User:CogitoErgoCogitoSum/sandbox2

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This is a continuation of the discussion started on my main sandbox page. In particular, this page is intended to discuss the specifics of constructing a circle center, and restoring the conditions to the Poncelet–Steiner theorem, when provided two non-intersecting circles devoid of their centers, but having a common centerline between them provided, or when provided three non-intersecting circles devoid of their centers.

Fixed Points of Involution

Given is a line $m$ (in black) and four arbitrary points $A$ , $B$ , $C$ , and $D$ on the line. Given also is an arbitrary circle in the plane; observe the center is not indicated as it is not utilized in this construction.

We may construct points $E$ and $F$ , known as the fixed points of involution. These points are defined by the four points on the line $m$ , and are independent of the circle. The order of these points is important as the relationship is particular. The six points hold a proportionality relationship: ${\frac {{\overline {AB}}\cdot {\overline {EC}}}{{\overline {AC}}\cdot {\overline {EB}}}}={\frac {{\overline {CD}}\cdot {\overline {BF}}}{{\overline {BD}}\cdot {\overline {CF}}}}$ .

Taking points $A$ , $B$ , $C$ , and $D$ out of order, the fixed points $E$ and $F$ would be constructed in different locations. Though, with the appropriate relabeling of points, the proportionality would still hold.

The fixed points may be constructed thus:

Choose a point $S$ arbitrarily on the circle.
Draw a line, four in total (each in red), passing through point S and each of the four points on the line: A, B, C, and D.
- These four lines intersect the circle at points $a$ , $b$ , $c$ , and $d$ , respectively.
Construct the line XY (in pink):
- This is the polar of the point $Q$ defined by the intersections of lines $ab$ and $cd$ (none of which are depicted).
- To find point $X$ , intersect lines $bd$ and $ac$ .
- To find point $Y$ , intersect (or find the concurrent) between lines $bc$ and $ad$ .
Intersect line XY with the circle, arriving at points e and f.
- Label the points such that $e$ is between $a$ and $b$ ; $f$ is between $c$ and $d$ .
Project points e and f back onto the line m from perspective point S:
- Draw lines $Se$ and $Sf$ (each in blue).
- Intersect these lines with line $m$ , arriving at points $E$ and $F$ , respectively.

In the above construction, it is assumed that segments $AB$ and $CD$ are exterior one another; i.e. they do not overlap. This has the consequence of placing points $E$ and $F$ interior their respective line segments. If segments $AB$ and $CD$ overlap or one is interior the other, then points $E$ and $F$ may have a less predictable and exterior placement.

It is the case that points $E$ and $F$ are both projective harmonic conjugates of one another with respect to both segments $AB$ and $CD$ , simultaneously. Alternatively phrased, line segment $EF$ is the segment to which both $A$ and $B$ are projective conjugates of one another, and both $C$ and $D$ are conjugates of one another.

Projecting a point through a circle defined by five points

The purpose of this construction is to find the intersection points of a circle, and a line defined by a point in the plane and a point on the circle. In other words, from the perspective of a point in the plane, a line to any point on the circle intersects at a new point to be constructed. The point on the circle is projected through the circle, along the line, to the other side of the circle. In order to complete this construction the arc of the circle is not required, nor a circle of any kind in the plane, but five points on the arc of the circle.

In the image below, a light blue dashed circle is apparent. This circle is not constructed by compass; the arc does not exist in the plane. It is only shown for convenience of exposition. We may not intersect lines with it directly.

Provided are five points in space - points $A$ , $B$ , $C$ , $D$ , $E$ arranged counter clockwise about the circle - such that all five exist on the circumference of the same circle. There exists also a point $M$ in space.

We wish to project any point on the circle (point $B$ is chosen in the construction below) to the opposite side of the circle as viewed from the perspective of point $M$ . That is to say, we wish to find the intersection point between the circle defined by the five points, and the line $MB$ , where point $B$ must necessarily be one of the five points on the circle circumference.

Construct line $MB$ (in dark blue)
Construct line CD (in red)
- These lines intersect at a point $I$ .
Construct lines BE and AD (both in orange)
- These lines intersect at a point $J$ .
Construct lines IJ and CE (both in green)
- These lines intersect at a point $K$ .
Construct line $AK$ (in purple)
Lines AK and MB intersect at a point B'.
- This point $B'$ is the desired point.
- It represents the projection of point $B$ , from point $M$ , through to the opposite side of the circle.
- It represents the intersection of line $MB$ with the circle defined by the five points.

From the perspective of point $M$ , both points $B$ and $B'$ are colinear with point $M$ , and existing on the circle. Thus point $B'$ is the intersection point between the circle and line $MB$ . This was achieved without the need of the arc; only five points on the circle were required.

Creating the polar of a point to a circle defined by five points

In the image below, a light blue dashed circle is apparent. This circle is not constructed by compass; the arc does not exist in the plane. It is only shown for convenience of exposition. We may not intersect lines with it directly. The circle is strictly identified by five unique points on its circumference, as per the previous construction, but only two of which - $B$ and $C$ - are depicted.

The polar is the line connecting points of tangency. In the image below, the tangent lines that pass through point $M$ to the circle defined by five points (points not depicted) will intersect the circle at the points of tangency. A single line passing through the points of tangency is known as the polar.

Choose two (of the five) points on the circle arbitrarily. In this example points $B$ and $C$ have been selected. The remaining three points $A$ , $D$ , and $E$ have been hidden in this construction, but are still necessary for the referenced sub-constructions.

Construct point $B'$ as the projection of point $B$ , per the previous construction.
Construct point $C'$ as the projection of point $C$ , per the previous construciton.
Draw lines BC' and B'C (both in red).
- These lines intersect at a point $X$ .
Draw lines BC and B'C' (both in green).
- These lines intersect at a point $Y$ , if they intersect.
Draw line XY (in blue).
- If point $Y$ does not exist then draw the concurrent through point $X$ .
- This is the desired line, the polar of point $M$ to the circle.

Observe that this construction is not significantly different than the one provided on in my first sandbox, specifically the construction of tangent lines to a circle. Polars, tangent lines and points of tangency are all intimately interrelated. Any line passing through point $M$ and passing through the circle suffices for that construction, but necessitates the use of both intersection points to the circle. Here, one of the intersections must be constructed as a projection of another. Additionally here we are using only one pair of lines through point $M$ rather than two - a variant discussed in the referenced construction.

Intersection of a line and the circle using fixed points and polars

In the image below, a light blue dashed circle is apparent. This circle is not constructed by compass; the arc does not exist in the plane. It is only shown for convenience of exposition. We may not intersect lines with it directly.

This construction will be used for circles defined by five points on the arc, or any circle in which line-circle intersection points cannot be directly found. The five points are not depicted, but will be required for the referenced sub-constructions. The referenced sub-constructions also require the use of a provided circle somewhere in the plane (its center point is not necessary).

Given a line $MN$ which passes through the circle defined by five points, we may find the points of intersection between the circle and the line.

From point M construct its polar to the circle (in purple).
- The polar intersects the line $MN$ at point $M'$ .
From point N construct its polar to the circle (in magenta).
- The polar intersects the line $MN$ at point $N'$ .
Points $M$ , $M'$ , $N'$ , and $N$ , in that order, define two fixed points of involution, points $I$ (on segment $MM'$ ) and $J$ (on segment $NN'$ ).
Points I and J exist on the circle and on the line, and are thus the desired points of intersection of the line with the circle.
- Points $I$ and $J$ are not illustrated in the image.

Finding additional points on a circle

The goal of this section is to construct points on the circle which is only identified by the presence of other points also already on the circle. The circle is not compass-drawn. A minimum of three points on the arc are required to identify a unique circle, but in order to construct additional points on the circle using straightedge only, more information than three points is required. Various scenarios are discussed below.

From five points on the circle

Given five distinct points on the arc of a circle, wherein the arc is not compass-drawn, additional points of the circle may be constructed. A variety of techniques are available.

Using the projection of a line through a circle

This is a trivial implementation of an earlier construction, projecting a point through a circle defined by five points. We may choose arbitrarily or construct any point in the plane. Through this point we may project any of the five points already on the circle, the pair of which define a line, through to the other side of the circle colinear with the other two points. In this way, additional points may be added to the set of points known to be on the circle.

Construct or arbitrarily choose a point $P$ in the plane.
Choose one of the five points on the circle arbitrarily.
- Point $B$ is chosen for example.
From the perspective of point P, project the point B through the circle.
- Point $B'$ is constructed on the circle, and is a sixth point.

by intersecting a line

Using the previous construction to intersect an arbitrary line with a circle of five points, any arbitrary line through the circle may have both of its intersection points with the circle constructed. This is a trivial implementation of a previous construction.

Draw an arbitrary line $m$ through the circle of five points.
Intersect line m with the circle.
- Points $I$ and $J$ are the points of intersection, and are two additional points of the circle.

It should be noted that this approach is unnecessarily complicated, and requires the implementation of other, simpler constructions such as the aforementioned projections through a circle in order to complete, which results in further point constructions on the circle, thus resulting in more than just two additional points of the circle. This construction is offered only as a possibility, not as a practical solution.

by the polar construction

This construction utilizes the polar construction, both from three cutting lines and from two cutting lines. First, using any of four points on the circle, cutting lines may be constructed which intersect at a point which we may call a pole. From the pole and using the two cutting lines, the polar may be constructed. Utilizing the fifth point, the pole, and the known polar, we may establish a third cutting line and thus a sixth point of intersection.

Using the properties of an inscribed hexagon (Pascal's theorem)

This is an implementation of Pascal's theorem, in particular as it is used with circles. The theorem states that whenever a hexagon is inscribed in a circle (conic section, more generally), opposite sides may be extended to form three distinct intersection points, even if "at infinity", all of which are colinear. The line of these intersection points is known as a Pascal line. This construction makes use of this property.

Given five points on a circle, four sides of a hexagon are established; that is, vertices may be arbitrarily paired to define edges. Accordingly, one pair of known opposing sides are definable. A degree of freedom left for the geometer is in the order in which the five points are selected to determine hexagonal sides. A hexagon may have a side omitted between any pair of the established points, and furthermore, the hexagon need not be simple; Pascal's theorem allows for self-intersecting (non-simple) hexagons. The arbitrariness of these selections contribute to the placement of the new point we aim to construct.

The known opposite sides, which were arbitrarily chosen, may be used to find a single point of intersection on Pascal's line. Any arbitrary line may be chosen through this point to be the Pascal line, as no known second point on the line is determinable from the provided information; this is a second degree of freedom left to the geometer also contributing to the arbitrariness of the final new point. The remaining two unpaired sides of the hexagon may be extended to Pascal's line, intersecting it and establishing its opposite side. The two new constructed opposite sides intersect one another on the circle at the hexagons sixth vertex.

Given are points $A$ , $B$ , $C$ , $D$ , and $E$ , all on the arc of the circle which is not here compass-drawn. The line segments $AB$ , $BC$ , $CD$ , and $DE$ (all in black), which are interior the circle, arbitrarily define five of the six sides of an inscribed hexagon. Further, sides $AB$ and $DE$ are arbitrarily chosen to be opposite one another.

Extend lines AB and DE (in red).
- These lines intersect at a point $X$ .
Draw an arbitrary Pascal line (in magenta) through point $X$ .
Through each of the other sides - BC and CD - extend the lines (each in green).
- These lines intersect Pascals line at points $Y$ and $Z$ , respectively.
Draw lines AZ and YE (each in blue).
- These lines intersect one another at a point $F$ .
- Point $F$ establishes the sixth point of the hexagon on the circle.
- Segments $FE$ and $FA$ (in black) are emphasized to complete the internal hexagon, which happens to be self-intersecting.

From four points on a circle and a tangent line

This may be viewed as the reverse of a tangent line construction, referencing the section Construction of a Tangent/Polar to a Circle or Arc by way of subsection from a point on the arc of a circle. In a previous construction, five points on a circle were used in order to create a tangent line to the circle through one of the points.

In that construction, these five points were chosen arbitrarily on an arc of a compass-drawn circle in the plane. The arc does not need to exist for the tangent through a point to be constructed, as the arc was never used, but as a curve upon which to arbitrarily place five points. If the points are provided and known to be on the arc already, the same construction may be achieved without the circle arc being drawn.

The construction in this section, however, attempts to construct one of the five points on the arc of the circle, using four other points which includes a point of tangency and a tangent line. This construction utilizes the geometric properties of the pentagram, and constructs the same set of lines in a different order, from a different initial set of conditions to arrive at a different end goal. In this case, we start with a tangent line (and the point of tangency) and three additional points on the arc of the circle, to arrive at a fifth point.

In the image, a point $A$ exists on the circle, and is a point or tangency for the tangent line (drawn in black). The circle is not compass-drawn, but is illustrated (in dashed gray) for expositional purposes; we may not directly intersect with the arc of the circle. Also on the circle are points $B$ , $C$ , and $E$ . We wish to construct a point $D$ between the latter two points. The required segments of the pentagram are drawn (in black): segments $AC$ , $CE$ , and $BE$ .

Each of the pentagrams sides must be extended as lines (in red).
- Line $BE$ intersects the tangent at a point $P$ .
Choose a point $M$ arbitrarily on line $AC$ .
Line $PM$ (in magenta) intersects line $CE$ at point $N$ .
Line BM and AN (both in blue) intersect at point D.
- Point $D$ completes the pentagram, and thus exists on the circle.

From three points on a circle and two tangent lines

Using a modified approach to the construction intersecting a line with the circle of an arc, it is possible using the same set of lines constructed in reverse order, to find two additional points (a fourth and a fifth) on a circle when only three are provided, two of which having tangent lines through them.

Finding points on a circle defined by a point in a coaxial system

This section allows us to construct points on the arc of a circle that is only identified by a single point in the plane, in a coaxial system. A coaxial system is a set of circles, even if they are yet to be constructed, any pair of which share the same radical axis. They share the same centerline as well, though this is not the defining attribute. Two compass-drawn circles are provided by hypothesis; these establish a radical axis and therefore define a coaxial system. For any point in space - excepting those on the radical axis - there exists a single circle passing through it which belongs to the same coaxial system. This section will demonstrate methods for constructing additional points on this circle, given the one initial point and the two provided circles.

Provided below are two compass-drawn circles, arbitrarily labeled $l$ and $r$ , for their left and right relative positions.

Using the tangent line and projective harmonic conjugate

From any point C in space, construct a point of tangency to circle l through which a line through C is tangent.
- The point of tangency is called point $A$ .
- The polar of point $C$ , line $c$ (dashed, gray) is provided for exposition.
- The second point of tangency, $T$ , is not needed for this construction but may still be used in a reiteration of these steps to establish a different additional point.
Construct the tangent line $AC$ (in orange).
From tangent point $A$ , construct the polar, line $a$ (in red), to circle $r$ .
The polar, line a, intersects the line AC at a point B.
- If point $C = B$ then the construction will fail (see next steps).
With respect to segment AB, find the projective harmonic conjugate of point C.
- This is point $D$ , the desired construction, and is on the same circle of the coaxial system as is point $C$ .
- Points $C$ and $D$ should be unique for a successful construction.
If point C=D then the construction has failed:
- It is the case that points $C = B$ , resulting in the harmonic conjugate also being coincident.
- Point $C$ is a point of tangency on the circle in question, defined by point $C$ in the coaxial system.
- Line $AC$ is a common point of tangency between circle $l$ and the circle defined by point $C$ .

Additional new points on the circle may be systematically constructed in sequence. Each newly constructed point $D$ may be relabeled the new point $C$ in the next iteration of the application of this construction, thereby producing a sequence of new points on the circle. Each constructed point becomes a new launch point, leading to one additional new point. This process may continue until and unless points $C = D$ , at which point the reiterative process may terminate. If insufficient points have been constructed, alternative solutions must be considered.

In each iteration of the construction, one of the tangent lines to circle $l$ from the point $C$ had already been previously constructed, and leads back to a previously constructed point. Only one tangent line is useful and has the potential to lead to new points. This is true for all constructed points on the circle, except the first. The original point $C$ , which defined the circle in the coaxial system, offered two tangents to circle $l$ , neither of which had previously been constructed, thus offering two unique new points. This may be where the original point $T$ becomes useful, and may be relabeled point $A$ . Switching the roles of points $A$ and $T$ of the first iteration will produce a new set of iterations, which may continue until again points $C = D$ .

Alternatively, we may also resolve the problem by swapping the roles of circles $l$ and $r$ , and construct tangent lines to the alternate circle. Placing points $A$ on circle $r$ , and finding its polar to circle $l$ instead, may also resolve some limitations.

It should also be noted that, in addition to the original provided point $C$ , only four new points need to be constructed, in the general case, or three new points and a known tangent line, or two new points and two tangent lines, before the constructions in the previous section Finding additional points on a circle may be utilized instead, which do not require a second circle or a coaxial system to work.

Furthermore, the next subsection offers a completely unique method based on projective points, of constructing new points on this circle in the coaxial system, that does not suffer the same limitations.

Using projective points

Given is a point point $C$ in the plane.

Through point C, draw an arbitrarily line m through circle r.
- Line $m$ intersects circle $r$ at two points $A$ and $B$ .
Choose an arbitrary point $S$ on circle $l$ .
Draw line SA.
- The line intersects circle $l$ at a point $a$ .
Draw line SB.
- The line intersects circle $l$ at a point $b$ .
Draw line ab.
- The line intersects line $m$ at a point $X$ .
- It is acceptable if point $X$ exists "at infinity".
Draw line SC.
- The line intersects circle $l$ at a point $c$ .
Draw line Xc.
- It is acceptable to use concurrent line constructions, if necessary.
- The line intersects circle $l$ at a point $d$ .
Draw line Sd.
- The line intersects line $m$ at a point $D$ .
- If these lines do not intersect then they are parallel; no circle defined by point C in the coaxial system exists because point C is on the radical axis.
  - This serves as a method of constructing two arbitrary parallel lines given two arbitrary circles devoid of their centers and a point on the radical axis.
Point $D$ is the desired constructed point, existing on the same circle as, and defined by, point $C$ in the coaxial system.

Additional points may be constructed as necessary. This construction may be repeated, using many arbitrary but unique lines $m$ through point $C$ . It is also valid to substitute point $C$ in the plane with any other point also on the same circle, such as any of the points previously constructed in the sequence of such points.

Only three points on the circle and two tangent lines, four points on the circle and one tangent line, or five points on the circle, are required before the constructions of the previous section Finding additional points on the circle may be utilized, which do not require a second circle or a coaxial system.

Finding the Centers of Two Circles with a Common Centerline

Given are two non-intersecting circles in the plane - $l$ and $r$ - through which a common centerline, $c$ , is drawn. The centers of these circles may be constructed. In the construction below, it is presumed that the provided circles are external one another, though this is not required.

Arbitrarily draw a line m in the plane such that it passes through each circle at least one intersection points each.
- Label the intersection points on circle $l$ as points $A$ and $B$ .
- Label the intersection points on circle $r$ as points $C$ and $D$ .
- If either pair of points are coincident then line $m$ is tangent to the associated circle; this simplifies the scenario. Refer to comments below.
Find the fixed points of involution on line m given the points A, B, C, and D, respectively.
- Fixed point $P$ corresponds to circle $l$ .
- Fixed point $P$ shall be on line segment $AB$ , if and only if circles $l$ and $r$ are exterior one another, and on the line but not the line segment otherwise.
- Fixed point $Q$ corresponds to circle $r$ .
- Fixed point $Q$ shall be on line segment $CD$ , if and only if circles $l$ and $r$ are exterior one another.
- Fixed points P and Q define two distinct circles in the coaxial system defined by circles l and r.
  - Let us call the circles $p$ and $q$ , respectively.
  - Due to the involution, each point represents a point of tangency on their circles, with line $m$ the common tangent to both circles $p$ and $q$ .
  - Circle $p$ is contained entirely within circle $l$ , and $q$ within $r$ , if and only if circles $l$ and $r$ are exterior one another.
Find the intersection points of each of circles p and q with their common centerline, by way of previous constructions:
- Construct four or more additional points on each circle.
- Construct or arbitrarily choose two distinct points on the centerline and their polars in the circles.
- These centerline points, along with where their polars intersect the centerline, define fixed points of an involution.
- The fixed points are the desired intersection points of the circle with the centerline.
- Denote the intersection points of circle $p$ as $U_{p}$ and $V_{p}$ , and the corresponding points of circle $q$ as $U_{q}$ and $V_{q}$ .
Draw lines PU_p and QU_q.
- These lines are parallel.
- Alternatively, the two points $V$ may be used instead. Only one corresponding set are required. Lines $PV p$ and $QV q$ suffice.
Construct the circle centers:
- Two new parallel lines in the plane now exist, which are not orthogonal to the common centerline.
- A new, distinct centerline may be constructed through each circle using previous techniques.
- Each new centerline intersects the common centerline at the circles center.

In such a scenario where the line $m$ is tangent to one or both circles, the construction simplifies greatly. If, for example, the line $m$ is tangent to circle $l$ , the following may be done. The point $A = B$ may be labeled $P$ , circle $l$ may be labeled circle $p$ . The intersection points of circle $p$ with the centerline may be labeled $U_{p}$ and $V_{p}$ . Steps 2 and 3 may be skipped, for the associated circle $l$ . Line $PU p$ is trivially constructed. The alternate circle $r$ for which line $m$ is not tangent must still undergo the full construction to derive line $QU q$ , after which steps 4 and 5 may be followed. A similar argument can be made for each circle in kind; the simplest scenario is if line $m$ is tangent to both circles.