The disk with which the folding is performed shall be represented by a circle x 2 + y 2 = r 2 with radius r > 0 and center C = ( 0 , 0 ) . The point D inside the circle and different from the center C shall be described as D = ( d , 0 ) with | d | < r . The fold performed by laying a point A 0 of the circumference of the disk on top of D can be described by the perpendicular bisector of the line segment A 0 D ¯ , where A 0 = ( ξ 0 , η 0 ) fulfills ξ 0 2 + η 0 2 = r 2 . The straight line running through D = ( d , 0 ) and A 0 has the form

η 0 x − ( ξ 0 − d ) y = η 0 d . (1)

The perpendicular bisector of line segment A 0 D ¯ is given by

( ξ 0 − d ) x + η 0 y = 1 2 ( r 2 − d 2 ) . (2)

The point of intersection of both straight lines (1) and (2) is M 0 = ( 1 2 ( ξ 0 + d ) , 1 2 η 0 ) . They are perpendicular to each other because their normal vectors are orthogonal: η 0 ( ξ 0 − d ) − ( ξ 0 − d ) η 0 = 0 . See Figure 1.

The term “sequential folding” was established by Gorkin and Shaffer [1] and means: After the first fold, proceed in such a way as to start the next fold where the previous fold ended. In contrast to Gorkin and Shaffer [1] in this paper after the first fold, the second fold and the third fold are starting at the two different ends of the first fold on the circle x 2 + y 2 = r 2 , at ( x 1 , y 1 ) and ( x 2 , y 2 ) respectively. This shall be named “bidirectional folding”. In the second fold, A 1 = ( ξ 1 , η 1 ) has to be chosen on the circle x 2 + y 2 = r 2 in such a way that ( x 1 , y 1 ) is a point on the perpendicular bisector of A 1 D ¯ :

( ξ 1 − d ) x 1 + η 1 y 1 = 1 2 ( r 2 − d 2 ) ,     ξ 1 2 + η 1 2 = r 2 . (3)

In the third fold, a point A 2 = ( ξ 2 , η 2 ) has to be chosen on the circle x 2 + y 2 = r 2 such that ( x 2 , y 2 ) is a point on the perpendicular bisector of A 2 D ¯ :

( ξ 2 − d ) x 2 + η 2 y 2 = 1 2 ( r 2 − d 2 ) ,     ξ 2 2 + η 2 2 = r 2 . (4)

The final problem will be, to show that perpendicular bisectors of A 1 D ¯ and A 2 D ¯

( ξ 1 − d ) x + η 1 y = 1 2 ( r 2 − d 2 ) ( ξ 2 − d ) x + η 2 y = 1 2 ( r 2 − d 2 ) (5)

have a point of intersection ( x T , y T ) on the circle x 2 + y 2 = r 2 . Because of the quadratic Equations in (3) and (4) involved in the calculation of ( ξ 1 , η 1 ) and ( ξ 2 , η 2 ) we get two solutions ( ( ξ 1 ) ± , ( η 1 ) ∓ ) and ( ( ξ 2 ) ± , ( η 2 ) ∓ ) respectively. The number of solutions can be reduced, however, looking at two straight lines: one passing through ( ( ξ 1 ) + , ( η 1 ) − ) and ( ( ξ 1 ) − , ( η 1 ) + ) the other through ( ( ξ 2 ) + , ( η 2 ) − ) and ( ( ξ 2 ) − , ( η 2 ) + ) . It will be shown that the point of intersection ( x S , y S ) of both straight lines is located on the circle x 2 + y 2 = r 2 and is equal to A 0 = ( ξ 0 , η 0 ) .

The procedure described above constructs an arbitrary triangle circumscribing an ellipse and having vertices on a circle. This is in contrast to the approach in [1], where a special triangle was chosen to circumscribe an ellipse and have vertices on a circle. The generalization to an arbitrary triangle is accomplished by employing Poncelet’s Theorem [2]. In a remark in [1], a different way of proof is indicated with a Blaschke product of degree three, treated more extensively in [3].

Next we want to find the points of intersection of (2) with the circle x 2 + y 2 = r 2 . Assuming η 0 ≠ 0 Equation (2) can be rewritten as

y = − ξ 0 − d η 0 x + r 2 − d 2 2 η 0 (6)

which gives after substitution into the equation of the circle and multiplication with η 0 2

η 0 2 x 2 + ( − ( ξ 0 − d ) x + 1 2 ( r 2 − d 2 ) ) 2 − η 0 2 r 2 = 0.

So we get

( ( ξ 0 − d ) 2 + η 0 2 ) x 2 − ( ξ 0 − d ) ( r 2 − d 2 ) x + 1 4 ( r 2 − d 2 ) 2 − η 0 2 r 2 = 0. (7)

The highest coefficient in (7) is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 because

( ξ 0 − d ) 2 + η 0 2 = r 2 − 2 ξ 0 d + d 2 ≥ r 2 − 2 r | d | + d 2 = ( r − | d | ) 2 > 0. (8)

So we get

x 2 − ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 x + 1 4 ( r 2 − d 2 ) 2 − η 0 2 r 2 ( ξ 0 − d ) 2 + η 0 2 = 0. (9)

Before writing down the solutions of (9) we are introducing the expression

R ( ξ 0 , η 0 ) = 4 r 2 ( ( ξ 0 − d ) 2 + η 0 2 ) − ( r 2 − d 2 ) 2 . (10)

Lemma 1

R ( ξ 0 , η 0 ) is positive for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 .

Proof: The expression ( ξ 0 − d ) 2 + η 0 2 is bounded below according to (8). Thus

R ( ξ 0 , η 0 ) ≥ 4 r 2 ( r − | d | ) 2 − ( r − | d | ) 2 ( r + | d | ) 2 ≥ ( r − | d | ) 2 ( 4 r 2 − ( r + | d | ) 2 ) .

For the expression 4 r 2 − ( r + | d | ) 2 holds

4 r 2 − ( r + | d | ) 2 = ( 2 r − ( r + | d | ) ) ( 2 r + ( r + | d | ) ) = ( r + | d | ) ( 3 r + | d | ) > 0.

Therefore R ( ξ 0 , η 0 ) is positive for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 . □

With expression (10) the solutions of (9) have the form

x 1 , 2 = 1 2 ( ( ξ 0 − d ) 2 + η 0 2 ) ( ( ξ 0 − d ) ( r 2 − d 2 ) ± η 0 R ( ξ 0 , η 0 ) ) , (11)

where the index 1 of x shall correspond to the +sign and the index 2 to the −sign. Substituting (11) into (6) we obtain

y 1,2 = 1 2 ( ( ξ 0 − d ) 2 + η 0 2 ) ( η 0 ( r 2 − d 2 ) ∓ ( ξ 0 − d ) R ( ξ 0 , η 0 ) ) , (12)

where the index 1 of y corresponds to the −sign and the index 2 to the +sign. For i ∈ { 1,2 } this makes for x i 2 + y i 2 = r 2 .

Although the derivation of (11) was carried out under the assumption η 0 ≠ 0 , the case η 0 = 0 can be recovered from (11). For η 0 = 0 follows ξ 0 = ± r . For ξ 0 = + r we get x 1 = 1 2 ( r + d ) and for ξ 0 = − r we have x 2 = − 1 2 ( r − d ) . In each case we are getting from x 1 , 2 2 + y 1 , 2 2 = r 2 two y -values: y 1 = ± 1 2 ( 3 r + d ) ( r − d ) and y 2 = ± 1 2 ( 3 r − d ) ( r + d ) .

From (11) and (12) we can deduce for i ∈ { 1,2 } :

( ξ 0 − d ) x i + η 0 y i = 1 2 ( r 2 − d 2 ) . (13)

The term “Gardner ellipse” has been used in [1] for the ellipse constructed by folding disks, going back to Gardner’s publication [4]. In order to calculate the point of intersection ( x B , y B ) of the straight line through C = ( 0 , 0 ) and A 0 = ( ξ 0 , η 0 ) and straight line (2) we are looking at the linear system

η 0 x B − ξ 0 y B = 0 ( ξ 0 − d ) x B + η 0 y B = 1 2 ( r 2 − d 2 ) . (14)

For the determinant D B of the linear system (14) holds:

D B = η 0 2 + ( ξ 0 − d ) ξ 0 = η 0 2 + ξ 0 2 − ξ 0 d = r 2 − ξ 0 d

and we have the following positive lower bound

D B = r 2 − ξ 0 d ≥ r 2 − | ξ 0 | | d | ≥ r 2 − r | d | = r ( r − | d | ) > 0.

The solutions of the linear system (14) are

x B = ξ 0 ( r 2 − d 2 ) 2 ( r 2 − ξ 0 d ) ,       y B = η 0 ( r 2 − d 2 ) 2 ( r 2 − ξ 0 d ) . (15)

Since B = ( x B , y B ) is lying on straight line (2), the perpendicular bisector of A 0 D ¯ , the line segments D B ¯ and A 0 B ¯ have the same length. Thus

D B ¯ + B C ¯ = A 0 B ¯ + B C ¯ = A 0 C ¯ = r . (16)

See Figure 2.

Letting point A 0 = ( ξ 0 , η 0 ) move on circle x 2 + y 2 = r 2 we obtain a sequence of points B = ( x B , y B ) which have a constant sum of distances from two fixed points C and D. According to (16) this sum of distances is equal to r.

Therefore the so generated points are forming an ellipse, the so-called “Gardner

ellipse”. It has foci C and D and midpoint ( 1 2 d ,0 ) . The semi axes of the Gardner ellipse are given by a = r 2 and b = 1 2 r 2 − d 2 . The formula of the ellipse is

( x − d 2 ) 2 a 2 + y 2 b 2 = 1. (17)

The coordinates of B = ( x B , y B ) given in (15) fulfill Equation (17), if the values for the semi axes a and b are substituted.

The tangent to the ellipse (17) at the point ( x B , y B ) has the form

( x B − d 2 ) ( x − d 2 ) a 2 + y B y b 2 = 1. (18)

Proposition 1

The tangent to ellipse (17) at the point ( x B , y B ) , given by (18), is equivalent to the perpendicular bisector (2) of line segment A 0 D ¯ .

Proof: From (15) we get

x B − d 2 = ξ 0 ( r 2 − d 2 ) − d ( r 2 − ξ 0 d ) 2 ( r 2 − ξ 0 d ) = r 2 ( ξ 0 − d ) 2 ( r 2 − ξ 0 d )

and

x B − d 2 a 2 = 2 ξ 0 − d r 2 − ξ 0 d ,         y B b 2 = 2 η 0 r 2 − ξ 0 d .

This leads to

( x B − d 2 ) ( x − d 2 ) a 2 + y B y b 2 = 2 ξ 0 − d r 2 − ξ 0 d ( x − d 2 ) + 2 η 0 r 2 − ξ 0 d y = 1 ,

( ξ 0 − d ) ( x − d 2 ) + η 0 y = 1 2 ( r 2 − ξ 0 d ) ,

( ξ 0 − d ) x + η 0 y = 1 2 ( r 2 − ξ 0 d ) + d 2 ( ξ 0 − d ) = 1 2 ( r 2 − d 2 ) ,

which is the perpendicular bisector (2) of line segment A 0 D ¯ .

Remark: Because of the relations for the semi axes of the Gardner ellipse a = r 2 and b = 1 2 r 2 − d 2 an arbitrary ellipse with semi axes a and b can be interpreted as Gardner ellipse, choosing r = 2 a as radius of the surrounding circle and d = r 2 − ( 2 b ) 2 = 2 a 2 − b 2 as distance of the foci.

Having obtained the points ( x 1 , y 1 ) and ( x 2 , y 2 ) according to (11) and (12) the second fold is starting from ( x 1 , y 1 ) and the third fold from ( x 2 , y 2 ) , described by Equations (3) and (4) respectively.

The calculation of the points A 1 = ( ξ 1 , η 1 ) and A 2 = ( ξ 2 , η 2 ) is combined by looking for A i = ( ξ i , η i ) for i ∈ { 1,2 } with

( ξ i − d ) x i + η i y i = 1 2 ( r 2 − d 2 ) ,     ξ i 2 + η i 2 = r 2 . (19)

For y i = 0 , because of x i = ± r , we obtain from (19) for i ∈ { 1,2 }

ξ i = d ± 1 2 r ( r 2 − d 2 )       and       η i = ± r 2 − ξ i 2 .

In the case y i ≠ 0 we use the expansion

ξ i 2 = ( ξ i − d + d ) 2 = ( ξ i − d ) 2 + 2 d ( ξ i − d ) + d 2

and multiply the second Equation of (19) with y i 2

( ( ξ i − d ) 2 + 2 d ( ξ i − d ) + d 2 ) y i 2 + η i 2 y i 2 = r 2 y i 2 . (20)

Solving the first Equation of (19) for η i y i and substituting in (20) gives

( ξ i − d ) 2 y i 2 + 2 d y i 2 ( ξ i − d ) + ( − ( ξ i − d ) x i + 1 2 ( r 2 − d 2 ) ) 2 − ( r 2 − d 2 ) y i 2 = 0 ,

( ξ i − d ) 2 ( x i 2 + y i 2 ) + ( 2 d y i 2 − x i ( r 2 − d 2 ) ) ( ξ i − d ) + 1 4 ( r 2 − d 2 ) 2 − ( r 2 − d 2 ) y i 2 = 0.

Because of x i 2 + y i 2 = r 2 we obtain a quadratic equation for ξ i − d :

( ξ i − d ) 2 − 2 r 2 ( x i 2 ( r 2 − d 2 ) − d y i 2 ) ( ξ i − d ) + 1 r 2 [ ( 1 2 ( r 2 − d 2 ) ) 2 − ( r 2 − d 2 ) y i 2 ] = 0. (21)

Before writing down the solutions of (21) we are introducing the expression

S ( x i ) = r 4 − ( d x i + 1 2 ( r 2 − d 2 ) ) 2 . (22)

Lemma 2

For i ∈ { 1,2 } and x i 2 + y i 2 = r 2 the expression S ( x i ) is positive.

Proof: For i ∈ { 1,2 } the expression S ( x i ) can be split up in the factors

S ( x i ) = [ r 2 − ( d x i + 1 2 ( r 2 − d 2 ) ) ] [ r 2 + ( d x i + 1 2 ( r 2 − d 2 ) ) ] = 1 2 ( r 2 − 2 d x i + d 2 ) 1 2 ( 2 ( r 2 + d x i ) + r 2 − d 2 ) .

S ( x i ) is positive for x i 2 + y i 2 = r 2 with i ∈ { 1,2 } since

r 2 − 2 d x i + d 2 ≥ r 2 − 2 | d | r + d 2 = ( r − | d | ) 2 > 0

and

2 ( r 2 + d x i ) + r 2 − d 2 ≥ 2 r ( r − | d | ) + r 2 − d 2 > 0.

□

With expression (22) the solutions of (21) are

( ξ i − d ) ± = 1 r 2 [ x i 2 ( r 2 − d 2 ) − d y i 2 ± y i S ( x i ) ] . (23)

From (23) we obtain

( ξ i ) ± = 1 r 2 [ x i ( d x i + 1 2 ( r 2 − d 2 ) ) ± y i S ( x i ) ] . (24)

Substituting (23) into the first Equation of (19) yields

( η i ) ∓ = 1 r 2 [ y i ( d x i + 1 2 ( r 2 − d 2 ) ) ∓ x i S ( x i ) ] (25)

or expressed otherwise: (23) and (25) combine to

( ξ i − d ) ± x i + ( η i ) ∓ y i = 1 2 ( r 2 − d 2 ) . (26)

In addition ( ξ i ) ± 2 + ( η i ) ∓ 2 = r 2 can be verified.

For i ∈ { 1,2 } we are looking at the straight lines passing through

( ( ξ i ) + , ( η i ) − )       and       ( ( ξ i ) − , ( η i ) + ) . (27)

Proposition 2

For i ∈ { 1,2 } the straight line

x i x + y i y = x i ( ξ i ) + + y i ( η i ) − (28)

is passing through points (27).

Proof: It is clear that ( ( ξ i ) + , ( η i ) − ) is fulfilling (28). In order to prove

x i ( ξ i ) − + y i ( η i ) + = x i ( ξ i ) + + y i ( η i ) −

we are going to show

x i ( ( ξ i ) − − ( ξ i ) + ) + y i ( ( η i ) + − ( η i ) − ) = 0. (29)

According to (24) and (25) we have:

( ξ i ) − − ( ξ i ) + = 1 r 2 [ x i ( d x i + 1 2 ( r 2 − d 2 ) ) − y i S ( x i ) ]     − 1 r 2 [ x i ( d x i + 1 2 ( r 2 − d 2 ) ) + y i S ( x i ) ] = − 2 r 2 y i S ( x i ) ,

( η i ) + − ( η i ) − = 1 r 2 [ y i ( d x i + 1 2 ( r 2 − d 2 ) ) + x i S ( x i ) ]   − 1 r 2 [ y i ( d x i + 1 2 ( r 2 − d 2 ) ) − x i S ( x i ) ] = 2 r 2 x i S ( x i ) .

Thus (29) follows from

x i ( ( ξ i ) − − ( ξ i ) + ) + y i ( ( η i ) + − ( η i ) − ) = 2 r 2 ( − x i y i + y i x i ) S ( x i ) = 0.

□

Next we are looking at the point of intersection ( x S , y S ) of the straight lines (28), i.e. the solution of the following linear system

x 1 x S + y 1 y S = x 1 ( ξ 1 ) + + y 1 ( η 1 ) − x 2 x S + y 2 y S = x 2 ( ξ 2 ) + + y 2 ( η 2 ) − . (30)

See Figure 3.

Proposition 3

For the the determinant D S of the linear system (30) holds

D S = x 1 y 2 − x 2 y 1 = r 2 − d 2 2 ( ( ξ 0 − d ) 2 + η 0 2 ) R ( ξ 0 , η 0 ) , (31)

which is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 .

Proof: According to (11) and (12) we have

x 1 y 2 = 1 4 ( ( ξ 0 − d ) 2 + η 0 2 ) ( 4 r 2 ( ξ 0 − d ) η 0 + ( r 2 − d 2 ) R ( ξ 0 , η 0 ) ) ,

x 2 y 1 = 1 4 ( ( ξ 0 − d ) 2 + η 0 2 ) ( 4 r 2 ( ξ 0 − d ) η 0 − ( r 2 − d 2 ) R ( ξ 0 , η 0 ) ) .

This gives the result (31).

D S is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 because by Lemma 1 R ( ξ 0 , η 0 ) is positive and r 2 − d 2 > 0 holds. □

Theorem 1

For i ∈ { 1,2 } holds:

  Either       ( ( ξ i ) + , ( η i ) − ) = ( ξ 0 , η 0 )       or       ( ( ξ i ) − , ( η i ) + ) = ( ξ 0 , η 0 ) . (32)

Proof: Because of (13) we have for i ∈ { 1,2 }

x i ξ 0 + y i η 0 = d x i + 1 2 ( r 2 − d 2 ) . (33)

Furthermore we have for i ∈ { 1,2 }

x i ( ξ i ) + + y i ( η i ) − = d x i + 1 2 ( r 2 − d 2 ) , (34)

because (24) and (25) yield

x i ( ξ i ) + + y i ( η i ) − = x i r 2 [ x i ( d x i + 1 2 ( r 2 − d 2 ) ) + y i S ( x i ) ]         + y i r 2 [ y i ( d x i + 1 2 ( r 2 − d 2 ) ) − x i S ( x i ) ] = 1 r 2 [ ( x i 2 + y i 2 ) ( d x i + 1 2 ( r 2 − d 2 ) ) + ( x i y i − y i x i ) S ( x i ) ] = d x i + 1 2 ( r 2 − d 2 ) .

Combining (33) and (34) we get for i ∈ { 1,2 }

x i ξ 0 + y i η 0 = x i ( ξ i ) + + y i ( η i ) − .

This means

x 1 ξ 0 + y 1 η 0 = x 1 ( ξ 1 ) + + y 1 ( η 1 ) − x 2 ξ 0 + y 2 η 0 = x 2 ( ξ 2 ) + + y 2 ( η 2 ) − . (35)

According to Proposition 3 the determinant D S is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 . Therefore the linear systems (30) and (35) have one and only one solution, which leads to ( x S , y S ) = ( ξ 0 , η 0 ) . Point A 0 = ( ξ 0 , η 0 ) is lying on both straight lines (28) and on the circle x 2 + y 2 = r 2 . One of the points of intersection of both straight lines (28) with the circle, ( ( ξ i ) + , ( η i ) − ) or ( ( ξ i ) − , ( η i ) + ) for i ∈ { 1,2 } , must coincide with ( ξ 0 , η 0 ) what means (32). □

Corollary 1

With the result of Theorem 1 for the points A i = ( ξ i , η i ) with i ∈ { 1,2 } calculated in Section 4 therefore remains the following attribution:

If for i ∈ { 1,2 } : ( ξ 0 , η 0 ) = ( ( ξ i ) + , ( η i ) − ) , then ( ξ i , η i ) = ( ( ξ i ) − , ( η i ) + ) .

If for i ∈ { 1,2 } : ( ξ 0 , η 0 ) = ( ( ξ i ) − , ( η i ) + ) , then ( ξ i , η i ) = ( ( ξ i ) + , ( η i ) − ) . □

With this information and formulas (23) and (25) the following expressions can be evaluated for i ∈ { 1,2 } :

( ξ 0 − d ) ( ξ i − d ) = r 2 − d 2 4 r 2 ( − 4 y i 2 + r 2 − d 2 ) (36)

η 0 η i = r 2 − d 2 4 r 2 ( − 4 x i ( x i − d ) + r 2 − d 2 ) . (37)

Both expressions (36) and (37) combine to

( ξ 0 − d ) ( ξ i − d ) + η 0 η i = − r 2 − d 2 2 r 2 ( ( x i − d ) 2 + y i 2 ) . (38)

We are introducing for i ∈ { 1,2 } the straight lines

( x i − d ) x + y i y = − 1 2 ( ( x i − d ) 2 + y i 2 ) , (39)

which will have a correspondence with the perpendicular bisectors in (5).

Proposition 4

( x 2 , y 2 ) fulfills ( x 1 − d ) x + y 1 y = − 1 2 ( ( x 1 − d ) 2 + y 1 2 ) .

( x 1 , y 1 ) fulfills ( x 2 − d ) x + y 2 y = − 1 2 ( ( x 2 − d ) 2 + y 2 2 ) .

Proof: We have to show

( x 1 − d ) x 2 + y 1 y 2 = − 1 2 ( ( x 1 − d ) 2 + y 1 2 ) ( x 2 − d ) x 1 + y 2 y 1 = − 1 2 ( ( x 2 − d ) 2 + y 2 2 ) . (40)

Transforming the right hand sides of (40) according to ( x i − d ) 2 + y i 2 = r 2 − 2 d x i + d 2 with i ∈ { 1,2 } both statements of (40) can be equivalently transformed to

x 1 x 2 + y 1 y 2 − d ( x 1 + x 2 ) = − 1 2 ( r 2 + d 2 ) . (41)

It suffices to prove (41). From (11) and (12) we obtain

x 1 x 2 = 1 4 ( ( ξ 0 − d ) 2 + η 0 2 ) ( ( r 2 − d 2 ) 2 − 4 r 2 η 0 2 )

y 1 y 2 = 1 4 ( ( ξ 0 − d ) 2 + η 0 2 ) ( ( r 2 − d 2 ) 2 − 4 r 2 ( ξ 0 − d ) 2 )

leading to

x 1 x 2 + y 1 y 2 = 1 4 ( ( ξ 0 − d ) 2 + η 0 2 ) [ 2 ( r 2 − d 2 ) 2 − 4 r 2 ( ( ξ 0 − d ) 2 + η 0 2 ) ] = ( r 2 − d 2 ) 2 2 ( ( ξ 0 − d ) 2 + η 0 2 ) − r 2 .

Together with

d ( x 1 + x 2 ) = 2 d ( ξ 0 − d ) ( r 2 − d 2 ) 2 ( ( ξ 0 − d ) 2 + η 0 2 )

we get

x 1 x 2 + y 1 y 2 − d ( x 1 + x 2 ) = ( r 2 − d 2 ) 2 − 2 d ( ξ 0 − d ) ( r 2 − d 2 ) 2 ( ( ξ 0 − d ) 2 + η 0 2 ) − r 2 . (42)

The numerator of the fraction in statement (42) can be transformed

( r 2 − d 2 ) ( r 2 − d 2 − 2 d ( ξ 0 − d ) ) = ( r 2 − d 2 ) ( r 2 − 2 ξ 0 d + d 2 ) = ( r 2 − d 2 ) ( ( ξ 0 − d ) 2 + η 0 2 ) .

Thus we obtain

x 1 x 2 + y 1 y 2 − d ( x 1 + x 2 ) = 1 2 ( r 2 − d 2 ) − r 2 = − 1 2 ( r 2 + d 2 ) ,

which is (41). □

Next the point of intersection ( x T , y T ) of the straight lines (39) shall be calculated, this means solving the following linear system

( x 1 − d ) x T + y 1 y T = − 1 2 ( ( x 1 − d ) 2 + y 1 2 ) ( x 2 − d ) x T + y 2 y T = − 1 2 ( ( x 2 − d ) 2 + y 2 2 ) . (43)

Proposition 5

For the determinant D T of the linear system (43) holds

D T = ( x 1 − d ) y 2 − ( x 2 − d ) y 1 = 1 2 R ( ξ 0 , η 0 ) , (44)

which is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 .

Proof: Because of (31) we have

x 1 y 2 − x 2 y 1 = r 2 − d 2 2 ( ( ξ 0 − d ) 2 + η 0 2 ) R ( ξ 0 , η 0 )

and because of (12)

− d ( y 2 − y 1 ) = − 2 d ( ξ 0 − d ) 2 ( ( ξ 0 − d ) 2 + η 0 2 ) R ( ξ 0 , η 0 ) .

This yields for D T

D T = r 2 − d 2 − 2 d ( ξ 0 − d ) 2 ( ( ξ 0 − d ) 2 + η 0 2 ) R ( ξ 0 , η 0 ) = r 2 − 2 d ξ 0 + d 2 2 ( ( ξ 0 − d ) 2 + η 0 2 ) R ( ξ 0 , η 0 )

and because of r 2 − 2 d ξ 0 + d 2 = ( ξ 0 − d ) 2 + η 0 2 formula (44). That R ( ξ 0 , η 0 ) is positive for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 was shown in Lemma 1. Therefore also D T is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 . □

Proposition 6

The solutions of the linear system (43) are as follows

x T = d − ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 ,       y T = − η 0 ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 . (45)

Proof: Because D T is different from zero for ( ξ 0 , η 0 ) with ξ 0 2 + η 0 2 = r 2 the linear system (43) has one and only one solution. It is sufficient to show for i ∈ { 1,2 }

( x i − d ) ( x T − d ) + y i y T = − 1 2 ( ( x i − d ) 2 + y i 2 ) − d ( x i − d ) . (46)

Substitution of x T − d and y T on the left side of (46) yields

( x i − d ) ( x T − d ) + y i y T = − ( x i − d ) ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 − y i η 0 ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 = − r 2 − d 2 ( ξ 0 − d ) 2 + η 0 2 ( ( x i − d ) ( ξ 0 − d ) + y i η 0 ) .

Because of (13) we have

( x i − d ) ( ξ 0 − d ) + y i η 0 = x i ( ξ 0 − d ) + y i η 0 − d ( ξ 0 − d ) = 1 2 ( r 2 − d 2 ) − d ξ 0 + d 2 = 1 2 ( r 2 − 2 d ξ 0 + d 2 ) = 1 2 ( ( ξ 0 − d ) 2 + η 0 2 )

Therefore we get

( x i − d ) ( x T − d ) + y i y T = − 1 2 ( r 2 − d 2 ) = − 1 2 ( r 2 − 2 d x i + d 2 + 2 d x i − 2 d 2 ) = − 1 2 ( ( x i − d ) 2 + y i 2 ) − d ( x i − d ) .

which is (46). □

Proposition 7

( x T , y T ) is located on the circle x 2 + y 2 = r 2 .

Proof: From (45) we get

( x T − d ) 2 + y T 2 = ( r 2 − d 2 ) 2 ( ( ξ 0 − d ) 2 + η 0 2 ) ( ( ξ 0 − d ) 2 + η 0 2 ) 2 = ( r 2 − d 2 ) 2 ( ξ 0 − d ) 2 + η 0 2 .

Because of x T 2 = ( x T − d ) 2 + 2 d x T − d 2 we have

x T 2 + y T 2 = ( x T − d ) 2 + y T 2 + 2 d x T − d 2 = ( r 2 − d 2 ) 2 ( ξ 0 − d ) 2 + η 0 2 + 2 d x T − d 2 .

Substituting x T from (45) yields

x T 2 + y T 2 = ( r 2 − d 2 ) 2 ( ξ 0 − d ) 2 + η 0 2 + 2 d [ d − ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 ] − d 2 = ( r 2 − d 2 ) 2 − 2 d ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 + d 2 = ( r 2 − d 2 ) ( r 2 − d 2 − 2 d ( ξ 0 − d ) ) ( ξ 0 − d ) 2 + η 0 2 + d 2 = ( r 2 − d 2 ) ( r 2 − 2 d ξ 0 + d 2 ) ( ξ 0 − d ) 2 + η 0 2 + d 2 = r 2 − d 2 + d 2 = r 2 .

□

See Figure 4.

Next we want to show that ( x T , y T ) is a point on the perpendicular bisectors of D A i ¯ for i ∈ { 1,2 } given in (5).

Proposition 8

For i ∈ { 1,2 } holds:

( ξ i − d ) x T + η i y T = 1 2 ( r 2 − d 2 ) . (47)

Proof: Equation (47) can be equivalently rewritten as

( ξ i − d ) ( x T − d ) + η i y T = − d ( ξ i − d ) + 1 2 ( r 2 − d 2 ) = 1 2 ( r 2 − 2 d ξ i + d 2 ) . (48)

Substituting x T − d and y T according to (45) into the left hand side of (48) we get:

( ξ i − d ) ( x T − d ) + η i y T = − ( ξ i − d ) ( ξ 0 − d ) ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 − η i η 0 ( r 2 − d 2 ) ( ξ 0 − d ) 2 + η 0 2 = − r 2 − d 2 ( ξ 0 − d ) 2 + η 0 2 [ ( ξ i − d ) ( ξ 0 − d ) + η i η 0 ] (49)

The expression in square brackets of (49) is substituted by (38); this yields

= ( r 2 − d 2 ) 2 ( ( x i − d ) 2 + y i 2 ) 2 r 2 ( ( ξ 0 − d ) 2 + η 0 2 ) = ( r 2 − d 2 ) 2 ( r 2 − 2 d x i + d 2 ) 2 r 2 ( r 2 − 2 d ξ 0 + d 2 ) . (50)

Statement (48) is proven if we can show for the expression on the right hand side of (50)

( r 2 − d 2 ) 2 ( r 2 − 2 d x i + d 2 ) 2 r 2 ( r 2 − 2 d ξ 0 + d 2 ) = 1 2 ( r 2 − 2 d ξ i + d 2 ) .

This means

( r 2 − d 2 ) 2 ( r 2 − 2 d x i + d 2 ) = r 2 ( r 2 − 2 d ξ 0 + d 2 ) ( r 2 − 2 d ξ i + d 2 ) . (51)

Statement (51) will be equivalently transformed. The right hand side of (51) can be expanded

r 2 ( r 2 − d 2 − 2 d ( ξ 0 − d ) ) ( r 2 − d 2 − 2 d ( ξ i − d ) ) = r 2 [ ( r 2 − d 2 ) 2 − 2 d ( r 2 − d 2 ) ( ξ 0 − d + ξ i − d ) + 4 d 2 ( ξ 0 − d ) ( ξ i − d ) ] . (52)

Subtracting r 2 ( r 2 − d 2 ) 2 from the left hand side of (51) and from (52) we get

( r 2 − d 2 ) 2 ( − 2 d x i + d 2 ) = r 2 [ − 2 d ( r 2 − d 2 ) ( ξ 0 + ξ i − 2 d ) + 4 d 2 ( ξ 0 − d ) ( ξ i − d ) ] . (53)

Substituting on the right hand side of (53)

ξ 0 + ξ i − 2 d = 2 r 2 ( x i ( d x i + 1 2 ( r 2 − d 2 ) ) − r 2 d ) = 2 r 2 ( 1 2 x i ( r 2 − d 2 ) − d y i 2 )

according to Corollary 1 and (24) as well as ( ξ 0 − d ) ( ξ i − d ) according to (36) we get

− 4 d ( r 2 − d 2 ) ( 1 2 x i ( r 2 − d 2 ) − d y i 2 ) + d 2 ( r 2 − d 2 ) ( − 4 y i 2 + r 2 − d 2 ) (54)

The expression − 4 d 2 ( r 2 − d 2 ) y i 2 vanishes in (54). So we have for (53)

( r 2 − d 2 ) 2 ( − 2 d x i + d 2 ) = − 4 d ( r 2 − d 2 ) 1 2 x i ( r 2 − d 2 ) + d 2 ( r 2 − d 2 ) 2 = − 2 d ( r 2 − d 2 ) 2 x i + d 2 ( r 2 − d 2 ) 2 = ( r 2 − d 2 ) 2 ( − 2 d x i + d 2 ) . (55)

□

Theorem 2

The ellipse created by folding of a disk is enclosed by an arbitrary triangle of tangents, the vertices of which are lying on the circle representing the disk.

Proof: According to Proposition 8 for i ∈ { 1,2 } the perpendicular bisectors of D A i ¯

( ξ i − d ) x + η i y = 1 2 ( r 2 − d 2 )

are passing through ( x T , y T ) . Because of their construction in Section 4 they are also going through ( x i , y i ) . Thus there is a coincidence with the straight lines (39) in the following way: the straight lines

( ξ 1 − d ) x + η 1 y = 1 2 ( r 2 − d 2 ) ( x 2 − d ) x + y 2 y = − 1 2 ( ( x 2 − d ) 2 + y 2 2 ) , (56)

having the same points ( x 1 , y 1 ) and ( x T , y T ) , are thus identical; the straight lines

( ξ 2 − d ) x + η 2 y = 1 2 ( r 2 − d 2 ) ( x 1 − d ) x + y 1 y = − 1 2 ( ( x 1 − d ) 2 + y 1 2 ) , (57)

having the same points ( x 2 , y 2 ) and ( x T , y T ) , are thus also identical.

Therefore the ellipse with foci C = ( 0 , 0 ) and D = ( d ,0 ) and large semi axis a = r 2 is enclosed by a triangle of tangents

( ξ 0 − d ) x + η 0 y = 1 2 ( r 2 − d 2 ) ( ξ 1 − d ) x + η 1 y = 1 2 ( r 2 − d 2 ) ( ξ 2 − d ) x + η 2 y = 1 2 ( r 2 − d 2 ) (58)

with vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) and ( x T , y T ) lying on the circle x 2 + y 2 = r 2 . □

It was the intention to show by elementary means of analytic geometry the enclosure of the Gardner ellipse, created by folds of a disk, with an arbitrary triangle of tangents, the vertices of which are lying on the surrounding circle representing the disk.

Remark: The Figures in this paper were constructed with Mathematica using the initial data r = 4 , d = − 3 , ξ 0 = − 0.9 and η 0 = + r 2 − ξ 0 2 .

The author declares no conflicts of interest regarding the publication of this paper.

Klein, P.P. (2022) Enclosing Ellipses by Folding Disks. Applied Mathematics, 13, 147-162. https://doi.org/10.4236/am.2022.132012