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STEINER’S HAT: A CONSTANT-AREA DELTOID

ASSOCIATED WITH THE ELLIPSE

RONALDO GARCIA, DAN REZNIK, HELLMUTH STACHEL, AND MARK HELMAN

Abstract.

The Negative Pedal Curve (NPC) of the Ellipse with respect to

a boundary point

M

is a 3-cusp closed-curve which is the aﬃne image of the

Steiner Deltoid. Over all

M

the family has invariant area and displays an array

of interesting properties.

Keywords

curve, envelope, ellipse, pedal, evolute, deltoid, Poncelet, osculating,

orthologic.

MSC 51M04 and 51N20 and 65D18

1. Introduction

Given an ellipse

E

with non-zero semi-axes

a, b

centered at

O

, let

M

be a point in

the plane. The Negative Pedal Curve (NPC) of

E

with respect to

M

is the envelope

of lines passing through points

P

(

t

)on the boundary of

E

and perpendicular to

[

P

(

t

)

−M

][

4

, pp. 349]. Well-studied cases [

7

,

14

] include placing

M

on (i) the

major axis: the NPC is a two-cusp “ﬁsh curve” (or an asymmetric ovoid for low

eccentricity of

E

); (ii) at

O

: this yielding a four-cusp NPC known as Talbot’s Curve

(or a squashed ellipse for low eccentricity), Figure 1.

Figure 1. The Negative Pedal Curve (NPC) of an ellipse Ewith respect to a point Mon the plane

is the envelope of lines passing through P(t)on the boundary, and perpendicular to P(t)−M.Left:

When Mlies on the major axis of E, the NPC is a two-cusp “ﬁsh” curve. Right: When Mis at

the center of E, the NPC is 4-cusp curve with 2-self intersections known as Talbot’s Curve [12].

For the particular aspect ratio a/b = 2, the two self-intersections are at the foci of E.

Date: May, 2020.

1

2 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 2. Left: The Negative Pedal Curve (NPC, purple) of Ewith respect to a boundary point

Mis a 3-cusped (labeled P0

i) asymmetric curve (called here “Steiner’s Hat”), whose area is invariant

over M, and whose asymmetric shape is aﬃnely related to the Steiner Curve [12]. ∆u(t)is the

instantaneous tangency point to the Hat. Right: The tangents at the cusps points P0

iconcur at

C2, the Hat’s center of area, furthermore, Pi, P 0

i, C2are collinear. Video: [10, PL#01]

As a variant to the above, we study the family of NPCs with respect to points

M

on the boundary of

E

. As shown in Figure 2, this yields a family of asymmetric,

constant-area 3-cusped deltoids. We call these curves “Steiner’s Hat” (or ∆), since

under a varying aﬃne transformation, they are the image of the Steiner Curve (aka.

Hypocycloid), Figure 3. Besides these remarks, we’ve observed:

Main Results:

•

The triangle

T0

deﬁned by the 3 cusps

P0

i

has invariant area over

M

, Figure 7.

•

The triangle

T

deﬁned by the pre-images

Pi

of the 3 cusps has invariant area

over

M

, Figure 7. The

Pi

are the 3 points on

E

such that the corresponding

tangent to the envelope is at a cusp.

•

The

T

are a Poncelet family with ﬁxed barycenter; their caustic is half the

size of E, Figure 7.

•

Let

C2

be the center of area of ∆. Then

M, C2, P1, P2, P3

are concyclic,

Figure 7. The lines Pi−C2are tangents at the cusps.

•

Each of the 3 circles passing through

M, Pi, P 0

i

,

i

= 1

,

2

,

3, osculate

E

at

Pi

, Figure 8. Their centers deﬁne an area-invariant triangle

T00

which is a

half-size homothety of T0.

The paper is organized as follows. In Section 3we prove the main results. In

Sections 4and 5we describe properties of the triangles deﬁned by the cusps and

their pre-images, respectively. In Section 6we analyze the locus of the cusps. In

Section 6.1 we characterize the tangencies and intersections of Steiner’s Hat with

the ellipse. In Section 7we describe properties of 3 circles which osculate the ellipse

at the cusp pre-images and pass through

M

. In Section 8we describe relationships

between the (constant-area) triangles with vertices at (i) cusps, (ii) cusp pre-images,

and (iii) centers of osculating circles. In Section 9we analyze a ﬁxed-area deltoid

obtained from a “rotated” negative pedal curve. The paper concludes in Section 10

with a table of illustrative videos. Appendix Aprovides explicit coordinates for

cusps, pre-images, and osculating circle centers. Finally, Appendix Blists all symbols

used in the paper.

STEINER’S HAT: A CONSTANT-AREA DELTOID 3

Figure 3. Two systems which generate the 3-cusp Steiner Curve (red), see [2] for more methods.

Left: The locus of a point on the boundary of a circle of radius 1rolling inside another of radius

3.Right: The envelope of Simson Lines (blue) of a triangle T(black) with respect to points P(t)

on the Circumcircle [12]. Q(t)denotes the corresponding tangent. Nice properties include (i) the

area of the Deltoid is half that of the Circumcircle, and (ii) the 9-point circle of T(dashed green)

centered on X5(whose radius is half that of the Circumcircle) is internally tangent to the Deltoid

[13, p.231].

2. Preliminaries

Let the ellipse Ebe deﬁned implicitly as:

E(x, y) = x2

a2+y2

b2−1=0, c2=a2−b2

where

a > b >

0are the semi-axes. Let a point

P

(

t

)on its boundary be parametrized

as P(t)=(acos t, b sin t).

Let

P0

= (

x0, y0

)

∈R2

. Consider the family of lines

L

(

t

)passing through

P

(

t

)

and orthogonal to

P

(

t

)

−P0

. Its envelope ∆is called antipedal or negative pedal

curve of E.

Consider the spatial curve deﬁned by

L(P0) = {(x, y, t) : L(t, x, y) = L0(t, x, y)=0}.

The projection

E

(

P0

) =

π

(

L

(

P0

)) is the envelope. Here

π

(

x, y, t

) = (

x, y

). In general,

L(P0)is regular, but E(P0)is a curve with singularities and/or cusps.

Lemma 1. The envelope of the family of lines L(t)is given by:

x(t) = 1

w[(ay0sin t−ab)x0−by2

0cos t−c2y0sin(2t)

+b

4((5a2−b2) cos t−c2cos(3t))]

y(t) = 1

w[−ax2

0sin t+ (by0cos t+c2sin(2t))x0−aby0

−a

4((5a2−b2) sin t−c2sin(3t)](1)

where w=ab −bx0cos t−ay0sin t.

4 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Proof. The line L(t)is given by:

(x0−acos t)x+ (y0−bsin t)y+a2cos2t+b2sin2t−ax0cos t−by0sin t= 0

Solving the linear system

L

(

t

) =

L0

(

t

) = 0 in the variables

x, y

leads to the result.

Triangle centers will be identifed below as

Xk

following Kimberling’s Encyclopedia

[6], e.g., X1is the Incenter, X2Barycenter, etc.

3. Main Results

Proposition 1.

The NPC with respect to

Mu

= (

acos u, b sin u

)a boundary point

of Eis a 3-cusp closed curve given by ∆u(t)=(xu(t), yu(t)), where

xu(t) = 1

ac2(1 + cos(t+u)) cos t−a2cos u

yu(t) =1

bc2cos tsin(t+u)−c2sin t−a2sin u

(2)

Proof. It is direct consequence of Lemma 1with P0=Mu.

Expressions for the 3 cusps P0

iin terms of uappear in Appendix A.

Remark 1.As

a/b →

1the ellipse becomes a circle and ∆shrinks to a point on the

boundary of said circle.

Remark 2.Though ∆can never have three-fold symmetry, for

Mu

at any ellipse

vertex, it has axial symmetry.

Remark 3.The average coordinates

¯

C

= [

¯x

(

u

)

,¯y

(

u

)] of ∆

u

w.r.t. this parametriza-

tion are given by:

¯x(u) = 1

2πZ2π

0

xu(t)dt =−(a2+b2)

2acos u

¯y(u) = 1

2πZ2π

0

yu(t)dt =−(a2+b2)

2bsin u(3)

Theorem 1.

∆

u

is the image of the 3-cusp Steiner Hypocycloid

S

under a varying

aﬃne transformation.

Proof. Consider the following transformations in R2:

rotation: Ru(x, y) = cos usin u

−sin ucos u! x

y!

translation: U(x, y) = (x, y) + ¯

C

homothety: D(x, y) = 1

2(a2−b2)(x, y).

linear map: V(x, y) = (x

a,y

b)

The hypocycloid of Steiner is given by

S

(

t

) = 2(

cos t, −sin t

)+(

cos

2

t, sin

2

t

)[

7

].

Then:

(4) ∆u(t)=[xu(t), yu(t)] = (U◦V◦D◦Ru)S(t)

Thus, Steiner’s Hat is of degree 4 and of class 3 (i.e., degree of its dual).

STEINER’S HAT: A CONSTANT-AREA DELTOID 5

Figure 4. The iso curves of signed area for the negative pedal curve when Mis interior to the

ellipse are closed algebraic curves of degree 10. These are shown in gray for an NPC with two

cusps (left), and 4 cusps (right).

Corollary 1.

The area of

A

(∆) of Steiner’s Hat is invariant over

Mu

and is given

by:

(5) A(∆) = (a2−b2)2π

2ab =c4π

2ab

Proof.

The area of

S

(

t

)is

RSxdy

= 2

π

. The Jacobian of (

U◦S◦D◦Ru

)given by

Equation 4is constant and equal to c4/4ab.

Noting that the area of

E

is

πab

, Table 1shows the aspect ratios

a/b

of

E

required

to produce special area ratios.

a/b approx. a/b A(∆)/A(E)

p2 + √3 1.93185 1

ϕ= (1 + √5)/2 1.61803 1/2

√2 1.41421 1/4

11 0

Table 1. Asp ect ratios yielding special area ratios of main ellipse Eto Steiner’s Hat ∆.

It is well known that if

M

is interior to

E

then the NPC is a 2-cusp or 4-cusp

curve with one or two self-intersections.

Remark 4.It can be shown that when

M

is interior to

E

the iso-curves of signed

area of the NPC are closed algebraic curves of degree 10, concentric with

E

and

symmetric about both axes, see Figure 4.

It is remarkable than when

M

moves from the interior to the boundary of

E

, the

iso-curves transition from a degree-10 curve to a simple conic.

Remark 5.It can also be shown that when

M

is exterior to

E

, the NPC is a

two-branched open curve, see Figure 5.

6 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 5. Left: When Mis exterior to Ethe NPC is a two-branched open curve. One branch is

smooth and non-self-intersecting, and the other has 2 cusps and one self-intersection. Right: Let

t1, t2be the parameters for which MP (t)is tangent to E. At these positions, the NPC intersects

the line at inﬁnity in the direction of the normal at P(t1), P (t2), i.e., the lines through P(t)

perpendicular to P(t)−Mare asymptotes.

Proposition 2. Let C2be the center of area of ∆u. Then C2=¯

C.

Proof. The center of area is deﬁned by

C2=1

A(∆) Zint(∆)

x dx dy, Zint(∆)

y dx dy!.

Using Green’s Theorem, evaluate the above using the parametric in Equation

(1)

.

This yields the expression for

¯

C

in Equation 3. Alternatively, one can obtain the

same result from the aﬃne transformation deﬁned in Theorem 1.

Referring to Figure 6(left):

Corollary 2.

The locus of

C2

is an ellipse always exterior to a copy of

E

rotated

90◦about O.

Proof.

Equation 3describes an ellipse. Since

a2

+

b2≥

2

ab

the claim follows directly.

Let

T

denote the triangle of the pre-images

Pi

on

E

of the Hat’s cusps, i.e.

P

(

ti

)

such that ∆

uti

is a cuspid. Explicit expressions for the

Pi

appear in Appendix A.

Referring to Figure 7:

Theorem 2. The points M, C2, P1, P2, P3are concyclic.

Proof.

∆

u

(

t

)is singular at

t1

=

−u

3

,

t2

=

−u

3−2π

3

and

t3

=

−u

3−4π

3

. Let

Pi= [acos ti, b sin ti], i = 1,2,3. The circle Kpassing through these is given by:

(6) K(x, y) = x2+y2−c2cos u

2ax+c2sin u

2by−1

2(a2+b2)=0.

Also, K(M) = K(acos u, b sin u) = 0.

STEINER’S HAT: A CONSTANT-AREA DELTOID 7

Figure 6. Left: The area center C2of Steiner’s Hat coincides with the barycenter X0

2of the

(dashed) triangle T0deﬁned by the cusps. Over all M, both the Hat and T0have invariant area.

C2’s locus (dashed purple) is elliptic and exterior to a copy of Erotated 90◦about its center (dashed

black). Right: Let P0

i(resp. Pi), i= 1,2,3denote the Hat’s cusps (resp. their pre-images on E),

colored by i. Lines PiP0

iconcur at C2.

The center of

K

is (

M

+

C2

)

/

2. It follows that

C2∈ K

and that

MC2

is a

diameter of K.

In 1846, Jakob Steiner stated that given a point

M

on an ellipse

E

, there exist 3

other points on it such that the osculating circles at these points pass through

M

[8, page 317]. This property is also mentioned in [4, page 97, Figure 3.26].

It turns out the cusp pre-images are said special points! Referring to Figure 8:

Proposition 3.

Each of the 3 circles

Ki

through

M, Pi, P 0

i

,

i

= 1

,

2

,

3, osculate

E

at Pi.

Proof. The circle K1passing through M,P1and P0

1is given by

K1(x, y) =2 ab(x2+y2)−4bc2cos3u

3x−4ac2sin3u

3y

+ab 3c2cos 2u

3−a2−b2= 0.

Recall a circle osculates an ellipse if its center lies on the evolute of said ellipse,

given by [4]:

(7) E∗(t) = c2cos3t

a,−c2sin3t

b

8 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 7. The cusp pre-images Pideﬁne a triangle T(orange) whose area is invariant over M.

Its barycenter X2is stationary at the center of E, rendering the latter its Steiner Ellipse. Let C2

denote the center of area of Steiner’s Hat. The 5 points M, C2, P1, P2, P3lie on a circle (orange),

with center at X3(circumcenter of T). Over all M, the Tare a constant-area Poncelet family

inscribed on Eand tangent to a concentric, axis-aligned elliptic caustic (dashed orange), half the

size of E, i.e., the latter is the (stationary) Steiner Inellipse of the T. Note also that Mis the

Steiner Point X99 of Tsince it is the intersection of its Circumcircle with the Steiner Ellipse.

Furthermore, the Tarry Point X98 of Tcoincides with C2, since it is the antipode of M=X99 [6].

Video: [10, PL#02,#05].

It is straightforward to verify that the center of

K1

is

P00

1

=

E∗

(

−u

3

). A similar

analysis can be made for K2and K3.

Since the area of E∗is A(E∗) = 3πc4

8ab , and the area of ∆is given in Equation 5:

Remark 6.The area ratio of ∆and the interior of E∗is equal to 4/3.,

3.1.

Why is

∆

aﬃne to Steiner’s Curve.

Up to projective transformations, there

is only one irreducible curve of degree 4 with 3 cusps. In a projective coordinate

frame (

x0

:

x1

:

x2

)with the cusps as base points (1:0:0),(0:1:0)and (0:0:1)

and the common point of the cusps’ tangents as unit point (1:1:1), the quartic has

the equation

x2

0x2

1+x2

0x2

2+x2

1x2

2−2x0x1x2(x0+x1+x2)=0

STEINER’S HAT: A CONSTANT-AREA DELTOID 9

Figure 8. The circles passing through a cusp P0

i, its pre-image Pi, and Mosculate Eat the Pi.

The centers P00

iof said circles deﬁne a triangle T00 (dashed black) whose area is constant for all

M.X00

2denotes its (moving) barycenter. Video: [10, PL#03,#05].

At Steiner’s three-cusped curve, the cusps form a regular triangle with the

tangents passing through the center. Hence, whenever a three-cusped quartic has

the meeting point of the cusps’ tangents at the center of gravity of the cusps, it

is aﬃne to Steiner’s curve, since there is an aﬃne transformation sending the four

points into a regular triangle and its center.

4. The Cusp Triangle

Recall T0=P0

1P0

2P0

3is the triangle deﬁned by the 3 cusps of ∆.

Proposition 4.

The area

A0

of the cusp triangle

T0

is invariant over

M

and is

given by:

A0=27√3

16

c4

ab

Proof. The determinant of the Jacobian of the aﬃne transformation in Theorem 1

is

|J|

=

c4

4ab

. Therefore, the area of

T0

is simply

|J|Ae

, where

Ae

is the area of an

equilateral triangle inscribed in a circle of radius 3with side 3√3.

Referring to Figure 6:

Proposition 5.

The barycenter

X0

2

of

T0

coincides with the center of area

C2

of

∆.

Proof. Direct calculations yield X0

2=C2.

Referring to Figure 9:

Proposition 6.

The Steiner Ellipse

E0

of

T0

has constant area and is a scaled

version of Erotated 90◦about O.

10 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 9. Left: The Steiner Ellipse E0of triangle T0deﬁned by the P0

iis a scaled-up and 90◦-

rotated copy of E.Right: At a/b = (1 + √10)/3'1.38743,E0and Ehave the same area.

Proof. E0

passes through the vertices of

T0

and is centered on

C2

=

X0

2

. Direct

calculations yield the following implicit equation for it:

a2x2+b2y2+a2+b2(acos ux +bsin uy)−a2−2b22a2−b2= 0

Its semi-axes are

b0

=

3c2

2a

and

a0

=

3c2

2b

. Therefore

E0

is similar to a 90

◦

-rotated

copy of E.

Remark 7.This proves that

T0

can never be regular and ∆has never a three-fold

symmetry.

Corollary 3.

The ratio of area of

E0

and

E

is given by

9

4

c4

a2b2

, and at

a/b

=

(1 + √10)/3, the two ellipses are congruent.

Proposition 7. The Steiner Point X0

99 of T0is given by:

X0

99(u) = "a2−2b2

acos u, −2a2−b2

bsin u#

Proof.

By deﬁnition

X99

is the intersection of the circumcircle of

T

(

K

) with the

Steiner ellipse. The Circumcircle K0of the triangle T0={P0

1, P 0

2, P 0

3}is given by:

K0(x, y) =8 a2b2x2+y2

+2 acos u3a4−2a2b2+ 7 b4x+ 2 bsin u7a4−2a2b2+ 3 b4y

−a2+b2c2a2+b2cos 2 u+ 5 a4−14 a2b2+ 5 b4= 0.

With the above, straightforward calculations lead to the coordinates of

X0

99

.

STEINER’S HAT: A CONSTANT-AREA DELTOID 11

5. The Triangle of Cusp Pre-Images

Recall

T

=

P1P2P3

is the triangle deﬁned by pre-images on

E

to each cusp of ∆.

Proposition 8.

The barycenter

X2

of

T

is stationary at

O

, i.e.,

E

is is Steiner

Ellipse [12].

Proof.

The triangle

T

is an aﬃne image of an equilateral triangle with center at 0

and Pi=E(ti) = E(−u

3−(i−1)2π

3). So the result follows.

Remark 8.Mis the Steiner Point X99 of T.

Proposition 9.

Over all

M

, the

T

are an

N

= 3 Poncelet family with external

conic

E

with the Steiner Inellipse of

T

as its caustic [

12

]. Futhermore the area of

these triangles is invariant and equal to 3√3ab

4.

Proof.

The pair of concentric circles of radius 1and 1

/

2is associated with a Poncelet

1d family of equilaterals. The image of this family by the map (

x, y

)

→

(

ax, by

)

produces the original pair of ellipses, with the stated area. Alternatively, the ratio

of areas of a triangle to its Steiner Ellipse is known to be 3

√3/

(4

π

)[

12

, Steiner

Circumellipse] which yields the area result.

6. Locus of the Cusps

We analyze the motion of the cusps

P0

i

of Steiner’s Hat ∆with respect to

continuous revolutions of Mon E. Referring to Figure 10:

Remark 9.The locus C(u)of the cusps of ∆is parametrized by:

C(u) : 3c2

21

acos u

3,1

bsin u

3−a2+b2

21

acos u, 1

bsin u

(8)

This is a curve of degree 6, with the following implicit equation:

−4a6x6−4b6y6−12 a2x2b2y2a2x2+b2y2

+12 a4a4−a2b2+b4x4+ 12 b4a4−a2b2+b4y4+ 24 a2b2a4−a2b2+b4x2y2

−3a22a2−b2a2+b2 2a4−5a2b2+ 5 b4x2

+3 b2a2−2b2a2+b2 5a4−5a2b2+ 2 b4y2

+2a2−b22a2−2b22a2+b22

= 0

Proposition 10.

It can be shown that over one revolution of

M

about

E

,

C2

will

cross the ellipse on four locations Wj, j = 1,··· ,4given by:

Wj=1

2√a2+b2±apa2+ 3b2,±bp3a2+b2

At each such crossing, C2coincides with one of the pre-images.

Proof.

From the coordinates of

C2

given in equation

(3)

in terms of the parameter

u

,

one can derive an equation that is a necessary and suﬃcient condition for

C2∈ E

to

happen, by substituting those coordinates in the ellipse equation

x2/a2

+

y2/b2−

1 = 0.

Solving for

sin u

and substituting back in the coordinates of

C2

, one easily gets the

four solutions W1, W2, W3, W4.

12 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Now, assume that

C2∈ E

. The points

M, P1, P2, P3, C2

must all be in both the

ellipse

E

and the circumcircle

K

of

P1P2P3

. Since the two conics have at most 4

intersections (counting multiplicity), 2 of those 5 points must coincide. It is easy to

verify from the previously-computed coordinates that

M

can only coincide with the

preimages

P1, P2, P3

at the vertices of

E

. In such cases, owing to the symmetry of

the geometry about either the x- or y-axis, the circle

K

must be tangent to

E

at

M

. Thus, that intersection will count with multiplicity (of at least) 2, so another

pair of those 5 points must also coincide. Since

P1, P2, P3

must all be distinct,

C2

will coincide with one of the preimages. However, this can never happen, since if

M

is on one of the vertices of

E

,

C2

won’t be in

E

. In any other case, since

P1, P2, P3

must be distinct and

C2

is diametrically opposed to

M

in

K

,

C2

must coincide with

one of the preimages.

Remark 10.

C2

will visit each of the preimages cyclically. Moreover, upon 3

revolutions (with 12 crossings in total), each

Pi

will have been visited four times

and the process repeats.

Figure 10. The loci of the cusps of ∆(dashed line) is a degree-6 curve with 2 internal lobes

with either 2, 3, or 4 self-intersections. From left to right, a/b ={1.27,√2,1.56}. Note that at

a/b =√2the two lobes touch, i.e., the cusps pass through the center of E. Also shown is the

elliptic locus of C2(purple). Points Zi(resp. Wi) mark oﬀ the intersection of the locus of the

cusps (resp. of C2) with E. These never coincide Video: [10, PL#04].

6.1. Tangencies and Intersections of the Deltoid with the Ellipse.

Deﬁnition 1

(Apollonius Hyperbola)

.

Let

M

be a point on an ellipse

E

with

semi-axes

a, b

. Consider a hyperbola

H

, known as the Apollonius Hyperbola of

M

[5]:

H:h(x, y)−M, (y/b2,−x/a2)i= 0.

Notice that for

P

on

E

, only the points for which the normal at

P

points to

M

will lie on H. See also [4, page 403].

Additionally,

H

passes through

M

and

O

, and its asymptotes are parallel to the

axes of E.

Proposition 11.

∆is tangent to

E

at

E ∩ H

, at 1, 2 or 3 points

Qi

depending on

whether Mis exterior or interior to the evolute E∗.

STEINER’S HAT: A CONSTANT-AREA DELTOID 13

Figure 11. Steiner’s Hat ∆(purple, top cusp not shown) is tangent to Eat the intersections

Qiof the Apollonius Hyperbola H(olive green) with E, excluding M. Notice Hpasses through

the center of E.Left: When Mis exterior to the evolute E∗(dashed purple), only one tangent

Q1is present. Right: When Mis interior to E∗, three tangent points Qi, i = 1,2,3arise. The

intersections of E∗are given in Equation 10. Note: the area ratio of ∆-to-E∗is always 4/3.

Proof.

∆is tangent to

E

at some

Qi

if the normal of

E

at

Qi

points to

M

= (

Mx, My

),

i.e., when

H

intersects with

E

. It can be shown that their

x

coordinate is given by

the real roots of:

(9) Q(x) = c4x3−c2Mx(a2+b2)x2−a4(a2−2b2)x+a6Mx= 0

The discriminant of the above is:

−4c4a6(a2−M2

x)[(a2−b2)(a2+b2)3M2

x+a4(a2−2b2)3].

Let

±x∗

denote the solutions to (

a2−b2

)(

a2

+

b2

)

3M2

x

+

a4

(

a2−

2

b2

)

3

= 0.

Assuming

a>b

, Equation 9has three real solutions when

|x|< x∗.

The intersections

of the evolute

E∗

with the ellipse

E

are given by the four points (

±x∗,±y∗

), where:

(10) x∗=a2√a4−b4a2−2b2

3

2

(a4−b4) (a2+b2), y∗=b2√a4−b42a2−b2

3

2

(a4−b4) (a2+b2)

For

M∈ E ∩ E ∗

two coinciding roots result in a 4-point contact between ∆and the

ellipse.

Let

M

= (

Mx, My

)be a point on

E

and

J

(

x

)denote the following cubic polyno-

mial:

(11) J(x)=(a2+b2)2x2−2Mxc2(a2+b2)x−4a4b2+M2

x(a2+b2)2

Proposition 12.

∆intersects

E

at

Q

(

x

)

J

(

x

) = 0, in at least 3and up to 5locations

locations, where Qis as in Equation 9.

14 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Proof.

As before,

M

= (

Mx, My

) = (

acos u, b sin u

)

∈ E

and

P

= (

x, y

) =

(

acos t, b sin t

). The intersection ∆

u

(

t

)with

E

is obtained by setting

E

(∆

u

(

t

)) = 0.

Using Equation 1, obtain the following system:

F(x, y) =b2a2−2M2

xa2+b2c4x4+ 2 a2MxMya2+b2c4x3y

+2 a2b2Mxc2a4+b4x3−2a4Myc2a4+b4x2y

+−a4b2a2+b23a4−4a2b2+ 2 b4+a2b2c2M2

x3a2−b2a2+b2x2

−2a6MxMyc2a2+b2xy −2b2Mxa6a4−a2b2+b4x

+2 a12Myy+a8b22a4−(a2+b2)M2

x= 0

E(x, y) = x2

a2+y2

b2−1=0

By Bézout’s theorem, the system

E

(

x, y

) =

F

(

x, y

)=0has eight solutions, with

algebraic multiplicities taken into account. ∆has three points of tangency with

ellipse, some of which may be complex, which by Proposition 11 are given by the

zeros of

Q

(

x

). Eliminating

y

by computing the resultant we obtain an Equation

G

(

x

) = 0 of degree 8 over

x

. Manipulation with a Computer Algebra System yields

a compact representation for G(x):

G(x) = Q(x)2J(x)

with

J

as in Equation 11. If

|Mx| ≤ a

, the solutions of

J

(

x

) = 0 are real and given

by Jx= [c2Mx±2abpa2−M2

x]/(a2+b2)and |Jx| ≤ a.

Referring to Figure 10:

Proposition 13.

When a cusp

P0

i

crosses the boundary of

E

, it coincides with its

pre-image Piat Zi=1

√a2+b2(±a2,±b2).

Proof.

Assume

P0

i

is on

E

. Since

P0

i

is on the circle

Ki

deﬁned by

M

,

Pi

and

P0

i

which osculates

E

at

Pi

, this circle intersects

E

at

Pi

with order of contact 3 or 4.

By construction, we have

MPi⊥PiP0

i

, so

M

and

P0

i

are diametrically opposite in

Ki

. Thus,

M

and

P0

i

must be distinct. Since two conics have at most 4 intersections

(counting multiplicities), we either have

P0

i

=

Pi

or

Pi

=

M

. The second case will

only happen when

M

is on one of the four vertices of the ellipse

E

, in which case

the osculating circle

Ki

has order of contact 4, so

P0

i

could not also be in the ellipse

in the ﬁrst place. Thus, P0

i=Pias we wanted.

Substituting the parameterization of

Pi

in the equation of

E

, we explicitly ﬁnd

the four points Ziat which P0

ican intersect the ellipse E.

7. A Triad of Osculating Circles

Recall

Ki

are the circles which osculate

E

at the pre-images

Pi

, see Figure 8.

Deﬁne a triangle

T00

by the centers

P00

i

of the

Ki

. These are given explicit coordinates

in Appendix A. Referring to Figure 12:

Proposition 14.

Triangles

T0

and

T00

are homothetic at ratio 2 : 1, with

M

as the

homothety center.

STEINER’S HAT: A CONSTANT-AREA DELTOID 15

Figure 12. Lines connecting each cusp P0

ito the center P00

iof the circle which osculates Eat

the pre-image Piconcur at M. Note these lines are diameters of said circles. Therefore Mis the

perspector of T0and T00 , i.e., the ratio of their areas is 4. This perspectivity implies C2, X 00

2, M

are collinear. Surprisingly, the X00

2coincides with the circumcenter X3of the pre-image triangle T

(not drawn).

Proof.

From the construction of ∆, for each

i

= 1

,

2

,

3we have

MPi⊥PiP0

i

, that

is,

∠MPiP0

i

= 90

◦

. Hence,

M P 0

i

is a diameter of the osculating circle

Ki

that goes

through

M

,

Pi

, and

P0

i

as proved in Proposition 3. Thus, the center

P00

i

of

Ki

is the

midpoint of

M P 0

i

and therefore

P0

i

is the image of

P00

i

under a homothety of center

Mand ratio 2.

Corollary 4. The area A00 of T00 is invariant over all Mand is 1/4that of T0.

Proof.

This follows from the homothety, and the fact that the area of

T0

is invariant

from Proposition 4.

Proposition 15.

Each (extended) side of

T0

passes through an intersection of two

osculating circles. Moreover, those sides are perpendicular to the radical axis of said

circle pairs.

Proof.

It suﬃces to prove it for one of the sides of

T0

and the others are analogous.

Let

M1

be the intersection of

K2

and

K3

diﬀerent than

M

and let

M1/2

be the

16 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

midpoint of

M

and

M1

. Since

MM1

is the radical axis of

K2

and

K3

, the lines

P00

2P00

3

and

MM1

are perpendicular and their intersection is

M1/2

. Applying an

homothety of center

M

and ratio 2, we get that the lines

P0

2P0

3

and

MM1

(the

radical axis) are perpendicular and their intersection is M1, as desired.

Corollary 5.

The Steiner ellipse

E00

of triangle

T00

is similar to

E0

. In fact,

E00 =1

2E0.

Proof. This follows from the homothety of T0and T00.

8. Relations between T , T 0, T 00

As before we identify Triangle centers as

Xk

after Kimberling’s Encyclopedia [

6

].

Proposition 16.

The circumcenter

X3

of

T

coincides with the barycenter

X00

2

of

T00.

Proof.

Follows from direct calculations using the coordinate expressions of

Pi

and

P00

i. In fact,

X00

2=X3=c2

4cos u

a,−sin u

b.

Corollary 6. The homothety with center Mand factor 2sends X00

2to X0

2=C2.

Proposition 17.

The lines joining a cusp

P0

i

to its preimage

Pi

concur at ∆’s

center of area C2.

Proof.

From Theorem 2, points

M

and

C2

both lie on the circumcircle of

T

and

form a diameter of this circle. Thus, for each

i

= 1

,

2

,

3, we have

∠MPiC2

= 90

◦

.

By construction, ∠MPiP0

i= 90◦, so Pi,P0

i, and C2are collinear as desired.

Referring to Figure 14(left):

Corollary 7. C2=X0

2is the perspector of T0and T.

Lemma 2.

Given a triangle

T

, and its Steiner Ellipse Σ, the normals at each

vertex pass through the Orthocenter of T, i.e., they are the altitudes.

Proof.

This stems from the fact that the tangent to Σat a vertex of

T

is parallel

to opposide side of T[12, Steiner Circumellipse].

Referring to Figure 14(right):

Proposition 18.

The orthocenter

X4

is the perspector of

T

and

T00

. Equivalently,

a line connecting a vertex of

T

to the respective vertex of

T00

is perpendicular to the

opposite side of T.

Proof.

Since

T

has ﬁxed

X2

,

E

is its Steiner Ellipse. The normals to the latter at

Pi

pass through centers

P00

i

since these are osculating circles. So by Lemma 2the

proof follows.

Deﬁnition 2.

According to J. Steiner [

3

, p. 55], two triangles

ABC

and

DEF

are

said to be orthologic if the perpendiculars from

A

to

EF

, from

B

to

DF

, and from

C

to

DE

are concurrent. Furthermore, if this holds, then the perpendiculars from

D

to

BC

, from

E

to

AC

, and from

F

to

AB

are also concurrent. Those two points

of concurrence are called the centers of orthology of the two triangles [9].

STEINER’S HAT: A CONSTANT-AREA DELTOID 17

Figure 13. Consider a reference triangle T(blue), and its pedal T0(red) and antipedal T00 (green)

triangles with respect to some point Z. Construction lines for both pedal and antipedal (dashed

red, dashed green) imply that Zis an orthology center simultaneousy for both T,T0and T,T00,

i.e. these pairs are orthologic. Also shown are H0and H00, the 2nd orthology centers of said pairs

(construction lines omitted). Non-transitivity arises from the fact that perpendiculars dropped

from the vertices of T0to the sides of T00 (dashed purple, feet are marked X) are non-concurrent

(purple diamonds mark the three disjoint intersections), i.e., T0,T00 are not orthologic.

Note that orthology is symmetric but not transitive [

9

, p. 37], see Figure 13 for

a non-transitive example involving a reference, pedal, and antipedal triangles.

Lemma 3.

Let

−Pi

denote the reﬂection of

Pi

about

O

=

X2

for

i

= 1

,

2

,

3. Then

the line from

M

to

−P1

is perpendicular to the line

P00

2P00

3

, and analogously for

−P2

,

and −P3.

Proof.

This follows directly from the coordinate expressions for points

M

,

Pi

=

E

(

ti

)

and P00

i=E∗(ti). It follows that hM+P1, P 00

2−P00

3i= 0.

Referring to Figure 14(right):

Theorem 3.

Triangles

T

and

T0

are orthologic and their centers of orthology are

the reﬂections X671 of Mon X2and on X4.

Proof.

We denote by

X671

the reﬂection of

M

=

X99

on

O

=

X2

. From Lemma 3,

the line through

M

and

−P1

is perpendicular to

P00

2P00

3

. Reﬂecting about

O

=

X2

,

the line

P1X671

is also perpendicular to

P00

2P00

3

. Since

P00

2P00

3kP0

2P0

3

from the

homothety, we get that

P1X671 ⊥P0

2P0

3

. This means the perpendicular from

P1

to

P0

2P0

3

passes through

X671

. Analogously, the perpendiculars from

P2

to

P0

1P0

3

and

from

P3

to

P0

1P0

2

also go through

X671

. Therefore

T

and

T0

are orthologic and

X671

is one of their two orthology centers.

Let

Xh

be the reﬂection of

M

on

X4

. From Proposition 18, the line through

P1

and

P00

1

passes through the orthocenter

X4

of

T

, that is,

P00

1

is on the

P1

-altitude of

T

. This means that the perpendicular from

P00

1

to the line

P2P3

passes through

X4

.

Applying the homothety with center

M

and ratio 2, the perpendicular from

X0

1

to

X2X3

passes through

Xh

. Analogously, the perpendiculars from

X0

2

to

X1X3

and

18 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 14. Left: Tand T0are perspective on X2. They are also orthologic, with orthology

centers X671 and the reﬂection of X99 on X4.Right: Tand T00 are perspective on X4. They are

also orthologic, with orthology centers X4and X671.

from

X0

3

to

X1X2

also pass through

Xh

. Hence,

Xh

is the second orthology center

of Tand T0.

Theorem 4.

Triangles

T

and

T00

are orthologic and their centers of orthology are

X4and the reﬂection X671 of Mon X2.

Proof.

From Proposition 18, the perpendiculars from

P00

1

to

P2P3

, from

P00

2

to

P1P3

,

and from

P00

3

to

P1P2

all pass through

X4

. Thus, triangles

T

and

T00

are orthologic

and X4is one of their two orthology centers.

As before, we denote by

X671

the reﬂection of

M

=

X99

on

O

=

X2

. Again, from

Lemma 3, the line through

M

and

−P1

is perpendicular to

P00

2P00

3

, so reﬂecting it at

X2

, we get that

P1X671 ⊥P00

2P00

3

. Since the triangles

T00

and

T0

have parallel sides,

we get

P1X671 ⊥P0

1P0

3kP00

1P00

3

. Thus,

X671

is the second orthology center of

T

and

T00.

Theorem 5

(Sondat’s Theorem)

.

If two triangles are both perspective and orthologic,

their centers of orthology and perspectivity are collinear. Moreover, the line through

these centers is perpendicular to the perspectrix of the two triangles [11,9].

Referring to Figure 15:

Theorem 6. The perspectrix of Tand T0is perpendicular to the Euler Line of T.

Proof.

Since

T

and

T0

are both orthologic and perspective from Corollary 7and

Theorem 3, by Sondat’s Theorem, their perspectrix is perpendicular to the line

through their orthology centers (reﬂections of

M

at

X2

and

X4

) and perspector

(

X0

2

=

C2

=

X98

=reﬂection of

M

at

X3

). By applying a homothety of center

M

and ratio 1/2, this last line is parallel to the line through

X2

,

X3

, and

X4

, the Euler

line of

T

. Therefore the perspectrix of

T

and

T0

is perpendicular to the Euler line

of T.

Proposition 19.

The perspectrix of

T

and

T00

is perpendicular to the line

X4X671

(which is parallel to the line through Mand X376, the reﬂection of X2at X3).

Proof.

Since

T

and

T00

are both orthologic and perspective from Proposition 18

and Theorem 4, by Sondat’s Theorem, their perspectrix is perpendicular to the line

STEINER’S HAT: A CONSTANT-AREA DELTOID 19

Figure 15. The perspectrix of T, T 00 is perpendicular to the line through X4and X671 . Compare

with Figure 13: the perspectrix of T, T 0is perpendicular to the Euler Line of T.

through their orthology centers

X4

and

X671

. Reﬂecting this last line at

X2

, we

ﬁnd that it is parallel to the line through

M

and the reﬂection of

X4

at

X2

, which

is the same as the reﬂection of X2at X3.

Table 2lists a few pairs of triangle centers numerically found to be common over

T , T 0or T, T 00.

T T 0T00

X3–X00

2

X4–X00

671

X5–X00

115

X20 –X00

99

X76 –X00

598

X98 X0

2–

X114 X0

230 –

X382 –X00

148

X548 –X00

620

X550 –X00

2482

Table 2. Triangle Centers which coincide T, T 0or T , T 00 .

9. Addendum: Rotated Negative Pedal Curve

The Negative Pedal Curve is the envelope of lines

L

(

t

)passing through

P

(

t

)and

perpendicular to

P

(

t

)

−M

. Here we consider the envelope ∆

∗

θ

of the

L

(

t

)rotated

clockwise a ﬁxed θabout P(t); see Figure 16.

Proposition 20.

∆

∗

θ

is the image of the NPC ∆under the similarity which is the

product of a rotation about

M

through

θ

and a homothety with center

M

and factor

cos θ.

20 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

Figure 16. From left to right: for a ﬁxed M, the line passing through P(t)and p erpendicular to

the segment P(t)−M(dashed) purple is rotated clockwise by θ= 30,45,60 degrees, respectively

(dashed olive green). For all P(t)these envelop new constant-area crooked hats ∆∗(olive green)

whose areas are cos(θ)2that of ∆. For the θshown, these amount to 3/4,1/2,1/4of the area of ∆

(purple). As one varies θ, the center of area C∗

2of ∆∗sweeps the circular arc between C2and M

with center at angle 2θ. The same holds for the cusps running along the corresp onding osculating

circles (shown dashed red, green, blue), which are stationary and independent of θ.Video: [10,

PL#06]

Proof.

For variable parameter

t

, the lines

L

(

t

)and

L∗

θ

(

t

)perform a motion which

sends

P

along

E

, while the line through

P

orthogonal to

L

(

t

)slides through the ﬁxed

point

M

. Due to basic results of planar kinematics [

1

, p. 274], the instantaneous

center of rotation

I

lies on the normal to

E

at

P

and on the normal to

MP

at

M

.

We obtain a rectangle with vertices

P

,

M

and

I

. The fourth vertex is the enveloping

point

C

of

L

(

t

). The enveloping point

C∗

of

L∗

θ

is the pedal point of

I

. Since the

circumcircle of the rectangle with diameter

MC

also passes through

C∗

, we see that

C∗is the image of Cunder the stated similarity, Figure 17.

This holds for all points on ∆, including the cusps, but also for the center

C2

. At

poses where

C

reaches a cusp

P0

i

of ∆, then for all lines

L∗

θ

(

t

)through

P

the point

C∗

is a cusp of the corresponding envelope. Then the point is the so-called return

pole, and the circular path of Cthe return circle or cuspidal circle [1, p. 274].

Corollary 8. The area of ∆∗

θis independent of Mand is given by:

A=c4cos2θ π

2ab .

Note this is equal to cos2θof the area of ∆, see Equation 5.

Remark 11.For variable

θ

between

−

90

◦

and 90

◦

, the said similarity deﬁnes an

equiform motion where each point in the plane runs along a circle through

M

with

the same angular velocity. For each point, the conﬁguration at

θ

= 0 and

M

deﬁne

a diameter of the trajectory.

Recall the pre-images

Pi

of the cusps of ∆have vertices at

E

(

ti

), where

t1

=

−u

3

,

t2=−u

3−2π

3,t3=−u

3−4π

3, see Theorem 2.

Corollary 9.

The cusps

P∗

i

of ∆

∗

θ

have pre-images on

E

which are invariant over

θand are congruent with the Pi, i = 1,2,3.

Corollary 10. Lines PiP∗

iconcur at C∗

2.

STEINER’S HAT: A CONSTANT-AREA DELTOID 21

P(t)

M

C

C*

I

Figure 17. The construction of points C,C∗of the envelopes ∆(red) and ∆∗(blue) with the help

of the instant center of rotation Ireveals that the rotation about Mthrough θand scaling with

factor cos θsends ∆to ∆∗(Proposition 20).

Corollary 11. C∗

2

is a rotation of

C2

by 2

θ

about the center

X3

of

K

. In particular,

When θ=π/2,C∗

2=M, and ∆∗

θdegenerates to point M.

10. Conclusion

Before we part, we would like to pay homage to eminent swiss mathematician

Jakob Steiner (1796–1863), discoverer of several concepts appearing herein: the

Steiner Ellipse and Inellipse, the Steiner Curve (or hypocycloid), the Steiner Point

X99

. Also due to him is the concept of orthologic triangles and the theorem of 3

concurrent osculating circles in the ellipse. Hats oﬀ and vielen dank, Herr Steiner!

Some of the above phenomena are illustrated dynamically through the videos on

Table 3.

22 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

PL# Title Narrated

01 Constant-Area Deltoid no

02 Properties of the Deltoid yes

03 Osculating Circles at the Cusp Pre-Images yes

04 Loci of Cusps and C2no

05 Concyclic pre-images, osculating circles,

and 3 area-invariant triangles

no

06 Rotated Negative Pedal Curve yes

Table 3. Playlist of videos. Column “PL#” indicates the entry within the playlist.

Appendix A. Explicit Expressions for the Pi, P 0

i, P 00

i

P1=hacos u

3,−bsin u

3,i

P2=h−asin(u

3+π

6),−bcos(u

3+π

6)i

P3=h−acos(u

3+π

6), b sin(u

3+π

6)i

P0

1="3c2

2acos u

3−a2+b2

2acos u, 3c2

2bsin u

3−a2+b2

2bsin u#

P0

2="−3c2

4acos u

3−3√3c2

4asin u

3−a2+b2

2acos u, 3√3c2

4bcos u

3−3c2

4bsin u

3−a2+b2

2bsin u#

P0

3="−3c2

4acos u

3+3√3c2

4asin u

3−a2+b2

2acos u, −3√3c2

4bcos u

3−3c2

4bsin u

3−a2+b2

2bsin u#

P00

1=3c2

4acos u

3+c2

4acos u, 3c2

4bsin u

3+c2

4bsin u

P00

2="−3c2

8acos u

3−3√3c2

8asin u

3+c2

4acos u, 3√3c2

8bcos u

3−3c2

8bsin u

3−c2

4bsin u#

P00

3="−3c2

8acos u

3+3√3c2

8asin u

3+c2

4acos u, −3√3c2

8bcos u

3−3c2

8bsin u

3−c2

4bsin u#

Appendix B. Table of Symbols

STEINER’S HAT: A CONSTANT-AREA DELTOID 23

symbol meaning note

Emain ellipse

a, b major, minor semi-axes of E

chalf the focal length of Ec2=a2−b2

Ocenter E

M, Mua ﬁxed point on the boundary of E[acos u, b sin u],

perspector of T0, T 00,=X99

P(t)a point which sweeps the boundary of E[acos t, b sin t]

L(t)Line through P(t)perp. to P(t)−M

∆,∆uSteiner’s Hat, negative pedal curve

of Ewith respect to M

invariant area

∆∗

θenvelope of L(t)rotated θabout P(t)invariant area

¯

Caverage coordinates of ∆ = C2=X0

2

C2, C∗

2area center of ∆,∆∗C2=X0

2=X98

P0

i, Pi, P 00

iThe cusps of ∆, their pre-images,

and centers of Ki(see below)

P∗

icusps of ∆∗PiP∗

iconcur at C∗

2

T , T 0, T 00 triangles deﬁned by the Pi, P 0

i, P 00

iinvariant area over M

A, A0, A00 areas of T , T 0, T 00 A0/A00 = 4 for any M, a, b

SSteiner’s Curve aka. Hypocycloid and Triscupoid

E0Steiner Circumellipse

of cusp (P0

i) triangle

centered at C2

a0, b0major, minor semi-axes of E0invariant, axis-parallel

and similar to 90◦-rotated E

KCircumcircle of Tcenter X3, contains M, Pi, C2, C∗

2

K0Circumcircle of T0

KiCircles osculating Eat the Picontain Pi, P 0

i, M

E∗evolute of Ethe Kilie on it

HApollonius Hyperbola of Ewrt M∆is tangent to Eat H∩E

X3circumcenter of T=X00

2

X4perspector of Tand T00

X99 Steiner Point of T=M

X98 Tarry Point of T=C2

X0

2centroid of T0=C2, and perspector of T , T 0

X0

99 Steiner Point of T0

X00

2centroid of T00 =X3

Table 4. All Symbols used.

24 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN

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