# Harmonic division

Harmonic division: definition

In geometry, the harmonic division denotes a special positional relationship of four points on a straight line. So four points lie harmoniously if the line is divided by two points inside and outside (see picture) in such a way that the relationship for the sections ${\ displaystyle A, B, S, T}$ ${\ displaystyle [AB]}$${\ displaystyle S, T}$

• ${\ displaystyle | AS |: | SB | = | AT |: | TB |}$ is satisfied.

The right side can never become 1. So the center of must never be. Is to the right of , then is to the right of . Is to the left of , then is to the left of . ${\ displaystyle S}$${\ displaystyle M}$${\ displaystyle A, B}$
${\ displaystyle S}$${\ displaystyle M}$${\ displaystyle T}$${\ displaystyle B}$
${\ displaystyle S}$${\ displaystyle M}$${\ displaystyle T}$${\ displaystyle A}$

The above equation and the assumption that the line divides inside and outside means that the two partial ratios and have the same amount and the double ratio is equal to −1. ${\ displaystyle S}$${\ displaystyle [A, B]}$${\ displaystyle T}$ ${\ displaystyle (A, B; S)}$${\ displaystyle (A, B; T)}$ ${\ displaystyle (A, B; S, T)}$

Since the above equation is also so

${\ displaystyle | AS |: | AT | = | SB |: | TB |}$

lets write, the points also divide the route harmoniously. The harmonic division describes a symmetrical relation between pairs of points on a straight line. ${\ displaystyle A, B}$${\ displaystyle [S, T]}$

## Drawing determination of the partial points

### With the ray sentences

Harmonic division: construction with theorems of rays

If the segment and the partial point are given, the fourth harmonic point (more precisely: the fourth point, which together with these 3 points results in a harmonic division) can be found with the help of the ray sets according to the adjacent drawing: ${\ displaystyle [A, B]}$${\ displaystyle S}$ ${\ displaystyle T}$

1. The point is chosen arbitrarily, the straight lines and are parallel.${\ displaystyle C}$${\ displaystyle AC}$${\ displaystyle BD}$
2. The point results from the connection of with the given partial point .${\ displaystyle D}$${\ displaystyle C}$${\ displaystyle S}$
3. ${\ displaystyle D}$will be transferred after . The routes and are the same length.${\ displaystyle D '}$${\ displaystyle | BD |}$${\ displaystyle | BD '|}$
4. The partial point results from the intersection of the straight line with the straight line .${\ displaystyle T}$${\ displaystyle CD '}$${\ displaystyle AB}$

If the partial point is given, the same procedure is followed in reverse order. ${\ displaystyle T}$

If the partial ratio is given, you have to choose the point so that it is fulfilled. then results as the intersection of the straight line with . ${\ displaystyle \ lambda}$${\ displaystyle D}$${\ displaystyle | AC |: | DB | = \ lambda}$${\ displaystyle S}$${\ displaystyle CD}$${\ displaystyle AB}$

### With bisector of a triangle

Harmonic division: Construction with bisector of a triangle

If the points are not an isosceles triangle, the inside and outside bisectors cut two points from the straight line so that the points divide the line harmoniously in the ratio of the adjacent sides of the triangle (see picture). The proof uses the theorem about the circle of Apollonios . Note that it must be, s. above. ${\ displaystyle A, B, C}$${\ displaystyle C}$${\ displaystyle S, T}$${\ displaystyle AB}$${\ displaystyle S, T}$${\ displaystyle [A, B]}$${\ displaystyle b: a}$${\ displaystyle C}$${\ displaystyle a \ neq b}$

Further graphic methods for determining the 4th harmonic point can be found here .

## Mathematical determination of the partial points

Mathematically , the length of the route , if and the partial point are given, results from the formula: ${\ displaystyle [AT]}$${\ displaystyle A, B}$${\ displaystyle S}$

• ${\ displaystyle | AT | = {\ frac {| AS || AB |} {2 | AS | - | AB |}}}$if the denominator is (to the right of )${\ displaystyle> 0}$${\ displaystyle T}$${\ displaystyle B}$
${\ displaystyle | AT | = - {\ frac {| AS || AB |} {2 | AS | - | AB |}}}$if the denominator is (to the left of )${\ displaystyle <0}$${\ displaystyle T}$${\ displaystyle A}$

If one introduces on the straight line through coordinates in such a way that is, then the uniform formula results ${\ displaystyle A, B}$${\ displaystyle x}$${\ displaystyle A = 0, B = 1, S = s, T = t}$

• ${\ displaystyle t = {\ frac {s} {2s-1}}.}$

Examples of harmonious numbers:

${\ displaystyle {\ text {1)}} \ {\ color {blue} 0}, {\ color {red} {\ tfrac {3} {4}}}, {\ color {blue} 1}, {\ color {red} {\ tfrac {3} {2}}} \, \ quad {\ text {2)}} \ {\ color {blue} 0}, {\ color {red} {\ tfrac {2} { 3}}}, {\ color {blue} 1}, {\ color {red} 2} \, \ quad {\ text {3)}} \ {\ color {blue} 0}, {\ color {red} {\ tfrac {3} {5}}}, {\ color {blue} 1}, {\ color {red} 3} \, \ quad {\ text {4)}} \ {\ color {red} - { \ tfrac {1} {2}}}, {\ color {blue} 0}, {\ color {red} {\ tfrac {1} {4}}}, {\ color {blue} 1}}$

## Relationship to the harmonic mean of two numbers

The last equation can be reformulated like this:

• ${\ displaystyle {\ frac {1} {2}} \ left ({\ frac {1} {s}} + {\ frac {1} {t}} \ right) = 1}$

This means that the harmonic mean of the two coordinates is equal to 1. ${\ displaystyle s, t}$

## generalization

• Four points of an affine or projective straight line over a body of the characteristic lie harmoniously if the double ratio is.${\ displaystyle A, B, S, T}$ ${\ displaystyle K}$ ${\ displaystyle \ neq 2}$ ${\ displaystyle (A, B; S, T) = - 1}$

Terms such as between, inside, outside, lengths, distances , which are typical for an arranged body with a metric, are not required for this definition. In particular, the harmonic position is also defined for the affine / projective straight line over the complex numbers or a finite field.

The above coordination ( ) is also possible in the affine case over any body, so that the relationship continues to apply. ${\ displaystyle A = 0, B = 1, S = s, T = t}$${\ displaystyle t = {\ tfrac {s} {2s-1}} \}$

If one ends the affine straight line projectively with the symbol and calculates in the "usual" way, the formula between and the four points are harmonious also applies in this case . H. . ${\ displaystyle \ infty}$${\ displaystyle \ infty}$${\ displaystyle s, t}$${\ displaystyle {\ color {blue} 0}, {\ color {red} {\ tfrac {1} {2}}}, {\ color {blue} 1}, {\ color {red} \ infty}}$${\ displaystyle ({\ color {blue} 0}, {\ color {blue} 1}; {\ color {red} {\ tfrac {1} {2}}}, {\ color {red} \ infty}) = -1}$

Harmonic pairs of points: double ratio = -1

The importance of the harmonic position of four collinear points is that there is always an involutive projective mapping of the straight line that leaves two (of the four points) fixed and exchanges the other two. In the illustration above, the linear mapping that leaves fixed and maps onto creates such an involution. In inhomogeneous coordinates it causes: (mirroring at the zero point). That means: are fixed and are swapped. ${\ displaystyle {\ vec {u}}}$${\ displaystyle {\ vec {v}}}$${\ displaystyle - {\ vec {v}}}$${\ displaystyle \ infty \ to \ infty, x \ to -x}$${\ displaystyle A, B}$${\ displaystyle S, T}$

The general rule is:

• The fourth harmonic point of three affine points, with one being the center of the remaining pair of points, is always the far point (see construction of the fourth harmonic point).${\ displaystyle \ infty}$

And:

• The harmonic position of four points of a projective straight line is the analogue of the affine term center of two points .

Further harmonic pairs of points:

For , is the double ratio . ${\ displaystyle A = \ langle {\ vec {u}} \ rangle, \ B = \ langle {\ vec {v}} \ rangle, \ S (s) = \ langle s {\ vec {u}} + { \ vec {v}} \ rangle, \ T (s) = \ langle s {\ vec {u}} - {\ vec {v}} \ rangle \, s \ neq 0}$${\ displaystyle (A, B; S (s), T (s)) = - 1}$

The following applies:

• From follows: . This means that the harmonic position depends only on the two pairs of points and not on their arrangement.${\ displaystyle (A, B; S, T) = - 1}$${\ displaystyle (A, B; T, S) = (B, A; S, T) = (S, T; A, B) = - 1}$

### Construction of the 4th harmonic point

Construction of the 4th harmonic point
Construction of the 4th harmonic point: is the far point${\ displaystyle P_ {1}}$
Affine variant of the construction of the 4th harmonic point: Construction of the center M of A, B. (A, B, T are given)

If there are three points on a straight line of a projective plane, the fourth harmonic point can be constructed as follows: ${\ displaystyle A, B, S}$${\ displaystyle (A, B; S, T) = - 1}$

1. Do n't pick up a point .${\ displaystyle P_ {1}}$${\ displaystyle g}$
2. Draw the straight lines .${\ displaystyle AP_ {1}, BP_ {1}, SP_ {1}}$
3. Pick a point on the straight line .${\ displaystyle P_ {2}}$${\ displaystyle SP_ {1}}$
4. The straight line intersects the straight line at a point . The straight line intersects the straight line at a point .${\ displaystyle BP_ {2}}$${\ displaystyle AP_ {1}}$${\ displaystyle P_ {3}}$${\ displaystyle AP_ {2}}$${\ displaystyle BP_ {1}}$${\ displaystyle P_ {4}}$
5. The straight line intersects at the fourth harmonic point .${\ displaystyle P_ {3} P_ {4}}$${\ displaystyle g}$${\ displaystyle T}$

Note: The construction takes place in a projective plane, ie every two straight lines intersect.

Comment:

1. If a far point is chosen as the point and not on the long line, the lines are parallel in the plane of the drawing (affine part) (see figure).${\ displaystyle P_ {1}}$${\ displaystyle A, B, S}$${\ displaystyle AP_ {3}, SP_ {2}, BP_ {4}}$
2. If you want to construct as a fourth harmonic point , you choose freely, on the straight line and construct . is then the intersection of the straight line with .${\ displaystyle S}$${\ displaystyle A, B, T}$${\ displaystyle P_ {3}}$${\ displaystyle P_ {4}}$${\ displaystyle P_ {3} T}$${\ displaystyle P_ {1}, P_ {2}}$${\ displaystyle S}$${\ displaystyle P_ {1} P_ {2}}$${\ displaystyle g}$
3. If there are given and far points, then the affine construction of the center of two points shown in the picture results . ( form a parallelogram!)${\ displaystyle A, B, T}$${\ displaystyle P_ {1}, T}$ ${\ displaystyle M}$${\ displaystyle A, B}$${\ displaystyle A, B, P_ {3}, P_ {4}}$

The proof of the independence of the construction of the fourth harmonic point from the choice of auxiliary points results in the first affine variant from the ray theorems or more simply in the second affine variant (construction of the center point) from the fact that 1) the diagonals bisect in a parallelogram and that 2) in the case of parallel projection, the center point of a line merges into the center point of the image line. This is independent of the choice of points . ${\ displaystyle M}$${\ displaystyle P_ {3}, P_ {4}}$

### Construction of the 4th harmonic point with the help of a circle

Construction of the 4th harmonic point: with a circle

Another affine variant of the construction of the 4th harmonic point uses a circle (compass) and the plumb bob (set square ):
Let the three affine collinear points be given in such a way that initially lies between . Find the 4th harmonic point (outside). ${\ displaystyle A, B, S}$${\ displaystyle S}$${\ displaystyle A, B}$${\ displaystyle T}$

1. Draw the circle by its center and the center of the points is.${\ displaystyle k}$${\ displaystyle A, B}$${\ displaystyle M}$${\ displaystyle A, B}$
2. Set up the plumb line and cut it with the circle . Be an intersection .${\ displaystyle S}$${\ displaystyle l}$${\ displaystyle k}$${\ displaystyle Q}$
3. Construct the tangent to the circle at the point . ( ).${\ displaystyle t}$${\ displaystyle k}$${\ displaystyle Q}$${\ displaystyle t \ perp MQ}$
4. ${\ displaystyle t}$intersects g at the 4th harmonic point .${\ displaystyle T}$

If one of the points approaches , so too . Is , so is and the far point of the straight line . ${\ displaystyle S}$${\ displaystyle A, B}$${\ displaystyle T}$${\ displaystyle S = M}$${\ displaystyle t \ parallel g}$${\ displaystyle T}$${\ displaystyle g}$

The proof is given by the similarity of the triangles . (Note that you only have to prove the equation . The double ratio is then automatically −1, since it lies inside and outside the line !) The equation follows from the similarity: ${\ displaystyle MSQ, MQT}$${\ displaystyle | AS || TB | = | SB || AT |}$${\ displaystyle S}$${\ displaystyle T}$${\ displaystyle [A, B]}$

• ${\ displaystyle | MS || MT | = r ^ {2}}$, where r is the radius of the circle.

This equation and the design rule (see picture) also appear when mirroring a circle . (The reflection on the unit circle is described with complex numbers by .) When the reflection on the circle (see picture) the points are exchanged and are fixed points (every point of the circle remains fixed!). ${\ displaystyle z \ to {\ tfrac {1} {\ overline {z}}}}$${\ displaystyle k}$${\ displaystyle S, T}$${\ displaystyle A, B}$

If the point does not lie between the points , use the Thales circle to construct the point of contact of the tangent through to the circle . The perpendicular from to provides the 4th harmonic point . (In the picture you just have to swap and .) ${\ displaystyle S}$ ${\ displaystyle A, B}$${\ displaystyle Q}$${\ displaystyle t}$${\ displaystyle S}$${\ displaystyle k}$${\ displaystyle l}$${\ displaystyle Q}$${\ displaystyle g}$${\ displaystyle T}$${\ displaystyle T}$${\ displaystyle S}$

The method described here for the construction of the 4th harmonic point is an affine special case of the following statement:

• If a straight line intersects a non-degenerate projective conic section in two points and is a different point of the straight line , then the associated 4th harmonic point is the intersection of the polar to (with respect to ) .${\ displaystyle g}$ ${\ displaystyle k}$${\ displaystyle A, B}$${\ displaystyle S}$${\ displaystyle A, B}$${\ displaystyle g}$${\ displaystyle A, B, S}$${\ displaystyle T}$${\ displaystyle S}$${\ displaystyle k}$${\ displaystyle g}$