11718 *Given positive real numbers a _{1},a_{2},..,a_{n} with n≥2, minimize Σx_{i} subject to the condition that x_{1},x_{2},..,
x_{n} are positive and that Πx_{i} = Σa_{i}x_{i}.*

__Solution:__

Since, by Lagrage's rule, there must be some scalar λ such that Π(i≠k)x_{i} - a_{k} = λ for all k, and hence
Πx_{i} - a_{k}x_{k} = λx_{k} for all k, we find, using Πx_{i} = Σa_{i}x_{i}:

(-λ)x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = 0,

a_{1}x_{1} + (-λ)x_{2} + ... + a_{n}x_{n} = 0,

... ,

a_{1}x_{1} + a_{2}x_{2} + ... + (-λ)x_{n} = 0.

(As a consequence, λΣx_{i} = (n-1)Σa_{i}x_{i} = (n-1)Πx_{i} and λ must be positive.)

So we can find x_{1}, ... , x_{n} by first finding a positive λ that makes the determinant of the matrix of coefficients 0,
then solving the system of linear equations, and finally using λΣx_{i} = (n-1)Πx_{i}.

Here, the determinant is 0 iff

(*) λ^{n} = λ^{n-2}Σa_{i}a_{j} + 2λ^{n-3}Σa_{i}a_{j}a_{k} +
... + tλ^{n-t-1}Σa_{i1}a_{i2}..a_{i(t+1)} + ... +(n-1)a_{1}a_{2}..a_{n}

Of course we can find λ up to machine uncertainty with the procedure

begin λ:=0; repeat λ := λ + 0.00...01 until (*) end.

For n=2 we find: λ = √(a_{1}a_{2}) and x_{1} = a_{2} + √(a_{1}a_{2}), x_{2} = a_{1} + √(a_{1}a_{2}).

For n=3 we find, using Cardano's formula:

λ = ∛(a_{1}a_{2}a_{3} + √(a_{1}^{2}a_{2}^{2}a_{3}^{2} -
(a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3})^{3}/27)) +
∛(a_{1}a_{2}a_{3} - √(a_{1}^{2}a_{2}^{2}a_{3}^{2} -
(a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3})^{3}/27)).

In general we find with a bit of sweeping:

x_{1} = x,

x_{m} = x(λ+a_{1})/(λ+a_{m}) for m > 1.

Then, using λΣx_{i} = (n-1)Πx_{i}, we find:

x_{m} = ((λ/(n-1)).(Σ(k from 1 to n) Π(j not k) (λ+a_{j})))^{1/(n-1)}/(λ+a_{m}).

Σx_{m} = ((n-1)/λ)Πx_{m} = (λ/(n-1))^{1/(n-1)}(Σ(k from 1 to n) Π(j not k) (λ+a_{j}))^{n/(n-1)}/Π(λ+a_{i}).

Now, using P(λ) := Π(λ+a_{i}) and its derivative P'(λ) = Σ(k from 1 to n) Π(j not k) (λ+a_{j}), and (*), we find after a bit of calculation

Σx_{m} = ((n-1)/λ) P(λ)^{1/(n-1)}.

Notice that if all a_{m} are equal to a, then the solution is given by λ = (n-1)a and x_{m} = (na)^{1/(n-1)}.

11719 *Let f be a twice-differentiable function from [0,∞) into (0,∞) such that f "(x)/(f(x)(1 + f '(x) ^{2})^{2}) tends to ∞ if x tends to ∞.
*

Show that the product of the integrals ∫_{0}^{x} √(1 + f '(t)^{2})/f(t) dt and ∫_{x}^{∞} √(1 + f '(t)^{2})f(t) dt tends to 0 if x tends to ∞.

__Partial solution:__

We restrict ourselves to the special case f(x) = e^{p(x)}, where p(x) is a polynomial.

In this case, the condition is satified iff p(x) has degree n ≥ 2 and its coefficient of x^{n} is negative, which we henceforth assume.

Then √(1 + f '(t)^{2}) has a maximum on [0,∞), so we need only prove that the product of the integrals ∫_{0}^{x} 1/f(t) dt and
∫_{x}^{∞} f(t) dt tends to 0 if x tends to ∞.

To do this, we apply the theorems of l'Hôpital several times, and find eventually that the assertion is true if 1/(-p"(x)+p'(x)^{2}) tends to 0 if x tends to ∞.

Since this holds under our assumptions, we are finished in our special case.

11720 *Let E _{n}(t) be the Eulerian polynomial defined by E_{n}(t) := (1-t)^{n+1} Σ_{k=0}^{∞} (k+1)^{n}t^{k},
and let B_{n} be the n-th Bernouilli number.
*

Show that B_{n}(E_{n+1}(t) - (1-t)^{n}) is a polynomial with integer coefficients.

__Solution:__

It is well known that for odd n > 1 holds B_{n} = 0. So in these cases the assertion is true.

Starting with Σ_{k=0}^{∞} t^{k} = 1/(1-t), by repeatedly multiplying with t and differentiating the resulting power series, we find:

E_{n+1} = (1+nt)E_{n} + (t-t^{2})E_{n}',
so if E_{n} = 1 + a_{1}t + a_{2}t^{2} + ... + a_{2}t^{n-3} + a_{1}t^{n-2} + t^{n-1} then
E_{n+1} = 1 + (2a_{1}+n)t + (3a_{2}+(n-1)a_{1})t^{2} + (4a_{3}+(n-2)a_{2})t^{3} + ...
+ (4a_{3}+(n-2)a_{2})t^{n-3} + (3a_{2}+(n-1)a_{1})t^{n-2} + (2a_{1}+n)t^{n-1} + t^{n},
and:

E_{n} = 1 for n=1, 1+t for n=2, 1+4t+t^{2} for n=3, 1+11t+11t^{2}+t^{3} for n=4, 1+26t+66t^{2}+26t^{3}+t^{4} for n=5,
1+57t+302t^{2}+302t^{3}+57t^{4}+t^{5} for n=6,
1+120t+1191t^{2}+2416t^{3}+1191t^{4}+120t^{5}+t^{6} for n=7, ...

Furthermore we use the defining formula B_{n+1} = Σ_{j=0}^{n+1} (n+1 over j) B_{j}.

We find (n+1)B_{n} = - Σ_{j=0}^{n-1} (n+1 over j) B_{j}, and hence:
B_{1}= -1/2, B_{2} = 1/6, B_{3} = 0, B_{4} = -1/30, B_{5} = 0, B_{6} = 1/42, B_{7} = 0, ...

I read in the Wolfram site the denominator of B_{n} with even n is the product of the primes p for which p-1 divides n.

Now we readily find B_{n}(E_{n+1}(t) - (1-t)^{n}) = -t for n=1, t for n=2, 0 for n=3, -t-2t^{2}-t^{3} for n=4, 0 for n=5,
3t+28t^{2}+58t^{3}+28t^{4}+3t^{5} for n=6, 0 for n=7, ...

We see the coefficients of t^{k} and t^{n-k} in E_{n+1} are equal, for k=0 they are 1 and for k=1 they are Σ_{k=0}^{n} k*2^{n-k} = 2^{n+1}-n-2.

So the coefficients of t^{k} and t^{n-k} in E_{n+1} - (1-t)^{n} are equal too, for k=0 they are 0 and for k=1 they are 2^{n+1}-2.

It turns out that the coefficients of t^{k} and t^{n-k} in E_{n+1} all have the form
(-1)^{k}(n over k) + Σ_{j=2}^{k+1} (j^{n+1}-j) p_{j}(n), where p_{j}(n) is a polynomial in n with integer coefficients.

Hence, because of Fermat's little theorem, these coefficients are divisible by the primes p for which p-1 divides n.

11722 *Let ABC be an acute triangle in the plane, and let M be a point inside ABC. Let O1, O2, O3 be the circumcenters of BCM, CAM and ABM, repectively.
Let c be the circumcircle of ABC. Let D, E, F be the points opposite A, B and C, respectively, at which AM, BM and CM meet c.
Prove that DO1, EO2 and FO3 are concurrent at a point P that lies on c.*

__Solution:__

We give coordinates such that c has equation x^{2} + y^{2} = 1 and A(cos(ρ),sin(ρ)), B(cos(σ),sin(σ)), C(cos(τ),sin(τ)) and M(r,0).

After a bit of calculation we find D(d1/d, d2/d) and O1(o11/o1, o12/o1) where

d1 = 2r - (1+r^{2})cos(ρ), d2 = (r^{2}-1)sin(ρ), d = r^{2} - 2r cos(ρ) + 1, and

o11 = (1-r^{2})(sin(σ) - sin(τ)), o12 = (1-r^{2})(cos(τ) - cos(σ)), o1 = 2r(sin(τ) - sin(σ)) - 2sin(τ - σ).

Now we easily find the coordinates of E, O2, F, O3, too.

We may also calculate the second intersection points P, Q, R of c and, respectively, DO1, EO2 and FO3. (We use d1^{2} + d2^{2} = d^{2}, and likewise for E and F).

By the symmetry between ρ, σ and τ in the coordinates of P, we see already that Q and R must coincide with P.

For instance, after cancellations using goniometrical formulas, the coefficient of r^{6} in the numerator of the first coordinate turns out to be cos(ρ+σ+τ).

11723 *Let A,B,C be three points in the plane, and let D,E,F be points lying on BC,CA,AB, repectively.
Show there exists a conic tangent to BC,CA,AB at D,E,F, respectively, if and only if AD,BE,CF are concurrent.*

__Solution:__

The assertion is clearly true if A,B,C are collinear, so henceforward we suppose they aren't.

Let a:=BC, b:=AC and c:=AB.

Let j be the conic of lines (conic of tangents) to which a,b,c belong and with contact points D on a and E on b.

Let X be the contact point of j on c.

According to the theorem of Pascal (or Brianchon), applied to ((abc)(cab)), the lines a.a-b.c = DA, b.b-c.a = EB and c.c-a.b = XC are concurrent at a point P.

This point P must be DA.EB, so X is the intersection point of PC and c.

Now if DA,EB and FC are concurrent, then PC.c = F and the conic J with corresponding conic of tangents j satisfies the conditions.

On the other hand, if there is a conic J that meets the conditions, its corresponding conic of tangents must be j, and according to Pascal (Brianchon), the lines DA,EB,FC are concurrent.

11724 *Let f(n) = Σ _{k=1}^{n} k^{k} and let g(n) = Σ_{k=1}^{n} f(k).
*

Find lim_{n → ∞} g(n+2)/g(n+1) - g(n+1)/g(n).

__Partial solution:__

I ran the following pascal program:

program p11724;

var k:integer; f,g0,g1,r0,r1:real;

begin

g0:=1; g1:=6;

k:=2; f:=5;

r1:=6;

while true do

begin

r0:=r1; k:=k+1;

f:=f+exp(k*ln(k));

g0:=g1;

g1:=g1+f;

r1:=g1/g0;

writeln(r1-r0:13:5);

readln

end

end.

It seems the outcome tends to the well known transcendental number e = 2.71828...

11727 *Let R be a circle with center M. Let R _{2} and R_{3} be circles with centers M_{2} and M_{3} inside R, such that R_{2} and R_{3}
are externally tangent and both are internally tangent to R.
*

Give a straightedge and compass construction of the circle R_{1} that is internally tangent to R and externally tangent to R_{2} and R_{3}.

__Solution:__

See the following figure. Here a and b and γ are given, and we have to construct c.

This can be done as follows:

1) Using the cosine rule we find cos(α) = (2b+2c-2-bc)/(bc), cos(β) = (2a+2c-2-ac)/(ac), cos(γ) = (2a+2b-2-ab)/(ab).

2) Since α + β + γ = 2*π, we have 2 cos(α) cos(β) cos(γ) = cos^{2}(α) + cos^{2}(β) + cos^{2}(γ) -1.

3) Combining 1) and 2), we find the following quadratic equation with unknown c:

c^{2}(-a^{2}b^{2}+4a^{2}b+4ab^{2}-4a^{2}-4b^{2}-10ab+8a+8b-5)
+ c(4a^{2}b^{2}-10a^{2}b-10ab^{2}+8a^{2}+8b^{2}+20ab-14a-14b+8)
+ (-4a^{2}b^{2}+8a^{2}b+8ab^{2}-5a^{2}-5b^{2}-14ab+8a+8b-4) = 0.

4) Hence we can solve the equation in 3) and construct c.

The discriminant is
16(a^{4}b^{3}+a^{3}b^{4}-7a^{3}b^{3}-3a^{4}b^{2}-3a^{2}b^{4}+3a^{4}b+3ab^{4}+15a^{3}b^{2}
+15a^{2}b^{3}-a^{4}-b^{4}-13a^{3}b-13ab^{3}-27a^{2}b^{2}+4a^{3}+4b^{3}+21a^{2}b+21ab^{2}-6a^{2}
-6b^{2}-15ab+4a+4b-1), which is not the square of a polynomial p(a,b).

So for both possible solutions a construction procedure involves the construction of the square root of the discriminant.

11729 *An integer is called b-normal if all digits 0,1,..,b-1 appear the same number of times in the base-b expansion of n. Let N(b) be the set of all b-normal integers. Determine those b for which
Σ _{n ∈ N(b)} 1/n < ∞*.

__Solution:__

Any b-normal integer in which the digits appear t times lies between Σ_{k=0}^{b-1} (b-k-1)b^{tk}(1+b+...+b^{t-1}) + b^{tb-1} - b^{tb-t-1}
and Σ_{k=0}^{b-1} kb^{tk}(1+b+...+b^{t-1}).

Hence we see after a bit of calculation that any b-normal integer in which the digits appear t times is greater than b^{tb-1} and smaller than b^{tb}.

There are (tb)!/(t!)^{b} such integers, given t and b.

So Σ_{n ∈ N(b)} 1/n is greater than Σ_{t=1}^{∞} ((tb)!/(t!)^{b})b^{-tb} and smaller than
Σ_{t=1}^{∞} ((tb)!/(t!)^{b})b^{1-tb}.

According to Stirling, the general term of each of these last two series is comparable to t^{(1-b)/2}.

Therefore the answer is: for b > 3.

11731 *The integer simplex with dimension d and side-length m is the graph T(d,m) whose vertices are the non-negative (d+1)-tuples summing to m, with two vertices adjacent when they differ by 1 in two
places and are equal in all other places.
Determine the connectivity, the chromatic number, and the edge-chromatic number of T(d,m) (the latter when m > d).*

__First approach (after studying T(2,3)):__

Any such (d+1)-tuple in which k coordinates are equal to 0 and d+1-k coordinates positive is connected with k(d+1-k) + 2(d+1-k over 2) = d(d+1-k) other vertices.

I guess the connectivity is the minimum d of d(d+1-k), which is attained for k=d, and the edge-chromatic number is the maximum d(d+1) of d(d+1-k), attained for k=0.

11732 *Let a and b be real numbers with 1 < a < b and let m and n be real numbers and m not 0.
Find all continuous real functions f with domain [0,∞) such that for all x ≥ 0 holds f(a ^{x}) + f(b^{x}) = mx+n.*

__Solution:__

Let b = a^{s}, so s = ln(b)/ln(a) > 1.

Let's first restrict ourselves to the case that f(a^{x}) is a linear function, so we're looking for some real numbers c and d such that
f(a^{x}) = c + dx and f(b^{x}) = c + dsx for all x≥0.

In this case, f(a^{x}) + f(b^{x}) = mx + n for all nonnegative x if and only if 2c = n and d(s+1) = m, so c = n/2 and d = m ln(a)/ln(ab).

Next we examine the general case: we're seeking all functions g(x) such that f(a^{x}) = n/2 + m ln(a)/ln(ab) x + g(x) satisfies the requirements.

Such a g:[0,∞) → ℜ has to be continuous and, with our s>1, it has to satisfy g(sx) = - g(x) for all x in [0,∞).

But then g(x) = 0 for all x in [0,∞), because if there would exist a c for which g(c) is not 0, then (-1)^{k}g(c) = g(c/s^{k}) would tend to g(0) if k → ∞, which is
clearly a contradiction.

We conclude that any continuous function f with domain [0,∞) will do if and only if for all t = a^{x} ≥ 1 we have f(t) = n/2 + m ln(t)/ln(ab).

11734 *Find all lists (a,k,m,n) of positive integers such that a ^{m+n} + a^{n} - a^{m} - 1 = 15^{k}.*

__Partial solution:__

The solutions with k=1 are (4,1,1,1) and (2,1,2,2).

With Pascal I didn't find other solutions (for k:=1 to 8 do for a:=1 to 20 do for n:=1 to k+2 do for m:=1 to k+3-n do)

Suppose (a,k,m,n) is a solution with k>1.

Then the right hand side has exactly k prime factors 3 and exactly k prime factors 5 and no other prime factors, so this holds for the left hand side as well.

Now the left hand side is equal to (a - 1)(1 + a + a^{2} + ... + a^{n-1})(a^{m} + 1), so for some nonnegative integers i,j we have a = 1 + 3^{i}5^{j}.

Then the left hand side is equal to 3^{i}5^{j}(n + (n over 2) 3^{i}5^{j} + (n over 3) 3^{2i}5^{2j} + ... + 3^{(n-1)i}5^{(n-1)j})
(2 + m 3^{i}5^{j} + (m over 2) 3^{2i}5^{2j} + ... + 3^{mi}5^{mj}).

If i and j are both positive, then (2 + m 3^{i}5^{j} + (m over 2) 3^{2i}5^{2j} + ... + 3^{mi}5^{mj})
has a prime factor different from 3 and different from 5, so this isn't possible.

If i is positive and j=0 then we find
3^{i}(n + (n over 2) 3^{i} + (n over 3) 3^{2i} + ... + 3^{(n-1)i})
(2 + m 3^{i} + (m over 2) 3^{2i} + ... + 3^{mi}) = 3^{k}5^{k}, so (2 + m 3^{i} + (m over 2) 3^{2i} + ... + 3^{mi}) is a power of 5.

If i=0 and j is positive then we find
5^{j}(n + (n over 2) 5^{j} + (n over 3) 5^{2j} + ... + 5^{(n-1)j})
(2 + m 5^{j} + (m over 2) 5^{2j} + ... + 5^{mj}) = 3^{k}5^{k}, so (2 + m 5^{j} + (m over 2) 5^{2j} + ... + 5^{mj}) is a power of 3.

If i and j are both 0, then a=2 and (2^{n} - 1)(2^{m} + 1) = 15^{k}.

Because (a^{n}-1)(a^{m}+1) = (-1)^{k} modulo 16 and a = 1 + 3^{i}5^{j}, we see by considering the powers of 3 (mod 16) and those of 5 (mod 16)
that in all remaining cases where i and j aren't both positive, k must be odd.

(If a=2 and n=1 we have 2^{m} + 1 = 15^{k}, which is impossible since 2^{m} + 1 can't be divisible both by 3 and by 5.)

11735 *Let P be a point inside triangle ABC. Let d _{A} and d_{B} and d_{C} be the distances from P to A and B and C, respectively.
*

Let r_{A} and r_{B} and r_{C} be the radii of the circumcircles of PBC and PCA and PAB, respectively.

Prove that 1/d_{A} + 1/d_{B} + 1/d_{C} ≥ 1/r_{A} + 1/r_{B} + 1/r_{C} .

__Partial solution:__

Give coordinates P(0,0), A(d_{A} ,0), B(d_{B} cos(s), d_{B} sin(s)), C(d_{C} cos(t), -d_{C} sin(t)), with 0 < min(s,t) ≤ max(s,t) < π < s+t.

Then we find r_{A} = √(d_{B}^{2} + d_{C}^{2} - 2d_{B} d_{C} cos(s+t)) / |2 sin(s+t)|,
r_{B} = √(d_{A}^{2} + d_{C}^{2} - 2d_{A} d_{C} cos(t)) / (2 sin(t)),
r_{C} = √(d_{A}^{2} + d_{B}^{2} - 2d_{A} d_{B} cos(s)) / (2 sin(s)).

We could check the inequality by comparing its left side and its right side for, say, 180^{2}100^{3} elements
(s,t,d_{A},d_{B},d_{C}) ∈ {0 < min(s,t) ≤ max(s,t) < π < s+t, 0 < d_{A},d_{B},d_{C} ≤ 1}
with the help of a computer program.

There is equality if d_{A} = d_{B} = d_{C} and s = t = 2π/3.

11737 *Given an acute triangle ABC, let O be its circumcenter, let M be the intersection of the lines AO and BC, and let D be the other intersection of AO with the circumcircle of ABC.
Let E be that point on AD such that M is the midpoint of ED. Let F be the point at which the perpendicular to AD at M meets AC.
Prove that EF is perpendicular to AB.*

__Solution:__

Give coordinates O(0,0), A(1,0), B(cos(σ),sin(σ)), C(cos(τ),sin(τ)), with 0 < σ < π and π < τ <σ + π.

Then we find M(sin(σ-τ)/(sin(σ)-sin(τ)),0), D(-1,0), E(1 + 2 sin(σ-τ)/(sin(σ)-sin(τ)),0),
F(sin(σ-τ)/(sin(σ)-sin(τ)), ((sin(σ-τ) - sin(σ) + sin(τ)) sin(τ))/((sin(σ) - sin(τ))(cos(τ) - 1))).

Now, using trigonometric identities, it's straightforward to show that the inner product of the direction vectors of EF and AB is zero.

11738 *Three point particles are constrained to move without friction along a unit circle. Three ideal massless springs of stiffness k _{1}, k_{2} and k_{3} connect the particles
pairwise. *

Show that an equilibrium in which the particles occupy three distinct positions exists if and only if 1/k_{1}, 1/k_{2} and 1/k_{3} can be the lengths of the sides of a triangle.

Show also that if this happens, the equilibrium length L_{1} of the spring with stiffness k_{1} is given by L_{1} = √(k_{2}k_{3}) √((1/k_{2} + 1/k_{3})^{2} - 1/k_{1}^{2}).

__Partial solution:__

From the last assertion it is clear what L_{2} and L_{3} should be. We will try and prove only this second assertion, because it clearly implies the first one.

To find a symmetric relation involving k_{i} and L_{i} (i=1,2,3) that is as simple as possible and doesn't contain square roots, we intuitively calculate
k_{1}L_{1}^{2} + k_{2}L_{2}^{2} + k_{3}L_{3}^{2} with the proposed values of L_{i}, and find it is equal to
k_{1}k_{2}k_{3}(1/k_{1} + 1/k_{2} + 1/k_{3})^{2}.

So, although I don't know the physics implied in this problem, I guess these physics demand that exactly this relation be satisfied:
k_{1}L_{1}^{2} + k_{2}L_{2}^{2} + k_{3}L_{3}^{2} =
k_{1}k_{2}k_{3}(1/k_{1} + 1/k_{2} + 1/k_{3})^{2}.

Then it follows that L_{i} is as proposed (i=1,2,3).
(I think the proposer of the problem forgot to multiply the three equilibrium lengths with a constant so as to get L_{1} + L_{2} + L_{3} = 2π,
or else it isn't clear how he defines a 'unit circle'.)

11739 *Let B(x) be the 2x2 matrix whose diagonal elements are 1 and whose offdiagonal elements are x. Evaluate the infinite matrix product M = Π _{p} B(p^{-2}), where the product runs
over all primes in increasing order.*

__Partial solution:__

M is the 2x2 matrix whose diagonal elements are equal to 1 + Σ(p_{1}p_{2})^{-2} + Σ(p_{1}p_{2}p_{3}p_{4})^{-2} + ...

and whose offdiagonal elements are equal to Σp_{1}^{-2} + Σ(p_{1}p_{2}p_{3})^{-2} +
Σ(p_{1}p_{2}p_{3}p_{4}p_{5})^{-2} + ... ,

where each sum is for some fixed nonnegative integer k taken over all k-tuples of distinct primes p_{1} < p_{2} < ... < p_{k}.

I think the values of these sums, involving zeta functions, can be found somewhere in the mathematical literature.

** Earlier solutions 1:** click here .

** Earlier solutions 2:** click here .

** Earlier solutions 3:** click here .

** Earlier solutions 4:** click here .

** Earlier solutions 5:** click here .

** Earlier solutions 6:** click here .

** Earlier solutions 7:** click here .