11747 *Determine all natural numbers n such that ⌊ n/k ⌋ divides n for 1 ≤ k ≤ n.
Similarly, determine all natural numbers n such that ⌈ n/k ⌉ divides n for 1 ≤ k ≤ n*

__Partial solution:__

I ran a pascal program and think the following lists are complete: 1,2,3,4,6,8,12,24 and 1,2,4,6,12.

11748 *Is there a sequence of positive real numbers a _{1} , a_{2} , a_{3} , ... such that Π_{k=1}^{n} a_{k} < n^{n}
and Σ_{k=1}^{∞} a_{k} converges?*

__First consideration:__

If Π_{k=1}^{n} a_{k} = n^{n}, then a_{k} = k^{k} / (k-1)^{(k-1)}, and then 1/a_{k} is comparable to 1/(e(k-1)) if k→∞.

So in this case Σ_{k=1}^{∞} a_{k} does not converge.

If Π_{k=1}^{n} a_{k} < n^{n}, then a_{k} must be even smaller than k^{k} / (k-1)^{(k-1)} for some k,
so at first sight it seems Σ_{k=1}^{∞} a_{k} can't converge. But there's more room for many a_{k} to be larger ...

11751 *In a triangle with angles of radian measures A,B and C, prove that
(csc(A)+csc(B)+csc(C))/2 ≥ 1/(sin(B)+sin(C)) + 1/(sin(C)+sin(A)) + 1/(sin(A)+sin(B)) with equality iff the triangle is equilateral.*

__Numerical solution:__

I checked this with a pascal program for all A,B,C corresponding to an integer number of degrees: for a:=1 to 60 do for b:=a to 90 - (a div 2) do begin c:=180-a-b ...

11753 *Let f be a continuous map from [0,1] to ℜ that is differentiable on (0,1), with f(0)=0 and f(1)=1.
Show that for each positive integer n there exist distinct numbers c _{1} , ... , c_{n} in (0,1) such that Π_{k=1}^{n} f '(c_{k} ) = 1.*

__Graphical solution:__

According to the mean value theorem, there exists a number c ∈ (0,1) with f '(c) = 1.

Furthermore, the following is clear if we think of the graph of f:

There must exist disjunct intervals I and J within (0,1) such that if a ∈ I and f '(a) = tan(α), there is b ∈ J with f '(b) = tan(π/2 - α) = 1/f '(a).

(Since a and b both run through an interval, the assertion of the problem is rather weak.)

11756 *Let f be a function from [-1,1] to (-∞,∞) with continuous derivatives of all orders up to 2n+2. Given f(0) = f "(0) = ... = f ^{(2n)}(0) = 0, prove
(1/2)((2n+2)!)^{2}(4n+5) (∫_{-1}^{1} f(x) dx)^{2} ≤ ∫_{-1}^{1} (f ^{(2n+2)}(x))^{2} dx.*

__Partial solution:__

First we restrict ourselves to the case that f is a polynomial of degree at most 2n+2.

Then the coefficients of even powers of x (up to exponent 2n) are 0 and the terms with odd powers of x vanish if we integrate them from -1 to 1.

So we have (2n+2)! (∫_{-1}^{1} f(x) dx) = ∫_{-1}^{1} f ^{(2n+2)}(0) x^{2n+2} dx = (2/(2n+3)) f ^{(2n+2)}(0).

Furthermore we have f ^{(2n+2)}(x) = f ^{(2n+2)}(0), so ∫_{-1}^{1} (f ^{(2n+2)}(x))^{2} dx = 2 (f ^{(2n+2)}(0))^{2}.

So ((2n+3)^{2}/2) (2n+2)!^{2} (∫_{-1}^{1} f(x) dx)^{2} = ∫_{-1}^{1} (f ^{(2n+2)}(x))^{2} dx.
Then the inequality amounts to 4n+5 ≤ (2n+3)^{2}.

11757 *Prove that the coefficient of x ^{n}y^{n} in the power series of 1/((1-3x)(1-(y+3x-x^{2}))) is 9^{n}.*

__Partial solution:__

We use 1/(1-r) = 1 + r + r^{2} + ... (with r=3x and with r=y+3x-x^{2}).

Then we find the coefficient of x^{n}y^{n} is equal to

Σ_{0 ≤ k ≤ n} 3^{k} Σ_{(n-k)/2 ≤ j ≤ n-k} ((n+j over j)(j over n-k-j)(-1)^{n-k-j}3^{j}).

For any (small) fixed n we may verify this equals 9^{n}.

11758 *Acute triangle ABC has several ancillary points and properties, shown in the figure.
Segment AX _{A} is perpendicular to BC and segments marked with the same symbol have the same length.
The angle at C is less than the angle at B.
Lines PN and RM are parallel and perpendicular, respectively, to BC.
*

*
*

*(a) Prove that |RO|/|OM| = 2.
(b) Show that PQ is not parallel to BC.
(c) Letting D be the intersection of PQ and BC, show that if AD is perpendicular to BX*

__Solution:__

Give coordinates in such a way that B(0,0), X_{A}(1,0), Y_{A}(2,0), C(a,0) with a > 2, A(1,t) with t > 0.

Then we find X_{B}((2+a)/3,2t/3), Y_{B}((1+2a)/3,t/3) and P(1,2t/(2+a)), R((4+2a)/(4+a),4t/(4+a)), N((2+2a)/(2+a),2t/(2+a)), Q((1+2a)/(1+a),t/(1+a)),
M((4+2a)/(4+a),t(8+a)/((2+a)(4+a))) and D(3,0).

(a) Since R,O,M have the same x-coordinate, the distances are differences of y-coordinates and we quickly check |RO|/|OM| = 2.

(b) Since P and Q have distinct y-coordinates, PQ is not parallel to BC.

(c)(d) AD is perpendicular to BX_{B} iff a = t^{2} - 2.

Then we have P(1,2/t), R(2t^{2}/(2+t^{2}),4t/(2+t^{2})), M(2t^{2}/(2+t^{2}),(6+t^{2})/(2t+t^{3})), N((2t^{2}-2)/t^{2},2/t).

Now we can check that P,R,N and M are on a common circle by using the cosine rule in triangles PRN and PMN:
it turns out the cosines of the angles at R and M add up to 0, so the angles at R and M add up to 180 degrees.

11761 *For each positive integer n, determine the least integer m such that lcm{1,2,...,m} = lcm{n,n+1,...,m}.*

__Solution:__

For all i ∈ {1,2,..,n-1}, the maximal multiplicity of a prime power occurring in the representation of i as a product of prime powers must also occur
in the representation as a product of prime powers of some integer j ∈ {n,n+1,..,m}.

This is true iff m = 2p^{k}, where p^{k} is the greatest prime power smaller than n.

11763 *Characterize the twice-differentiable and bounded functions f mapping the set of positive reals into itself and satisfying
x g"(x) + (1 + x g'(x))g'(x) ≥ 0 for all (positive) x, where g = log(f). *

__Graphical characterization:__

The inequality can be reduced to h' + h/x ≥ 0, where h = f '.

So let α be the angle which the line through (0,0) and (x,h(x)) makes with the positive x-axis, and β the angle which the tangent
in (x,h(x)) to the graph of h(x) (for x > 0) makes with the positive x-axis.

Then the inequality is equivalent to tan(α) + tan(β) ≥ 0.

We could also give an economical characterization:

If x is input and h(x) the corresponding profit, marginal profit is never less than average loss.

11771 *Let n!! = Π _{i=0}^{[(n-1)/2]}(n-2i). Find lim_{n → ∞} ((2n-1)!!)^{1/n}*(tan(π(n+1)!^{1/(n+1)}/(4(n!)^{1/n})) - 1). *

__Solution:__

I ran a pascal program which yields a slowly ascending sequence of outcomes for n=1 through n=150 with last term 1.14.. .

Using the formula of Stirling, and that (2n-1)!!=(2n-1)(2n-3)..5.3.1 is between (n-1)!2^{n-1} and n!2^{n}, we find that the limit must be equal to

lim_{n → ∞} (2n/e)*(tan((π/4)(1+1/n)) - 1).

Substituting t=1/n and using the theorem of l'Hôpital, we find the limit is π/e = 1.159..

11774 *Let ω be the circumscribed circle of triangle ABC.
The A-mixtilinear incircle of ABC and ω is the circle that is internally tangent to ω, AB and AC, and similarly for B and C.
Let A', P*

*
*

*Prove that the triangles C'P _{B}B and CP_{C}B' are similar.*

__Partial solution:__

Give coordinates such that ω: x^{2} + y^{2} = 1,

A: (cos(u),sin(u)), B: (cos(w),sin(w)), C: (cos(v),sin(v)),

(see figure) A': (cos(p),sin(p)), B': (cos(t),sin(t)), C': (cos(s),sin(s)),

(see figure) O_{A}: (r cos(p), r sin(p)),

(see figure) P_{B}: (cos(u),sin(u)) + λ((cos(w),sin(w)) - (cos(u),sin(u))), P_{C}: (cos(u),sin(u)) + μ((cos(v),sin(v)) - (cos(u),sin(u))).

We are going to express p,t,s,λ and μ in u,v,w.

Thereafter we can calculate the angles of triangles C'P_{B}B and CP_{C}B', with the cosine rule or simply as angles at the circumference of a circle.

To calculate p,t,s,λ and μ, we use four conditions:

1) O_{A}A' = O_{A}P_{B}, 2) O_{A}A' = O_{A}P_{C}, 3) O_{A}P_{B} is perpendicular to AB,
4) O_{A}P_{C} is perpendicular to AC.

We find r = λ(1-λ)(1-cos(w-u))/(1-λcos(w-p)-(1-λ)cos(p-u)) = μ(1-μ)(1-cos(v-u))/(1-μcos(p-v)-(1-μ)cos(p-u)) =
(2λ-1)(1-cos(w-u))/(cos(w-p)-cos(p-u)) = (2μ-1)(1-cos(v-u))/(cos(p-v)-cos(p-u)).

Hence with λ^{2}/(2λ-1) = (1-cos(p-u))/(cos(w-p)-cos(p-u)) = A and μ^{2}/(2μ-1) = (1-cos(p-u))/(cos(p-v)-cos(p-u)) = B, we find

λ = A - √(A^{2}-A) = 2sin((p-u)/2)(sin(p-u)/2)-sin((w-p)/2))/(cos(w-p)-cos(p-u)) = (sin((p-u)/4) cos((p-u)/4)/((sin((w-u)/4)(cos(2p-u-w)/4)) and

μ = B - √(B^{2}-B) = 2sin((p-u)/2)(sin(p-u)/2)-sin((p-v)/2))/(cos(p-v)-cos(p-u)) = (sin((p-u)/4) cos((p-u)/4)/((sin((v-u)/4)(cos(2p-u-v)/4)) and
and λ/μ = sin((v-u)/2)/sin((w-u)/2) and sin((v-u)/4)/cos((w-u)/4) = sin((2p-u-v)/4)/cos((2p-u-w)/4).

Using goniometric formulas we find sin(p) = 2X/(1+X^{2}) and cos(p) = (1-X^{2})/(1+X^{2}) with X = tan(p/2) =
(sin((v-u)/4)cos((u+w)/4 + sin((u+v)/4)cos((w-u)/4))/(cos((u+v)/4)cos((w-u)/4 - sin((v-u)/4)sin((u+w)/4)).

Furthermore, we find from the coordinates of A' those of C' by substituting for u,v,w (simultaneously) v,w,u (respectively) and those of B'
by substituting for u,v,w (simultaneously) w,u,v (respectively).

So now we have the coordinates of the vertices of the triangles, and could in principle finish the job, but that would require another day of calculation.

I think this problem can be more easily solved by applying the theory of *inversion*, but I don't know enough of this theory.

In the picture it seems the lines BB', CC' and P_{B}P_{C} are concurrent, so maybe the triangles are mapped to each other by a reflexion in the point of intersection.

11782 *A signed binary representation of an integer m is a finite list a _{0} , a_{1} , ... of elements of {-1,0,1} such that Σ a_{i} 2^{i} = m.
*

A signed binary representation is sparse if no two consecutive entries in the list are nonzero.

(a) Prove that every integer has a unique sparse representation.

(b) Prove that for all integer m, every non-sparse signed binary representation of m has at least as many nonzero terms as the sparse representation.

__Solution:__

(a) First we prove for every (positive) integer the existence of a sparse representation.

This can be done with induction:

First, we make a list of sparse representations for m with 1 ≤ m ≤ 2^{k}, for instance for k=1 we have the list (1) (0,1).

Next, we see we can derive from it a list of sparse representations for m equal to 2^{k+1}+x with -2^{k-1} ≤ x ≤ 2^{k-1}
and hence a list of sparse representations for m with 1 ≤ m ≤ 2^{k+1}.

As for the uniqueness: suppose some m would have two distinct sparse representations Σ a_{i} 2^{i} = Σ b_{i} 2^{i}.

Now first cancel a_{i} and b_{i} (same i) if they are equal, next move all remaining negative terms to the other side of the equality sign, and use 2*2^{j} = 2^{j+1}.

Since the representations were sparse and distinct, we would get two distinct (unsigned) binary representations of some nonzero positive integer.

(b) Suppose there would be a non-sparse signed binary representation Σ b_{i} 2^{i} with less nonzero terms
than the sparse signed binary representation Σ a_{i} 2^{i} of the same nonzero number.

We may also assume there are not more negative terms at the left hand side of Σ a_{i} 2^{i} = Σ b_{i} 2^{i} (or else multiply by -1).

Now first cancel a_{i} and b_{i} (same i) if they are equal.

Next move all remaining negative terms to the other side of the equality sign, and use 2*2^{j} = 2^{j+1} where possible.

(At the right hand side it may then be necessary to repeat this last step, but not at the left hand side, so the number of nonzero terms stays less at the right hand side.)

Then again we would at last find two distinct (unsigned) binary representations of some nonzero positive integer.

11783 *Given a tetrahedron, let r denote the radius of its inscribed sphere. For 1 ≤ k ≤ 4, let h _{k} denote the distance from the k-th vertex to the plane of the opposite face.
Prove that Σ_{1 ≤ k ≤ 4} (h_{k}-r)/(h_{k}+r) ≥ 12/5.*

__Partial solution:__

Of course, by scaling, we may assume r=1. If the tetrahedron is regular we can give
coordinates so that A(0,0,3), B(-√2,-√6,-1), C(-√2,√6,-1), D(2√2,0,-1). Then we find h_{k}=4 for all k with 1 ≤ k ≤ 4 and the sum is 12/5.

It is plausible that the minimum of the sum occurs for regular tetrahedra, but we still have to prove it.

We may start with the four faces at distance 1 from (0,0,0) and equations

sin(θ)cos(α)X + sin(θ)sin(α)Y + cos(θ)Z = 1, sin(θ)cos(β)X + sin(θ)sin(β)Y + cos(θ)Z = 1,
sin(θ)cos(γ)X + sin(θ)sin(γ)Y + cos(θ)Z = 1, sin(σ)cos(τ)X + sin(σ)sin(τ)Y + cos(σ)Z = 1.

Then we still have to calculate the coordinates of the vertices and the distances h_{k} and the minimum of the sum.

It may be easier to prove that the sum is a *local* minimum for a regular tetrahedron.

Indeed, let's start with a regular tetrahedron whose inscribed sphere has radius r=1 and whose distances from a vertex to the opposite face are h_{1} = h_{2} = h_{3} =
h_{4} = h (= 4).

If we stretch or press some side(s) of this tetrahedron while keeping the same inscribed sphere, the sum changes with
approximately (2/(h+1)^{2})(dh_{1} + dh_{2} + dh_{3} + dh_{4}), where the last factor between brackets seems to be always positive.

11784 *Let ABC be an equilateral triangle with center O and circumradius r. Given R > r, let ρ be a circle about O with radius R. All points named P are on ρ.
(a) Prove that |PA| ^{2} + |PB|^{2} + |PC|^{2} = 3(R^{2} + r^{2}).
(b) Prove that min_{P ∈ ρ} |PA|*|PB|*|PC| = R^{3} - r^{3}, and max_{P ∈ ρ} |PA|*|PB|*|PC| = R^{3} + r^{3}.
(c) Prove that the area of a triangle with sides of length |PA|,|PB|,|PC| is (√(3)/4)*(R^{2} - r^{2}).
(d) Prove that if H,K,L are the respective projections of P onto AB, AC, and BC, then the area of triangle HKL is (3√(3)/16)(R^{2} - r^{2}).
(e) With the same notation, prove that |HK|^{2} + |KL|^{2} + |HL|^{2} = (9/4)(R^{2} + r^{2}).
*

__Solution:__

Give coordinates O(0,0), A(-r√3/2,-r/2), B(r√3/2,-r/2), C(0,r), P(R cos(φ),R sin(φ)).

(a) We find |PA|^{2} = R^{2} + r^{2} + 2rR sin(φ+π/3), |PB|^{2} = R^{2} + r^{2} + 2rR sin(φ-π/3),
|PC|^{2} = R^{2} + r^{2} -2rR sin(φ).

(b) The minimum is found for P nearest to a vertex, for example φ = π/2, the maximum for P at equal distance from two vertices, for example φ = 3π/2.

(c) The easiest way to check this is perhaps using Heron's formula for the area of a triangle with sides of length a,b,c:
(√((a^{2} + b^{2} + c^{2})^{2} - 2(a^{4} + b^{4} + c^{4})))/4.

(d)+(e): We find

H(R cos(φ),-r/2),

K((R/4)cos(φ) + (R√3/4)sin(φ) - r√3/4, (R√3/4)cos(φ) + (3R/4)sin(φ) + r/4),

L((R/4)cos(φ) - (R√3/4)sin(φ) + r√3/4, (-R√3/4)cos(φ) + (3R/4)sin(φ) + r/4).

Hence |HK|^{2} = (3/4)(R^{2} + r^{2}) + (3rR√3/4)cos(φ) +(3rR/4)sin(φ),
|HL|^{2} = (3/4)(R^{2} + r^{2}) - (3rR√3/4)cos(φ) +(3rR/4)sin(φ), |KL|^{2} = (3/4)(R^{2} + r^{2}) -(3rR/2)sin(φ).

So we find (e) at once, and for (d) we can again use Heron's formula.

11786 *Let x _{1} , x_{2} , ... be a sequence of positive numbers such that lim _{n → ∞} x_{n} = 0 and
lim _{n → ∞} log(x_{n} )/(x_{1} + ... + x_{n} ) is a negative number.
*

Prove that lim _{n → ∞} log(x_{n} )/log(n) = -1.

__First observation:__

If x_{n} = 1/n^{p} with p > 0, it's well known the conditions are satisfied iff p=1. And in this case p=1 the conclusion is valid too.

11788 *Let n be a positive integer, and suppose that 0 < y _{i} ≤ x_{i} < 1 for 1 ≤ i ≤ n.
*

Prove that (log x_{1} + ... + log x_{n})/(log y_{1} + ... + log y_{n}) ≤ √((1-x_{1})/(1-y_{1}) + ... + (1-x_{n})/(1-y_{n})).

__First try:__

If n=1 and x = x_{1} = 1-u and y = y_{1} = 1-v, then we have to prove log(1-u)/log(1-v) ≤ √(u/v) for 0 < u ≤ v < 1.

Using the power series for log, we see this amounts to proving ((u + u^{2}/2 + u^{3}/3 + ...)/(v + v^{2}/2 + v^{3}/3 + ...))^{2} ≤ u/v for 0 < u ≤ v < 1.

But this is straightforward, so then we are finished with the case n=1.

In general, it's sufficient to prove (Σ_{i=1}^{n} Σ_{k=1}^{∞} u_{i}^{k}/k)^{2}/(n u_{1}...u_{n}) ≤
(Σ_{i=1}^{n} Σ_{k=1}^{∞} v_{i}^{k}/k)^{2}/(v_{1}...v_{n})
if 0 < u_{i} ≤ v_{i} < 1 for 1 ≤ i ≤ n.

11789 *Let a and k be positive integers. Prove that for every positive integer d there exists a positive integer n such that d divides k*a ^{n} + n.*

__First exploration:__

I've checked the claim for all k,a,d that don't exceed 100, with the help of the following pascal program:

program modulo;

var n,k,a,d,p:integer; klaar:boolean;

begin

for k:=1 to 100 do for a:=1 to 100 do for d:=1 to 100 do

begin

n:=0; p:=1; klaar:=false;

repeat

if (((k*p+n) mod d) = 0) then

begin klaar:=true; writeln(k:6,a:6,d:6,n:6) end;

if not klaar then begin n:=n+1; p:=(p*a mod d) end

until (klaar or (n>10000));

if not klaar then begin writeln(k:6,a:6,d:6,'faalt'); readln end

end; readln

end.

11790 *Given a triangle with semiperimeter s, inradius r, and medians of length m _{a} , m_{b} , and m_{c} , prove that
m_{a} + m_{b} + m_{c} ≤ 2s - 3(2√3-3)r.*

__Solution (numerical):__

If we denote the side lengths with a,b,c and the angles with α, β, γ, we have by definition 2s = a+b+c and by the sine rule a/sin(α) = b/sin(β) = c/sin(γ) ( = h).

Furthermore, in the literature we find formulas 2m_{a} = √(2b^{2} + 2c^{2} - a^{2}), etc, and 2r = √((b+c-a)(a+c-b)(a+b-c)/(a+b+c)).

The inequality then amounts to

√(sin(α) + sin(β) + sin(γ))*(√(2sin^{2}(β) + 2sin^{2}(γ) - sin^{2}(α)) +
√(2sin^{2}(α) + 2sin^{2}(γ) - sin^{2}(β)) + √(2sin^{2}(α) + 2sin^{2}(β) - sin^{2}(γ))) ≤

2(sin(α) + sin(β) + sin(γ))*√(sin(α) + sin(β) + sin(γ)) - 3(2√3-3)√((sin(β) + sin(γ) - sin(α))(sin(α) + sin(γ) - sin(β))(sin(α) + sin(β) - sin(γ))).

We can check with mathematica that this inequality holds for all positive α, β, γ with α + β + γ = π.

There is equality iff α = β = γ = π/3.

11792 *Show that every infinite-dimensional Banach space V contains a closed subspace of infinite dimension and infinite codimension.*

__First guess:__

Let e_{1}, e_{2}, e_{3}, ... be a basis of V, consisting of vectors whose norm ||e_{i}|| is equal to 1.

Then I think {x ∈ V / ||x - e_{1}|| ≤ ||x - e_{2}||} could be such a subspace, but I don't know much about Banach spaces.

11794 *Find every twice differentiable function f on the real numbers such that for all nonzero x and y holds x f(f(y)/x) = y f(f(x)/y).*

__Partial solution:__

If we try f(t) = a + bt + ct^{2}, dividing by x-y and using a polynomial is zero iff the coefficients are all zero, we find c=0 and a = b^{2}.

Likewise, if we try f(t) = Σ_{k=0}^{n} a_{k}t^{k} for some natural number n, looking at the coefficient for the term with the highest exponent of x, we find a_{n}=0 unless n=1.

So the only polynomial solutions are f(t) = b^{2} + bt.