Download THE FOURTH TEST Problem 1. Show that, for all positive real

THE FOURTH TEST
Problem 1. Show that, for all positive real numbers a, b, c such that abc = 1, the inequality
1
1 + a2 + (b + 1)
holds.
Solution. We have
1
1+
a2
+ (b + 1)
LHS
≤
=
=
2
=
2
+
a2
1
1 + b2 + (c + 1)
+
b2
2
+
1
1 + c2 + (a + 1)
2
≤
1
2
1
1
≤
. Then
+ 2b + 2
2 (1 + b + ab)
1
1
1
1
+
+
=
2 1 + b + ab 1 + c + bc 1 + a + ac
1
1
ab
b
+
+
=
2 1 + b + ab ab + abc + ab2 c b + ab + abc
1
1
ab
b
1
+
+
= .
2 1 + b + ab ab + 1 + b b + ab + 1
2
Problem 2. Let us choose arbitrarily n vertices of a regular 2n-gon and color them red. The remaining
vertices are colored blue. We arrange all red-red distances into a nondecreasing sequence and do the same with
the blue-blue distances. Prove that the two sequences thus obtained are identical.
10th Vojtech Jarnik Mathematical Competition, Ostrava, 2000
Solution. We divide the segments determined by the 2n vertices into groups of segments having the same
length. We prove that, in each group, the number of red-red segments is equal to the number of the blue-blue
segments.
Let a be the number of the red-red segments from a certain group, b the number of the blue-blue segments
from that same group, and c the number of the segments from the group that join differently colored vertices.
Counting the total number of vertices joined by these segments, we have 2a + c red vertices and 2b + c blue
vertices. But each vertex was counted twice within this group, with the exception of the group consisting of
diameters in the circumcircle of the 2n-gon in which case each vertex is counted once.
As there are n red and n blue vertices, it follows that 2b + c = 2a + c, hence a = b.
In conclusion each group contains and equal number of red-red and blue-blue segments, therefore the sequence
of the red-red distances and the sequence of the blue-blue sequences coincide.
Problem 3. Let ABC be an arbitrary triangle, and let M , N , P be any three points on the sides BC, CA,
AB such that the lines AM , BN , CP concur. Let the parallel to the line AB through the point N meet the
line M P at a point E, and let the parallel to the line AB through the point M meet the line N P at a point F .
Then, the lines CP , M N and EF are concurrent.
MOP 1997
Solution 1. We exclude the trivial case M N k AB, where M ≡ E and N ≡ F. Let R be the intersection
point of the lines N E and BC. Suppose that E ∈ (N R). Then N ∈ (P F ) . If R ∈ (EN ), we have F ∈ (P N ) and
this case is similar to the previous one.
Let O be the intersection of the lines AM , BN , and CP, Q the intersection of the lines M N and EF , and S the
intersection of the lines CP and M N. We prove that Q coincides with S.
Applying Menelaus’ Theorem in triangle BM N with transversal O − S − C, and in triangle ABN with
transversal P − O − C, and using Ceva’s Theorem in triangle ABC, we obtain:
1
SN OB CM
·
·
= 1, SM ON CB
(·) SN M B CA
ON CA P B
SN
BC AN
·
·
= 1, ⇒
·
·
= 1, hence
=
·
.
OB CN P A
SM CB AN
SM
BM AC
PA MB NC
·
·
= 1, PB MC NA
Because of the similarity between triangles N EQ and M F Q, and also the one between P EN and P M F ,
according to Thales’ Theorem, we obtain:
QN
NE
PE
BR
BR BC
AN BC
SN
=
=
=
=
·
=
·
=
.
QM
MF
PM
BM
BC BM
AC BM
SM
As Q, S belong to the line segment M N, it follows that Q and S coincide, hence the lines M N, EF and CP are
concurrent.
Solution 2. We use the fact that in a trapezoid, the midpoints of the bases, the intersection point of the
diagonals, and the intersection point of the extensions of the legs are collinear.
In trapezoid M EN F, the intersection point of the extensions of the legs is P. If we prove that CP contains
one of the midpoints of the line segments N E and M F, then CP is the line passing through the four points
mentioned above, hence CP also contains the intersection point of the diagonals M N and EF of the trapezoid
M EN F.
Considering E ∈ (N R) , let T be the intersection point of lines CP and N E. We prove that T is the midpoint
of the line segment N E.
Applying Menelaus’ Theorem in triangle CT R with the transversal M − E − P, yields
M R ET P C
·
·
= 1.
M C ER P T
PT
AN
=
and from ER parallel to BP it follows that the triangles M ER
PC
AC
ER
MR
and M P B are similar, therefore
=
.
BP
MB
AN M C
MB PA NC
Multiplying the equalities above, we obtain ET =
·
· BP, and, as
·
·
= 1, it follows
AC M B
MC PB NA
NC
NT
NC
NT
that ET =
· AP =
· AP = N T, because
=
, hence T is the midpoint of the line segment N E.
AC
AP
AC
AP
As N T and AB are parallel, we get
Problem 4. A positive integer is called lonely if the sum of the inverses of its positive divisors (including 1
and itself) is not equal with the some of the inverses of the positive divisors of any other positive integer.
a) Show that any prime number is lonely.
b) Prove that there are infinitely many numbers that are not lonely.
Solution. We denote σ−1 (n) =
X1
d|n
d
and σ(n) =
X
d|n
d. As σ−1 (n) =
X1
d|n
d
=
1Xn
1X
1
=
e = σ(n),
n
d
n
n
d|n
e|n
and the function σ satisfies σ(a · b) = σ(a) · σ(b), for all positive co-prime integers a, b) (σ is multiplicative), it
follows that σ−1 (ab) = σ(a) · σ(b) (1), for all positive co-prime integers a, b (σ−1 is multiplicative).
1
p+1
. Let us suppose that there exists a positive
a) If p ≥ 2 is a prime number, then σ−1 (p) = 1 + =
p
p
p+1
integer n 6= p such that σ−1 (n) =
. We have p σ(n) = n(p + 1), hence p | n. If n 6= p, then σ−1 (n) =
p
X1
1
1
≥ 1 + + > σ−1 (p), which is a contradiction.
d
p n
d|n
b) From (1) it follows that if σ−1 (a) = σ−1 (b) then σ−1 (na) = σ−1 (nb) for all positive integer n that is
co-prime with a and b. Therefore, if a is not lonely, then the same is true for na for infinitely many choices of
n. In order to provide an example of a number that is not lonely, it is enough to remember that for a perfect
1
number n (i.e. one for which σ(n) = 2n) we have σ−1 (n) = σ(n) = 2. We can consider the first two perfect
n
2
numbers, 6 and 28. These numbers are not lonely, hence neither are 6n with n co-prime with 6 and 28.
Remarks. (Dan Schwarz)
1. The existence of even perfect numbers is related to the Mersenne primes (numbers of the form 2p − 1, with p
prime), of which it is not known whether they are infinitely many or not, but if 2p − 1 and p are both primes,
then 2p−1 (2p − 1) is a perfect number. Moreover, these are the only even perfect numbers, while odd perfect
numbers are not known. For p = 2, 3, 5, 7 we obtain 6, 28, 496, and 8128 which are the first four perfect numbers.
2. There also exist k-perfect numbers, with k ≥ 3, defined by σ(n) = kn. Obviously, they can be used similarly
to solve our problem. For example, numbers 120 and 672 are ,,triperfect”.
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