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Abstract and Figures

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.
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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 PiC2are 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
b21=0, c2=a2b2
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 tab)x0by2
0cos tc2y0sin(2t)
+b
4((5a2b2) cos tc2cos(3t))]
y(t) = 1
w[ax2
0sin t+ (by0cos t+c2sin(2t))x0aby0
a
4((5a2b2) sin tc2sin(3t)](1)
where w=ab bx0cos tay0sin t.
4 R. GARCIA, D. REZNIK, H. STACHEL, AND M. HELMAN
Proof. The line L(t)is given by:
(x0acos t)x+ (y0bsin t)y+a2cos2t+b2sin2tax0cos tby0sin 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 ta2cos u
yu(t) =1
bc2cos tsin(t+u)c2sin ta2sin 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(a2b2)(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)] = (UVDRu)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(∆) = (a2b2)2π
2ab =c4π
2ab
Proof.
The area of
S
(
t
)is
RSxdy
= 2
π
. The Jacobian of (
USDRu
)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
90about 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
32π
3
and
t3
=
u
34π
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+y2c2cos u
2ax+c2sin u
2by1
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 90about 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
3x4ac2sin3u
3y
+ab 3c2cos 2u
3a2b2= 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 Eis A(E) = 3πc4
8ab , and the area of is given in Equation 5:
Remark 6.The area ratio of and the interior of Eis 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
22x0x1x2(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=273
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 33.
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 90about 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)a22b22a2b2= 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) = "a22b2
acos u, 2a2b2
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 u3a42a2b2+ 7 b4x+ 2 bsin u7a42a2b2+ 3 b4y
a2+b2c2a2+b2cos 2 u+ 5 a414 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(i1)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 33ab
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
3a2+b2
21
acos u, 1
bsin u
(8)
This is a curve of degree 6, with the following implicit equation:
4a6x64b6y612 a2x2b2y2a2x2+b2y2
+12 a4a4a2b2+b4x4+ 12 b4a4a2b2+b4y4+ 24 a2b2a4a2b2+b4x2y2
3a22a2b2a2+b2 2a45a2b2+ 5 b4x2
+3 b2a22b2a2+b2 5a45a2b2+ 2 b4y2
+2a2b22a22b22a2+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
2a2+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 Eare given in Equation 10. Note: the area ratio of -to-Eis 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) = c4x3c2Mx(a2+b2)x2a4(a22b2)x+a6Mx= 0
The discriminant of the above is:
4c4a6(a2M2
x)[(a2b2)(a2+b2)3M2
x+a4(a22b2)3].
Let
±x
denote the solutions to (
a2b2
)(
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=a2a4b4a22b2
3
2
(a4b4) (a2+b2), y=b2a4b42a2b2
3
2
(a4b4) (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)2x22Mxc2(a2+b2)x4a4b2+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) =b2a22M2
xa2+b2c4x4+ 2 a2MxMya2+b2c4x3y
+2 a2b2Mxc2a4+b4x32a4Myc2a4+b4x2y
+a4b2a2+b23a44a2b2+ 2 b4+a2b2c2M2
x3a2b2a2+b2x2
2a6MxMyc2a2+b2xy 2b2Mxa6a4a2b2+b4x
+2 a12Myy+a8b22a4(a2+b2)M2
x= 0
E(x, y) = x2
a2+y2
b21=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±2abpa2M2
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
MPiPiP0
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
MPiPiP0
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
2P00
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
X3X00
2
X4X00
671
X5X00
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
Cis 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
32π
3,t3=u
34π
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,Cof 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=hasin(u
3+π
6),bcos(u
3+π
6)i
P3=hacos(u
3+π
6), b sin(u
3+π
6)i
P0
1="3c2
2acos u
3a2+b2
2acos u, 3c2
2bsin u
3a2+b2
2bsin u#
P0
2="3c2
4acos u
333c2
4asin u
3a2+b2
2acos u, 33c2
4bcos u
33c2
4bsin u
3a2+b2
2bsin u#
P0
3="3c2
4acos u
3+33c2
4asin u
3a2+b2
2acos u, 33c2
4bcos u
33c2
4bsin u
3a2+b2
2bsin u#
P00
1=3c2
4acos u
3+c2
4acos u, 3c2
4bsin u
3+c2
4bsin u
P00
2="3c2
8acos u
333c2
8asin u
3+c2
4acos u, 33c2
8bcos u
33c2
8bsin u
3c2
4bsin u#
P00
3="3c2
8acos u
3+33c2
8asin u
3+c2
4acos u, 33c2
8bcos u
33c2
8bsin u
3c2
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=a2b2
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
Eevolute of Ethe Kilie on it
HApollonius Hyperbola of Ewrt Mis 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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