Solution of a geometric problem of L. Lyusternik
I. M. Kamenetskiı̆
Abstract
This paper is a translation of Kamenetskiı̆, I. M.: 1947, Rešenie geometriceskoi zadači
L. Lyusternika, Uspekhi matematicheskikh nauk 2(2), 199–202.
In “Uspekhi Matematicheskikh Nauk”, Vol. 1, Nos. 3–4 (13–14), pp. 104–105, L. Lyusternik
posed two problems. We shall give here a solution to the first of them. We state this
problem: “Do there exist, apart from an inscribed circle, figures ϕ around which it is possible
to rotate a given triangle such that its sides remain supports for ϕ?” The point is to prove
the impossibility of such figures for triangles in which at least one angle is incommensurable
with π. (For triangles in which all the angles are commensurable with π, such figures exist.)
The solution to this problem is given by the following theorem:
Theorem. If at least one angle of a triangle ABC is incommensurable with π and the figure
ϕ possesses the property that the triangle ABC can be rotated such that its sides remain
supports for ϕ, then ϕ is a circle inscribed in ABC.
I shall refer to a bounded Jordan domain as a figure. For such a figure there exists a support
function h(ϕ) which is continuous everywhere, the proof of which I omit. If the coordinate
origin lies inside a figure Q, the function h(ϕ) is positive everywhere. The function h(ϕ) has
period 2π.
Suppose that a figure Q has the property that it is possible to rotate around it a fixed triangle
ABC, one of whose angles, for example the angle at the vertex A, is incommensurable with
π.
We place the coordinate origin at a point O lying inside the figure Q. The sides of our triangle
serve as support lines for Q. We drop perpendiculars OM , ON and OP to them from the
1
point O. We denote the angles between these perpendiculars as shown in the figure. Let us
denote the argument of the ray OM in our coordinate system by ϕ; then
arg(ON ) = ϕ + α
arg(OP ) = ϕ + α + β = ϕ + σ
where σ = α + β. We note that since α and β complement the angles A and B to π, we have
0 < α < π,
σ <α+π
(1)
Moreover, we can write
OM
= h(ϕ)
ON
= h(ϕ + α)
OP
= h(ϕ + σ)
Considering the areas of the triangles AOB, AOC and BOC, and denoting the lengths of the
sides of the triangle ABC as a, b and c, we can write
h(ϕ)b + h(ϕ + α)c + h(ϕ + σ)a = 2S,
where S is the area of the triangle ABC. The angles α, β and γ complement the corresponding
angles of the triangle ABC to π, and therefore
a = R sin(α)
b = R sin(β)
c = R sin(γ)
where R is the radius of the circle circumscribed about the triangle ABC. Substituting these
values into the previous relation, we obtain
h(ϕ) sin(β) + h(ϕ + α) sin(γ) + h(ϕ + σ) sin(α) = K
or, bearing in mind that γ = 2π − σ,
h(ϕ) sin(β) − h(ϕ + α) sin(σ) + h(ϕ + σ) sin(α) = K
(2)
where K = 2S/R = const. Equation (2) is a functional equation for the problem. We must
show that if α is incommensurable with π and if 0 < α < π < σ < α + π, which corresponds
to the conditions of the problem, then the everywhere continuous periodic function h(ϕ) with
period 2π satisfying this equation must have the form
h(ϕ) = d + a1 cos(ϕ) + b1 sin(ϕ)
(3)
where d, a1 , b1 are constants (the support function of the circle). The figure whose support
function has the form (2) is a circle.
2
We return to the proof. Let us write the Fourier series for the function h(ϕ):
h(ϕ) ∼
a0 X
+
an cos(nϕ) + bn sin(nϕ)
2
(4)
To prove the validity of the relation (2) it is sufficient to show that an = 0 and bn = 0 for
n > 1. The Fourier series for the function h(ϕ + α) can be written in the form
h(ϕ + α) ∼
where
a0 X ′
+
an cos(nϕ) + b′n sin(nϕ)
2
a′n = an cos(nα) + bn sin(nα),
b′n = bn cos(nα) − bn sin(nα).
Accordingly,
h(ϕ + σ) ∼
(5)
a0 X ′′
+
an cos(nϕ) + b′′n sin(nϕ)
2
a′′n = an cos(nσ) + bn sin(nσ),
b′′n = bn cos(nσ) − bn sin(nσ).
(5a)
Comparing (2), (4), (5) and (5a), we obtain
an sin(β) − a′n sin(σ) + a′′n sin(α) = 0,
bn sin(β) − b′n sin(σ) + b′′n sin(α) = 0,
(n = 1, 2, 3, · · ·)
(6)
Substituting into (6) the expressions for a′n , a′′n , b′n , b′′n from (4) and (5), we obtain
an Pn + bn Qn = 0,
bn Pn − an Qn = 0,
where
(n = 1, 2, 3, · · ·)
Pn = sin(β) − cos(nα) sin(σ) + cos(nσ) sin(α),
Qn = sin(nσ) sin(α) − sin(nα) sin(σ).
(7)
(7a)
From (7) we obtain
bn (Pn2 + Q2n ) = 0,
an (Pn2 + Q2n ) = 0,
(n = 1, 2, 3, · · ·)
(8)
It follows from this that if Pn2 + Q2n 6= 0, then an = bn = 0. We shall show that for n = 2, 3, . . .
the equalities Pn = 0 and Qn = 0 cannot hold simultaneously if, as is the case here, α is
incommensurable with π and 0 < α < π < σ < α + π [by virtue of (1)].
Bearing in mind that β = σ − α, we can write these relations in the form
sin(σ) cos(α) − cos(σ) sin(α) = sin(σ) cos(nα) − cos(nσ) sin(α)
sin(nσ) sin(α) − sin(nα) sin(σ) = 0.
(9)
For n = 1 these relations reduce to identities. Assuming n > 1, we shall consider the relations
(9) as a system of equations for α and σ, and we shall show that solutions of this system can
only be of the following three types: 1) the obvious solutions for which α = kπ or σ = lπ
(k, l = 0, ±1, ±2, . . .); 2) solutions different from the preceding ones but which represent pairs
of numbers commensurable with π; 3) solutions for which σ = α + kπ (k = 0, ±1, ±2, . . .).
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In fact, rejecting the solutions of type 1, we can assume that sin α 6= 0, sin σ 6= 0, in which
case the system (9) can be rewritten in the form
sin(nα)
sin(nσ)
=
;
sin(α)
sin(σ)
cos(α) − cos(nα)
cos(σ) − cos(nσ)
=
;
sin(α)
sin(σ)
(10)
Squaring the first equation in (10) and subtracting unity from each side of the resulting
equation, we obtain the equation
sin2 (nα) − sin2 (α)
sin2 (nσ) − sin2 (σ)
=
sin2 (α)
sin2 (σ)
or
cos2 (α) − cos2 (nα)
cos2 (σ) − cos2 (nσ)
=
.
sin2 (α)
sin2 (σ)
Thus, we obtain the system of equations
cos2 (α) − cos2 (nα)
cos2 (σ) − cos2 (nσ)
=
,
sin2 (α)
sin2 (σ)
cos(σ) − cos(nσ)
cos(α) − cos(nα)
=
sin(α)
sin(σ)





(11)




satisfied by all the solutions of the system (10). The system (11) splits into the two systems
of equations
cos(α) − cos(nα)
cos(σ) − cos(nσ)
=
=0
(11a)
sin(α)
sin(σ)
and
cos(α) + cos(nα)
cos(σ) + cos(nσ)
=
,
sin(α)
sin(σ)
cos(σ) − cos(nσ)
cos(α) − cos(nα)
=
.
sin(α)
sin(σ)




(11b)



The values of α satisfying the system (11a) are commensurable with π. From (11b) we obtain
cot(α) = cot(σ);
σ = α + kπ,
(k = 0, ±1, · · ·)
Thus, for n > 1 the system (10) has no solutions for which α is incommensurable with π and
0 < α < π < σ < α + π.
Consequently, for n > 1 we have an = 0 and bn = 0, and the function h(ϕ) has the form (3),
which completes the proof.
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