Isoptic point and Simson line of a quadrilateral

# The perpendicular bisectors construction, the isoptic point and the Simson line of a quadrilateral

## 1. Introduction

The perpendicular bisector construction that we investigate in this paper arises very naturally in an attempt to find a replacement for a circumcenter in the case of a noncyclic quadrilateral . Indeed, while there is no circle going through all four vertices, for every triple of vertices there is a unique circle (called the triad circle) passing through them. The centers of these four triad circles can be taken as the vertices of a new quadrilateral, and the process can be iterated to obtain a sequence of noncyclic quadrilaterals:

To reverse the iterative process, one finds isogonal conjugates of each of the vertices with respect to the triangle formed by the remaining vertices of the quadrilateral.

It turns out that all odd generation quadrilaterals are similar, and all even generation quadrilaterals are similar. Moreover, there is a point that serves as the center of spiral similarity for any pair of odd generation quadrilaterals as well as for any pair of even generation quadrilaterals. The angle of rotation is or depending on whether the quadrilateral is concave or convex, and the ratio of similarity is a constant that is negative for convex noncyclic quadrilaterals, zero for cyclic quadrilaterals, and for concave quadrilaterals. If , the same special point turns out to be the limit point for the iterative process or for the reverse process.

The main goal of this paper is to prove the following

###### Theorem.

For each quadrilateral there is a unique point that has any (and, therefore, all) of the following properties:

1. is the center of the spiral similarity for any two odd (even) generation quadrilaterals in the iterative process;

2. Depending on the value of the ratio of similarity in the iterative process, there are the following possibilities:

1. If , the quadrilaterals in the iterated perpendicular bisectors construction converge to ;

2. If , the iterative process is periodic (with period or ); is the common center of rotations for any two odd (even) generation quadrilaterals;

3. If , the quadrilaterals in the reverse iterative process (obtained by isogonal conjugation) converge to ;

3. is the common point of the six circles of similitude for any pair of triad circles , , where , , , .

4. (isoptic property) Each of the triad circles is visible from at the same angle.

5. (generalization of circumcenter) The (directed) angle subtended by any of the quadrilateral’s sides at equals to the sum of the angles subtended by the same side at the two remaining vertices.

6. (isodynamic property) The distance from to any vertex is inversely proportional to the radius of the triad circle determined by the remaining three vertices.

7. is obtained by inversion of any of the vertices of the original quadrilateral in the corresponding triad-circle of the second generation:

 W=Invo(2)1(A)=Invo(2)2(B)=Invo(2)3(C)=Invo(2)4(D),

where , , , .

8. is obtained by composition of isogonal conjugation of a vertex in the triangle formed by the remaining vertices and inversion in the circumcircle of that triangle.

9. is the center of spiral similarity for any pair of triad circles (of possibly different generations). That is, for all .

10. The pedal quadrilateral of is a (nondegenerate) parallelogram. Moreover, its angles equal to the angles of the Varignon parallelogram.

Many of these properties of were known earlier. In particular, several authors (G.T. Bennett in an unpublished work, De Majo [2], H.V. Mallison [3]) have considered a point that is defined as the common center of spiral similarities. Once the existence of such a point is established, it is easy to conclude that all the triad circles are viewed from this point under the same angle (this is the so-called isoptic property). Since it seems that the oldest reference to the point with such an isoptic property is to an unpublished work of G.T. Bennett given by H.F. Baker [1] in his “Principles of Geometry” (vol. 4), in 1925, we propose to call the center of spiral similarities in the iterative process Bennett’s isoptic point.

C.F. Parry and M.S. Longuet-Higgins [4] showed the existence of a point with property 7 using elementary geometry.

Mallison [3] defined using property 3 and credited T. McHugh for observing that this implies property 5.

Several authors, including Wood [5] and De Majo [2], have looked at the properties of the isoptic point from the point of view of the unique rectangular hyperbola going through the vertices of the quadrilateral, and studied its properties related to cubics. For example, P.W. Wood [5] considered the diameters of the rectangular hyperbola that go through . Denoting by the other endpoints of the diameters, he showed that the isogonal conjugates of these points in triangles , ,, coincide. Starting from this, he proved properties 4 and 7 of the theorem. He also mentions the reversal of the iterative process using isogonal conjugation (Also found in [5], [6], [15]). Another interesting property mentioned by Wood is that is the Fregier point of the center of the rectangular hyperbola for the conic , where O is the center of the rectangular hyperbola.

De Majo [2] uses the property that inversion in a point on the circle of similitude of two circles transforms the original circles into a pair of circles whose radii are inversely proportional to those of the original circles to show that that there is a common point of intersection of all circles of similitude. He describes the iterative process and states property 1, as well as several other properties of (including 8). Most statements are given without proofs.

Scimemi [6] describes a Möbius transformation that characterizes : there exists a line going through and a circle centered at such that the product of the reflection in the line with the inversion in the circle maps each vertex of the first generation into a vertex of the second generation.

The fact that the third generation quadrilateral is similar to the original quadrilateral and the expression for the ratio of similarity has appeared in the form of a problem by V.V. Prasolov in [7, 8]. Independently, the same question was formulated by J. Langr [9] , and the expression for the ratio (under certain conditions) was obtained by D. Bennett [10] (apparently, no relation to G.T. Bennett mentioned above), and J. King [11]. A paper by G.C. Shepard [12] found an expression for the ratio as well. (See [13] for a discussion of these works).

Properties 9 and 10 appear to be new.

For the convenience of the reader, we give a complete and self-contained exposition of all the properties in the Theorem above, as well as proofs of several related statements.

In addition to investigating properties of , we show that there is a unique point for which the feet of the perpendiculars to the sides lie on a straight line. In analogy with the case of a triangle, we call this line the Simson line of a quadrilateral and the point — the Simson point. The existence of such a point is stated in [16] where it is obtained as the intersection of the Miquel circles of the complete quadrilateral.

Finally, we introduce a version of isogonal conjugation for a quadrilateral and show that the isogonal conjugate of is a parallelogram, and that of the Simson point is a degenerate quadrilateral whose vertices are at infinity, in analogy with the case of the points on the circumcircle of a triangle.

## 2. The iterative process

Let be a quadrilateral. If is cyclic, the center of the circumcircle can be found as the intersection of the perpendicular bisectors to the sides of the quadrilateral.

Assume that is a noncyclic quadrilateral1. Is there a point that, in some sense, plays the role of the circumcenter? Let be the quadrilateral formed by the intersections of the perpendicular bisectors to the sides of . The vertices of the new quadrilateral are the circumcenters of the triad-triangles , , and formed by vertices of the original quadrilateral taken three at a time.

Iterating this process (i.e., constructing the vertices of the next generation quadrilateral by intersecting the perpendicular bisectors to the sides of the current one), we obtain the successive generations, , and so on, see Figure 2.1.

The first thing we note about the iterative process is that it can be reversed using isogonal conjugation. Recall that given a triangle and a point , the isogonal conjugate of with respect to (denoted by ) is the point of intersection of the reflections of the lines , and in the angle bisectors of , and respectively. One of the basic properties of isogonal conjugation is that the isogonal conjugate of is the circumcenter of the triangle obtained by reflecting in the sides of (see, e.g., [15] for more details). This property immediately implies

###### Theorem 1.

The original quadrilateral can be reconstructed from the second generation quadrilateral using isogonal conjugation:

 A1=Iso△D2A2B2(C2),B1=Iso△A2B2C2(D2),C1=Iso△B2C2D2(A2),D1=Iso△C2D2A2(B2).

The following theorem describes the basic properties of the iterative process.

###### Theorem 2.

1. degenerates to a point if and only if is cyclic.

2. If is not cyclic, the corresponding angles of the first and second generation quadrilaterals are supplementary:

 ∠A1+∠A2=∠B1+∠B2=∠C1+∠C2=∠D1+∠D2=π.
3. If is not cyclic, all odd generation quadrilaterals are similar to each other and all the even generation quadrilaterals are similar to each other:

 Q(1)∼Q(3)∼Q(5)∼…, Q(2)∼Q(4)∼Q(6)∼….
4. All odd generation quadrilaterals are related to each other via spiral similarities with respect to a common center.

5. All even generation quadrilaterals are also related to each other via spiral similarities with respect to a common center.

6. The angle of rotation for each spiral similarity is (for a convex quadrilateral) or a (for a concave quadrilateral). The ratio of similarity is

 (2.1) r=14(cotα+cotγ)⋅(cotβ+cotδ),

where , , and are the angles of .

7. The center of spiral similarities is the same for both the odd and the even generations.

###### Proof.

The first and second statements follow immediately from the definition of the iterative process. To show that all odd generation quadrilaterals are similar to each other and all even generation quadrilaterals similar to each other, it is enough to notice that both the corresponding sides and the corresponding diagonals of all odd (even) generation quadrilaterals are pairwise parallel.

Let be the center of spiral similarity taking into . Similarly, let be the center of spiral similarity taking into . Denote the midpoints of segments and by and . (See fig. 2.2). To show that and coincide, notice that . Since the corresponding sides of these triangles are parallel, they are related by a spiral similarity. Since and , it follows that . Let now be the center of spiral similarity that takes into . By the same reasoning, , which implies that . Continuing by induction, we conclude that the center of spiral similarity for any pair of odd generation quadrilaterals coincides with that for any pair of even generation quadrilaterals. We denote this point by . ∎

Properties (2) and (3) of the theorem above imply the following

###### Corollary 3.

The even and odd generation quadrilaterals are similar to each other if and only if is a trapezoid.

The ratio of similarity takes values in and characterizes the shape of in the following way:

1. if and only if is convex. Moreover, if and only if is cyclic.

2. if and only if is concave. Moreover, if and only if is orthocentric (that is, each of the vertices is the orthocenter of the triangle formed by the remaining three vertices. Alternatively, an orthocentric quadrilateral is characterized by being a concave quadrilateral for which the two opposite acute angles are equal).

For convex quadrilaterals, can be viewed as a measure of how noncyclic the original quadrilateral is. Recall that since the opposite angles of a cyclic quadrilateral add up to , the difference

 (2.2) |(α+γ)−π|=|(β+δ)−π|

can be taken as the simplest measure of noncyclicity. This measure, however, treats two quadrilaterals with equal sums of opposite angles as equally noncyclic. The ratio provides a refined measure of noncyclicity. For example, for a fixed sum of opposite angles, , , where , the convex quadrilateral with the smallest is the parallelogram with , .

Similarly, for concave quadrilaterals, measures how different the quadrilateral is from being orthocentric.

Since the angles between diagonals are the same for all generations, it follows that the ratio is the same for all pairs of consecutive generations:

 Area(Q(n))Area(Q(n−1))=|r|.

Assuming the quadrilateral is noncyclic, there are the following three possibilities:

1. When (which can only happen for convex quadrilaterals), the quadrilaterals in the iterative process converge to .

2. When , the quadrilaterals in the inverse iterative process converge to .

3. When , all the quadrilaterals have the same area. The iterative process is periodic with period for all quadrilaterals with , except for the following two special cases. If is either a parallelogram with angle (so that ) or forms an orthocentric system (so that ), we have , , and the iterative process is periodic with period .

By setting in formula (2.1), we obtain the familiar relations between the sides and diagonals of a cyclic quadrilateral :

Since the vertices of the next generation depend only on the vertices of the previous one (but not on the way the vertices are connected), one can see that and for the (self-intersecting) quadrilaterals and coincide with those for . This observation allows us to prove the following

###### Corollary 4.

The angles between the sides and the diagonals of a quadrilateral satisfy the following identities:

 (cotα+cotγ)⋅(cotβ+cotδ) = (cotα1−cotβ2)⋅(cotδ2−cotγ1), (cotα+cotγ)⋅(cotβ+cotδ) = (cotδ1−cotα2)⋅(cotβ1−cotγ2)

where , are the directed angles formed between sides and diagonals of a quadrilateral (see Figure 2.3).

###### Proof.

Since the (directed) angles of are and the directed angles of are , the identities follow from formula (2.1) for the ratio of similarity. ∎

## 3. Properties of the Center of Spiral Similarity

We will show that , defined as the limit point of the iterated perpendicular bisectors construction in the case that (or of its reverse in the case that ), is the common center of all spiral similarities taking any of the triad circles into another triad circle in the iterative process.

First, we will prove that any of the triad circles of the first generation quadrilateral can be taken into another triad circle of the first generation by a spiral similarity centered at (Theorem 8). This result allows us to view as a generalization of the circumcenter for a noncyclic quadrilateral (Corollary 9 and Corollary 12), to prove its isoptic (Theorem 10), isodynamic (Corollary 13) and inversive (Theorem 14) properties, as well as to establish some other results. We then prove several statements that allow us to conclude (see Theorem 23) that serves as the center of spiral similarities for any pair of triad circles of any two generations.

Several objects associated to a configuration of two circles on the plane will play a major role in establishing properties of . We will start by recalling the definitions and basic constructions related to these objects.

### 3.1. Preliminaries: circle of similitude, mid-circles and the radical axis of two circles.

Let and be two (intersecting2) circles on the plane with centers and and radii and respectively. Let and be the points of intersection of the two circles. There are several geometric objects associated to this configuration (see Figure 3.1):

1. The circle of similitude is the set of points on the plane such that the ratio of their distances to the centers of the circles is equal to the ratio of the radii of the circles:

 |PO1||PO2|=R1R2.

In other words, is the Apollonian circle determined by points , and ratio .

2. The radical axis can be defined as the line through the points of intersection.

3. The two mid-circles (sometimes also called the circles of antisimilitude) and are the circles that invert into , and vice versa:

 InvMCi(o1,o2)(o1)=o2,i=1,2.

Here are several important properties of these objects (see [16] and [17] for more details):

1. is the locus of centers of spiral similarities taking into . For any , there is a spiral similarity centered at that takes into . The ratio of similarity is and the angle of rotation is .

2. Inversion with respect to takes centers of and into each other:

 InvCS(o1,o2)(O1)=O2.
3. Inversion with respect to any of the mid-circles exchanges the circle of similitude and the radical axis:

 InvMCi(o1,o2)(CS(o1,o2))=RA(o1,o2),i=1,2.
4. The radical axis is the locus of centers of all circles that are orthogonal to both and .

5. For any , inversion in a circle centered at takes the circle of similitude of the original circles into the radical axis of the images, and the radical axis of the original circles into the circle of similitude of the images:

 CS(o1,o2)′=RA(o′1,o′2),
 RA(o1,o2)′=CS(o′1,o′2).

Here denotes the image of an object under the inversion in a circle centered at .

6. Let be points on the circles respectively. Then

 (3.1) ∠AMB=∠AKB+∠ALB,

where the angles are taken in the sense of directed angles.

7. Let be a chord of a circle and be a chord of a circle . Then are on a circle if and only if

It is also useful to recall the construction of the center of a spiral similarity given the images of two points. Suppose that and are transformed into and respectively. Let . The center of the spiral similarity can be found as the intersection , where here and below stands for the circle going through .

We will call point in this construction the joint point associated to two given points and their images under spiral similarity.

There is another spiral similarity associated to the same configuration of points. Let be the joint point for the spiral similarity taking and into and respectively. A simple geometric argument shows that the center of this spiral similarity, determined as the intersection of the circles , coincides with . We will call such a pair of spiral similarities centered at the same point associated spiral similarities.

Let be the spiral similarity centered at that takes into . The following Lemma will be useful when studying properties of the limit point of the iterative process (or of its inverse):

###### Lemma 5.

Let and be two circles centered at and respectively and intersecting at points and . Let be points on the circle of similitude such that and are symmetric to each other with respect to the line of centers, . Then the joint points corresponding to taking , by and taking , by coincide. The common joint point lies on .

###### Proof.

Perform inversion in the mid-circle. The image of is the radical axis (i.e., the line through and ). The images of and lie on the line and are symmetric with respect to . Similarly, the images of and are symmetric with respect to and lie on the line of centers. By abuse of notation, we will denote the image of a point under inversion in the mid-circle by the same letter.

The lemma is equivalent to the statement that lies on the circle . To show this, note that since and W lie on a circle, we have Since and it follows that which implies that lie on a circle. After inverting back in the mid-circle, we obtain the result of the lemma. ∎

Notice that the lemma is equivalent to the statement that

 RR1,2∩SS2,1=(WRO1)∩(WSO2)∈O1O2.

### 3.2. W as the center of spiral similarities for triad circles of Q(1).

Denote by and the triad circles , , and respectively.3 For triad circles in other generations, we add an upper index indicating the generation. For example, denotes the first triad-circle in the rd generation quadrilateral, i.e., circle . Let and be the triad triangles , , and respectively.

Consider two of the triad circles of the first generation, and , . The set of all possible centers of spiral similarity taking into is their circle of similitude . If is a nondegenerate quadrilateral, it can be shown that and intersect at two points (i.e. are not tangent). Let be the other point of intersection of and .4

Let be the spiral similarity centered at that takes into for any .

###### Lemma 6.

Spiral similarities have the following properties:

1. .

2. .

###### Proof.

Assume that . Let be the joint point of the spiral similarity (centered at ) taking into and into . Since points lie on a circle (see Lemma 5), it follows that . Thus, is a diameter of Since is centered at , the circles and are tangent at . It is easy to see that the converse is also true: if and are tangent at , then .

Since lie on a circle, it follows that . Since and under , . This implies that the circles and are tangent at . It is easy to see that is tangent to if and only if is tangent to .

Similarly to the above, let be the joint point of the spiral similarity centered at and taking into . Then . Similarly to the argument above, is tangent to if and only if is tangent to . This is equivalent to .

The second statement follows since . (Here and below the compositions of transformations are read right to left). ∎

Notice that circles and have two common vertices, and . The next Lemma shows that takes (i.e., the third vertex on ) to (the third vertex on ). This property is very important for showing that any triad circle from the first generation can be transformed into another triad circle from the first generation by a spiral similarity centered at . Similar properties hold for and . Namely, we have

, , .

###### Proof.

Lemma 6 shows that implies . Assume that . To find the image of under , represent the latter as the composition . First, , where is as in Lemma 6, see Figure 3.3. For brevity, let . (The indices refer to the fact that is the image of under spiral similarity taking into ).

Now we construct . By Lemma 5, , where is as in Lemma 6. Applying Lemma 5 to the circle , we conclude that it passes through . Since by assumption , it follows that . Thus, . The other statements in the Lemma can be shown in a similar way. ∎

The last Lemma allows us to show that lies on all of the circles of similitude .

for all .

###### Proof.

By definition, . We will show that for any .

Recall that . Let be the second point in the intersection , so that . By Lemma 7, . Since , it follows that , which implies that . Therefore, is the common point for all the circles of similitude , . ∎

### 3.3. Properties of W

The angle property (3.1) of the circle of similitude implies

###### Corollary 9.

The angles subtended by the quadrilateral’s sides at are as follows (see Figure 3.4):

 ∠AWB = ∠ACB+∠ADB, ∠BWC = ∠BAC+∠BDC, ∠CWD = ∠CAD+∠CBD, ∠DWA = ∠DBA+∠DCA.

This allows us to view as a replacement of the circumcenter in a certain sense: the angle relations above are generalizations of the relation between the angles in a cyclic quadrilateral with circumcenter . (Of course, in this special case, ).

Since for all , can be used as the center of spiral similarity taking any of the triad circles into another triad circle. This implies the following

###### Theorem 10.

(Isoptic property) All the triad circles subtend equal angles at

In particular, is inside of all of the triad circles in the case of a convex quadrilateral and outside of all of the triad circles in the case of a concave quadrilateral. (This was pointed out by Scimemi in [6]). If is inside of a triad circle, the isoptic angle equals to , where and are the points on the circle so that goes through and . (See Figure 3.5, where and are halves of the isoptic angle in and respectively). If is outside of a triad circle centered at and is the tangent line to the circle, so that is point of tangency, is half of the isoptic angle. Inverting in a triad circle of the second generation, we get that the triad circles are viewed at equal angles from the vertices opposite to their centers (see Figure 3.5).

Recall that the power of a point with respect to a circle centered at with radius is the square of the length of the tangent from to the circle, that is,

 h=|PO|2−R2.

The isoptic property implies the following

###### Corollary 11.

The powers of with respect to triad circles are proportional to the squares of the radii of the triad circles.

This property of the isoptic point was shown by Neville in [14] using tetracyclic coordinates and the Darboux-Frobenius identity.

Let be sides of the quadrilateral. For any , let be the foot of the perpendicular bisector of side on the opposite side. (E.g., is the intersection of the perpendicular bisector to the side and the side ). The following corollary follows from Lemma 7 and expresses as the point of intersection of several circles going through the vertices of the first and second generation quadrilaterals, as well as the intersections of the perpendicular bisectors of the original quadrilateral with the opposite sides (see Fig. 3.6).

###### Corollary 12.
 (3.2) W=(A1FbB2)∩(A1FcD2)∩(B1FcC2)∩(B1FdA2)∩(C1FdD2)∩(C1FaB2)∩(D1FaA2)∩(D1FbC2),

This property can be viewed as the generalization of the following property of the circumcenter of a triangle:

Given a triangle with sides opposite to vertices , let denote the feet of the perpendicular bisector to side on the side (or its extension), where . Then the circumcenter is the common point of three circles going through vertices and feet of the perpendicular bisectors in the following way:5 (3.3)

The similarity between (3.3) and (3.2) supports the analogy of the isoptic point with the circumcenter.

The last corollary provides a quick way of constructing . First, construct two vertices (e.g., and ) of the second generation by intersecting the perpendicular bisectors. Let be the intersection of the lines and . Then is obtained as the second point of intersection of the two circles and .

Recall the definition of isodynamic points of a triangle. Let be a triangle with sides opposite to the vertices . For each , where , consider the circle centered at and going through . The circle of similitude of two distinct circles and is the Apollonian circle with respect to points with ratio . It is easy to see that the three Apollonian circles intersect in two points, and , which are called the isodynamic points of the triangle.

Here are some properties of isodynamic points (see, e.g., [16], [17] for more details):

1. The distances from (and ) to the vertices are inversely proportional to the opposite side lengths:

 (3.4) |SA1|:|SA2|:|SA3|=1a1:1a2:1a3.

Equivalently,

 |SAi|:|SAj|=sinαj:sinαi,i≠j∈{1,2,3},

where is the angle in the triangle. The isodynamic points can be characterized as the points having this distance property. Note that since the radii of the circles used to define the circles of similitude are the sides, the last property means that distances from isodynamic points to the vertices are inversely proportional to the radii of the circles.

2. The pedal triangle of a point on the plane of is equilateral if and only if the point is one of the isodynamic points.

3. The triangle whose vertices are obtained by inversion of with respect to a circle centered at a point is equilateral if and only if is one of the isodynamic points of .

It turns out that has properties (Corollary 13, Theorem 29, Theorem 26) similar to properties 1–3 of .

###### Corollary 13.

(Isodynamic property of ) The distances from to the vertices of the quadrilateral are inversely proportional to the radii of the triad-circles going through the remaining three vertices:

 |WA1|:|WB1|:|WC1|:|WD1|=1R3:1R4:1R1:1R2,

where is the radius of the triad-circle . Equivalently, the ratios of the distances from to the vertices are as follows:

 |WA1|:|WB1| = |A1C1|sinγ:|B1D1|sinδ, |WA1|:|WC1| = sinγ:sinα, |WB1|:|WD1| = sinδ:sinβ.

From analysis of similar triangles in the iterative process, it is easy to see that the limit point of the process satisfies the above distance relations. Therefore, (defined at the beginning of this section as the second point of intersection of and ) is the limit point of the iterative process.

One more property expresses as the image of a vertex of the first generation under the inversion in a triad circle of the second generation. Namely, we have the following

###### Theorem 14.

(Inversive property of W)

 (3.5) W=Invo(2)1(A1)=Invo(2)2(B1)=Invo(2)3(C1)=Invo(2)4(D1).
###### Proof.

To prove the first equality, perform inversion in a circle centered at . The image of a point under the inversion will be denoted by the same letter with a prime. The images of the circles of similitude , and are the perpendicular bisectors of the segments , and respectively. By Theorem