32nd VTRMC, 2010, Solutions
- It is easily checked that 101 is a prime number (divide 101 by the
primes whose square is less than 101, i.e. the primes
1≤r≤100, we may choose a positive integer
q such that
rq≡1 mod 101. Since
(I + A + ... + A100)(I - A) = I - A101, we see that
A101 = I, in
particular A is invertible with inverse A100. Suppose
1 < n≤100 and set
r = 101 - n. Then
An + ... + A100 is invertible if and only if
I + ... + Ar-1 is
invertible. We can think of
I + ... + Ar-1 as
(I - Ar)/(I - A), which should have inverse
(I - A)/(I - Ar). However
A = (Ar)q and so
(I - A)/(I - Ar) = I + Ar + ... + (Ar)q-1. It
is easily checked that
(I + ... + Ar-1)(I + Ar + ... + (Ar)q-1) = I.
It follows that
An + ... + A100 is invertible for all
n≤100. We conclude that
An + ... + A100 has determinant
±1 for all positive integers
n < 100.
- First we will calculate
fn(75)mod 16. Note that if a, b are
odd positive integers and
a≡b mod 16, then
aa≡bbmod 16. Also
33≡11 mod 16 and
1111≡3 mod 16. We now prove by induction on n that
f2n-1(75)≡11 mod 16 for all
n∈N. This is clear for n = 1 so
f2n-1(75)≡11 mod 16 and set
k = f2n-1(75)
m = f2n(75). Then
||≡3 mod 16
||≡11 mod 16
and the induction step is complete. We now
fn(a)≡fn+2(a)mod 17 for all
a, n∈N with a prime to 17 and n even.
In fact we have
fn+1(a)≡a3mod 17, fn+2(a)≡(a3)11≡a mod 17.
f100(75)≡f2(75)mod 17. Therefore
f100(75)≡711≡14 mod 17.
- First note the
1 + x + ... + x6 = 0,
so by taking the real part, we obtain
∑n=0n=6cos 2nπ/7 = 0.
cos 2π/7 = cos 12π/7,
cos 4π/7 = cos 10π/7 = - cos 3π/7 and
cos 6π/7 = cos 8π/7 = - cosπ/7,
we see that
1 - 2 cosπ/7 + 2 cos 2π/7 - 2 cos 3π/7 = 0.
Observe that if
1 - 2 cosθ +2 cos 2θ -2 cos 3θ = 0, then
cos 2θ = 2 cos2θ - 1 and
cos 3θ = 4 cos3θ -3 cosθ, we find that
8x3 -4x2 - 4x + 1 = 0. Thus in particular
this equation. Next note that
1 - 2 cos 3π/7 + 2 cos 6π/7 - 2 cos 9π/7 = 1 - 2 cos 3π/7 - 2 cosπ/7 + 2 cos 2π/7, so
cos 3π/7 is also a root of
8x3 -4x2 - 4x + 1. Finally
since the sum of the roots of this equation is 1/2, we find that
-cos 2θ is also a root. Thus the roots of
8x3 -4x2 - 4x + 1 are
- The equation
4A + 3C = 540o tells us that A = 3B. Let D on
BC such that
∠ADC = 3B, and then let E on BD such that
∠AED = 2B.
Since triangles ABD and AED are similar, we
BD/AD = AD/ED = AB/AE.
Also BE = AE because
B = ∠BAE, and
BE = BD - BE. We
BD2 = AD2 + AB·AD. Since
BD = BC - CD, we conclude that
(BC - CD)2 = AD2 + AB·AD.
Next triangles ABC and ADC are similar, consequently
BC/AC = AB/AD = AC/CD.
AD = AB·AC/BC and
CD = AC2/BC. We deduce that
(a - b2/a)2 = c2b2/a2 + c2b/a.
(a2 - b2)2 = bc2(a + b) and the result follows.
- Let X denote the center of A, let Y denote the center of
B, let Z be where A and B touch (so X, Y, Z are collinear),
θ = ∠PXY. Note that YQ makes an angle
2θ downwards with respect to the horizontal, because
∠QYZ = 3θ.
Choose (x, y)-coordinates such that X is the origin and
XP is on the line y = 0. Let (x, y) denote the coordinates of
Q. Then we have
||= 2 cosθ + cos 2θ
||= 2 sinθ - sin 2θ.
By symmetry the area above the x-axis equals the area below the
x-axis (we don't really need this observation, but it may make
things easier to follow).
θ goes from
2π to 0 as circle B goes
round circle A.
Therefore the area enclosed by the locus of Q is
||= 2∫π0(2 sinθ - sin 2θ)(- 2 sinθ -2 sin 2θ) dθ
||= 2∫0π(4 sin2θ +2 sinθsin 2θ -2 sin22θ) dθ
||= ∫0π(4 - 4 cos 2θ +2 cosθ -2 cos 3θ -2 + 2 cos 4θ) dθ = 2π.
- Note that if
0 < x, y < 1, then
0 < 1 - y/2 < 1 and
0 < x(1 - y/2) < 1, and it follows that (an) is a positive monotone
decreasing sequence consisting of numbers strictly less that 1. This
sequence must have a limit z where
0≤z≤1. In particular
an+2 - an+1 = anan+1/2 has limit 0, so
limn-> ∞anan+1 = 0. It follows that z = 0.
bn = 1/an. Then
bn+2 = bn+1/(1 - an/2) = bn+1(1 + an/2 + O(an2)). Therefore
bn+2 - bn+1 = bn+1(an/2 + O(an2)). Also
an/an+1 = (1 - an-1/2)-1 = 1 + O(an)
and we deduce that
bn+2 - bn+1 = bn(1 + O(an))(an/2 + O(an2)). Therefore
bn+2 - bn+1 = 1/2 + O(an). Thus
ε > 0, there exists
N∈N such that
| bn+1 - bn -1/2| < ε for all n > N. We deduce that if
k is a positive integer, then
| bn+k/k - bn/k - 1/2| < ε.
Thus for k sufficiently large,
| bn+k/(n + k) - 1/2| < 2ε.
We conclude that
limn-> ∞bn/n = 1/2
limn-> ∞nan = 2.
- It will be sufficient to prove that
∑n=1∞n2/(1/a12 + ... +1/an2) is convergent. Note we may assume that (an) is
monotonic decreasing, because rearranging the terms in series
∑an does not affect its convergence, whereas the terms of the above
series become largest when (an) is monotonic decreasing. Next
observe that if
∑n=1∞an = S, then
all positive integers n.
(2n)2/(1/a12 + ... +1/a2n2). This is
≤(2n)2S/(1/a1 +2/a2 + ... +2n/a2n)≤4n2S/(n2/an) = 4San. The result follows.