Harmonic conjugate

Sorry, your browser is not Java compliant

With respect to 2 lines

Given two lines d1 and d2. Two points M and N are harmonic conjugates (of each other) with respect to d1 and d2 says that M and N are harmonic conjugates with respect to intersection points A,B of MN with d1 and d2 : MA/MB = -NA/NB.
Really, we should also consider the case when MN is parallel to d1 or d2, or even to both (if d1//d2). To express the conjugate property in just one sentence in all cases, we have to consider projective geometry, or consider "points at infinity" as if they were ordinary points.

Sorry, your browser is not Java compliant

With respect to a circle

Given a circle Γ. Two points M and N are harmonic conjugates with respect to Γ says that M and N are harmonic conjugates with respect to intersection points of MN with Γ. MA/MB = -NA/NB.
Really, we should consider the case when MN doesn't intersect Γ. The previous definition should then require considering "imaginary" intersection points. That is in ² instead of ². Then, for instance, the circle x² + y² = 9 intersects the line x = 5 in y = ±4i : 25 - 16 = 9, points (5, ±4i) satisfy algebraically both equations, but they are not "real". The real points (5, ±4) are harmonic conjugates of points (5, ±4i), on the line : 4² = (-4)² = (4i)×(-4i), where (5,0) is midpoint of (5, ±4)
If we want to stay in the "classical" euclidean geometry, we have to use another equivallent definition :

Perpendicular circles

Two circles C and Γ are perpendicular if they intersect at a right angle, that is if the tangents in an intersection point are perpendicular.
This is equivallent to power of O to Γ is OT² = R² that is, if A and B are intersection points of a diameter MN of C with Γ, OM² = ON² = R² = OA.OB, (ABMN) is an harmonic range, M and N are harmonic conjugates with respect to Γ.

This allows to extend the definition of conjugate points, even if line MN doesn't intersect Γ :

 Two points M and N are conjugates with respect to Γ iff circle with diameter MN is perpendicular to Γ 

Pole and polar

With respect to 2 lines

Consider two secants MAB and MA'B'. Let N conjugate of M with respect to AB. Line ON intersects MA'B' in N'. Let's prove that N' is the harmonic conjugate of M with respect to A'B'.
The parallel lines to OM in N and N' intersect d1 and d2 in CD and C'D'.
Triangles MOB and NDB are similar : hence MO/ND = MB/NB
Triangles MOA and NCA are similar : hence MO/NC = MA/NA
As MB/NB = MA/NA (M,N conjugates), ND = NC and N is the midpoint of CD.
Dilation with center O implies then N' midpoint of C'D', and a similar calculation as the previous one, but in reverse order, using triangles MOB', N'D'B' and MOA', N'C'A', results into MB'/N'B' = MA'/N'A'. Hence N' is conjugate of M with respect to A'B'.
Moving the secant MA'B', we get then :

 The locus of all conjugate points of M with respect to d1,d2 is a line : 
 The polar of M with respect to lines d1 and d2

M being conjugate of N, OM is the polar of N, hence another property/definition :

Two points M and N are conjugates iff one lies on the polar of the other 

The pencil of lines OM,ON,d1,d2 define then on any secant MN an harmonic range :
It is an  harmonic pencil  of lines (The property doesn't depend on the secant, but is an intrinsic property of the pencil).

Construction of the polar

The previous proof gives a first construction of the polar :
A parallel to OA intersects d1 and d2 in C and D. Let N the midpoint of CD, then the polar of M is line ON.
If d1//d2, O is "at infinity" and OA is parallel to d1 and d2, hence C and D don't exist (are "at infinity") and this construction doesn't work.

Another construction, more efficient as it is done with the straightedge alone :
Let two secants MAB and MA'B', N and N' the harmonic conjugates of M with respect to AB and A'B'
They are on the polar of M with respect to d1 and d2. Let I the intersection point of AB' and A'B. Considering the secants MAB and MA'B' intersecting lines IAB' and IA'B, N and N' are on the polar of M with respect to lines IAB' and IA'B, and this polar goes through I.
Line NN' (polar of M with respect to d1 and d2) then goes through I, hence is line OI, without the need to effectively construct N or N' before.

 

 

If d1//d2, this construction also works : line "OI" is the parallel to d1 and d2 from I.
To avoid constructing a parallel, we may draw a 3rd secant to get in the same way a point J, the polar being then line IJ, constructed using straightedge alone.

With respect to a circle

Sorry, your browser is not Java compliant From the definition as :
"M and N conjugates iff the circle with diameter MN is perpendicular to Γ",
or in other words, midpoint O of MN has power OT² = OM² to Γ.
Locus of O is the radical axis of Γ and the circle-point (M) : locus of all points with equal power to Γ and (M).
N is the dilation of O from center M in ratio 2 O.
Hence locus of N is a line : the polar of M.
Of course, as the radical axis, it is perpendicular to O'M.
The "foot" H of the polar is such as O'U² = O'V² = O'X² = O'M.O'H.
If M is outside the circle, MX is then tangent to circle Γ (which seems obvious : A B N are then the same point X)
Conversedly, M is on the polar of N, hence if M is inside the circle, HX is tangent to circle.

This gives a construction of the polar, by constructing the tangents to circle Γ from M (if outside), or the tangents to Γ in the intersection points of the perpendicular in M to O'M (if inside).
If M is on the circle, the polar is the tangent in M.
If M is at center of circle, the polar is the "line at infinity".

Consider two secants to circle from M, intersecting the circle in AB and A'B'. The conjugates of M being N and N' with respect to AB and A'B', these are also conjugates of M with respect to lines AA' and BB', hence the polar NN' of M with respect to circle Γ is also the polar of M with respect to lines AA' and BB'. Hence I is on this polar.
Which gives a construction with straightedge only of the polar of M, by drawing three secants.
Also results immediately a construction with straightedge only of the de la tangents from M to circle, because the touch points are the intersection points of polar with Γ.

Exercice : construct with straightedge only the tangent in a given point T on Γ
Hint

Consider a line PQ. Let I the intersection of the polar of P with the polar of Q.
Both P and Q being conjugate of I, PQ is the polar of I.
I is called the pole of PQ, the only point conjugate to all points of PQ.

Generalization

We can define the pole and polar properties in a pure algebraic way, in a projective plane, with respect to any conic section.
Two lines, or a circle, being specific cases of conic section, even degenerated.
Not going so far, a central projection transforms a circle into conic section, or reversal. (consider a cone intersected by a plane in a circle, and by another plane in any conic section). The constructions "with straightedge only" of pole and polar are kept the same through the projection and show the definitions and properties of pole/polar with respect to any conic section.

 

 

Home Arithmetic Geometric Misc Topics Scripts Games Exercices Mail Version Française Previous topic Next topic