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rloginunix
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Posts: 1026
 Inversion Tutorial. Part 1.   « on: Sep 7th, 2014, 10:07am » Quote Modify

Inversion with Respect to a Circle in a Real Plane. Part 1. The Basics.

Purpose.

The purpose of this small tutorial is to remove the paywall between an inquisitive mind and a fundamental academic information and to provide a gentle introduction into an engaging idea that dovetails into geometric transformations in general, complex analysis, group theory and hyperbolic geometry. Inversion with respect to a circle will add to your arsenal of problem solving methods a deterministic technique of finding simple, elegant and comprehensible solutions to a number of geometric problems.

Whether you are a grade school student looking ahead, a college student studying the inversion now or a grownup who wants to step out of one's comfort zone and learn something new a few basic Euclidean geometry facts is all you will need to comprehend the upcoming material.

To benefit from this tutorial the most do all the constructions yourself, whether on real paper or with your favorite graphing software (I use Cinderella and GeoGebra). After you are done with the basic construction abuse it in a creative, exploratory way. Move the points about to see when things break, when they don't, what the overall patterns are. Go slow. Do not move on to the next step unless you fully comprehended the current one.

Before you read my construction steps or proofs try to come up with your own. If you find a better way - reference your improvement and post it!

Context.

The context of this tutorial is Euclidean geometry only. Though a lot of the upcoming material can be set in the context of complex numbers we shall avoid them completely. As such it should be understood from the outset that even though I will be omitting the qualifier "real" - real numbers and real spaces, R2 and R3, is what I really mean and skip only for the sake of avoiding the repetition.

Style.

Rigor is a requirement in math but too much of it tends to kill the organic connection. Gut feel alone, however, may carry you in the wrong direction. Both are needed to produce correct results. In this tutorial I strove for a tasteful blend of both favoring the latter. That is why I qualified the upcoming observations as Properties, not Theorems. As such, the abbreviation ICP1 will stand for "Inversion with respect to a Circle Property 1", etc.

In a similar fashion the abbreviation ICE1 will stand for Inversion with respect to a Circle Exercise 1 and so on.

Notation.

Line(A, B) means a straight line passing through the points A and B.
LineSegment(A, B) means a line segment between the points A and B.
Ray(O, A) means a ray originated at the point O and passing through the point A.
Circle(O, R) means a circle centered at the point O with radius R.

If you see the above notation for the first time during a particular construction it means take action and construct the corresponding object. If this notation is repeated it means a reference to the previously constructed object.

I will refer to Euclid's book Elements in an abbreviated form: B1.P1, for example, means Book 1 Proposition 1. B1.D1 means Book 1 Definition 1 and so on.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #1 on: Sep 7th, 2014, 10:25am » Quote Modify

Prologue. A Point Far Far Away.

To ease our way into the inversion with respect to a circle let us consider something that is trivial to visualize - a sphere of radius R (R-sphere) touching a plane P at a point S - the south pole of the R-sphere. A straight line through S perpendicular to the plane P will pass through the center of the sphere and will also intersect the sphere at a point N, the north pole of the R-sphere, diametrically opposite to the point S.

At random pick a point A anywhere on the sphere but its north pole N. Draw a ray from N through A. That ray will pierce the R-sphere through the point A and the plane P through the point A':

If you keep repeating the above process for all the points on the R-sphere you will create a stereographic projection of the sphere (the space, if you vary R) onto the plane. As a topologist you would want the things to be nice and wholesome - if you were able to carry out the above operation for any point on the sphere then you would be able to say that topologically the sphere and the plane are equivalent.

However, as things sit right now there's a bit of trouble going on. Let us do two very simple operations: cut our above 3D formation with a plane through the points N, S and A' and turn that plane so that it is parallel to the plane of our monitors:

Consider the triangles NAS and NSA'. The NSA' triangle is right by construction. The point A is on the R-sphere's great circle passing through N and S. From Euclid's B3.P31 we know that in a circle the angle in the semicircle is right. Hence, the angle NAS is 90 degrees. The angle at N is common. From B1.P32 we know that the sum of the internal angles in a triangle is 180 degrees. Hence, the remaining angles NSA and NA'S are equal and, finally, by AAA, these triangles are similar.

From B6.P4 we know that in similar triangles the corresponding sides are in the same proportion:

SA'/NS = AS/AN

AN = NS*AS/SA' = 2R*AS/SA'

Now let us move A along the surface of the R-sphere arbitrarily close to N. In that case AS > R and

AN > 2R2/SA'

which means that as the length of the line segment AN arbitrarily diminishes the length of the line segment SA' arbitrarily grows and as is N can not have its image (N') on the plane P. To complete the picture and include N in the set of points on the sphere that have a corresponding point (image) on the plane under the stereographic projection we have to introduce a special point - a point at infinity. Hopefully this example will help you remember and somehow visualize this special point which we will meet it again soon.

Double That.

In the previous section we used the similarity of triangles to justify the need for a special point - a point at infinity. Let us reuse the same similarity of the same triangles but in a different way - by constructing a different proportion. Instead of using only the sides of the right triangles let us use their hypotenuses:

NA'/NS = NS/NA, where NS = 2R

NA'*NA = 4R2

We see that under the stereographic projection the product of the distances between the north pole, the point on the R-sphere and its image on the plane is the square of the R-sphere's diameter.

Let us add another sphere to the picture above. Make the radius of the new sphere 2R, place its center at N and name it the 2R-sphere:

From the drawing above it should be clear that even though our point A is now inside the 2R-sphere all the previous ratios stand. So, by replacing 2R with r we get:

NA'*NA = r2

By manipulating our spheres in a simple way we observe the following. The ray NA connects the center of the 2R-sphere with A. The line AS is perpendicular to NA. The line that is perpendicular to the ray NS intersects the ray NA at A'. Keep this picture in mind.

What we have just done is called the inversion (of the space) with respect to a sphere. It carries a fair amount of similarity with the inversion (of the plane) with respect to a circle which we will now take a closer look at.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #2 on: Sep 7th, 2014, 10:40am » Quote Modify

Inversion with Respect to a Circle in a Real Plane.

An inversion with respect to a circle is a type of plane transformation with the following rule.

Assume that in a given plane P a circle c centered at O with radius r is fixed. A point A' is an inverse of a point A in the plane P with respect to the circle c if:

Condition ICC1). A and A' are different from O and a point at infinity.

Condition ICC2). O, A and A' are collinear with A and A' on the same side of O.

Condition ICC3). OA*OA' = r2.

The circle c is called the circle of inversion, the point O is called the center of inversion and the radius r is called the radius of inversion.

Under the inversion with respect to a circle the center of inversion O is transformed into a point at infinity and a point at infinity is transformed into the center of inversion O. This follows from ICC3 directly when we diminish the length of the line segment OA arbitrarily. These two special points are usually excluded from the properties that follow.

Points.

Property ICP1. Under the inversion with respect to a circle c(O, r) any point A on c is transformed into itself.

Such points are called fixed points.

From ICC3 we have:

OA*OA' = r2
OA = r
r*OA' = r2
OA' = r

From ICC2 it follows that A' does coincide with A.

Property ICP2. Under the inversion with respect to a circle c(O, r) the circle of inversion is transformed into itself.

Such figures are called invariant figures and such transformations are called identity transformations.

This follows directly from ICP1.

Property ICP3. Under the inversion with respect to a circle c(O, r) a point A inside c different from O is transformed into the point A' outside c.

From ICC3 we have:

OA*OA' = r2
OA < r
OA' = r2/OA > r2/r = r
OA' > r

Exercise ICE1. Given the circle of inversion c(O, r) and a point A inside c other than O construct A's inverse with respect to c.

Recall the spheres image I was asking you to remember. It suggests the following construction steps:

1). Ray(O, A).
2). Perpendicular to Ray(O, A) through A until it intersects c to locate B.
3). Ray(O, B).
4). Perpendicular to Ray(O, B) through B until it intersects Ray(O, A) to locate A'.

By construction the triangles OAB and OBA' are right. The angle at O is common. Hence, by AAA, these triangles are similar and their corresponding sides are in the same proportion:

OA/OB = OB/OA'
OB = r
OA*OA' = r2

We see that the idea of similar triangles works again, as it did in the situation with the spheres. We will employ it again soon.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #3 on: Sep 7th, 2014, 10:53am » Quote Modify

Property ICP4. Under the inversion with respect to a circle c(O, r) a point A outside c other than a point at infinity is transformed into a point A inside the circle of inversion other than O.

From ICC3 we have:

OA*OA' = r2
OA > r
OA' = r2/OA < r2/r = r
OA' < r

Exercise ICE2. Given the circle of inversion c(O, r) and a point A outside c other than a point at infinity construct A's inverse with respect to c.

Here we have to reconstruct a tangent to a circle (c) through a point (A):

1). Ray(O, A).
2). Bisect LineSegment(O, A) to locate B: BA = BO.
3). Circle(B, BO = BA) until it intersects c(O, r) to locate D.
4). Perpendicular through D to Ray(O, A) to locate A'.

From B3.P31 (in a circle the angle in the semicircle is right) it follows that the angle ODA is right by construction. And so is the triangle OA'D. The angle at O is common. By AAA these triangles are similar and their corresponding sides are in the same proportion:

OA/OD = OD/OA'
OA*OA' = OD2 = r2

For completeness sake let us do two more constructions.

Exercise ICE3. Given two distinct points A and A' that are inverses of each other with respect to a circle with radius r locate the center of inversion O.

From the definition of inversion with respect to a circle c(O, r) it follows that if we name OA = x and AA' = a then:

x*(x + AA') = r2
x*(x + a) = r2

Solving it for x we get:

x = (sqrt(a2 + 4r2) - a)/2

Rewriting the above as

x + a/2 = sqrt( (a/2)2 + r2)

we see that the square root on the right hand side of the equation is the hypotenuse of a right triangle with known sides. That suggests the following construction:

1). Line(A, A').
2). Bisect LineSegment(A, A') to locate B: BA = BA'.
3). Perpendicular p to Line(A, A') through A (or A').
4). Circle(A, r) until it intersects p to locate D.
5). Circle(B, BD) until it intersects Line(A, A') to locate O and O'.

Exercise ICE4. Given two distinct points A and A' that are inverses of each other with respect to a circle centered at O construct the radius of inversion r.

This is almost an exact copy of the previous exercise with just a couple of the last steps changed:

1). Line(A, A').
2). Bisect LineSegment(A, A') to locate B: BA = BA'.
3). Perpendicular p to Line(A, A') through A (or A').
4). Circle(B, BO = d + a/2) until it intersects p to locate D.
5). LineSegment(A, D) == r.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #4 on: Sep 7th, 2014, 10:59am » Quote Modify

Property ICP5. A point A' inverse to a point A under the inversion with respect to a circle c(O, r) is unique.

This follows directly from ICC3. Algebraically, if we put x = OA' and take OA as a known, we get:

x = r2/OA

a first degree equation yielding a unique solution.

Property ICP6. Under the inversion with respect to a circle c(O, r) an even number of inversions with respect to c transforms a point A into itself.

Two consecutive inversions is all we need to look at. Assume that A is inverted into A' and A' is inverted into A'' with respect to the same circle. From ICC3 we get:

OA*OA' = r2
OA'*OA'' = r2

Since OA' is one and the same (ICP5) we get:

r2/OA = r2/OA''
OA = OA''

From ICC2 it follows that A'' does coincide with A.

Another way to look at this property is to state that if a point A' is the inverse of a point A under the inversion with respect to a circle c then A is the inverse of A' with respect to c.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #5 on: Sep 7th, 2014, 11:07am » Quote Modify

Straight Lines.

Property ICP7. Under the inversion with respect to a circle c(O, r) a straight line l passing through O is transformed into itself.

Barring both intersection points of l with c (which are fixed) the points on l are not transformed into themselves. However, l as a whole - is, which is another example of an invariant figure and an identity transformation.

The proof of this property follows directly from the definition of an inversion with respect to a circle. Pick two arbitrary points P1 and P2 other than O or a point at infinity on either side of O. P'1, by definition, is on l - O, P1, P'1 are collinear. And so is P'2. It follows then that a line through P'1 and P'2 coincides with l.

Property ICP8. Under the inversion with respect to a circle c(O, r) a straight line l that does not pass through O and has exactly two common points with c, F1 and F2, is transformed into a circle q' passing through O and F1 and F2.

Construct a perpendicular to l through O to locate the point of their intersection P. Invert P with respect to c (see ICE1). Invert an arbitrary point A on l inside c with respect to c to locate A'. Consider the triangles OPA and OA'P'. They share a common angle at O. From ICC3 we have:

OP*OP' = r2 = OA*OA'
OP/OA = OA'/OP'

which means that two triangles have two sides about the equal angles in the same proportion - the essence of B6.P6. Hence, the triangles OPA and OA'P' are similar and their corresponding angles, OPA and OA'P', are equal. But OPA is right by construction and hence OA'P' is right.

Since A was picked at random the locus traced by A' must be a circumference of a circle with diameter OP' (you should verify this using a different point, B, on l picked outside c).

This suggests the following ruler and compass construction:

1). Perpendicular to l through O to locate their intersection at P.
2). Invert P with respect to c to locate P'.
3). Bisect LineSegment(F1, P') until it intersects Line(O, P) at Q'.
4). Circle(Q', Q'P' = Q'O) = q'.

Property ICP9. Under the inversion with respect to a circle c(O, r) a straight line l that has exactly one common point with c, P, is transformed into a circle q' passing through O and P.

Since P is on the circumference c - it is fixed. From B3.P16 it follows that an arbitrary point A on l different from P will be outside c. Its inverse A' will then be thrown inside c - ICP4. By construction (ICE2) the angle OA'P is right. Hence, the triangle OA'P is right. Hence, from B3.P31 (in a circle the angle in the semicircle is right) it follows that the locus traced by A' must be a circumference of a circle with diameter OP.

The ruler and compass construction steps are especially easy this time:

1). Line(O, P).
2). Bisect LineSegment(O, P) to locate Q': Q'O = Q'P.
3). Circle(Q', Q'O = Q'P = r/2) = q'.

Property ICP10. Under the inversion with respect to a circle c(O, r) a straight line l that has no common points with c is transformed into a circle q' passing through O having no common points with c.

Construct a perpendicular to l through O to locate the point of their intersection, P. Invert P with respect to c (see ICE2). Invert an arbitrary point A other than P on l (which will be outside c) with respect to c to locate A'. Consider the triangles OPA and OA'P'. They share a common angle at O. From ICC3 we have:

OP*OP' = r2 = OA*OA'
OP/OA = OA'/OP'

which means that the two triangles have two sides about the equal angles in the same proportion - the essence of B6.P6. Hence, the triangles OPA and OA'P' are similar and their corresponding angles, OPA and OA'P', are equal. But OPA is right by construction and hence OA'P' is right.

Since A was picked at random the locus traced by A' must be a circumference of a circle with diameter OP' (you should verify this using a different point, B).

This suggests the following ruler and compass construction:

1). Perpendicular to l through O to locate their intersection at P.
2). Invert P with respect to c to locate P'.
3). Bisect LineSegment(O, P') to locate Q': Q'O = Q'P'.
4). Circle(Q', Q'O = Q'P') = q'.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #6 on: Sep 7th, 2014, 11:09am » Quote Modify

From the properties 8, 9 and 10 we see that for the most part inversion with respect to a circle transforms a straight line into a circle that passes through the center of inversion.

It should also be obvious that the converse statements of the properties 8, 9 and 10 are true. I believe you have enough gut feel developed by now to prove these on your own. I will only formulate these properties here for the record:

Property ICP11. Under the inversion with respect to a circle c(O, r) a circle q that passes through O and has exactly two common points, F1 and F2, with c is transformed into a straight line l' passing through F1 and F2.

Property ICP12. Under the inversion with respect to a circle c(O, r) a circle q that passes through O and has exactly one common point P with c is transformed into a straight line l' tangent to c at P.

Property ICP13. Under the inversion with respect to a circle c(O, r) a circle q that passes through O and has no common points with c is transformed into a straight line l' that has no common points with c.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #7 on: Sep 7th, 2014, 11:34am » Quote Modify

Circles.

Property ICP14. Under the inversion with respect to a circle c(O, r) a circle q concentric with and located inside c is transformed into a circle q' concentric with and located outside c.

By definition q is a set of points equidistant from O. According to ICP3 any point on q, A for example, will be tossed outside c in such a way that:

OA' = r2/OA = r2/R = const = R', where OA = R.

All such distances, R', will remain the same for all the images of the points on q - a definition of a circle.

Property ICP15. Under the inversion with respect to a circle c(O, r) a circle q concentric with and located outside c is transformed into a circle q' concentric with and located inside c.

This property is the converse of the previous one, ICP14. Swap the roles of the circles and the resulting points. Rename q as q', q' as q, A as A', A' as A, etc.

Note here that as the length of R approaches the length of r arbitrarily so does the length of R'. The circle of inversion c gets locked in between q and q' and in their limiting position when R = r = R' q and q' coincide with c.

Property ICP16. Under the inversion with respect to a circle c(O, r) a circle q that does not pass through O and touches c internally is transformed into a circle q' that does not pass through O and touches c externally.

Let T be the single common point of c and q. According to ICP1 T is fixed. Draw Ray(O, T) to locate A on q - a point, according to B3.P8, closest to O. Invert A with respect to c to obtain A' - a point, according to ICC3, furthest from O, and, according to ICP3, outside c. At random pick a point B on q other than A and T and invert it with respect to c to obtain B' also outside c.

The angle OTB' is exterior for the triangle TB'A' and according to B1.P32 it is equal to the sum of the two interior opposite angles:

angle OTB' = beta + alpha

From ICC3 we have:

OB*OB' = r2
r = OT
OB*OB' = OT2 = OT*OT or
OB/OT = OT/OB'

which means that in the triangles BOT and TOB' two sides about the equal (shared) angle at O are in the same proportion. According to B6.P6 these triangles are similar and have their corresponding angles equal: the angle OBT equals the angle OTB'. But the angle OBT = gamma + 90 (where the angle ABT = 90 degrees as a circumference angle in a semicircle) and hence:

angle OTB' = angle OBT = gamma + 90 = beta + alpha

Now consider the triangles OAB and OB'A' with a common angle at O. According to ICC3:

OA*OA' = r2 = OB*OB' or
OA/OB = OB'/OA'

which, based on the same logic as above, means that these triangles are similar with the corresponding angles equal: gamma = alpha and hence:

alpha + 90 = beta + alpha
beta = 90

And since we've picked B arbitrarily the locus traced by B' must be a circumference of a circle with diameter TA'. This suggests the following ruler and compass construction:

1). Ray(O, T) until it intersects q at A.
2). Invert A with respect to c to locate A'.
3). Bisect LineSegment(T, A') to locate Q': Q'T = Q'A'.
4). Circle(Q', Q'T = Q'A') = q'.
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rloginunix
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Posts: 1026
 Re: Inversion Tutorial. Part 1.   « Reply #8 on: Sep 7th, 2014, 11:38am » Quote Modify

Property ICP17. Under the inversion with respect to a circle c(O, r) a circle q that does not pass through O and touches c externally is transformed into a circle q' that does not pass through O and touches c internally.

This property is the converse of the previous one, ICP16. Swap the roles of the circles by renaming them and the resulting points. Rename q' as q and q as q'. The point A', outside c, becomes A. And its image, formerly A, becomes A', inside c, etc.

Property ICP18. Under the inversion with respect to a circle c(O, r) a circle q that does not pass through O, has zero common points with c and is located inside c is transformed into a circle q' that does not pass through O, has zero common points with c and is located outside c.

The proof of this property is similar to the proof of ICP16. The point T from ICP16 is now split into two - B inside and B' outside c.

Let us highlight some key observations. Using the fact that the angle at O is common and ICC3 we can prove that the following triangles are similar pairwise: OBP with OP'B' and OAP with OP'A'. From the similarity of the first pair we conclude that angle OPB = angle OB'P'.

From the similarity of the second pair we conclude that angle OPA = angle OA'P' = alpha.

The angle OB'P' is exterior for the triangle B'P'A' and hence:

angle OB'P' = angle OPB = alpha + 90 = beta + alpha
beta = 90

Property ICP19. Under the inversion with respect to a circle c(O, r) a circle q that does not pass through O, has zero common points with c and is located outside c is transformed into a circle q' that does not pass through O, has zero common points with c and is located inside c.

This property is the converse of the previous one, ICP18. Again, swap the roles of the circles q and q', rename the resulting points, etc.

Before we move on to angles do keep in mind that the center of the original circle and the center of the inverse circle are not inverses of each other with respect to the circle of inversion.
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 Re: Inversion Tutorial. Part 1.   « Reply #9 on: Sep 7th, 2014, 11:57am » Quote Modify

Angles.

Property ICP20. Under the inversion with respect to a circle c(O, r) the angle between two straight lines l1 and l2, both of which pass through O, is equal to the angle between their images l'1 and l'2.

According to ICP7 the straight lines l1 and l2 are transformed into themselves and, hence, the angle between their images, l'1 and l'2, is equal to the angle between l1 and l2.

Property ICP21. Under the inversion with respect to a circle c(O, r) the angle between two straight lines l1 and l2, of which exactly one passes through O, is equal to the angle between their images l'1 and l'2.

Without the loss of generality let us assume that l1 passes through O and, from ICP7, it follows that its image with respect to c also passes through O and coincides with l1. From ICP8,9,10 it follows that l2 is transformed into a circle l'2 that passes through O. The angle between l'1 and l'2 then is the angle between the straight line l'1 and the tangent t to l'2 at O.

Let us consider three cases: l1 and l2 are parallel, perpendicular, neither.

When l1 and l2 are parallel the angle between them is zero degrees and the tangent t to l'2 at O is parallel to l2 (B1.P27). We then have two parallel straight lines, l'1 and t, pass through one point O. Which means that they must coincide. Which makes the angle between them also zero, thus preserving it:

When l1 and l2 are perpendicular O, A, A' and Q', by ICP8,9,10, are all on l'1. Hence, the diameter of l'2 is also on l'1 and, by definition, both tangents to l'2, t at O and t' at A', are parallel to l2 (B1.P27) and, by B1.P29, are also perpendicular to l'1, thus preserving the angle:

When l1 and l2 are neither parallel nor perpendicular take the smaller angle between them as alpha:

The tangent t to l'2 at O is parallel to l2 (B1.P27). It follows then that a straight line l'1 is falling on two parallel straight lines t and l2. From B1.P29 it follows that the angle BOA is equal to alpha, thus preserving the angle between l'1 and l'2 at O.

Consider the angle between the second tangent, t', and l'1. By construction (and the definition of a tangent) the angles Q'AB and Q'OB are right and Q'A = Q'O (being the radii of l'2). Hence, the right triangles Q'AB and Q'OB, by Pythagoras, have BA = BO (BQ' being a common hypotenuse). Hence, the triangle OBA is isosceles having its interior angles at the base OA equal (B1.P5). Hence, the angle BAO = alpha.
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 Re: Inversion Tutorial. Part 1.   « Reply #10 on: Sep 7th, 2014, 12:13pm » Quote Modify

Property ICP22. Under the inversion with respect to a circle c(O, r) the angle between two straight lines l1 and l2, none of which pass through O, is equal to the angle between their images l'1 and l'2.

Again, let us consider three cases: l1 and l2 are parallel, perpendicular, neither. Let P be the second point of intersection of l'1 and l'2. Let t1 and t'1 be tangents to l'1 at O and P. Let t2 and t'2 be tangents to l'2 at O and P.

When l1 and l2 are parallel then the angle between them is zero degrees. Then both tangents at O, t1 to l'1 and t2 to l'2, are parallel (B1.P27 and B1.P30) and pass through O - they must coincide making the angle between them also zero, thus preserving it:

When l1 and l2 are perpendicular then together with t1, which is parallel to l1 (B1.P27) and t2, which is parallel to l2 (B1.P27), they form a parallelogram. Hence, from B1.P34 it follows that t1 and t2 are also perpendicular. In the previous property, ICP21, we've proved that the triangles OQ'1P and OQ'2P are isosceles. Hence:

90 = alpha + beta

must be true at both vertexes, O and P, thus proving that the angle between t'1 and t'2 is preserved:

When l1 and l2 are neither parallel nor perpendicular take the smaller angle between them as alpha:

As in the previous (perpendicular) case we prove that the angle between t1 and t2 is alpha. However, this time the tangents to circles intersect not at their centers but elsewhere - at the points A and B. Still, by construction the triangles Q'1OB and Q'1PB are right. They have a common hypotenuse Q'1B and two equal sides, Q'1O = Q'1P, since these are the radii of l'1. By Pythagoras BP = BO. Hence, the triangle OBP is isosceles having its interior angles at the base OP equal, gamma. In a similar way we can prove that AP = AO, the triangle OAP is isosceles having its interior angles at the base OP equal, beta. Hence:

alpha = beta + gamma

proving that the angle between t'1 and t'2 is also preserved.

From the above straight line angle preservation properties two more follow:

Property ICP23. Under the inversion with respect to a circle c(O, r) the angle between two circles is equal to the angle between their images.

Property ICP24. Under the inversion with respect to a circle c(O, r) the angle between a circle and a straight line is equal to the angle between their images.
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rloginunix
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 Re: Inversion Tutorial. Part 1.   « Reply #11 on: Sep 7th, 2014, 12:30pm » Quote Modify

Points of Tangency.

Property ICP25. If two circles touch each other at a point T different from the center of inversion, then their inverses must also touch at a point T'.

Property ICP26. If a circle and a straight line touch each other at a point T different from the center of inversion, then their inverses must also touch at a point T'.

By definition a tangency point T is single. It belongs to two geometric objects at the same time - two circles or a circle and a straight line. From ICP5 it follows that its image under the inversion with respect to a circle, T', must be unique and it has to belong to the inverse figures, straight lines and/or circles, as well.

Orthogonal Circles.

Two curves are orthogonal at a point T or intersect orthogonally at a point T if their tangents at T are perpendicular. In practical terms, then, two circles are orthogonal if their radii form a right angle at any intersection point. From B3.P36 and B3.P37 two properties follow:

Property ICP27. Any circle q passing through two points A and A' inverse to each other with respect to a circle c(O, r) is orthogonal to c.

From ICC3 we have:

OA*OA' = r2

The locus of centers of circles passing through A and A' must, by definition, be on the perpendicular bisector of AA', p. Pick an arbitrary point Q on p. Construct a circle centered at Q with radius QA = QA'. Name one point where c and q intersect T. Since T is on c's circumference:

OT = r

and the ICC3 equation stands. Since T is on q's circumference B3.P37 is applicable: if OT2 = OA*OA' then OT touches q.  Since OT2 = OA*OA' is true it follows that OT is perpendicular to QT.

The converse statement is a bit awkward but also true:

Property ICP28. If two circles c(O, r) and q(Q, R) are orthogonal then two points of intersection of any secant of q through O are inverses of each other with respect to c and vice versa.

Here it is given that OT is tangent to q, O is outside q and by B3.P36 for any secant of q

OT2 = OA'*OA

but OT = r, O, A and A' are collinear with A and A' on the same side of O. Hence, A and A' are inverses of each other with respect to c.

Note here that because B3.P36 and B3.P37 apply to any secant q is invariant under inversion with respect to c which splits q into two mutually inverse arcs - the inner, inside c, and the outer, outside c. Barring the intersection points T and T' (which are fixed) all other points on the inner arc have their images on the outer arc and vice versa. So even though these points on q do not transform into themselves the circle as a whole does. This is yet another example of an invariant figure and an identity transformation.

Epilogue.

Now that we've seen some basic properties of inversion with respect to a circle we can identify certain patterns.

1). For the most part the absolute value of the radius of inversion didn't really matter. As such, quite often it is permissible to simply make it a unity. Algebraically, for two distances x and y we have then:

x*y = 1

Since each distance is just an inverse of the other it becomes more clear of where the name inversion may have come from. Earlier constructions also show yet another way to obtain the geometric reciprocals of given linear measures.

2). Similarity of triangles showed up and worked so often that we can branch it out into its own theorem:

If A and B are two distinct points different from the center of inversion O and a point at infinity, noncollinear with O and A' and B' are their images under the inversion with respect to a circle c(O, r) correspondingly, then the triangles OAB and OB'A' are similar.

You should have no problems coming up with a proof.

3). Reflection with respect to a straight line l transforms a point X not on it into a point X' such that XO = OX' where O is the intersection point of the perpendicular to l through X. Any point on l is transformed into itself. Reflecting X with respect to a straight line twice transforms X into itself.

We see that reflection with repsect to a straight line has some features that are very similar to those of inversion with respect to a circle. For that reason inversion with respect to a circle is sometimes called reflection in a circle.

Inversion with respect to a circle has many other interesting properties. But I think this is enough for now. You can try your hand at investigating the properties of inversion with respect to a sphere or move on to applying this material to problem solving in Part 2 of this tutorial.
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 Re: Inversion Tutorial. Part 1.   « Reply #12 on: Sep 29th, 2014, 5:21pm » Quote Modify

Inversion Tutorial Part 2 is now complete and it is located here.
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