @@
-'''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.  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.
+{{asy image|<asy>
+size(250);
+D(unitcircle,red);
+pair A=dir(20), B=dir(40),
+     C=dir(70), D=dir(120),
+     E=dir(210),F=dir(300);
+D(L(D(IP(D(L(A,B,0.5,7)),D(L(E,D,0.5,1.5)))),
+D(IP(D(L(C,B,0.5,10)),D(L(E,F,0.5,3.3)))),0.1),dashed);
+D(IP(D(L(A,F,1,0.5)),D(L(C,D,2,0.5))));
+D(A--B--C--D--E--F--A,black+linewidth(1));
+</asy>|right|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 <math> \displaystyle \pi </math>.
+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 <math>\pi </math>.
-'''Lemma.''' Let <math> \displaystyle \omega_1, \omega_2 </math> be two circles which intersect at <math> \displaystyle M, N </math>, let <math> \displaystyle AB </math> be a chord of <math> \displaystyle \omega_1 </math>, and let <math> \displaystyle C, D </math> be the second intersections of lines <math> \displaystyle AM, BN </math> with <math> \displaystyle \omega_2 </math>.  Then <math> \displaystyle AB </math> and <math> \displaystyle CD </math> are parallel.
+'''Lemma.''' Let <math>\omega_1, \omega_2 </math> be two circles which intersect at <math>M, N </math>, let <math>AB </math> be a chord of <math>\omega_1 </math>, and let <math>C, D </math> be the second intersections of lines <math>AM, BN </math> with <math>\omega_2 </math>.  Then <math>AB </math> and <math>CD </math> are parallel.
-''Proof.''  Since <math> \displaystyle ABNM, CDMN </math> are two sets of concyclic points and <math> \displaystyle A,M,C </math> and <math> \displaystyle 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>
-<math> \angle CAB \equiv \angle MAB \equiv \angle MNB \equiv \angle MND \equiv \angle MCD \equiv \angle ACD </math>.
+<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>
-This implies that <math> \displaystyle AB </math> and <math> \displaystyle CD </math> are parallel. {{Halmos}}
+Because alternate interior angles <math>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}}
-'''Theorem.''' Let <math> \displaystyle 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> \displaystyle 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.
-''Proof.'' Let <math> \displaystyle \omega_1 </math> be the circumcircle of <math> \displaystyle A_1A_2A_3A_4A_5A_6 </math>, and let <math> \displaystyle \omega_2 </math> be the circumcircle of triangle <math> \displaystyle A_2A_5P_2 </math>.  Let <math> \displaystyle B_1 </math> be the second intersection of <math> \displaystyle \omega_2 </math> with <math> \displaystyle A_1A_2 </math>, and let <math> \displaystyle B_2 </math> be the second intersection of <math> \displaystyle A_4A_5 </math> with <math> \displaystyle \omega_2 </math>.  By lemma, <math> \displaystyle A_1P_3 = A_1A_6 </math> is parallel to <math> \displaystyle B_1P_2 </math>, and <math> \displaystyle \displaystyle A_1A_4 </math> is parallel to <math> \displaystyle B_1B_2 </math>, and <math> \displaystyle P_3A_4 = A_4A_3 </math> is parallel to <math> \displaystyle P_2B_2 </math>.  It follows that triangles <math> \displaystyle P_3A_1A_4 </math> and <math> \displaystyle P_2B_1B_2 </math> are homothetic, so the line <math> \displaystyle P_3P_2 </math> passes through the intersection of lines <math> \displaystyle A_1B_1 </math> (which is the same as line <math> \displaystyle A_1A_2 </math>) and <math> \displaystyle A_4B_2 </math> (which is the same as line <math> \displaystyle A_4A_5 </math>), which interesect at <math> \displaystyle P_1 </math>.  {{halmos}}
+''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}}
 == 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> \displaystyle ABC </math> be a "hexagon" <math> \displaystyle 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> \displaystyle ABC </math> that pass through <math> \displaystyle 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> \displaystyle 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.
+==Small Pascal's theorem==
+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>
+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>
+Prove that the points <math>D, E</math> and <math>F</math> are collinear.
+<i><b>Proof</b></i>
+<i><b>1. Solution in Barycentric coordinates.</b></i>
+[[Barycentric coordinates | Small Pascal's theorem]]
+<i><b>2. Simplest case -  Lemoine point.</b></i>
+Let <math>P</math> be the Lemoine point.

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Difference between revisions of "Pascal's Theorem"

Latest revision as of 14:23, 6 July 2024

Proof

Notes

Small Pascal's theorem