Download 1. Colour the 3 × 3 cube in this manner: colour each corner sub

1. Colour the 3 × 3 cube in this manner: colour each corner sub-cube black.
Every other sub-cube is coloured black or white so that each sub-cube is a
different colour than all the other sub-cubes that it shares a face with.
Now, notice that the corner sub-cube is black, and the centre sub-cube is
white. But as the mouse goes through from sub-cube to sub-cube, the destination sub-cube is a different colour from his original cube. Since there are 13
white cubes and 14 black cubes, the mouse’s path must go BW BW . . . BW B.
The last cube must be black. Thus, it cannot finish in the centre sub-cube.
2. Suppose that an arbitrary MP has rating R. When he enters
P a group,
say group Gi , he carries the relative rating ri = SRi , where Si = xk ∈Gi Rk .
When he moves from group Gi to group Gj , the given condition translates into
the inequality
R
R
.
<
Si
Sj + R
It is obviously equivalent to R + Sj − Si < 0.
After the move from Gi to Gj , the new sums of absolute ratings are:
Si0 = Si − R,
Sj0 = Sj + R.
The idea is to associate to the Parliament a daily invariant:
X
L=
Si2 .
i
When an MP moves from one group to another, the invariant L changes value
to a new one
X
L0 =
Sk2 + (Si − R)2 + (Sj + R)2 .
k6=i,j
It is easy to see that
L0 − L = (Si − R)2 + (Sj + R)2 − Si2 − Sj2 = 2R(R + Sj − Si ) < 0.
Therefore L0 < L. Since there are only finitely many possibilities to arrange the
MPs in groups, the invariant L can take only a finite number of values. Thus it
cannot decrease infinitely.
Here is another argument. If S is the least possible sum of ranks in a group,
no PM can move from a group with the sum S. Assume that there are counter
examples to the statement of the problem. Of all such counter examples, let us
choose one of possibly several which have the minimal number of groups, the
maximal value of S and the minimal number of groups with sum S. In this
example all groups must participate in the process an infinite number of times
(else we can disregard the groups which participate a finite number of times and
get a smaller counter example). Then a group of sum S has to participate. The
only possibility is that a PM moves to this group. But then either the minimal
sum will get larger, or the number of groups of sum S will decrease.
3. We start with a lemma.
1
Lemma. Let p be a prime number and let a1 , a2 , . . . , ak be k < p (not
necessarily different) non-zero P
residues modulo p. For every non-empty subset
M ⊂ {1, 2, . . . , k} let SM =
(mod p). Then there exist at least k
i∈M ai
different non-zero residues SMj .
Proof. Induction on k. For k = 1, the sum of one element a1 is non-zero.
For the induction step, consider k + 1 < p non-zero residues a1 , a2 , . . . , ak , ak+1
and let 0 < SM1 < SM2 < . . . SMk < p be k different sums that we can obtain
(by the induction assumption) using the first k numbers only. First consider
the case when ak+1 = 1. If for some j we have SMj + 1 < SMj+1 , then the sum
SMj + 1 = SMj ∪{k+1} gives a new residue. Otherwise SMk − SM1 = k − 1 and
either SMk < p − 1 so that SMk + 1 = SMk ∪{k+1} is a new residue, or 1 < S1
and then ak+1 itself is a new residue (here we used that k + 1 < p).
Finally, consider the case when ak+1 6= 1. Since we can divide modulo a
ai
(mod p) and use the
prime number we can consider the k + 1 residues bi = ak+1
previous case to obtain k + 1 different sums (note that bk+1 = 1). Multiplying
those sums by ak+1 we get the desired result and the lemma is proved.
Now we are ready to prove the main result. Consider first the case when
n = p is a prime number. Let x1 ≤ x2 ≤ · · · ≤ x2p−1 be the ordered residues of
the given numbers. We have to choose p of them such that their sum is equal to
zero modulo p. If for some i we have xi = xp+i−1 , then xi = xi+1 = · · · = xp+i−1
and their sum is equal to zero modulo p. Otherwise the p − 1 numbers
a1
a2
a3
...
ap−1
= xp+1 − x2 ,
= xp+2 − x3 ,
= xp+3 − x4 ,
= ...
= xp+p−1 − xp
are non-zero and we can, according to the lemma, create all p − 1 non-zero
residues by summing them. If S = a1 + . . . + ap = 0 (mod p), we have found
the desired p elements. If it is not zero, we can add SM such that S + SM = 0
(mod p) and this sum is the desired one — it is equal to
X
X
x1 +
xi+1 +
xp+i .
i6∈M
i∈M
It remains to consider the case when n = ab is not a prime number. Here we
can use induction by the number of prime factors (and what we already proved).
From some 2a − 1 of the 2n − 1 numbers (2a − 1 < 2n − 1) we can choose a such
that their sum A1 is divisible by a. We can pick up the next 2a − 1 numbers
from the remaining 2n − a − 1 numbers and choose the next a from them with
sum A2 divisible by a. Continuing this process we stop when the number of
remaining numbers 2n − am − 1 gets less than 2a − 1; the sum corresponding
to the last step is Am . In other words,
2n − am − 1 < 2a − 1 ⇔ 2n < a(m + 2) ⇔ 2b < m + 2 ⇔ 2b − 1 ≤ m.
2
So we can get 2b − 1 numbers A1 = ax1 , . . . , A2b−1 = ax2b−1 . Using the induction assumption once again we choose b of the 2b − 1 indices such that the sum
of the corresponding xi numbers is divisible by b. This means that we get ab
numbers such that their sum is divisible by ab
Here is another solution. Consider the sum
X
(xσ(1) + xσ(2) + · · · + xσ(p) )p−1 ,
where the summation is over all ordered choices of p different indices between
1 and 2p − 1. The number of such choices is 2p−1
, a number that is not
p
divisible by p. If the statement is false, then by Fermat’s little theorem the sum
will not be divisible by p. On the other hand, each term Cxei11 xei22 . . . xeikk with
e1 + e2 + . . . + ek = p − 1 and k nonzero ei is included 2p−1−k
times (the
p−k
number of ways to choose the remaining p − k indices), which is divisible by p.
4. Denote the weights of the cows by x1 , x2 , . . . , x101 . These numbers then
satisfy a linear 101 × 101 system of equations with all coefficients along the
main diagonal equal to zero, 50 coefficients in each equation equal to 1, and 50
coefficients in each equation equal to −1. All right-hand-sides are equal to zero,
i.e. the system is homogeneous. For a homogeneous system, the sum of any two
solutions is a solution and any multiple of a solution is also a solution.
Obviously x1 = x2 = . . . = x101 = c solves the system for any real c. We need
to prove that these equalities occur for all (non-trivial) solutions. (Actually, to
make any sense, all solutions to the problem need to be strictly positive.)
Assume the contrary, i.e. assume that there exists a solution (y1 , y2 , . . . , y101 )
such that not all yi are equal to each other. Since the system is homogeneous, we can add any solution to (y1 , y2 , . . . , y101 ) and still get a solution; let
us add (c, c, . . . , c) in such a way that in the new solution (denoted again by
(y1 , . . . , y101 )) we have yk = 0 for some k. By our assumption above, this solution has to be non-trivial (i.e. at least one yi 6= 0). The system has integer
coefficients. By induction (the details can be found in the Lemma below), we
conclude that there is a rational non-trivial solution (y1 , y2 , . . . , y101 ) with the
above property (some yk = 0). Multiplying if necessary all equations by a
suitable natural number, we can further assume that all yi are integers. Also,
without loss of generality, we can assume that y101 = 0 and that the largest
common divisor of all yk is 1. Consider the last equation: since all coefficients are ±1, there must be an even number m of odd numbers among yk ,
k = 1, . . . , 100. We have assumed the solution to be non-trivial, hence m ≥ 2.
But then (y1 , y2 , . . . , y101 ) couldn’t satisfy any of the equations where 0 stands
as coefficient in front of an odd yk . Contradiction!
The contradiction means that x1 = x2 = · · · = x101 must hold for all
solutions.
Lemma. If a homogeneous linear system of equations in n variables with
rational coefficients has a non-trivial solution such that xn = 0, then it also has
a rational non-trivial solution with the same property.
Proof. Note that a solution is rational iff all free variables are rational
(because we get the basic variables from them by division with rational pivot
3
elements — see any book in Linear algebra for more details). For the same
reasons, the solution of a homogeneous system is trivial iff all free variables are
zero.
Thus to have a non-trivial solution, we have to have free variables. Note
that at least one of them is different from xn . Indeed, xn cannot be the only
free variable because otherwise the only solution with xn = 0 would be trivial.
So let xi with i 6= n be a free variable. Let xi = 1 and other free variables be
zero. The obtained solution is non-trivial and because xn was the last variable
we still have xn = 0 both in the cases when xn is free and basic.
5. Let A, B, C, D be the prices in cents. We have
 A
B
C
D
+ 100
+ 100
= 711
 100 + 100
100
,
 A
B
C
D
= 711
100 · 100 · 100 · 100
100
so that
A+B+C +D
A·B·C ·D
=
=
711
79 · 26 · 32 · 56 .
One of the numbers has to be divisible by 79, assume that A = 79a, a < 9.
Not all of them are divisible by 5, so one of them has to be divisible by 25, let
B = 25b. Evidently b < 25, so one of the numbers C, D must be divisible by 5,
let C = 5c. We get
79a + 25b + 5c + D =
711
a·b·c·D
= 26 · 32 · 53 .
Consider two cases. Suppose first that D ≡ 0 (mod 5), meaning that D = 5d.
Then a ≡ −1 (mod 5) and since a < 9, we get that a = 4. Now we have
5b + c + d =
79
b·c·d
= 24 · 32 · 52 .
Since b < 25 we can choose c = 5s, which implies d ≡ 4 (mod 5), so that
d = 4 + 5t and
b+s+t
=
15
b · s · (4 + 5t) = 24 · 32 · 5.
From the first equation we have t ≤ 13, so possible values of d = 4 + 5t are
4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69,
only 4, 9√and 24 being divisors of 24 · 32 · 5. If d ≤ 9, then b · s ≥ 16 · 5 and
b+s
≥ 16 · 5 > 8, which contradicts the first equation. If d = 24 we get
2
b + s = 11, b · s = 30, which leads to the solution
A = 79 · 4 = 316,
B = 25 · 6 = 150,
C = 25 · 5 = 125,
4
D = 5 · 24 = 120.
B
a
c
K
M
dB
dA
O
A
dC
R
C
L
b
Figure 1: Carnot’s theorem.
In the second case D 6≡ 0 (mod 5). Then c should be divisible by 5, c = 5s.
Using symmetry between s and b we can also assume that b is divisible by
5, b = 5t (since a < 9 and therefore cannot be divisible by 25). We get
79a + 125t + 25s + D =
711
a·t·s·D
= 26 · 32 · 5.
If a is not divisible by 5 then s = 5r and, using symmetry, we can suppose
that t ≥ r. Since 125(t + r) < 625, we have only three choices: r = 1 and
t = 1, 2, 3. None of them leads to an integer solution.
So a should be divisible by 5 and hence a = 5. Then D = 16 (mod 25) and
hence D is one of the numbers in the sequence 16, 41, 66, 91, 116, 141, . . ., only
one of those numbers being a divisor of 26 · 32 , namely D = 16. This possibility
does not give any integer solution either.
Hence, the only integer solution to the problem is
A = 79 · 4 = 316,
B = 25 · 6 = 150,
C = 25 · 5 = 125,
D = 5 · 24 = 120.
6. We shall first prove the Carnot’s theorem.
Lemma. Let the triangle ABC be inscribed in the circle of centre O and
radius R. Define the quantity dA as ± the distance from O to the side BC, the
plus sign in case O and A lie in the same half-plane with respect to the line
through B and C; the minus sign in case O and A lie in different half-planes
with respect to this line (dA = 0 in case O ∈ BC). Analogously, define dB and
dC . Denote by r the radius of the inscribed circle in 4ABC. Then
dA + dB + dC = R + r.
5
Proof. Denote, as usual, the side lengths of the triangle by a, b, c. Let
K, L, M be the orthogonal projections of O on BC, CA, AB respectively. Consider the quadrilateral ALM O. The points M, L lie on the circle with diameter
AO. Ptolemy’s theorem gives that
AL · M O + AM · LO = M L · AO.
In the introduced notations this means that
cdC
aR
bdB
+
=
2
2
2
and, analogously,
bdB
adA
cR
+
=
,
2
2
2
adA
cdC
bR
+
=
.
2
2
2
Whatever the signs of dA , dB , dC , for the area of 4ABC we have
S=
adA
bdB
cdC
(a + b + c)r
+
+
=
.
2
2
2
2
Summing all four obtained equalities and dividing both sides by a+b+c
, we
2
deduce that dA + dB + dC = R + r.
We can now return to the original problem. Denote by n the number of
vertices of the polygon. Observe that any triangulation by non-intersecting
diagonals results in triangles inscribed in the same circle as the polygon itself;
denote the centre and radius of this circle by O and R respectively. Choose one
triangulation; denote by r1 , r2 , . . . , rn−2 the radii of the circles inscribed in the
triangles (the number n − 2 of such triangles follows easily by computing the
sum of their angles and comparing that sum with the sum of the angles of the
polygon). Sum the equalities from the above lemma for all such triangles. In
this sum the distance from O to any side of the polygon will appear only once
with plus sign, whereas the distance from O to any of the chosen diagonals will
appear twice, once with plus sign and once with minus sign, since each such
diagonal is a side in two of the chosen triangles and these two triangles lie in
different half-planes with respect to the diagonal. Hence we get
d1 + d2 + . . . + dn = (n − 2)R + r1 + r2 + . . . + rn−2 ,
where d1 , d2 , . . . , dn are the distances from O to the n sides of the polygon. It
immediately follows that
r1 + r2 + . . . + rn−2
does not depend on the triangulation.
Alternative solution. It actually suffices to prove the statement for an inscribed quadrilateral ABCD and use induction to establish the result for arbitrary n. We shall use the following results: given a triangle of side lengths
6
a, b, c, angles α, β, γ (opposite to a, b, c respectively), area S, inradius r and
circumradius R, we have
r=
2S
,
a+b+c
a
b
c
=
=
= 2R.
sin α
sin β
sin γ
We shall use index 1 for quantities corresponding to 4ABC, index 2 for
4CDA, index 3 for 4ABD and index 4 for 4BCD. What we have to prove
is then r1 + r2 = r3 + r4 . Denote α = ∠BAC = ∠BDC; β = ∠CAD =
∠CBD; γ = ∠ABD = ∠ACD; δ = ∠ADB = ∠ACB. We then have
r1 =
2S1
,
2R(sin α + sin(β + γ) + sin δ)
r2 =
S2
,
R(sin β + sin γ + sin(α + δ))
r4 =
S4
.
R(sin β + sin(δ + γ) + sin α)
and similarly
r3 =
S3
,
R(sin γ + sin(α + β) + sin δ)
For the areas of the triangles we get expressions of the type
S1 =
2R sin δ · 2R sin α · sin(β + γ)
= 2R2 sin α sin δ sin(β + γ) .
2
Hence
r1 + r2 = 2R
sin α sin δ sin(β + γ)
sin β sin γ sin(α + δ)
+
sin α + sin δ + sin(β + γ) sin β + sin γ + sin(α + δ)
.
Since sin α, sin δ, sin(β + γ) are angles in a triangle we have (check!)
sin α + sin δ + sin(β + γ) = 4 cos
α
δ
β+γ
cos cos
,
2
2
2
which gives
α
δ
β+γ
β
γ
α+δ
sin sin
+ sin sin sin
).
2
2
2
2
2
2
Similar computations for r3 + r4 and the relation β + γ = π − (α + δ) (since
the quadrilateral is inscribed in a circle) yield the result.
The statement in the problem is called sometimes the Japanese theorem.
7. Note that if |x| ≥ 1 then |2x2 − 1| ≥ 1 and |8x4 − 8x2 + 1| = |2(2x2 − 1)2 −
1| ≥ 1 so we have no solutions in this case. Otherwise we can put x = cos α and
rewrite the equation as
r1 + r2 = 4R (sin
−8 cos α cos(2α) cos(4α) = 1.
Multiplying this with sin α we get
−8 sin α cos α cos(2α) cos(4α) = sin α ⇔
−4 sin 2α cos(2α) cos(4α) = sin α ⇔ sin 8α = sin(−α).
We get 8α = −α +2πm or 8α = π + α + 2πn. Therefore x = cos 2mπ
9
or x = cos (2n+1)π
. Because sin x = 0 cannot give us a solution we get 7
7
different solutions for m = 1, 2, 3, 4 and n = 1, 2, 3.
7