On the Existence of Poncelet-Morley Points in Euclidean Geometry

In this paper, we prove the existence of the Poncelet-Morley point for a given elliptic configuration. The paper ends with an application of such a point in angle trisection problem.


Introduction
Trisection is a classic problem of compass and straightedge constructions of ancient Greek mathematics.It concerns the construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge, and a compass.The problem as stated is generally impossible to solve, as shown by Pierre Wantzel in (Arnaudiès J. M. & Fraysse H., 1990).Note that the fact that there is no way to trisect an angle in general with just a compass and a straightedge does not mean that there is no trisecting angle: for example, it is relatively straightforward to trisect a right angle, that is, to construct an angle of measure 30 degrees.It is, however, possible to trisect an arbitrary angle by using tools other than straightedge and compass.We can also prove using trigonometry and algebraic tools that it is possible to trisect an angle; see solution of Morley problem in (Jean Fresnel, 1996).In plane geometry, Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisector form an equilateral triangle, called the first Morley triangle or simply the Morley triangle.The theorem was discovered in 1899 by Frank Morley.It has various generalizations; in particular, if all of the trisectors are intersected, one obtains four other equilateral triangles.A simple proof of the problem is given in (Jean Fresn, 1996) but it remains difficult to define a construction programme associated to an arbitrary triangle.
Our aim is to propose a new geometrical tool and use it to prove that there are angles which cannot be trisected.Of course, we also need the Poncelet result to prove the existence of our main tool called the Poncelet-Morley point for a given configuration.
We recall that it was in 1813 during his captivity as war prisoner that J. V. Poncelet discovered that if C 1 and C 2 are non degenerate conics in general position which neither meet nor intersect, if there is an n-sided polygon inscribed in C 1 and circumscribed about C 2 , then for any point P of C 1 there exist an n-sided polygon also inscribed in C 1 and circumscribed about C 2 which has P as one of its vertices.We can then define our main tool.Let (E l ) be a non degenerate ellipse with focus F and F ′ , X an external point of (E l ), T 1 and T 2 tangents on (E l ) respectively at We prove the existence of a Poncelet-Morley point for a given configuration.We also prove that for any elliptic configuration, there are two symmetric Poncelet-Morley points.We use the Poncelet-Morley point to prove that it is not possible to trisect all angles.

Morley Point Associated to an Ellipse Configuration
Let (E l ) be a non degenerate ellipse with focus F and F ′ and let X be an external point of (E l ).X is called a Morley point associated to (E l ) if there exist two points M and N on (E l ) such that MXF ′ = F ′ XF = FXN.
) and (T 2 ) are tangents to (E l ) respectively at M 1 and M 2 passing through X and (XM) ∩ (E l ) ∅, then (XM) is in interior region bounded by (T 1 ) and (T 2 ).We shall prove that (∆ ′ ) is interior with (T 1 ) and (T 2 ).Since M 1 = T 1 ∩ E l and M 2 = T 2 ∩ E l then, according to the second Poncelet theorem, (XA) and (XM) are symmetric relatively to Γ. Since (XM) is in the interior region bounded by (T 1 ) and (T 2 ) and (XA) is the symmetric of XM relatively to (Γ) then (XA) = (∆ ′ ) is in the interior bounded by (T 1 ) and (T 2 ).Therefore (∆ ′ ) ∩ (E l ) ∅. ) ) is in the interior region bounded by ) is in the interior Therefore X is a Morley point associated to (E l ).

Poncelet-Morley Points
Let (E l ) be an ellipse with focus F and F ′ , X an external point of (E l ), T 1 and T 2 tangent to (E l ) respectively at M 1 and Let us note that a Poncelet-Morley point of the ellipse E l is the intersection of two tangents to E l .Given any tangent T 1 to an ellipse E l , determining a Poncelet-Morley point associated to a giving configuration is equivalent to construct another tangent T 2 to E l such that X = T 1 ∩ T 2 is a Morley point associated to E l .
Proposition 2. Let (E l ) be an ellipse with focus F and F ′ ; M a point of (E l ).There exist a Poncelet-Morley point associated to M.
Proof.The proof includes a construction programme of Poncelet-Morney.
Let (E l ) be an ellipse with focus F and F ′ .M a point of (E l ).Let (T l ) be the tangent at M to (T 1 ).Consider A M a point of (T 1 ) such that AF ′ = FF ′ and MA is minimal.
is on the median line of the segment [AF].Let X be the intersection of the bisector of angle Therefore F ′ is on the median line of the segment [AF].Let X be the intersection of the bisector of angle FF ′ A and (T 1 ).As AF ′ = FF ′ , the triangle ∆AFF ′ is isosceles in F ′ thus (F ′ X) is the bisector of angle AF ′ F and median of the segment [AF].Therefore the triangle ∆XAF is isosceles in X and then mes MXF ′ = mes AXF ′ = mes F ′ XF.Let (T 2 ) be another tangent at N to (E l ) passing through X and different from (T 1 ).According to the second Poncelet theorem, mes MXF ′ = mes F ′ XF = mes FXN.Then X is the Poncelet-Morley point of (E l ) associated M.

Number of Poncelet-Morley points associated with a given point of an ellipse
In this section, (E l ) denotes a non degenerate ellipse of focus F ′ and F; (T 1 ) is a tangent at M to (E l ) and A is an arbitrary point of (T 1 ) different to M. We denote by [MA) the half-line of (T 1 ) containing A.
Proposition 3. Let (E l ) be an ellipse of focus F and F ′ and M ∈ (E l ).There exist at most two Poncelet-Morley points associated to M.
Proof.The proof also gives a construction programme.
The existence of a Poncelet-Morley point associated with M is given by the above proposition.We remark that there is no Poncelet-Morley point if the bisector of the angle AF ′ F is parallel to (T 1 ).
One can thus conclude that to each point of (E l ) we can associate at most two Poncelet-Morley points.
Theorem 4. If X is a Poncelet-Morley point associated with M ∈ (E l ) and if M ′ and X ′ are respective images of M and X relatively to a symmetrical line of (E l ), then X ′ is a Poncelet-Morley point associated with M ′ .
Proof.Since the orthogonal symmetry preserves the length and the measure of non oriented angle, the image of a tangent to (E l ) is another tangent and the image of the bisector of an angle is the bisector of the image of the angle.
Corollary 1.Given a non degenerated ellipse (E l ) with focus F and F ′ , there is an infinity Poncelet-Morley point associated to (E l ).

Poncelet-morley Point and Angle Trisection Problem
Angle trisection is a classic problem of compass and straightedge constructions of ancient Greek mathematics.It concerns the construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge, and a compass.The problem as stated is generally impossible to solve, as shown by Pierre-Laurent Wantzel in 1837; in (Pierre-LaurentWantzel, 1837).However, although there is no way to trisect an angle in general with just a compass and a straightedge, some special angles can be trisected; that is the case of a right angle.It is also possible to trisect an arbitrary angle by using tools other than straightedge and compass; that is the case of neusis construction; which is due to Archimedes.
Wantzel published his proof earlier in 1837 than Galois's, whose work was published in 1846.Therefore, his proof did not use the connection between field extensions and groups that is the subject of Galois theory itself.
According to Andrew M. Gleason in (Andrew M Gleason, 1988), the origin of this problem come from Gauss who discovered in (Carl Friedrich Gauss, 1966) how to construct a regular 17-gon using only ruler and compass.Gauss showed (Carl Friedrich Gauss, 1966) that regular polygons with 257 or 65537 sides can be constructed.He also stated without proof that no other regular polygons are constructible.
In what follows, we use the Poncelet-Morley point associated to any trisection problem of a given angle a cubic equation and use Galois theory to prove that the problem as stated is generally impossible to solve.If θ is a constructible angle, 3θ is also (it is the multiplication of the angles by three ).Conversely if 3θ is constructible is it always possible to construct θ?
Our aim is to see if we can use the Poncelet-Morley point and give a solution to the problem of the trisection.In other words if ABC is an angle is there any ellipse (E l ) such as B is a Poncelet-Morley point associated with A? The answer to this question consists in the construction of the focus F and F ′ of (E l ).Let ABC be an acute angle such that AB = BC.Let D be a point of segment [BC]  Let F ′ be the intersection point of (∆) and (C) such that DF ′ is maximal and we note β the angular measure ABF ′ .
When D covers [BC], β changes between 0 and α 2 .It can then take the value α 3 Let (∆) be the perpendicular line to (AB) passing through A and (∆ 1 ) the parallel line to (AB) passing through D. We have (∆ 1 ) ∩ (C) = F ′ since DF ′ is maximum.Let (∆ 2 ) be the perpendicular line to (BC) passing through D. We denote by (∆ ′ 1 ) the image of (∆ 1 ) by the orthogonal symmetry relatively to (∆ 2 ).The image of the straight line (AF ′ ) by orthogonal symmetry relatively to (∆) intersects (∆ ′ 1 ) at F. If k is the solution of the problem, then F and F ′ are focus of the ellipse tangent to (AB) and (BC) respectively at A and D such that angles ABF ′ , F ′ BF; and FBC are congruent.
and (C) the circle of center A passing through B.The parallel line (∆) with (AB) passing through D intersect (C) in two points.Indeed if H is the orthogonal projection of B on (∆) then BH < BD < BC = AB but ABC is not a right angle.Consequently d(A, (∆)) = BH < AB = r the radius of the circle (C).Then (∆) meet (C).
] which for each point D ∈ [BC] associate the measure of ABF ′ , is a bijection.We can then conclude that for all α one and only one point D ∈ [BC] such that mes ABF ′ = α 3 .This completes the algebraic proof of the existence of the trisection angle, but the trisection problem remains unsolved.That will be the case if we give the construction programme of the point D. Since D ∈ [BC] there exist k ∈]0, 1[ such that D is an image of C by homothety with center B and the ratio k.The trisection angle problem is then equivalent to the determination of ratio k.D is the image of C by the homothety of center B and ratio k;

Fγ
= 26.69• δ = 26.66•If k is the solution of the problem then BD = kAB, since AB = BC.Therefore, in the isosceles triangle ∆ ABF ′ one has: BF ′ = 2AB cos θ.We deduce that BD 2 = BF ′2 + DF ′2 − 2BF ′ .DF ′ cos θ.This last equality implies that k 2 AB 2 = Each non degenerate ellipse (E l ) has Morley points.Proof.If (E l ) is a non degenerate ellipse of focus F and F ′ then FF ′ 0. Therefore the circle (C) with center F ′ and radius FF ′ is non degenerate and it intersects