Fermat bases of conics

J. Geom.c© 2013 Springer BaselDOI 10.1007/s00022-013-0194-y Journal of Geometry

Fermat bases of conics

Zvonko Cerin

Abstract. The concepts of a Fermat base of a plane curve or a Fermatlocus of a quadrangle come from an old geometric problem by Pierre deFermat about a semicircle on a side of a rectangle with ratio of adja-cent sides equal to

√2, which was resolved by synthetic methods first

by Leonard Euler in 1750. An arbitrary quadrangle has a plane curveof order four as its Fermat locus. Conics are Fermat loci of trapeziums.Conversely, for every conic one can ask to find all of its Fermat bases.We give answers separately for parabolas, hyperbolas, ellipses and circlestreating standard classes of special trapeziums: parallelograms, rhombi,rectangles and squares.

Mathematics Subject Classification (2000). 54H01.

Keywords. Fermat geometric problem, rectangle, semicircle, Euler, conic,parabola, hyperbola, ellipse, circle, quadrangle, trapezium,parallelogram, rhombus, square.

1. Introduction

Let τ = (u, v) be a pair of positive real numbers. For a curve Γ and a quadrangleQ = ABB′A′ in the plane, we use the notation Q

τ� Γ provided for every pointP on Γ the relation |AD|2 + u |BC|2 = v |AB|2 holds, where C and D are theintersections of the line AB with the lines A′P and B′P (see Fig. 1). In thissituation, we say that Q is a (Fermat) τ -base for Γ or that Γ is a (Fermat)τ -locus of Q (since Γ is generated from Q). We use terms ”Q is a Fermat baseof Γ” and ”Γ is a Fermat locus of Q” if there exist real numbers u and v suchthat Q is a (u, v)-base for Γ.

For a rectangle R = ABB′A′ in the plane, let us call s(R) = |AB||AA′| its shape.

Among the numerous questions that Pierre de Fermat has formulated, thefollowing geometric problem is our main concern (see Fig. 1).

Z. Cerin J. Geom.

Figure 1 The configuration of the Fermat problem

Fermat Problem. Let R = ABB′A′ be a rectangle with the shape√

2 and letΓ denote the semicircle on the side AB as a diameter. If θ = (1, 1), prove that

Rθ� Γ .

The great Leonard Euler in [6] has provided the first rather long proof thatavoids analytic geometry (which offers rather simple proofs). Several othermore concise synthetic proofs are now known (see [1, pp. 168, 169], [7, pp. 602,603], [8, pp. 181, 264], and [11]).

An analytic proof was recently recalled in [9] where it was observed that theEuler’s theorem can be extended to the circle with the segment AB as adiameter. The author also has done some recent contributions in the Euclideanplane (see [2] and [3]) and in the Euclidean space (see [4] and [5]). See also thenice description of Euler’s proof in [12].

In this paper we consider the generalization of the Fermat problem to conics.Let τ = (u, v) be a pair of positive real numbers. Our idea is to put the quad-rangle Q = ABB′A′ at the center of the stage and try to determine the curveΓ such that Q

τ� Γ holds. Using analytic geometry, it is easy to show that Γis a curve of order 4. However, when AB and A′B′ are parallel (i. e., when Qis a trapezium), then Γ is a conic. The type of this conic depends on the pairτ and the quotient |A′B′|

|AB| .

In the rest of this paper, we study the problem of finding conditions when ageneral trapezium (parallelogram, rhombus, rectangle or square) is a Fermat(u, v)-base of a given (standard) conic. It turns out that the cases of parabola,hyperbola, ellipse and circle each have their own peculiarities so that we treatthem separately.

In order to give an insight into our results we list here only some of these casescovering very special classes of rectangles.


For any positive real number s and every parabola Γ there are exactly two rec-tangles with the shape s and symmetric with respect to the line of symmetryof Γ which are its Fermat (1, 2)-bases. Similarly, for every real number s andevery hyperbola Γ with the semi-axes h and k there are exactly four rectan-gles of the shape s which are its Fermat η-bases for η =

(1, 2 + 2 h2

k2 s2

). These

rectangles have two sides parallel to either line of symmetry of the hyperbola.

Finally, for every sufficiently large positive real number s and every ellipse Γwith semi-axes h and k, there are exactly four rectangles of the shape s whichare its Fermat μ-bases and exactly four rectangles of the shape s which are itsFermat ν-bases for μ =

(1, 2 − 2 h2

k2 s2

)and ν =

(1, 2 − 2 k2

h2 s2

). These rectangles

again have two sides parallel to either line of symmetry of the ellipse.

The case of the circle (when h = k = r, its radius) is somewhat different. Forevery line π through its center (line of symmetry), there are exactly four rec-tangles of the shape s symmetric with respect to the line π which are its Fermatλ-bases for λ =

(1, 2 − 2

s2

). Of course, the Fermat problem follows from this

result for the circle.

Hence, we can say in an imprecise way that a circle, (genuine) ellipse, hyperbolaand parabola have infinitely many, exactly eight, four and two rectangles of agiven shape symmetric with respect to axis as its Fermat bases, respectively.

2. Quadrangles are τ -bases of curves of order four

Theorem 1. Let τ = (u, v) be a pair of positive real numbers. Every planequadrangle Q = ABB′A′ is a τ -base of some curve of degree four.

Proof. We shall use analytic geometry, which offers a simple proof. Let theorigin of the rectangular coordinate system be the midpoint O of the sideAB so that the points A and B have coordinates (−a, 0) and (a, 0) for somepositive real number a. The coordinates of the points A′ and B′ are (−b,−c)and (d, e) for some real numbers b, c, d and e. Of course, we assume that thequadrangle Q is non-degenerate. Hence, no three of its vertices are collinear.

An arbitrary point P on the required locus has coordinates (p, q). From thesimilar right-angled triangles, we find that C

(p c−q b

q+c , 0)

and D(

q d−p eq−e , 0

).

The condition |AD|2 + u |BC|2 = v |AB|2 will hold if and only if

[(a + d)q − e p − a e]2 E + u [(a + b)q − c p + a c]2 F − 4 v a2 E F = 0,

where E = (q + c)2 and F = (q − e)2. Hence, the locus is a curve of degreefour. �Figure 2 shows (shaded) the

(12 , 1

)-base of the curve Γ with the equation

f(p, q) = 0, where f(p, q) = h p2 + k p + m has the following coefficients

m = 4q4 + 4q3 − 23q2 − 44q − 20, k = 12q3 + 30q2 + 24q + 8

and h = 9q2 + 20q + 12.

Z. Cerin J. Geom.

D

C

B'(1,-2)

A'(-1,-1)

B(1,0)A(-1,0)

P

Figure 2 Curve Γ of order four and its(

12 , 1

)-base

3. Conics are τ -loci of trapeziums

We continue under the assumption that the vertices of the quadrangle ABB′A′

and the point P have coordinates as described in the proof of Theorem 1.

When the lines AB and A′B′ are parallel, then e = −c so that we can dividethe above equation of the locus by a common factor (q + c)2. What remainsis the quadratic equation

[(a + d)q + c(p + a)]2 + u [(a + b)q − c(p − a)]2 − 4 v a2 (q + c)2 = 0.

Theorem 2. Let τ = (u, v) be a pair of positive real numbers. Then every trapez-ium T = ABB′A′ in the plane with parallel sides AB and A′B′ is a τ -base ofa conic. This conic is an ellipse, a parabola or a hyperbola provided the number

v is less than, equal to or greater than u1+u

(1 + |A′B′|

|AB|)2

, respectively.

Proof. The above equation has the familiar form Ξ = 0, with

Ξ = a11p2 + 2 a12p q + a22q

2 + 2 a13p + 2 a23q + a33,

where a11 = c2(1 + u), a12 = c[(a + d) − (a + b)u], a13 = a c2(1 − u),

a22 = (a + d)2 + (a + b)2u − 4 v a2, a23 = a c[(a + d) + (a + b)u − 4 a v]

and a33 = a2c2(1 + u − 4 v).

It is well-known (see [10, 2.4]) that the invariants

I = a11 + a22 = [c2 + (a + d)2] + [c2 + (a + b)2]u − 4 a2 v,

D =∣∣∣∣a11 a12

a12 a22

∣∣∣∣ = c2[(2 a + b + d)2 u − 4 a2 v(1 + u)],

and


A =

∣∣∣∣∣∣a11 a12 a13

a12 a22 a23

a13 a23 a33

∣∣∣∣∣∣= −4 c4 a2(b + d)2 u v

determine the properties of a conic which do not depend on its position on theplane.

Since A �= 0, the conic is non-degenerate. Since |AB| = 2 a and |A′B′| = b + d(we assume that ABB′A′ has negative orientation!), its type depends on thesign of D as stated in the theorem. �Corollary 1. Let Γ be a parabola and let T = ABB′A′ be a trapezium with thelengths of parallel sides |AB| = 2 t and |A′B′| = 2 b. If T

τ� Γ holds, thenT is either a parallelogram or a crossed trapezium with b = 3 t if and only ifv = 4 u

1+u .

Proof. Without loss of generality, we can take that B′(c + 2 e b,−d), B(t, 0),A(−t, 0) and A′(c,−d), where d is a positive real number, c is a real numberand e = ±1. Note that T is a parallelogram if and only if b = t and e = 1. Also,T is a crossed trapezium if and only if e = −1.

Repeating the above arguments (i. e., writing down the equation of the locusunder the current selection of coordinates), we conclude that the D-invariant

4 d2[u(t + e b)2 − t2(u + 1)v]

of the Fermat (u, v)-locus of T (i. e., of the parabola Γ) vanishes. Hence,v = u

1+u

(1 + e b

t

)2. Thus, u

1+u

(1 + e b

t

)2= 4 u

1+u if and only if either e bt = 1 or

e bt = −3. In the first case T is a parallelogram while in the second case it is a

crossed trapezium with b = 3 t. �By solving the equation I2 = 4D (or the system a12 = 0, a11 = a22) in vari-ables u and v, we can answer the question for which pairs τ will the trapeziumABB′A′ be in the relation

τ� with a circle.

Theorem 3. For every trapezium T = ABB′A′ in the plane with parallel sidesAB and A′B′, there is only one pair τ = (u, v) of positive real numbers suchthat T

τ� Γ holds for a circle Γ.

Proof. The required pair is τ =(

a+da+b ,

(2 a+b+d)[(a+b)d+a2+a b−c2]4 a2(a+b)

), where the

values a, b, c and d have been explained in the proof of Theorem 1. �

4. Fermat bases of a parabola

The main question now is to determine for a given conic Γ and a pair τ ofpositive real numbers all trapeziums T in the plane that are its τ -bases (i. e.,such that T

τ� Γ holds).

Z. Cerin J. Geom.

In view of Corollary 1, this task is the simplest for parabolas that we considernow.

Theorem 4. For every parabola Γ and every positive real number u there is atwo-parameters family of parallelograms Pc,e that are its

(u, 4 u

1+u

)-bases.

Proof. Since every parabola can be reduced to a form q = 2 g p2, where g is apositive real number, using rotations and translations, it suffices to prove theclaim for this standard parabola.

Let α = c − e and β = c + e. An arbitrary parallelogram ABB′A′ has verticesA (a, b), B (c, d), A′ (e, f) and B′ (c + e − a, d + f − b) for some real numbersa, b, c, d, e and f . By repeating the above procedure for v = 4 u

1+u and solvingthe system a11 = n, a12 = 0, a22 = 0, a13 = 0, a23 = − n

4 g , a33 = 0 in the vari-ables a, b, c, d, e, f and n, we conclude that the solutions are parallelogramsABB′A′ with the coordinates of vertices

(a−, g y1

2

),

(c, g y2

2

),

(a+, g y3

2

)and(

e,− g y42

), where a± = β±αu

2 , y1 = (c − 5e)αu + β2, y2 = 3α2u + (3c − e)β,y3 = (c + 3e)αu + β2 and y4 = α2u + (c − 3e)β. �Similarly, for every parabola Γ and every positive real number u there isa two-parameters family of crossed trapeziums Qc,e with the bottom three

times longer than the top that are its(u, 4 u

1+u

)-bases. Indeed, if α = c − e and

β = 3c − e, then the vertices of Qc,e have the coordinates(a−, g y1

2

),(c, g y2

2

),(

a+, g y32

)and

(e,− g y4

2

), where a− = β+αu

2 , a+ = β+3αu2 ,

y1 = (9c − 5e)αu + β2, y2 = 3α2u + (c + e)β,

y3 = (3c + e)αu + β2 and y4 = 9α2u + (3c − 5e)β.

Theorem 5. For every parabola Γ and every positive real number u there is aone-parameter family of rhombi Hh that are its

(u, 4 u

1+u

)-bases.

Proof. We shall again work with the standard parabola q = 2 g p2, where g isa positive real number.

An arbitrary rhombus ABB′A′ has vertices either

A(a, d+f

2

), B(c, d), B′

(2c − a, d+f

2

)and A′(c, f)

or

A (t+, b) , B (c, d) , B′ (t−, d + f − b) and A′(e, f)

for some real numbers a, b, c, d, e and f , where c �= e and

t± =c2 − e2 ± (d2 − f2) ± 2(f − d)b

2(c − e).

By repeating the above procedure for v = 4 u1+u and solving the system a11 = n,

a12 = 0, a22 = 0, a13 = 0, a23 = − n4 g , a33 = 0 in the variables a, b, c, d, e, f and

n, we conclude that for the first group of rhombi there are no solutions whilefor the second group (when c �= e) the solutions are rhombi Hh = ABB′A′


with the coordinates of vertices(

x1y1

, x2y2

),(

2x3y1

, x4y2

),(

hg , x5

y2

)and

(− 1

16gh , x6y2

),

where

x1 = (32h2 + 1)u − 1, y1 = 16gh(u + 1), x2 = 32h2(24h2 + 1)u + 1,

y2 = 128gh2(u + 1), x3 = 8h2(u − 1) − 1, x4 = 32h2(8h2u + 1) + 3,

x5 = 32h2(8h2u − 1) − 1 and x6 = 1 − 32h2(8h2 + 1)u.

�Theorem 6. Among the parallelograms Pc,e there are exactly two which arerectangles of the given shape s. They are symmetric with respect to the line ofsymmetry of the parabola Γ.

Proof. We shall again work with the standard parabola q = 2 g p2 and use itsparallelograms Pc,e = ABB′A′ described in the proof of the previous theorem.

In order that ABB′A′ is a rectangle of the shape s > 0, the sides AB and AA′

must be perpendicular and |AB| = s|AA′|. These two conditions hold onlywhen c = ± (3 u−1)(u−1)s

8gu(u+1) ± u+18gsu and e = ± (u−1)s

8gu ∓ u+18gsu . Substituting these

values back into the coordinates of the vertices A, B, A′ and B′ we inferthat the following rectangle ABB′A′ of the shape s is the

(u, 4 u

1+u

)-base of

the parabola Γ. The second such base is its reflection in the y-axis.

Let α = u − 1, β = u + 1, γ = u − 3 and δ = 3u − 1. The vertices of ABB′A′

have the coordinates(− x1

8 g ,− y132 g u

),

(x2

8 g u , y232 g u

),

(x38 g , y3

32 g u

)and

(x4

8 g u ,− y432 g u

),

where

x1 =αγs

β+

β

s, y1 =

γ(αs)2

β− 3β2

s2− 2αγ, x2 =

αδs

β+

β

s,

y2 =δ(αs)2

β+

3β2

s2+ 2αδ, x3 = αs +

β

s, y3 =

δ(αs)2

β− β2

s2+ 2αβ,

y4 =γ(αs)2

β+

β2

s2+ 2αβ and x4 = αs − β

s.

�The side AA′ of the above rectangle will be parallel to the y-axis (the line ofsymmetry of the standard parabola Γ) if and only if u = 1. Hence, we havethe following corollary.

Corollary 2. For every parabola Γ and every positive real number s there areexactly two rectangles R = ABB′A′ and R′ = BAA′B′ such that s(R) = s(R′)= s, the line of symmetry of Γ is the perpendicular bisector of the side ABand R and R′ are the (1, 2)-bases of Γ.

Note that, for a standard parabola, the above rectangle R has the pairs(− 1

4 g s , 38 g s

),

(1

4 g s , 38 g s

),

(1

4 g s ,− 38 g s

)and

(− 1

4 g s ,− 38 g s

)as the coordi-

nates of vertices. Hence, for s = 1, we get the following result (see Fig. 3).

Z. Cerin J. Geom.

D CA

B'

S

A'

B

P

Figure 3 |AD|2 + |BC|2 = 2 |AB|2

Corollary 3. The square ABB′A′ is the (1, 2)-base of the parabola which hasits center as the focus and the line A′B′ as the directrix.

5. Fermat bases of a hyperbola

In this section we study the problem of identifying the Fermat bases of ahyperbola. This task is much harder than the case of a parabola because uand v do not satisfy any relation (except that the invariant D is negative).

It suffices to consider the standard hyperbola Φ with the equation p2

h2 − q2

k2 = 1,for positive real numbers h and k.

Assume that a trapezium ABB′A′ is a (u, v)-base for Φ and that its verticeshave the coordinates (a, b), (c, d),

(g, f + (b−d)(e−g)

c−a

)and (e, f), for real num-

bers a, b, c, d, e, f and g with c �= a. Let

E = bc − ad, F = c − a, G = b − d, H = g − e, K = f − d,

L = c − e, M = a − g, N = b − f, O = g − e,

U = cf − de, V = fF + eG and W = gE − aV.

The points C and D are(

xz , y

z

)and

(sz , t

F z

), where z = Gp + Fq − V ,

x = (E − fF )p + e(Fq − E), y = fGp + (E − eG)q − fE,

s = (E − fF + GH)p + g(Fq − E)

and

t = (Gp − E)(fF − GH) + Fq(E − gG).

In this notation, AD2 + u BC2 − v AB2 is the quotient (F 2+G2)ΞF 2 z2 , where

a11 = F 2(K2u − G2v) + (FK + GH)2,a12 = F 2(KLu − FGv) − FM(FN + GO),


a22 = F 2(L2u − F 2v + M2),a13 = F 2(GV v − KUu) + W (FN + GO),a23 = F 2(FV v − LUu) + FMW and a33 = F 2(U2u − V 2v) + W 2.

In order to solve the system of the six equations a11 = nh2 , a12 = 0, a22 = − n

k2 ,a13 = 0, a23 = 0, a33 = −n in ten variables a, b, c, d, e, f , g, u, v and n, we solveany pair of equations in u and v, then substitute these values in the remainingfour equations and solve them in n and in a triple of the main variables a,. . . ,g. This procedure is quite straightforward but is difficult to write down in alldetails. In this way, we obtain the following six families of bases of Φ.

(1) Let a, b, c and d be real numbers. Define

E = h2(k2 + c2) − k2b2, F = h2c2(h2 − d2) + k2(h2 − bd)2,G = h2c2 + k2(h2 − 2bd + d2), H = h2c2 + k2(h2 − bd) andK = (h2c2 − k2b2)ad + h2k2(ab + cd − bc).

Assume that cd, E, K �= 0. Let

A

(h2cG − aF

cdE,k2

[h2c(b − d)H − abF

]h2c2dE

), B

(abd + h2(c − a)

cd, a

),

A′ (b, c) , B′(

d,k2(bd − h2)

h2c

), u =

k2F

h2c2Eand v =

h2k2(a − c)2FK2

.

(2) Let a, b and c be real numbers. Define G = h2(k2 + c2), E = G − k2b2

and F = G − k2ab. Assume that c, E, F �= 0. Let A′ (b, c), A(a, (ab−h2)k2

c h2

),

B(b + aE

G , c), B′

(0,−k2

c

), u = k2G

c2E and v = a2k2GF 2 .

(3) Let a, b and c be real numbers. Define E = h2− c2, F = a − c andG = h2 − ac. Assume that bc, E,G �= 0. Let A

(h2

c , b + k2FGbcE

), B′

(h2

c , k2Fbc

),

B (a, b), A′ (c, 0), u = h2

c2

(k2F 2

b2E − 1)

and v = − h2

G2

(b2Ek2 + F 2

).

(4) Let a, b and c be real numbers. Define

E = h2 − c2, F = a − b, G = k2E − c2b2 and H = k2E − c2ba.

Assume that c, E, F,G,H �= 0. Let

B

(h2

c, a

), B′(c, 0), A′

(h2

c, b

), A

(c[(k2 + ab)E − h2b2]

G,k2EF

G

),

u = −k2c2Eh2G and v = k2c2F 2E

H2 .


E = h2 − bc, F = h2 + c2 − bc and G = h2 + c2 − 2bc.

Z. Cerin J. Geom.

Assume that c, c − a, E, F − ac, G �= 0. Let

A

(a,∓ (h2 − ab)kc

hE

), B

(aG ∓ (b − c)F

E,k

(F 2 − h2c2 ∓ acG

)hcE

),

B′ (c,∓kch

), A′ (b,±kE

hc

), u = c2

G and v = c2(a−c)2

(F−ac)2 .

Note that in (5) there are two families depending on the choice of signs andthat the family in (1) depends on four parameters while in (2), (3), (4) and(5) only on three. In each of these six cases, if u, v > 0, then the trapeziumABB′A′ is the (u, v)-base of Φ.

When a = c, we get the following two additional families that depend on threeand on two parameters, respectively.

(6) Let a, b and c be real numbers. Let E and F be h2k2 + h2c2 − k2b2 andh2 − b2. Assume c, E, F �= 0. Let A

(h2c−aF

bc ,−k2aFh2c2

), A′ (b, c), B′

(b,−k2F

h2c

),

B(

h2c−aFbc , a

), u = k2F

h2c2 and v = h2k2(a−c)2Fa2b2E .

(7) Let a and b �= 0 be real numbers. Let A(a,−k2

b

), B (a, b), A′ (0, b),

B′(0,−k2

b

), u = k2

b2 and v = a2k2

h2(b2+k2) .

We can summarize the above arguments in the following theorem.

Theorem 7. For every hyperbola Γ there are eight families of trapeziums suchthat each trapezium T from any of these families is a Fermat base of Γ. Oneof these families depends on two, another on four and the remaining six onthree parameters.

Just as in the case of parabolas, we can now explore when will the abovetrapeziums be parallelograms, rhombi, rectangles or squares. Of course, thesespecial trapeziums lead to the reduction of a number of parameters on whichthey depend.

For example, if for a trapezium ABB′A′ in the family (1) we require thatthe sides AA′ and BB′ are parallel this condition allows to express the firstvariable a in terms of the other three variables b, c, d and the semi-axes h andk. This leads to the following three-parameters family of parallelograms whichare (u, v)-bases for Φ when u, v > 0.

Let a, b and c be real numbers such that b, E, G �= 0, where F = h2 − ac, E =h2b2c + k2aF , hx = h2 − x2 and G = h2b2 + k2ha. Let A

(x1E , k2x2

h2bE

), A′ (a, b),

B(

x3E ,− bx4

E

), B′

(c,−k2F

h2b

), u = k2 Z

h2b2G and v = 4h2k2 ZE2 , where

x1 = h2b2(h2 + hc) + k2F (F + h2), x2 = h2b2[h2(2a − c) − c2a] + k2aF 2,

x3 = h2b2(F + h2) + k2(F + ha), x4 = E + 2h2k2(c − a)

and Z = k2F 2 + h2b2hc.


Here is an example of a statement for rhombi when the base is unique.

Theorem 8. For every hyperbola Γ and every point A′ in the plane not on Γand on its axes there is a unique triple of points (A,B,B′) and a pair (u, v)of real numbers such that when u and v are both positive then the rhombusABB′A′ is a (u, v)-base of Γ.

Proof. We shall work with the standard hyperbola Φ again. Let the pointA′ has the coordinates a and b. Our assumptions imply a, b, E �= 0, whereE = (hb)2 − (ka)2 + (hk)2. Let G = (h2b)2 + (k2a)2 and F = h2 + k2. Repeat-ing the above procedure we infer that A

(x1

aFG , x2bFG

), B′

(h4

aF ,− k4

bF

), B

(y1G , y2

G

),

u = (hk)2H(abF )2E and v = (hk)2H

G2 are the required points and real values, where

x1 = h2[2(abF )2 + (F + k2)(k2a)2 − (h3b)2],x2 = 2(kabF )2 + (k4a)2 − (kh2b)2(F + h2),y1 = a[(F + k2)(hb)2 − (k2a)2 + 2(hk2)2],y2 = b[(F + h2)(ka)2 − (h2b)2 − 2(h2k)2]

and H, on which the signs of u and v depend, is (abF )2 + (k3a)2 − (h3b)2. �Another case when only few quadrangles appear as Fermat bases is coveredby the following result for special rectangles.

Recall that the line joining the foci of a central conic is its principal axis whileits other line of symmetry is regarded as the secondary axis.

Theorem 9. For every hyperbola with semi-axes h and k and every real numbers there are exactly four rectangles that have the shape s and are symmetricwith respect to its principal axis and which are its λ-bases, where λ is the pair(1, 2 + 2 k2

h2 s2

).

Proof. As in the previous proof, it suffices to prove the claim for the standardhyperbola Φ.

Let α =√

h2 s2 + k2 and β = h2 s2 + 2 k2. We can assume that A (a, −b),B (a, b), A′ (c,−b) and B′ (c, b) for a positive real number b and real numbersa and c. By repeating the above procedure and solving the system a11 = n

h2 ,a12 = 0, a22 = − n

k2 , a13 = 0, a23 = 0, a33 = −n in the variables a, b, c, u, v and

n, we conclude that the rectangle ABB′A′, where A(

βs α , k2

α

), B

(β

s α ,−k2

α

),

A′(

h2 sα , k2

α

)and B′

(h2 sα ,−k2

α

)is one solution for the pair λ. The second is

the reflection of this rectangle in the p-axis while the third and the fourth arethe reflections of the first and the second rectangle in the q-axis. �By an analogous argument one can prove that there is no rectangle symmetricwith respect to the secondary axis of a hyperbola that is its (u, v)-basis forany positive real numbers u and v.

When h = k, we have the following corollary (see Fig. 4).

Z. Cerin J. Geom.

D C

S

A' B'

A B

P

Figure 4 |AD|2 + |BC|2 = 4 |AB|2

Corollary 4. The square ABB′A′ is the (1, 4)-basis of the equilateral hyperbolawhich has its center S as the focus and the center in the reflection of S in theline A′B′.

6. Fermat bases of an ellipse

In this final section we study the problem of identifying the Fermat bases ofan ellipse. Of course, this case is analogous to the case of a hyperbola studiedin the previous section. Circles are covered also since they are special cases ofellipses (with eccentricity zero).

It suffices again to consider only the standard ellipse Ψ with the equationp2

h2 + q2

k2 = 1, for positive real numbers h and k.

In a similar way, we obtain the following five families of bases for Ψ.

(1) Let a, b, c and d be real numbers. Define

E = h2(c2 − k2) + k2b2, F = h2c2(d2 − h2) + k2(h2 − bd)2,G = h2c2 − k2(h2 − 2bd + d2), H = h2(c2 − k2) + k2bd

and K = (h2c2 + k2b2)ad + h2k2(bc − ab − cd). Assume that cd, E and K

are different from 0. Let A

(h2cG+aF

cdE ,k2[abF−h2c(b−d)H]

h2c2dE

), B

(abd+h2(c−a)

cd , a),

A′ (b, c), B′(d, k2(h2−bd)

h2c

), u = k2F

h2c2E and v = h2k2(a−c)2FK2 .


(2) Let a, b and c be real numbers. Define G = h2(c2 − k2), E = G +k2b2

and F = G + k2ab. Assume that c, E, F �= 0. Let A′ (b, c), A(a, (h2−ab)k2

c h2

),

B(b + aE

G , c), B′

(0, k2

c

), u = −k2G

c2E and v = −a2k2GF 2 .

(3) Let a, b and c be real numbers. Define E = h2− c2, F = c − a andG = h2 − ac. Assume that bc, E,G �= 0. Let A

(h2

c , b + k2FGbcE

), B′

(h2

c , k2Fbc

),

B (a, b), A′ (c, 0), u = −h2

c2

(k2F 2

b2E + 1)

and v = h2

G2

(b2Ek2 − F 2

).


E = h2 − c2, F = a − b, G = k2E + c2b2 and H = k2E + c2 a b.

Assume that c, E, F,G,H �= 0. Let A(

c[(k2−ab)E+h2b2]G , k2EF

G

), B

(h2

c , a),

A′(

h2

c , b), B′ (c, 0), u = w

h2G and v = w F 2

H2 , where w = −k2c2E.

(5) Let a, b and c be real numbers. Let E = c2 − k2, F = Ea + k2b andG = h2E + k2b2. Assume that a, b, c, E,G �= 0. Let B

(a, F

bc

), A

(h2Eak2b2 , F

bc

),

A′ (b, c), B′(−h2E

k2b , c), u = h2E

k2b2 and v = h2k2(a−b)2Ea2c2G .

Note that the family in (1) depends on four parameters while in (2), (3), (4)and (5) only on three. In each of these five cases, if u, v > 0, then the trapeziumABB′A′ is the (u, v)-base of Ψ.

When a = c, we get the following additional family that depends on threeparameters.


E = h2c2 − h2k2 + k2b2 and F = h2 − b2.

Assume that c, E, F �= 0. Let A′ (b, c), B′(b, k2F

h2c

), A

(h2c−aF

bc , −k2aFh2c2

),

B(

h2c−aFbc , a

), u = − k2F

c2h2 and v = −h2k2(a−c)2Fa2b2E .

We can summarize the above arguments in the following theorem.

Theorem 10. For every ellipse Γ there are one four-parameters family and fivethree-parameters families of trapeziums such that every trapezium in these sixfamilies is a Fermat base of Γ.

Just as in the case of parabolas and hyperbolas, we can now explore whenwill the above trapeziums be parallelograms, rhombi, rectangles or squares. Ofcourse, these special trapeziums lead to the reduction of a number of parame-ters on which they depend.

For example, if for a trapezium ABB′A′ in the family (1) above we requirethat the sides AA′ and BB′ are parallel this condition allows to express thefirst variable a in terms of the other three variables b, c, d and the semi-axesh and k. This leads to the following three-parameters family of parallelogramswhich are (u, v)-bases for Ψ when u, v > 0.

Z. Cerin J. Geom.

Let a, b and c be real numbers such that b, E, G �= 0, where E = h2b2c − k2aF ,F = h2 − ac, hx = h2 − x2 and G = h2b2 − k2ha. Let A

(x1E , k2x2

h2bE

), A′ (a, b),

B(

x3E ,− bx4

E

), B′

(c, k2F

h2b

), u = k2 Z

h2b2G and v = 4h2k2 ZE2 , where

x1 = h2b2(h2 + hc) − k2F (F + h2), x2 = h2b2[h2(c − 2a) + c2a] + k2aF 2,

x3 = h2b2(F + h2) − k2(F + ha), x4 = E − 2h2k2(c − a)

and Z = k2F 2 − h2b2hc.

Here is an example of a statement for rhombi when the base is unique.

Theorem 11. For every ellipse Γ and every point A′ in the plane not on Γ andon its axes there is a unique triple of points (A,B,B′) and a pair (u, v) of realnumbers such that when u and v are both positive then the rhombus ABB′A′

is a Fermat (u, v)-base of Γ.

Proof. We shall work with the standard ellipse Ψ again. Let the point A′ hasthe coordinates a and b. Our assumptions imply a, b, E �= 0, where E = (hb)2+(ka)2 − (hk)2. Let G = (h2b)2 + (k2a)2 and F = h2 − k2. Repeating the aboveprocedure we infer that A

(x1

aFG , x2bFG

), B

(y1G , y2

G

), B′

(h4

aF ,− k4

bF

), u = (hk)2H

(abF )2E

and v = (hk)2HG2 are the required points and real values, where

x1 = h2[2(abF )2 + (F − k2)(k2a)2 − (h3b)2],x2 = k2[(k3a)2 + (h2b)2(F + h2) − 2(abF )2],y1 = a

[2(hk2)2 + (F − k2)(hb)2 − (k2a)2

],

H = (h3b)2 + (k3a)2 − (abF )2 and y2 = b[2(h2k)2 − (F + h2)(ka)2 − (h2b)2].Note that the sign of v depends on the sign of H and the sign of u dependson the signs of E and H. �Of course, we can view the rhombi in the previous theorem as the two-parameters family of rhombi. Its parameters are the coordinates a and b ofthe vertex A′. It is interesting that when k > h, then there are the followingtwo additional one-parameter families of rhombi M = ABB′A′ and pairs (u, v)of positive real numbers such that M is a Fermat (u, v)-base of the standardellipse Ψ.

Let Y =√

k2 − h2 and Z = h Y . For a real number a �= 0, let U = 1 − Z2,V = 1 + Z2, B

(−a,±UY

), A′ (a,±V

Y

), A

((h Z)2

a ,±UY

), B′

(− (h Z)2

a , ±VY

),

u =(

h Za

)2and v = (2 h Z)2

V [a2+(h Z)2] .

Another case when only few quadrangles appear as Fermat bases is coveredby the following result for special rectangles.

Theorem 12. For every ellipse with semi-axes h and k with h �= k and every realnumber s > h

k there are exactly four rectangles ABB′A′ with the shape s which

are its Fermat λ-bases for λ =(1, 2 − 2 h2

s2 k2

)and the line AB is parallel with

the principal line of symmetry of the ellipse. Moreover, for every real number


DC

F'

F

A'B'

B A

P

Figure 5 |AD|2 + |BC|2 = 32 |AB|2

s > kh there are exactly four rectangles ABB′A′ with the shape s which are

its Fermat ν-bases for ν =(1, 2 − 2 k2

s2 h2

)and the line AB is parallel with the

secondary line of symmetry of the ellipse.

Proof. As in the previous proofs, it suffices to prove the claim for the standardellipse Ψ.

We again assume that A (−a, b), B (a, b), A′ (−a,−c) and B′ (a,−c) for a posi-tive real number a and real numbers b and c. By repeating the above procedureand solving the system a11 = n

h2 , a12 = 0, a22 = nk2 , a13 = 0, a23 = 0, a33 = −n

in the variables a, b, c, u, v and n, we conclude that the rectangle ABB′A′,where A

(−h2

T , h2−T 2

s T

), B

(h2

T , h2−T 2

s T

), A′

(−h2

T ,−k2 sT

), B′

(h2

T ,−k2 sT

)and

T =√

k2 s2 − h2 is one solution for λ =(1, 2 T 2

k2 s2

). The second is its reflection

in the p-axis while the third and the fourth are the reflections of the first andthe second rectangle in the q-axis.

We shall get another set of four Fermat ν-bases repeating the above argumentfor the points A (a,−b), B (a, b), A′ (c,−b) and B′ (c, b) with a positive realnumber b and real numbers a and c. One solution is u = 1, v = 2 − 2 k2

s2 h2 ,

A(

T 2−k2

s T , k2

T

), B

(T 2−k2

s T ,−k2

T

), A′

(h2 sT , k2

T

)and B′

(h2 sT ,−k2

T

)with T =√

h2 s2 − k2. The second solution is its reflection in the q-axis while the thirdand the fourth are the reflections of the first and the second rectangle in thep-axis. �Figure 5 shows the square ABB′A′ with the center in a focus of the ellipsex2 + y2

4 = 1 that is its Fermat(1, 3

2

)-base.

In the case of a circle γ, when h = k = r (the radius of the circle), the aboveargument implies that for every real number s > 1 and every line π through

Z. Cerin J. Geom.

the center of γ, there are exactly four rectangles that are Fermat μ-bases of γand that have two sides parallel with π for μ =

(1, 2 − 2

s2

).

The same conclusion can be formulated also as follows.

Corollary 5. For every circle γ, every line π in the plane and every positivereal number v smaller than

√2, there are exactly four rectangles ABB′A′ that

are the Fermat (1, v2)-bases of γ such that the lines AA′ and π are parallel.

The Fermat problem is hidden in the above corollary. Indeed, if we want thatthe second number 2 − 2

s2 is 1, then the shape s has to be√

2 and the sec-ond of the above rectangles has vertices A(−r, 0), B(r, 0), A′(−r,−r

√2) and

B′(r,−r√

2) so that the (semi)circle has the side AB as a diameter.

References

[1] Catalan, E.C.: Theoremes et problemes de Geometrie Elementaire (6e ed.).Dunod, Paris (1879)

[2] Cerin, Z.: On the Fermat geometric problem. Forum Geom. 13, 135–147 (2013)

[3] Cerin, Z.: On the modified Fermat problem. Missouri J. Math. Sci. (2013, toappear)

[4] Cerin, Z.: Rectangular hexahedrons as Fermat bases of quadrics. Sarajevo J.Math. 9(22), 1–9 (2013)

[5] Cerin, Z.: Generalization of a geometry problem posed by Fermat. J. Geom.Graph. 17(1), 1–5 (2013)

[6] Euler, L.: Variae demonstrations geometricae. Novi commentarii academiae sci-entiarum Petropolitanae 1, 49–66 (1750)

[7] Gabriel-Marie, F.: Exercices de Geometrie (6e ed.). J. de Gigord, Paris (1920)

[8] Hacken, M.L.: Sur un theoreme de Fermat. Mathesis 27, 181, 264 (1907)

[9] Hanjs, Z., Volenec, V.: O jednom (davno rijesenom) Fermatovom problemu.Matematicko-fizicki list LXI, 225 (2010–2011)

[10] Korn, G.A., Korn, T.M.: Mathematical Handbook. McGraw-Hill, New York(1961)

[11] Lionnet, E.: Solutions des questions proposees. Nouvelles Annales deMathematiques Series 2 9, 189–191 (1870)

[12] Sandifer, E.: A forgotten Fermat Problem. How Euler Did It. MAA, Washington(2007)

Zvonko CerinKopernikova 710010 ZagrebCroatiae-mail: [email protected]

Received: December 8, 2011.

Revised: June 12, 2013.

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