11248 *
Let n be a positive integer, and let f be a continuous real-valued function on [0,1] with the property that
INT(x from 0 to 1: x ^{k}f(x) dx) = 1 for k=0,1,...,n-1.
Prove that INT(x from 0 to 1: f(x)^{2} dx) is at least n^{2}.
*

__Solution:__

Using that for any reals c_{0}, c_{1}, ... , c_{n-1},
INT(x from 0 to 1: (f(x)-SUM(k from 0 to n-1: c_{k}x^{k}))^{2} dx) is at least 0,
we derive:

INT(x from 0 to 1: f(x)^{2} dx) is at least

c_{0}(2 - c_{0}/1 - c_{1}/2 - c_{2}/3 - c_{3}/4 - ... - c_{n-1}/n) +

c_{1}(2 - c_{0}/2 - c_{1}/3 - c_{2}/4 - c_{3}/5 - ... - c_{n-1}/(n+1)) +

c_{2}(2 - c_{0}/3 - c_{1}/4 - c_{2}/5 - c_{3}/6 - ... - c_{n-1}/(n+2)) +

... +

c_{n-1}(2-c_{0}/n-c_{1}/(n+1)-c_{2}/(n+2)-c_{3}/(n+3)-...-c_{n-1}/(2n-1)).

To maximize this sum, we set the partial derivatives equal to 0. This amounts to setting the n expressions between brackets
equal to 1.

We calculate c_{0}, c_{1}, ... , c_{n-1}, using the inverse of the well-known matrix
with coefficients 1/(i+j-1). Then the sum turns out to be equal to n^{2}.

11249 *
A node-labeled rooted tree is a tree such that any parent with label p has p+1 children, labeled 1,2,..,p+1, and such that the
root vertex (generation 0) has label 1. Find the population of generation n.
*

__Solution:__

Let g^{(k)}(m) be the number of labels m of generation k.

Initialize
g^{(k)}(0):=0 for all k, g^{(0)}(1):=1. Furthermore use the following recurrence relation:

For k=1,2,3,... successively, for m from 1 to k+1, calculate g^{(k)}(m):=SUM(p from m-1 to k: g^{(k-1)}(p)).

Clearly the population g^{(n)}_{total} of generation n is g^{(n+1)}(1)=g^{(n+1)}(2).

I did the calculations for k from 1 to 10, using a triangular matrix schedule (g^{(k)}(m)). I found

g^{(n)}_{total}=1,2,5,14,42,132,429,1430,4862,16800 for n=1,2,..,10.

Furthermore, looking at the matrix elements, I discovered an easier calculation scheme:

Initialize g^{(k)}(0):=0 and g^{(k)}(k+1):=1 for k=0,1,2,3,...

For k=1,2,3,... successively, for m:=k downto 1, calculate
g^{(k)}(m):=g^{(k)}(m+1)+g^{(k-1)}(m-1).

This reminds us of the Pascal triangle. But why should we express the results
g^{(n)}_{total} in binomial coefficients, if our triangle is as easy as Pascal's?

11250 *Show that if n is a positive integer and x _{1}, x_{2}, ... x_{n} are nonnegative
real numbers that sum to 1, then *

(SUM j: j from 1 to n: sqrt(x_{j}))*(SUM k: k from 1 to n: 1/(1+sqrt(1+2x_{k})) is at most n^{2}/(sqrt(n)+sqrt(n+2)).

__Partial solution:__

Since the right hand side of the inequality is increasing as a function of n, and the case n=1 is trivial,
we may assume without loss of generality that all x_{i} are positive and smaller than 1. Therefore the maximum
of the left hand side is attained in a point of Lagrange. With our restriction x_{1}+x_{2}+...+x_{n}=1,
this means that all partial derivatives must be equal. We want to deduce that, then, all x_{i} must be equal.
Then the maximum is attained for x_{i}=1/n (all i); thus we get the right hand side as maximum for the left hand side.

Suppose that among the x_{i} there exist x and y with x smaller than y.
We calculate the partial derivatives, and use they must be equal. Then we get:

(SUM k: k from 1 to n: 1/(1+sqrt(1+2x_{k}))*(1/(2*sqrt(x))-1/(2*sqrt(y))) =
(SUM j: j from 1 to n: sqrt(x_{j}))*(1/((sqrt(1+2x)*(1+sqrt(1+2x))^{2})-1/((sqrt(1+2y)*(1+sqrt(1+2y))^{2})).
Somehow, this must lead us to a contradiction.

11251 *Let a,b and c be positive real numbers, two of which are less than or equal to 1, satisfying ab+ac+bc=3.
Show that1/(a+b) ^{2}+1/(a+c)^{2}+1/(b+c)^{2}-3/4 is at least 3(a-1)(b-1)(c-1)/(2(a+b)(a+c)(b+c)).
*

__Practical solution:__

This amounts to checking whether 4(a^{2}+3)^{2}(b^{2}+3)^{2}+
4(a+b)^{4}(b^{2}+3)^{2}+4(a+b)^{4}(a^{2}+3)^{2} is at least
3(a+b)^{2}(a^{2}+3)(b^{2}+3)(15+5a^{2}+5b^{2}+8ab-8a-8b-a^{2}b^{2})
for a and b between 0 and 1 (inclusive). I did this with Maple.

11252 *Let n be an integer greater than 2, and let a _{1}, a_{2}, ... , a_{n} be positive
numbers. *

Let S=SUM(i from 1 to n: a_{i} ). Let b_{i}=S-a_{i} and let S' =SUM(i from 1 to n: b_{i} ). Show that

PRODUCT(i from 1 to n: a_{i} )/PRODUCT(i from 1 to n: S-a_{i} ) is less than or equal to PRODUCT(i from 1 to n: b_{i} )/PRODUCT(i from 1 to n: S'-b_{i} ).

__Provisional solution:__

Since S'= (n-1)S and S'-b_{i}=(n-2)S+a_{i}, the inequality is equivalent to

PRODUCT(i from 1 to n: a_{i} )*PRODUCT(i from 1 to n: (n-2)S+a_{i} ) is less than or equal to
PRODUCT(i from 1 to n: (S-a_{i} )*PRODUCT(i from 1 to n: (S-a_{i} ), or

PRODUCT(i from 1 to n: a_{i}^{2}-2*S*a_{i}+n*a_{i}S ) is less than or equal to
PRODUCT(i from 1 to n: a_{i}^{2}-2*S*a_{i}+S^{2} ).

Now the sum of the factors at the left hand side of this last inequality is the same as the sum of the factors at the right hand side,
but the factors at the left hand side have a greater variation. It is wellknown that in such a case the product at the left
hand side is less than or equal to the product at the right hand side.

11253 *Let n be a positive integer and A an n by n matrix with all entries a _{i,j} positive. Let P be the
permanent of A. Prove that P is at least n! times (PRODUCT(i,j from 1 to n: a_{i,j}))^{1/n}.*

__Solution:__

By definition, P is equal to (SUM s: s permutation of {1,2,..,n}: a_{s(1),1}a_{s(2),2}...a_{s(n),n}).

By the arithmetic-geometric mean inequality, this is at least n! times (PRODUCT: s permutation of {1,2,..,n}:
a_{s(1),1}a_{s(2),2}...a_{s(n),n}))^{1/(n!)}.

The right hand side is equal to n! times (PRODUCT(i,j from 1 to n: a_{i,j}))^{((n-1)!)/(n!)}.

11254 *For a prime p greater than 3, let S _{p} be the set of positive integers less than (p-1)/2 and
relative prime to p-1. Characterize the primes p for which there exists a primitive root g modulo p such that the
product of g^{a}, taken over all a in S_{p}, is also a primitive root modulo p.*

__Partial solution:__

Any root of unity modulo p is primitive iff it has order p-1, and any power of a primitive
root is also primitive iff the exponent is relative prime to p-1. Since a primitive root of unity g always exists, p
satisfies the demand iff the sum of the elements of S_{p} is relative prime to p-1.

The primes p between 3 and 30 which satisfy the demand are the following: 5,7,19,23.

I searched the internet and the book of Hardy and Wright for a formula that gives the sum of the elements of S_{p},
but in vain.

11256 *For complex numbers a,b and c, let f(x)=max(Re(a*e ^{ix}),Re(b*e^{ix},Re(c*e^{ix})).
*

Find INT(x from 0 to 2*pi: f(x) dx).

__Solution:__

Let a=a_{1}+ia_{2} (a_{1},a_{2} real), and likewise for b and c.
Re(a*e^{ix})=A=a_{1}cos(x)-a_{2}sin(x), and likewise for b and c.

Without loss of generality we assume a_{2} is at least b_{2}, and b_{2} at least c_{2}.

We find: A is larger than B iff ((x lies between -pi/2 and pi/2, and tan(x) is smaller than
(a_{1}-b_{1})/(a_{2}-b_{2})) or (x lies between pi/2 and 3*pi/2, and tan(x) is greater than
(a_{1}-b_{1})/(a_{2}-b_{2}))).

Let at(a,b)=arctan((a_{1}-b_{1})/(a_{2}-b_{2})), and likewise at(a,c) and at(b,c).

We find: A is larger than B iff ((x lies modulo 2*pi between -pi/2 and at(a,b)) or (x lies modulo 2*pi between pi+at(a,b)
and 3*pi/2)).

Now henceforth we assume, for instance, that at(a,c) lies between at(a,b) and at(b,c) (other cases are similar).

Then max(A,B,C)=B for x modulo 2*pi lying between
at(a,b) and at(b,c), max(A,B,C)=C for x modulo 2*pi lying between at(b,c) and pi+at(a,c), max(A,B,C)=A for x modulo 2*pi
lying between pi+at(a,c) and 2*pi+at(a,b).

So INT(x from 0 to 2*pi: f(x) dx) = INT(x from at(a,b) to at(b,c): B dx) + INT(x from at(b,c) to pi+at(a,c): C dx) +
INT(x from pi+at(a,c) to 2*pi+at(a,b): A dx).

If x=at(b,c) then
sin(x)=(b_{1}-c_{1})/sqrt((b_{1}-c_{1})^{2}+(b_{2}-c_{2})^{2}) =
(b_{1}-c_{1)})/sqrt(b,c), and cos(x)=(b_{2}-c_{2})/sqrt(b,c), etc.

So INT(x from at(a,b) to at(b,c): B dx) = [b_{1}sin(x)+b_{2}cos(x)] (x in at(b,c) minus x in at(a,b)) =
(b_{1}(b_{1}-c_{1})+b_{2}(b_{2}-c_{2}))/sqrt(b,c) -
(b_{1}(a_{1}-b_{1})+b_{2}(a_{2}-b_{2}))/sqrt(a,b).

Likewise we find: INT(x from at(b,c) to pi+at(a,c): C dx) =
-(c_{1}(a_{1}-c_{1})+c_{2}(a_{2}-c_{2}))/sqrt(a,c) -
(c_{1}(b_{1}-c_{1})+c_{2}(b_{2}-c_{2}))/sqrt(b,c); and:

INT(x from pi+at(a,c) to 2*pi+at(a,b) : A dx) = (a_{1}(a_{1}-b_{1})+a_{2}(a_{2}-b_{2}))/sqrt(a,b)
+ (a_{1}(a_{1}-c_{1})+a_{2}(a_{2}-c_{2}))/sqrt(a,c).

These terms sum up to sqrt(a,b)+sqrt(b,c)+sqrt(c,a), this is the circumference of triangle ABC.

11261 *A triangle of area 1 has vertices A _{1}, A_{2} and A_{3}. The sides
A_{2}A_{3}, A_{3}A_{1}, A_{1}A_{2} subtend angles of measure
alpha_{1}, alpha_{2}, alpha_{3}, respectively, at an internal point P. The triangle has
angles at A_{1}, A_{2}, A_{3} of measure a_{1}, a_{2}, a_{3},
respectively. The extensions of A_{1}P, A_{2}P, A_{3}P to their opposite sides meet those sides
at B_{1}, B_{2}, B_{3} respectively.
*

Express the area of triangle B_{1}B_{2}B_{3} in alpha_{1}, alpha_{2}, alpha_{3}, a_{1}, a_{2} and a_{3}.

__Partial solution:__

Give coordinates P(0,0), A_{2}(r,0), A_{3}(s*cos(alpha_{1}),s*sin(alpha_{1})),
A_{1}(t*cos(alpha_{3}),-t*sin(alpha_{3}), and

B_{2}(-u,0),
B_{3}(-v*cos(alpha_{1}),-v*sin(alpha_{1})),
B_{1}(-w*cos(alpha_{3}),w*sin(alpha_{3})).

Using that B_{2} lies on A_{1}A_{3} we get
u=s*t*sin(alpha_{2})/(s*sin(alpha_{1})+t*sin(alpha_{3})). Likewise:

v=r*t*sin(alpha_{3})/(r*sin(alpha_{1})+t*sin(alpha_{2})),
w=s*r*sin(alpha_{1})/(s*sin(alpha_{2})+r*sin(alpha_{3})).

So we can express the side lengths B_{2}B_{3}, B_{3}B_{1}, B_{1}B_{2}
in r,s,t and alpha_{1}, alpha_{2}, alpha_{3}.

Furthermore, using A_{1}A_{2}*A_{1}A_{3}*sin(a_{1})/2 = 1, etc, we find
A_{1}A_{2} = sqrt((2*sin(a_{3})/(sin(a_{1})*sin(a_{2}))), etc.

So, using Heron's formula for the area of triangle B_{1}B_{2}B_{3}, we can now reach our
aim if we can express r,s,t in the angles
alpha_{1},alpha_{2},alpha_{3} and the side lengths
A_{2}A_{3}, A_{3}A_{1}, A_{1}A_{2}.

This last task can be done, at least in principle, by using

A_{1}A_{2}^{2} = r^{2}+t^{2}-2*r*t*cos(alpha_{3}),
A_{2}A_{3}^{2} = r^{2}+s^{2}-2*r*s*cos(alpha_{1}),
A_{1}A_{3}^{2} = s^{2}+t^{2}-2*s*t*cos(alpha_{2}).

__Remark:__ If we use the formula for the area of a Cevian triangle with given trilinear coordinates, we
get

2*rst*sin(alpha_{1})*sin(alpha_{2})*sin(alpha_{3})/((r*sin(alpha_{1})+t*sin(alpha_{2}))*
(s*sin(alpha_{1})+t*sin(alpha_{3}))*(s*sin(alpha_{2})+r*sin(alpha_{3}))).

But then, again, we still have to express r,s,t in the angles
alpha_{1},alpha_{2},alpha_{3} and the side lengths
A_{2}A_{3}, A_{3}A_{1}, A_{1}A_{2}.

11309 *Let c and d be real numbers with sqrt(c ^{2}+d^{2}) smaller than pi/2. Prove that cos(c*sin(x)) is greater than sin(d*cos(x)) for all real x.*

__Solution:__

Note that c=r*cos(e) and d=r*sin(e) for some real e and r from [0,pi/2).

We distinguish cases a): c*sin(x) positive, b): c*sin(x) nonpositive.

In case a) we observe that cos(c*sin(x)) = sin(pi/2-c*sin(x)) and sin is strictly ascending on (-pi/2,pi/2);

since d*cos(x) is greater than -pi/2, pi/2-c*sin(x) smaller than pi/2, and
c*sin(x)+d*cos(x) = r*sin(e+x) smaller than pi/2, so pi/2-c*sin(x) greater than d*cos(x), we find that the inequality in the problem is true.

In case b) we observe that sin(d*cos(x)) = cos(pi/2-d*cos(x)) and cos(c*sin(x)) = cos(-c*sin(x)) and cos is strictly descending on [0,pi);

since -c*sin(x) is greater than or equal to 0, pi/2-d*cos(x) smaller than pi, and
-c*sin(x)+d*cos(x) = r*sin(e-x) smaller than pi/2, so pi/2-d*cos(x) greater than -c*sin(x), we find that the inequality in the problem is true.

11315 *Define f on the positive integers by letting f(n)=PRODUCT(k from 1 to r: a _{k}^{pk}) when n has prime factrorization
PRODUCT(k from 1 to r: p_{k}^{ak}), with the empty product yielding f(1)=1. Prove that for all n the sequence n, f(n), f(f(n)), ... , is eventually periodic, with period at
most 2.*

__Partial solution:__

The sequence becomes periodic, with period 1 or 2, as soon as we get a term whose prime factorization has only exponents that are prime and distinct.

But this will always happen at some moment, because the sequence SUM(k from 1 to r: a_{k}+p_{k}), derived from the prime factorizations of the terms, is mostly decreasing until it happens.

11325 *Let P be a point within a triangle ABC. Let the lines AP, BP and CP intersect the sides BC, CA and AB at L, M, N respectively. Show that AP*BP*CP is at least 8*LP*MP*NP, with equality iff P is
the centroid of ABC.*

__Solution:__

Give coordinates P(0,0), A(-a,0), L(1,0) and, for some t with sin(t) not equal to 0, B(1+b*cos(t),b*sin(t)), C(1-c*cos(t),-c*sin(t)) (a,b,c positive).

We calculate the coordinates of M and N and find, besides AP/LP = a, the ratios BP/MP = (b+c+ac)/(ab) and CP/NP = (b+c+ab)/ac.

Hence we have to prove that for positive a,b,c holds: (b+c+ac)(b+c+ab) - 8abc (= f) is at least 0, with equality iff the three ratios are equal to 2.

We may write f/(bc) = (a+(s-8)/2)^{2} + s-(s-8)^{2}/4 with s=(b+c)^{2}/(bc).

Now s-(s-8)^{2}/4 is positive for s in (4,16), and, since a is positive, for s at least 16 we find that f is positive.

Furthermore, s is at least 4 with equality iff b=c. If s=4 then f=(a-2)^{2}.

Hence, for positive a,b,c we find: f is at least 0 with equality iff (b=c and a=2).

Finally, it is straightforward to show that the three ratios are equal to 2 iff (b=c and a=2).

11328 *Let ABCD be a convex quadrilateral. Let P be a point outside ABCD such that angle APB is a right angle and P is equidistant from A and B. Let points Q, R and S be given by the same conditions
with respect to the other three edges of ABCD. Let J, K, L and M be the midpoints of PQ, QR, RS and SP, respectively. Prove that JKLM is a square.*

__Solution:__

Give coordinates A(0,0), B(1,0), C(c_{1},c_{2}), D(d_{1},d_{2}) with c_{2} and d_{2} positive and c_{1} greater than d_{1}.

Then we find P(1/2,-1/2), Q(1/2+c_{1}/2+c_{2}/2,1/2-c_{1}/2+c_{2}/2), R(c_{1}/2-c_{2}/2+d_{1}/2+d_{2}/2,c_{1}/2+c_{2}/2-d_{1}/2+d_{2}/2),
S(d_{1}/2-d_{2}/2,d_{1}/2+d_{2}/2)

and hence J(1/2+c_{1}/4+c_{2}/4,-c_{1}/4+c_{2}/4), K(1/4+c_{1}/2+d_{1}/4+d_{2}/4,
1/4+c_{2}/2-d_{1}/4+d_{2}/4), L(c_{1}/4-c_{2}/4+d_{1}/2,c_{1}/4+c_{2}/4+d_{2}/2),
M(1/4+d_{1}/4-d_{2}/4,-1/4+d_{1}/4+d_{2}/4).

Now it's straightforward to show that the vectors __JK__ and __KL__ are perpendicular, and likewise __KL__ and __LM__, __LM__ and __MJ__, __MJ__ and __JK__, and that the four lengths
|__JK__|, |__KL__|, |__LM__| and |__MJ__| are equal.

11330 *For a triangle with semiperimeter s, inradius r, circumradius R, and heigths h _{a}, h_{b}, h_{c}, show that*

h_{a}+h_{b}+h_{c} - 9r is at least 2s*√(2r/R) -6r*√3.

__Partial solution:__

As several times before, we use:

1) area D = ab*sin(γ)/2 = bc*sin(α)/2 = ac*sin(β)/2 = (1/2)*(abc)^{2/3}(sin(α)*sin(β)*sin(γ))^{1/3};

2) r = D/s = 2D/(a+b+c), R = abc/(4D);

3) h_{a}=c*sin(β), h_{b}=a*sin(γ), h_{c}=b*sin(α);

4) a/sin(α) = b/sin(β) = c/sin(γ) = h.

Then we find, again, that the proposed inequality is equivalent to a goniometric inequality that we may check using Pascal. This time it reads:

(α, β, γ positive and summing up to π, and)

(sin(α)+sin(β)+sin(γ))(sin(α)*sin(β)+sin(α)*sin(γ)+sin(β)*sin(γ)) ≥ 2√(sin(α)+sin(β)+sin(γ))*√(sin(α)*sin(β)*sin(γ))*(sin(α)+sin(β)+sin(γ)) + (9-6*√3)*sin(α)*sin(β)*sin(γ).

{The Pascal program includes the following central part:

for k:=0 to 10000 do for m:=k to (15000 - (k div 2)) do begin a:=k*pi/30000; b:=m*pi/30000; c:=pi-a-b; f:= .... ; if not (f ≥ 0) then writeline(k:6,m:6) end}

We have equality if α = β = γ = π/3.

11333 *Show that PRODUCT(n from 2 to inf: ((n ^{2}-1)/n^{2})^{2(n2-1)}((n+1)/(n-1))^{n}) = π.*

__Solution:__

PRODUCT(n from 2 to N: ((n^{2}-1)/n^{2})^{2(n2-1)}((n+1)/(n-1))^{n}) =
2*3^{2}*4^{2}*5^{2}*..*N^{2}*(N/(N+1)^{2})*(N+1)^{2N2+N}/N^{2N2+3N}.

We can replace (N+1)^{2N2+N}/N^{2N2+3N} by (e^{2N}/N^{2N}) asymptotically.
Furthermore, we can replace (N/(N+1)^{2}) by 1/N asymptotically.

Hence the infinite product is LIM(N to inf: (N!)^{2}(e^{2N}/N^{2N})/(2N)). According to Stirling, this becomes π.

11337 *Suppose in triangle ABC we have opposite sides of length a,b,c respectively, with a≤b≤c. Let w _{a} and w_{b} be the lengths of the bisectors of the angles A and
B respectively. Show that a+w_{a} ≤ b+w_{b}*.

__Partial solution:__

Without loss of generality we may assume: c=1; A(0,0), B(1,0), C(b*cos(α),b*sin(α)).

We find w_{a}=2b*cos(α/2)/(1+b), w_{b}=2a*cos(β/2)/(1+a).

So we have to prove: 2b(1+a)cos(α/2)-2a(1+b)cos(β/2) ≤ (b-a)(1+a)(1+b).

Now, according to the sine rule a/sin(α)=b/sin(β)= 1/sin(α+β). So the inequality becomes:

sin(α+β)(2sin(β)(sin(α+β)+sin(α))cos(α/2)-2sin(α)(sin(α+β)+sin(β))cos(β/2)) ≤ (sin(β)-sin(α))(sin(α+β)+sin(α))(sin(α+β)+sin(β)) for 0≤α≤β≤π-α-β.

This can be checked with Pascal.

11345 *Find all nondecreasing functions f from ℜ to ℜ such that f(x+f(y)) = f(f(x)) + f(y) for all x and y*.

__Solution:__

Suppose f(0)=A. Substituting x=y=0 yields f(A)=f(A)+A, so A=0. So f(0)=0.

Then substituting y=0 yields f(x)=f(f(x)) for all x.

So if f(x)=B then f(B)=B. So for any B in the range of f we have f(B)=B.

Furthermore, we find f(x+f(y)) = f(x) + f(y) for all x and y. Hence, if B and C belong to the range of f then f(x+C) = B+C for all x with f(x)=B.

Now suppose f(y)=C. Substituting x=-C yields 0 = f(0) = f(-C+C) = f(-C) + C. So f(-C) = -C. Hence, if C belongs to the range of f, then so does -C.

If B and C belong to the range of f, then f(B)=B and f(C)=C and f(B+C) = B+C, so B+C belongs to the range of f.

Conclusions:

1) The range f(ℜ) of f is a subgroup of (ℜ,+).

2) If f(ℜ) has a smallest positive element E then, for all D ∈ f(ℜ), f^{-1}(D) is a half open interval of length E containing D, obtained by translating
f^{-1}(0) a distance k times E for some integer k; so f is a kind of step function.

3) If f(ℜ) has no smallest positive element E then either f(x)=0 for all x or f(x)=x for all x.

11357 *Let I _{a} , I_{b} , I_{c} and r_{a} , r_{b} , r_{c} be respectively the excenters and exradii of the triangle ABC.
*

If ρ_{a} , ρ_{b} , ρ_{c} are the inradii of triangles I_{a}BC , I_{b}CA , I_{c}AB , show that ρ_{a} /r_{a} + ρ_{b} /r_{b} + ρ_{c} /r_{c} = 1.

__Solution:__

The picture below explains everything.

11397 *Let a,b,c,x,y,z be positive numbers such that a+b+c = x+y+z and abc = xyz. Show that if max{a,b,c} ≥ max{x,y,z}, then min{a,b,c} ≥ min{x,y,z}.*

__Solution:__

Suppose f = U^{3} + pU^{2} + qU + s = (U-a)(U-b)(U-c) and g = U^{3} + pU^{2} + rU + s = (U-x)(U-y)(U-z),
where a ≥ b ≥ c > 0 and x ≥ y ≥ z > 0.

Then g - f = (r-q)U = tU.

If a ≥ x then t ≥ 0 and we have the following picture:

So c ≥ z.

11398 *Suppose that acute triangle ABC has its middle-sized angle at A. Suppose further that the incenter I is equidistant from the circumcenter O and the orthocenter H.
Show that angle A has measure 60 degrees and that the circumradius of IBC is the same as that of ABC.*

__Partial solution:__

From Wolfram we learn: if D is the area of the triangle and R = abc/(4D) is the circumradius and r = 2D/(a+b+c) is the inradius, then
IO = √(R(R-2r)), IH = √(2r^{2} + 4R^{2} -(a^{2}+b^{2}+c^{2})/2).

Now let h := a/sin(α) = b/sin(β) = c/sin(γ). Then D = ab sin(γ)/2 = h^{2}sin(α)sin(β)sin(γ)/2.

We deduce R = h/2 and r = h*sin(α)sin(β)sin(γ)/(sin(α)+sin(β)+sin(γ)).

So IO = IH is equivalent to 1/4 - sin(α)sin(β)sin(γ)/(sin(α)+sin(β)+sin(γ)) = 2*(sin(α)sin(β)sin(γ)/(sin(α)+sin(β)+sin(γ)))^{2} + 1
- (sin^{2}(α) + sin^{2}(β) + sin^{2}(γ))/2.

With a Pascal program we seek all solutions of this last equation under the restraint α+β+γ = 180. We find: α=60, β and γ arbitrary such that β+γ
= 120.

If α = 60 then the angle at I in triangle IBC is 120. The circumradius of ΔABC is h/2 = a/(2*sin(60)), and the circumradius of ΔIBC is a/(2*sin(120)). So these two circumradii are always equal if α = 60.

11402 *Let f:[0,1] → [0,∞) be a continuous function such that f(0) = f(1) = 0 and f(x) is positive for x ∈ (0,1).
Show that there exists a square with two vertices in the interval (0,1) on the x-axis and the two other vertices on the graph of f.
*

__Partial solution:__

Here is the solution when f is increasing at the beginning of (0,1) and decreasing at the end.

Let c ∈ (0,1] be maximal such that f(x) is at most 1-x for all x ∈ [0,c].

Let g(x) := f(x+f(x)) - f(x).

Then g is continuous on [0,c].

Let s be maximal such that f is increasing on [0,s] and let t ∈ [0,s] be maximal such that f(x) is at most s-x on [0,t].

Then, for every δ ∈ (0,t), g(δ) is positive.

Analoguously, if c=1, there exists a positive ε (smaller than 1-t) such that g(1-ε) is negative (because f is decreasing on [u,1] for some u ∈ [s,1]).

Again, if c is smaller than 1, then for ε := 1-c (positive and smaller than 1-t) we find g(1-ε) is negative (because g(1-ε) = g(c) = f(c+f(c))-f(c) = f(1)-f(c) = -f(c)).

Now, since g(δ) is positive and g(1-ε) negative, there exists an u ∈ (δ,1-ε) such that g(u) = 0.

Since f(u+f(u)) = f(u), we find a square as required.

11404 * Any three non-concurrent cevians of a triangle create a subtriangle. Identify the set of non-concurrent cevians which create a subtriangle whose incenter coincides with the incenter of the primary triangle
(a cevian is a line segment joining a vertex to an interior point of the opposite edge.)*

__Partial solution:__

If we use coordinates such that the incenter is (0,0), and one of the sides of the primary triangle is y=-1, and the other two sides have direction coefficients m (positive) and n (negative), then the vertices have
coordinates

(-(√(m^{2}+1)+1)/m,-1) , (-(√(n^{2}+1)+1)/n,-1), and (U,V) =
((√(n^{2}+1)-√(m^{2}+1))/(m-n),(m(√(n^{2}+1)-1)-n(√(m^{2}+1)-1))/(m-n)).

Now if the cevians from these three vertices have direction coefficients (respectively) a ∈ (0,m), b ∈ (n,0) and c ∈ (n,m), then, because the distances from (0,0) to these three cevians
must be equal, we find two independent linear equalities that relate a,b,c to m,n (so there is a one-dimensional solution set for (a,b,c)). These inequalities read:

|((a/m)(1+√(m^{2}+1))-1)/√(a^{2}+1)| = |((b/n)(1+√(n^{2}+1))-1)/√(b^{2}+1)| = |(Uc-V)/√(c^{2}+1)|.

If we use A:=|a|/√(a^{2}+1), etc, and M:=(1+√(m^{2}+1))/|m|, etc, then the one-dimensional solution set for (A,B,C) has direction vector (|U|/M,|U|/N,1).

In each of the cases, we readily find the full solution set.

11405 *Let P be an interior point of a tetrahedron ABCD. When X is a vertex, let X' be the intersection of the opposite face with the line through X and P. Let XP denote the length of the line
segment from X to P.
(a) Show that PA.PB.PC.PD ≥ 81PA'.PB'.PC'.PD', with equality if and only if P is the centroid of ABCD.
(b) When X is a vertex, let X" be the foot of the perpendicular from P to the plane of the face opposite X. Show that PA.PB.PC.PD = 81PA".PB".PC".PD" if and only if the tetrahedron is regular and P is
its centroid.
*

__Solution:__

(a) Let D=O and, for any point X, denote the vector from O to X by __x__.

If P is an interior point of tetrahedron OABC, then __p__ = x__a__ + y__b__ + z__c__ for positive x,y,z whose sum is smaller than 1.

We readily calculate PA' = x/(1-x)PA, PB' = y/(1-y)PB, PC' = z/(1-z)PC, PO' = (1-x-y-z)/(x+y+z)PO.

So we have to find the minimum of ((1-x)/x)((1-y)/y)((1-z)/z)((x+y+z)/(1-x-y-z)) for x,y,z as above.

We calculate the partial derivatives and demand that they be zero, and find x(1-x) = y(1-y) = z(1-z) = (x+y+z)(1-x-y-z) and hence x=y=z=1/4.

(b) If the tetrahedron is regular and P is the centroid, then X"=X' for all vertices X, so the equality holds.

If the tetrahedron is not regular or P is not the centroid, then PX/PX" ≥ PX/PX' for all vertices X and PX/PX" > PX/PX' for at least one vertex X, so the equality does not hold.

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