Minimum-Area ellipse containing a finite set of points. IRAB/Backup Oct 22 2011/f_3_CalculationForWFIRSTML/Bob2.pdfMINIMUM-AREA ELLIPSE CONTAINING A FINITE SET OF POINTS. I 1117 -_-.)

Ukrainian Mathematical Journal, VoL 50, No. 7, 1998

MINIMUM-AREA ELLIPSE CONTAINING A FINITE SET OF POINTS. I

B. V. Rublev and Yu. I. Petunin UDC 514.17+513+681.3

From the geometric point of view, we consider the problem of construction of a minimum-area ellipse containing a given convex polygon. For an arbitrary triangle, we obtain an equation for the boundary of the minimum-area ellipse in explicit form. For a quadrangle, the problem of construction of a minimum- area ellipse is connected with the solution of a cubic equation. For an arbitrary polygon, we prove that if the boundary of the minimum-area ellipse has exactly three common points with the polygon, then this ellipse is the minimum-area ellipse for the triangle obtained.

Introduction

We consider the problem of construction of the minimum-area ellipse containing a given finite set of points. Passing to a convex hull o f a finite collection of points, we can easily verify that this problem is equivalent to the problem of construction of the minimum-area ellipse containing a convex polygon. We propose the exact solution of this problem with the use of classical geometric methods and without using the theory of convex programming [1]. One of the first results obtained in this field belongs to Rademacher and Trpl i tz , who solved an analogous problem for a circle [2].

Definition 1. A minimum-area ellipse for a bounded set �9 is defined as an ellipse e ( ~ ) that satisfies the

following conditions: (i) ~ C e ( ~ ) and (ii) for any other ellipse e 1 such that �9 C e 1, we have S (e(dO)) < S(e l ) , where S (M) is the area of the figure M.

It is clear that, for a bounded set ~ , the minimum-area ellipse for d~ and conv (qb) (a convex hull of ~ ) co-

incide. Thus, the only restriction imposed on the class of bounded sets �9 (they can even include infinitely many

points) is the requirement that conv (qb) be a convex polygon, i.e., its boundary must consist of finitely many segments.

By contradiction, one can easily prove the following statement with the use of affine mappings on the plane of homothety and contraction:

Theorem 1, For an arbitrary convex polygon M, at least three vertices of M lie on the boundary of the minimum-area ellipse.

1. Minimum-Area Ellipse for a Triangle

We begin our investigation from the following obvious result:

Theorem 2, For a regular triangle T, the minimum-area ellipse is a circle circumscribed about T.

Proof. The proof of the theorem is equivalent to the proof of the following statement: The maximum-area triangle inscribed into a given circle is regular [3].

Kiev University, Kiev. Translated from Ukrainskii Matematicheskii Zhurnal, Vol. 50, No. 7, pp. 980-988, July, 1998. Original article submitted November 4, 1996.

0041-5995/98/5007-1115 $20.00 �9 1999 Kluwer Academic/Plenum Publishers 1115

I I 16 B.V. RUBLEV AND YU. I. I>ETUNIN

z+ C

/ /

/ /

/

/ /

/

/ /

A I / / 1 0 / B X

C,

Y

Fig. 1

By virtue of Theorem 2, we can obtain the minimum-area ellipse for an arbitrary triangle T. For this purpose, it is necessary to perform the affine transformation of T into a regular triangle To, then circumscribe a circle K about T 0, and perform the inverse affine transformation. As a result, T o transforms into T and the required minimum-area ellipse for the triangle T is the image of the circle K.

For further investigation of the minimum-area ellipse, we must solve the following problem:

Problem 1. Find an equation of the minimum-area ellipse for an arbitrary triangle.

To solve this problem, we introduce a Cartesian coordinate system OXYZ in the space ]~3. Let ABC be an

arbitrary triangle. We place this triangle in the plane ZOX as follows: the side AB lies on the OX-axis so that its midpoint coincides with the origin of coordinates O, and the vertex C is located in the upper half-plane (i.e., z > 0) (Fig. 1).

Assume that the vertices of the triangle ABC thus located have the following coordinates in the space •3:

A(-a ,O,O) , B(a,O,O), and C(b,O,c).

Then, in the plane XOY, we construct a regular triangle ABC" so that the vertex C' lies in the upper half- plane of XOY (i.e., y > 0). It is clear that, in this case, the point C' belongs to the OY-axis and has the coordi-

nates (0, ~/-3a, 0).

We project the plane ZOX onto XOY along the line l = CC'. In this case, the triangle ABC is mapped into the regular triangle ABC'. By virtue of affinity of the mapping, ellipses in one plane are mapped into ellipses in another plane and, moreover, the ratio of the areas of the projected figures remains unchanged. Let us find the equation for the image of the circle circumscribed about the triangle ABC' under this mapping. It follows from the argument presented above that this image is the required minimum-area ellipse for the triangle ABC.

MINIMUM-AREA ELLIPSE CONTAINING A FINITE SET OF POINTS. I 1117

-_-.)

CC'= (b , -~/-3a, c), The vector along which the projection is performed has the coordinates and an arbitrary point M ( x o, O, Zo) in the plane Z O X is mapped into the point M" in the plane X O Y with the coordinates

�9 Zo b y, ~/-3 az0 , x - + x 0, - - - , z = 0. (1)

C c

In the plane XOY, the circle circumscribed about the triangle ABC" is determined by the equation

a -- 4 a 2 . (2) (x')2 + Y ' - ~ 3

Relations (1) imply that the equation of the ellipse that is the image of circle (2) under the above-mentioned projection has the form

/xo ob) 2 - - - + -- = a 2" c c - f3

(3)

Hence, the equation for the minimum-area ellipse that solves Problem 1 has the form

c 2 x 2 - 2 b c x o z o + ( b 2 + 3 a 2 ) z 2 - 2 a 2 c z o - a 2 c 2 = O.

2. Minimum-Area Ellipse Circumscribed about a Convex Quadrangle

The problem of construction of the minimum-area ellipse for a convex quadrangle is more complicated than the analogous problem for a triangle because all triangles are invariant under affine transformations (recall that two figures are called affine-invariant if they can be transformed in one another by affine mappings), whereas the same is not true for convex quadrangles. Convex quadrangles are separated into classes of aff'me-invariant figures (equiv- alence classes [4]). One of these classes is formed by all parallelograms. The set of all trapezoids consists of infinitely many classes, and each of these classes contains an isosceles trapezoid; it is convenient to regard the latter as a representative of the corresponding class.

For what follows, we need the criterion of affine invariance of convex quadrangles presented below without

proof.

Lemma 1. Any two convex quadrangles whose diagonals are divided at the point o f intersection in equal

ratios a : c and b" d are affine-invariant.

It follows from the properties of affine mappings that the ratios in which the diagonals are divided at the point of intersection are equal for affme-invariant quadrangles. According to Lemma 1, the problem of determination of the minimum-area ellipse for a given convex quadrangle reduces to the solution of the problem of construction of the minimum-area ellipse for a quadrangle with fixed ratios a : c and b : d of the segments of diagonals divided by the

point of intersection. Let ABCD be a convex quadrangle whose diagonals intersect at the point O and are divided in the ratios

AO 1 BO n 2 . . . . . (4)

OC k 2' OD 1

In this case, without loss of generality, we can assume that

1118 B . V . RUBLEV AND YU. I. PETUNIN

0 < k ~ 1, 0 < n 5 1. (5)

Denote by W(k, n) the class of all convex quadrangles that are affine-invariant with respect to the given quadrangle A B C D .

In the course of investigation of minimum-area ellipses for convex quadrangles, we must solve the following problems:

P r o b l e m 2. Construct the minimum-area ellipse for a quadrangle from the class W ( k , n).

P r o b l e m 2a. Construct the minimum-area ellipse for a quadrangle from the class W ( k , n ) if it is known that all vertices of the quadrangle lie on the boundary of this ellipse.

It is easy to show that Problem 2a is equivalent to the following problem:

P r o b l e m 3. Inscribe the maximum-area quadrangle from the class W(k, n) into a given circle.

To solve Problem 3, we use methods of analytic geometry. Since the quadrangle A B C D is inscribed into the

circle, we have A O . O C = B O . OD. We set A O = 1. Then it follows from (4) that OC = k 2. Furthermore, if D O =

q, then B O = n 2 q, which yields A O . OC = k 2 = B O . OD =q2n2. Therefore, q = n / k and

A O = 1, CO = k z, BO = nk , DO = k / n . (6)

Any quadrangle whose diagonals satisfy condition (6) is inscribed into the circle and belongs to the class W ( k ,

n), and the angle q0 between its diagonals satisfies the inequality

0 < cp< re. (7)

Consider an arbitrary quadrangle for which conditions (6) are satisfied. We place this quadrangle in the coordinate plane X O Y as shown in Fig. 2. The point of intersection of its diagonals is located at the origin of coordinates, and the other points have the following coordinates:

A(1 ,0) , B ( - k n c o s c p , - k n s i n c p ~ C ( - k 2, 0 ) , D(-k cos cp;-k sin cp]. \ n n

We seek the center F of the circle circumscribed about the quadrangle A B C D as the point equidistant from the vertices A, B, C, and D. If 11 and 12 are the perpendiculars dropped to the midpoints of the segments A C

and BD, respectively, then F = l 1 f'] 12- Moreover,

11: x = ~ ( 1 - k 2 ) , (8)

12: y = (-cot 9)x + b,

where the parameter b is determined from the condition that 12 passes through the midpoint of the segment B D

(the point E). Therefore,

b = k(1-n2)

2n sin cp


y

- k n cos q0 _ ) 2 0

B ~

11 D

q~

- k n sin q~

X

Fig. 2

and

l 2" y = ( -cot q0)x + k ' l - n2 ' ( ] 2n sin cp

(9)

The abscissa x F of the point F is determined from Eq. (9). To determine the ordinate YF, we solve system (8),

(9). As a result, we get

1 - k 2 k ( 1 - n 2 ) - ( c o s q ~ ) ( 1 - k 2 ) n X F = - - ' Y F =

2 2n sin q~ (10)

To simplify the notation, we set

I = (n2k 2 + 1)(/7 2 + k2) , J = 2 n k ( 1 - n 2 ) ( 1 - k 2 ) ,

L = 4n2k 2. (11)

The square of the radius R of the required circle can now be written in the form

R 2 = l - J c o s c p - L c o s Z c P 4n 2 sin 2 q0

(12)

The area of the quadrangle ABCD is expressed in terms of the diagonals and the angle between them as follows:

1120 B.V.-RUBLEV AND YU. I. PETUNIN

S(ABCD)= (l+k2)(l+n2)ksinq~ 2n

This enables us to represent the ratio of the area of the quadrangle to the area of the circle as a funct ion o f the angle

qo in the form

-r = S(ABCD) 2nk(1 + k2)(1 + n2) sin 3 qo

rcR 2 ~ ( I - J cos qo - L cos 2 qo)"

For f ixed values o f n to the problem of determinat ion o f the minimum of the funct ion

I - J cos q0 - L cos 2 q0 f(~o) = sin 3 cp

The critical values o f f (q0) are determined f rom the relation

and k, the problem o f determinat ion of the maximum of the function 7(q0) is equivalent

L c o s 3 ~ + 2 J c o s 2 ~ + ( 2 L - 3 1 ) c o s ~ + J = O,

f ' ( ~ ) = s i n 4 ~

which is equivalent to the condi t ion

g ( ~ ) = L cos 3 q0 + 2 J COS 2 (p + (2L - 31) cos q) + J = 0.

Since q0 satisfies condit ion (7), we have cos q0 ~ ( - 1, 1 ). Introducing the new variable t = cos q0, we obtain the equat ion

g l ( t ) = L t 3 + 2 J t 2 + ( 2 L - 3 I ) t + J = 0. (13)

To determine the roots o f Eq. (13), we note that

l im g i ( t ) = - ~ , t - -~ - - ~

g l ( - 1 ) = 3(I+ J - L ) = 3 ( n + k 2 ) ( 1 - n k 2) > O,

gl(O) = J = 2 n k ( 1 - n Z ) ( 1 - k 2) > O,

g l ( 1 ) = 3(L + J - I ) = - 3 ( l + n k ) Z ( n - k ) 2 < O,

l im gl( t ) = +~.

This implies that the cubic polynomial g 1 ( t ) has three real roots, which belong, respect ively, to the intervals

( - ~ , - 1 ), (0, 1 ), and ( 1, +co). Since [ t] < 1, the equat ion g 1 ( t) = 0 has only one root in the interval ( - 1 , 1 ).


Returning to the function g(tp), we get g(0) = g I ( 1 ) < 0 and g(rt/2) = g l (0) > 0. Hence, its derivative has

a unique root q0 0 ~ (0, n/2) and changes its sign at this point (from plus to minus). Therefore, the function f(q0)

has the minimum at this critical point. Thus, the solution of Problem 3 is reduced to the investigation of the cubic equation (13) and determination of

the root lying in the interval (0, 1 ). It is easy to see that Problem 2a reduces to Problem 3. Namely, by solving Problem 3 for a given class of quad-

rangles W(k, n), we can construct a quadrangle A1B 1C1D 1 e W(k, n) with the maximum ratio of its area to the

area of the circumscribed circle. Then we can circumscribe the circle K 1 about the quadrangle A1 B1Cl Dl" This

circle is mapped into the required ellipse e under the affine mapping that transforms AIB 1C 1D 1 into ABCD.

3. Minimum-Area Ellipse for an Arbitrary Polygon in the Case where Exactly Three Vertices of the Polygon Lie on the Boundary of the Ellipse

Let M = A 1A 2--- A n be an arbitrary convex polygon. We consider various triangles A iAjAk generated by the

vertices of this polygon (1 < i < j < k < n ) and choose the maximum-area triangle BCD ({B, C, D} c {Ai [i =

1, n }). Let us construct the minimum-area ellipse e for the triangle BCD. If, in this case, the entire polygon M C e, then e is the minimum-area ellipse for the entire polygon, because, otherwise, there exists the minimum-area ellipse e 1 = e(M) such that S(e l) < S(e). Since the triangle BCD C e 1, we conclude that e is not the minimum- area ellipse for the triangle BCD. This yields the following obvious statement:

Lemma 2. Let M = A 1A2 ... A n be a convex polygon, let BCD be the maximum-area triangle among all

triangles generated by the vertices of the polygon M, and let e (BCD) be the minimum-area ellipse for the triangle BCD. I f the polygon M C e(BCD), then e(BCD) is the minimum-area ellipse for the entire polygon M. I f AiAjA k is not the maximum-area triangle among all triangles with vertices from A 1A2 ... A n, then the

minimum-area ellipse for this triangle cannot be the minimum-area ellipse for the entire polygon M.

Theorem 3. The minimum-area ellipse for a convex polygon M = A 1A2... A n the boundary of which has

three common points with the polygon is the minimum-area ellipse for the triangle generated by these points.

Proof. As shown above (see Theorem 1), the boundary of the minimum-area ellipse for the polygon M has at least three common points with M. Let us prove the following assertion: If some ellipse contains the polygon M, has exactly three vertices of the polygon M lying on its boundary, and is not the minimum-area ellipse for the triangle generated by these vertices, then it is not the minimum-area ellipse for the polygon M. Therefore, one can construct an ellipse of smaller area that contains the polygon M. We prove this assertion by contradiction.

Assume that the boundary of the minimum-area ellipse e for the polygon M contains exactly three vertices A, B, and C of this polygon. We perform the affine transformation of the ellipse e into a circle K so that the polygon M is transformed into a polygon M" and the triangle ABC is transformed into a triangle A'B'C'.

Let B'C" be the greatest side of the triangle A'B'C'. We consider the Cartesian coordinate system with the

origin at the center of the circle K. We assume that the segment B'C" is parallel to the abscissa axis and is located in the lower half-plane, and K is the unit circle. In what follows, to simplify the notation, we omit the primes of all

corresponding quantities. The vertices B and C have the following coordinates:

B(a , -b ) , C ( - a , - b ) , ae ( 0, 1], b e [ 0, 1).

The case of an obtuse-angled triangle ABC is obvious. Let the triangle ABC be not obtuse-angled and let the third vertex A of it have the coordinates ( p, q), p >__ 0,

q > 0 (Fig. 3). In this case, the following conditions are satisfied:

1122 B.V. -RUBLEV AND YU. I. PETUNIN

r j

" - ] ~ A

, t I I I

0 I I

P

- b

c\ t1 m

2

B

X

F i g . 3

p 2 + q 2 = 1, a 2 + b 2 = 1, p<a , q>b. (14)

Furthermore, the following inequality holds:

1 b < - . (15)

2

Indeed, if b > 1/2, then the angle against the side BC is less than re/3. This contradicts the fact that the side

BC is the greatest side of the triangle ABC [5]. If the segment b = 1/2, then the angle BAC is equal to re/3, the triangle ABC is regular, and the circle K is the minimum-area ellipse for it [6], which contradicts the assumption.

Consider the ellipse e 6 centered at the point (0, fi), where ~ > 0, that passes through the vertices of the tri-

angle ABC and has axes parallel to the coordinate axes. The equation of the boundary of this ellipse has the form

( ~ 1, (16)

where U and V are unknown parameters, which are determined from the condition that the ellipse passes through

the vertices of the triangle ABC. 1 1

By setting t~ =~--~ and [3 = ~--2, we obtain

a 2 _ p2

( q - ~)2a2 _ p2(b + 5) 2,


= (q - 5) 2 - (b + ~)2

(q _ 8)2a2 _ p2(b + ~)2"

Note that, for sufficiently small 8 > 0, all vertices of the polygon M lie inside the ellipse e 5 or on its boundary.

Therefore, the parameters c~ and [3 are positive. Let us show that, for sufficiently small ~ > 0, we have S(K) > S(es) or

2 S2(K ) = 7g 2 > $ 2 ( e 5 ) = /g2U2V 2 = 71:.

Thus, it remains to prove that cx ~ > 1 for sufficiently small positive 5 > 0. This inequality is equivalent to

- 2 ( q 2 - b 2 ) ( q + b ) > - 4 ( b + q ) ( q 2 - b 2 ) ( 1 - q b )

or

2 ( q 2 - b 2 ) ( q + b ) ( 1 - 2 b q ) > O.

By virtue of conditions (14) and (15), the last inequality takes the form 2bq < 1, which obviously follows from the

inequalities 0 < q < 1 and 0 < b < 1/2 proved above.

Therefore, for sufficiently small 8 > 0, the area of the ellipse e 5 is smaller than the area of the circle K.

Moreover, as 5--) 0, we have

q2 _ b 2 a 2 _ p2 ~----> q2a2 _ p2b2 - 1, ~----~ q2a2 _ p2b2 - 1 .

Therefore, the ellipse e 5 tends to the circle K in the Hausdorff metric. By assumption, three vertices of the poly-

gon M lie on the boundary of the circle K, whereas the other vertices lie inside it. Therefore, for sufficiently small

5 > 0, exactly three vertices of the polygon M lie on the boundary of the ellipse e 5, and the other vertices lie

inside e 5- This implies that the circle K is not the minimum-area ellipse for the polygon M, which contradicts the

construction of the circle K. This contradiction completes the proof of the theorem.

Let us return to the solution of Problem 2. Let a certain quadrangle ABCD ~ W (k, n) be given. We consider

the triangles ABC, ABD, ACD, and BCD and choose the maximum-area triangle among them. For definiteness,

let it be the triangle ABC. For this triangle, we construct the minimum-area ellipse and denote it by e (ABC) If the

vertex D belongs to the ellipse e, the problem of construction of the minimum-area ellipse for the quadrangle

ABCD (and, hence, for the entire class W(k, n)) is solved, and the ellipse e is the required minimum-area ellipse. If the point D does not belong to e, then, by virtue of Lemma 2 and Theorem 3, the minimum-area ellipse for the

quadrangle ABCD is the ellipse circumscribed about this quadrangle. Hence, we arrive at Problem 2a, which has been solved in Sec. 2. Thus, Problem 2 is completely solved.

R E F E R E N C E S

1. L. Yu. Khachiyan, "Problems of optimal algorithms in convex programming, decompositions, and sorting," in: Computer and Problems of Choice [in Russian] (1983), pp. 161-205.

1124 B . V . RUBLEV AND YU. I. PETUNIN

2. H. Rademacher and O. T/Splitz, Numbers and Figures [Russian translation], Fizmatgiz, Moscow (1962). 3. W. Blaschke, Kreis undKugel, Velt, Bedin (1956). 4. I.A. Lavrov and L. L. Maksimova, Problems of Theory of Sets, Mathematical Logic, and the Theory of Algorithms [in Russian],

Nauka, Moscow (1975). 5. D.O. Shklyarskii, N. N. Chentsov, and I. M. Yaglom, Geometric Inequalities and Problems on Maxima and Minima [in Russian],

Nauka, Moscow (1970). 6. G. P61ya and G. Szeg/5, Isoperimetric Inequalities in Mathematical Physics, Princeton University, Princeton (1951).

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Minimum-Area ellipse containing a finite set of points. IRAB/Backup Oct 22 2011/f_3_CalculationForWFIRSTML/Bob2.pdfMINIMUM-AREA ELLIPSE CONTAINING A FINITE SET OF POINTS. I 1117 -_-.)