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Liceul Internaţional de Informatică Bucureşti
Societatea de Ştiinţe Matematice
Stelele Matematicii 2015, Seniori
Problema 1. Arătaţi că există un şir de numere
naturale impare m1 <m2
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Liceul Internaţional de Informatică Bucureşti
Societatea de Ştiinţe Matematice
Stars of Mathematics 2015, Senior Level — Solutions
Problem 1. Show that there are positive odd integers
m1 < m2 n, the difference m4−2n4 is a
perfect square, and it is readilychecked that m and n
are relatively prime. The conclusion follows.
Problem 2.
Let γ , γ 0, γ 1, γ 2
be coplanar circles such that γ i is
internally tangent to
γ at Ai,and γ i and
γ i+1 are externally tangent at Bi+2, i
= 0, 1, 2 (indices are reduced modulo 3). Thetangent
at Bi, common to γ i−1 and γ i+1,
meets γ at C i, located in the half-plane
opposite Aiwith respect to the line Ai−1Ai+1. Show that
the three lines AiC i are concurrent.
Flavian Georgescu
Solution 1. Let γ i cross the lines
AiC i+1 and AiC i+2 again at
X i+2 and Y i+1, respectively.Since the
homothety centred at Ai, transforming γ i
into γ ,
sends X i+2 to C i+1 and
Y i+1 to
C i+2, the lines X i+2Y i+1
and C i+1C i+2 are parallel,
so AiC i+1/AiC i+2 =
X i+2C i+1/Y i+1C i+2.On the other hand,
since C i has equal powers relative to
γ i+1
and γ i+2, C iX i+1
·C iAi+2 =
C iB2i = C iY i+2 ·
C iAi+1, it follows that C iX i+1 =
C iB2i /C iAi+2 and
C iY i+2 = C iB2i
/C iAi+1.Hence AiC i+1/AiC i+2 =
(Bi+1C i+1/Bi+2C i+2)
2(AiC i+2/AiC i+1), so
AiC i+1/AiC i+2 =Bi+1C i+1/Bi+2C i+2,
and consequently
2i=0(AiC i+1/AiC i+2) =
2i=0(Bi+1C i+1/Bi+2C i+2) =
1. Since the sines of the angles C iAiC j
are proportional to the lengths of the correspondingchords
AiC j, the conclusion follows by Ceva’s theorem in
trigonometric form in the triangleC 0C 1C 2.
A
A
A
D D
D
B B
B
C C
C
0
0
0
0 1
1
1
1
2
2
2
2
X
X
X
2
1
01
0
2
Y
Y
Y
T
T
T
P P
P
Z Z
Z
0
0
0 1
1
1
2
2
2
Fig. 1 Fig. 2
Solution 2. Let the tangents to γ at
Ai and Ai+1 meet at Di+2 (fig
1), and notice that thelatter has equal powers relative to
γ i and γ i+1 to deduce that it lies
on their radical axis,
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Bi+2C i+2. Consequently, the lines C iDi
are concurrent at the radical centre of the γ i,
and theconclusion follows by the lemma below (see Figure 2).
Lemma. Let P 0P 1P 2 be
a triangle, and let T i be the touchpoint
of the side P i+1P i+2 and the
incircle γ of the
triangle P 0P 1P 2. Let
further Z i be a point on the
arc T i+1T i+2
of γ
not containing T i. Then the
lines T iZ i are concurrent if and only
if the lines P iZ i are concurrent.
The lemma can be proved either projectively, by sending the
point of intersection of T 0Z 0and
T 1Z 1 to the incentre, or by a trigonometric
version of Ceva’s theorem.
Problem 3. Let n be a positive integer and let
a1, . . ., an be n positive
integers. Show that
nk=1
√ ak
1 + a1 + · · · + ak <n2k=1
1
k.
G. I. Natanson
Solution. Set b0 = 1 and bk = 1 +
a1 + · · · + ak, k = 1, . . . , n, to
obtain a strictly increasingstring of positive integers 1
= b0 < b1 n2} ≥ 1 — if there is no bk
> n2, let m = n + 1 —, to splitthe
above sum into
m−1k=1
bk − bk−1
bk+
nk=m
bk − bk−1
bk, (∗)
where empty sums are zero. We show that the first sum does not
exceed n2
k=2 1/k, and thesecond is always less than 1.
If k = 1, . . . , m − 1, write
(bk − bk−1)1/2/bk ≤ (bk −
bk−1)/bk = 1/bk + · · · + 1/bk
≤
bk j=bk−1+1
1/j, to deduce that the first sum in (∗) does not
exceed
m−1k=1
bk j=bk−1+1
1/j =
bm−1k=2 1/k ≤n2k=2 1/k.
Finally, if k = m, . . . , n, write
(bk − bk−1)1/2/bk < b−1/2k
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Problem 4. Let S be a finite planar
set no three points of which are collinear, and letD(S, r)
= {{x, y} : x, y ∈ S, dist(x, y)
= r}, where r is a positive real number, and
dist(x, y)is the Euclidean distance between the points
x and y. Show that
r>0
|D(S, r)|2 ≤ 3|S |2(|S | − 1)/4.
H. Lefmann, T. Thiele
Solution. Given a point x in
S and a real number r, let S (x,
r) = {y : y ∈ S, dist(x, y)
= r},and notice that the S (x, r),
r ≥ 0, partition S .
The number of non-degenerate isosceles triangles with vertices
in S and apex at x isr>0
|S (x,r)|2
, so the total number of non-degenerate isosceles triangles with
vertices in S is
N =
x∈S
r>0
|S (x,r)|2
, equilateral triangles with vertices in
S being counted three times
each. Now,
N =
x∈S r>0|S (x, r)|
2 =
r>0x∈S |S (x, r)|
2 ≥r>0
|S | 1
|S |
x∈S |S (x, r)|
2
=r>0
|S |2|D(S,r)|
|S |
2
= 2
|S |r>0
|D(S, r)|2 −r>0
|D(S, r)| = 2|S |r>0
|D(S, r)|2 −|S |
2
,
by Jensen’s inequality applied to the convex function
t → t2 = t(t − 1)/2, t ∈ R.On the
other hand, given two distinct points x and y
in S , there are at most two non-
degenerate isosceles triangles with vertices in S ,
base xy, and apex at a third point in S \{x, y}:the
apex must lie on the perpendicular bisector of the segment
xy, and since no three pointsin S are
collinear, there are at most two such. Hence N
≤ 2|S |2 .
Combining the lower and upper bounds for N and
rearranging terms yields the requiredinequality.
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