Solutions by H Reuvers, part 2

10969 A partition L of n is a nonincreasing sequence L=(L1,L2,...) of nonnegative integers with SUM(Li / i from 1 to infinity) = n. The zero sequence counts as a partition of zero. The conjugate L' of a partition L is the partition in which, for all i, the integer i occurs Li-Li+1 times.
Let p(n) denote the total number of partitions of n, and let t(n) denote the number of them satisfying O(L)=O(L') mod 4, where O(M) denotes the number of odd parts of a partition M.
Show that t(n)=(p(n)+f(n))/2, where SUM(f(n)xn) = PRODUCT((1+x2i-1)/((1-x4i)(1+x4i-2)2)).

Partial solution:

We can divide the number p(n) of partitions of n into those satisfying O(L)=O(L')(mod 4), number t(n), and those that don't satisfy this relation, number v(n); thus p(n)=t(n)+v(n).
We have to show f(n)=t(n)-v(n).
We write out the generating function of f(n), that is the product in the problem:
(1+x)(1+x3)(1+x5)(1+x7)...
(1+x4+x8+x12+x16+...) (1+x8+x16+x24+x32+...) (1+x12+x24+x36+x48+...) .....
(1-x2+x4-x6+x8+...) (1-x2+x4-x6+x8+...)
(1-x6+x12-x18+x24+...) (1-x6+x12-x18+x24+...)
(1-x10+x20-x30+x40+...) (1-x10+x20-x30+x40+...)
.....
To find all partitions we rewrite this as follows:
(1+x)(1+x3)(1+x5)(1+x7)...
(1+x4+x4x4+x4x4x4+...) (1+x8+x8x8+x8x8x8+...) (1+x12+x12x12+x12x12x12+...) .....
(1-x2+x2x2-x2x2x2+...) (1-x1x1+x1x1x1x1- x1x1x1x1x1x1+...)
(1-x6+x6x6-x6x6x6+...) (1-x3x3+x3x3x3x3- x3x3x3x3x3x3+...)
(1-x10+x10x10-x10x10x10+...) (1-x5x5+x5x5x5x5- x5x5x5x5x5x5+...)
.....
Hence we see: f(n)=s(n)-w(n) and p(n)=s(n)+w(n), where s(n) is the number of partitions that contribute +1 to f(n), and w(n) the number of partitions that contribute -1.
Studying which partitions contribute -1, we see that we have to distinguish blocks of equal parts of several types:
type A: 1+2m times 2+4k, for instance 666 or 2;
type B: 2+4m times 1+2k, for instance 55 or 777777;
type C: 3+4m times 1+2k, for instance 111.
We observe that w(n) is the number of partitions of n containing an odd number of blocks of types A, B or C, for instance 77777766653333111.
We still have to show that w(n)=v(n).

10970 Let ABC be an acute triangle and P a point in its interior. Denote by a,b,c the lengths of the triangle's sides, by alpha,beta,gamma its angles, by da,db,dc the distances from P to the triangle's sides, and by Ra,Rb,Rc the distances from P to the vertices A,B,C, respectively.
Show that da2+db2+dc2 is at least Ra2sin2(alpha/2)+ Rb2sin2(beta/2)+Rc2sin2(gamma/2) and this last number at least (1/3)(da+db+dc)2.
When is equality possible?

Solution:

There exist positive a1,a2,b1,b2,c1,c2 such that alpha=a1+a2, beta=b1+b2, gamma=c1+c2 and da=Rc*sin(c2)=Rb*sin(b2), db=Ra*sin(a2)=Rc*sin(c1), dc=Ra*sin(a1)=Rb*sin(b1).

Hence da2+db2+dc2=(1/2)(Ra2sin2(a1)+ Ra2sin2(a2)+Rb2sin2(b1)+Rb2sin2(b2)+ Rc2sin2(c1)+Rc2sin2(c2)).
To show that da2+db2+dc2 is at least Ra2sin2((a1+a2)/2)+Rb2sin2((b1+b2)/2)+ Rc2sin2((c1+c2)/2), it is sufficient to show that (1/2)(sin2(a1)+sin2(a2)) is at least sin2((a1+a2)/2).
Since (a1+a2)/2 is smaller than pi/4, this follows from the fact that the graph of sin2(x) is convex in [0,pi/4] and invariant under reflection in the point (pi/4,1/2).

Furthermore, da+db+dc=(1/2)(Ra*sin(a1)+Ra*sin(a2)+Rb*sin(b1)+Rb*sin(b2)+Rc*sin(c1)+Rc*sin(c2)).
To prove that (1/3)(da+db+dc)2 is at most Ra2sin2((a1+a2)/2) +Rb2sin2((b1+b2)/2)+Rc2sin2((c1+c2)/2), it is sufficient to prove that
(1/2)(Ra*sin(a1)+Ra*sin(a2))2+(1/2)(Rb*sin(b1)+Rb*sin(b2))2+ 2*Ra*Rb(sin(a1)+sin(a2)(sin(b1)+sin(b2)) is at most 6*Ra2sin2((a1+a2)/2) +6*Rb2sin2((b1+b2)/2).
Hence, since 4*Ra2sin2((a1+a2)/2)+4*Rb2sin2((b1+b2)/2) is at least 8*Ra*Rb*sin((a1+a2)/2)*sin((b1+b2)/2), it is also sufficient to prove that
(sin(a1)+sin(a2))2 is at most 4*sin2((a1+a2)/2) and (sin(b1)+sin(b2))2 is at most 4*sin2((b1+b2)/2) and (sin(a1)+sin(a2))*(sin(b1)+sin(b2)) is at most 4*sin((a1+a2)/2)*sin((b1+b2)/2).
Or again, it is sufficient to prove that sin(a1)+sin(a2) is at most 2*sin((a1+a2)/2).
But this follows from sin(a1)+sin(a2)=2*sin((a1+a2)/2)*cos((a1-a2)/2).

We see above that for each of both inequalities holds that it becomes an equality if and only if a1=a2 and b1=b2 and c1=c2, so if P is the incenter of the triangle.

10973 With Rk(n) defined as below, prove that lim(k to infinity) Rk(2)/Rk(3)=3/2.
Rk(n)=sqrt(2-sqrt(2+sqrt(2+sqrt(2+...+sqrt(2+sqrt(n))))))) (k squareroots).

First approach:

It's no proof, but if they didn't tell us the outcome, from the output of a little Pascal program I'd guess 3/2.
The core of this program is:
m:=sqrt(2);n:=sqrt(3);
for k:=1 to 10 do begin L:=sqrt((2-m)/(2-n));writeln(L:13:10);m:=sqrt(2+m);n:=sqrt(2+n) end

10974 The digital root R(n) of a positive integer n is the eventual image of n under the mapping that carries an integer n to the sum of its base-ten digits. Thus R(10974)=R(21)=3.
Find R(Fn), where Fn is the n-th Fibonacci number with F1= F2=1.

Solution:

For each natural number m, we have: R(m)=m MOD 9 if m is not divisible by 9, R(m)=9 if m is divisible by 9.
So R(m) can be 1,2,..,8,9.
For the Fibonacci numbers we have the recurrent relation Fn+2=Fn+1+Fn.
Since this relation holds modulo 9 as well, the function R(Fn) is known as soon as we find in the sequence R(F1), R(F2), ... a pair of two consecutive terms that we found earlier.
Since R(F1)=R(F2)=1=R(F25)=R(F26), we find: R(Fn)=R(Fn MOD 24).
Thus, to let people know the function R(Fn), we only have to mention the first 24 terms.
They are 1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9. Thereafter come the same 24 terms again and again ad infinity.

10980 Consider four distinct straight lines in the same plane, with the property that no two of them are parallel, no three are concurrent, and no three form an equilateral triangle. Prove that, if one of the lines is parallel to the Euler line of the triangle formed by the other three, then each of the four given lines is parallel to the Euler line of the triangle formed by the other three.

Solution:

Let three lines form triangle ABC.
Euler himself proved that, if the Euler line of triangle ABC intersects AB, AC and BC in C', B' and A' respectively, then the Euler line of triangle AB'C' is parallel to BC, etc.
See this website about the Gossard Perspector.
Hence, if the fourth line is parallel to the Euler line of triangle ABC, then it forms with AB and AC a triangle whose Euler line is parallel to BC, etc.

10983 Let T be the orthogonal projection of the orthocenter H of an acute triangle ABC on the line r bisecting the angle BAC at A. Let P be the orthogonal projection of T on BC, M the midpoint of BC, and A' the intersection of BC and r.
(a) Prove that TA' bisects the angle MTP.
(b) Prove that TM is parallel to AO, where O is the circumcenter of triangle ABC.

Solution:

Give coordinates O(0,0), A(-cos(s),-sin(s)), B(cos(s),-sin(s)), C(cos(t),sin(t)).
Then we find M((cos(t)+cos(s))/2,(sin(t)-sin(s))/2), H(cos(t),lamba*sin(t));
from the fact that BH is perpendicular to AC we find
lambda=(cos(2s)-cos(2t))/(sin2(t)+sin(s)*sin(t))-1.
Furthermore, since T lies on the bisectrix of the angle alpha=arc(BC)/2=(s+t)/2, we have:
T:(-cos(s),-sin(s))+mu*(cos((s+t)/4),sin((s+t)/4).
From the fact that AT is perpendicular to HT we find
mu=cos((s+t)/4)*(cos(s)+cos(t))+sin((s+t)/4)*(lambda*sin(t)+sin(s))=
(cos(s)+cos(t))(cos((s+t)/4)+sin((s+t)/4)*((cos(s)-cos(t))/(sin(s)+sin(t)).
For the vector from T to M we find
((sin(t)+sin(s))/2-mu*sin((s+t)/4),(cos(t)+3cos(s))/2-mu*cos((s+t)/4).
After some calculation this reduces to (cos(s),sin(s))*(1-cos((s+t)/2).
So TM is parallel to AO.
From the fact that TP is perpendicular to BC, we find for TP the direction vector
((sin(s)+sin(t),cos(s)-cos(t))=(cos((t-s)/2,sin(t-s)/2)*2sin((s+t)/2).
So the angles between the horizontal AB and (resp) TP,TA,TM are (resp) (t-s)/2, (s+t)/4, s.
Since (s+t)/4 = (1/2)*(s + (t-s)/2), we see that TA bisects angle MTP.

10990 The n+1 Bernstein polynomials of degree n are defined by
bn,k(x) = (n over k)*xk(1-x)n-k for k from 0 to n.
When all n+1 polynomials are plotted on the same graph for large fixed n over the interval [0,1], an 'upper envelope' begins to be seen.
Let b(x) = lim sqrt(n)*max(k from 0 to n: bn,k(x)). Find a closed-form expression for b(x).

Solution:

Differentiation learns that the maximum appears when x=k/n. We substitute k=xn, and use the formula of Stirling. After a bit of calculation we find:
b(x) = 1/sqrt(2*pi*x*(1-x)).

Remark:
professor van Lint found this earlier, but I found it independently.

10992 Let N and n be positive integers and (for natural numbers w1, ... ) let S={(w1,...,wn)/w1+..+wn is at most N} and T={(w1,...,wn)/w1, ... ,wn are distinct and at most N}.
Show that SUM(S, 1/(w1..wn)) = SUM(T, 1/(w1..wn)).

Partial solution:

We proceed with induction.
The cases n=1 and N=n are trivial.
Suppose that the assertion is true for n.
With respect to n+1, suppose that the assertion is true for N.
Then, by induction hypothesis, SUM(T,n+1,N+1) = SUM(T,n+1,N) + SUM(T,n,N)*(n+1)/(N+1) = SUM(S,n+1, N) + SUM(S,n,N)*(n+1)/(N+1).
On the other hand, SUM(S,n+1,N+1) = SUM(S,n+1,N) + SUM(w1+...wn=N+1, 1/(w1..w(n+1)).
To accomplish the induction step, we still have to prove that
SUM(S,n,N)*(n+1)/(N+1) = SUM(w1+...wn=N+1, 1/(w1..w(n+1)).
The left hand side is ((n+1)/(N+1))*SUM(k=1,..,N,SUM(w1+..+wn=k, 1/(w1..wn)).
The right hand side is SUM(k=1,..,N,SUM(w1+..+wn=k, 1/(w1..wn*(N+1-k))).
These two must be equal, but we still have to prove that.

10993 The excenter of a triangle T=ABC opposite A is the point (on the angle bisector at A) at which the normals through B and C to the bisectors of that angles meet.
Let T0 be a nondegenerate triangle and construct a sequence (Tn) of triangles by taking for the new vertices the excenters of the previous triangle Tn-1.
(a) Show that the shape of triangle Tn tends to equilateral as n goes to infinity.
(b) Obtain the same conclusion when the successive triangles arise by letting the new vertices be the points of contact of the incircle with the sides of the previous triangle.

Solution:

(a) Let the angles of Tn have measures an, bn, cn (in degrees).
It is straightforward to show that an+1 = (bn+cn)/2 = 90 - an/2, so abs(an+1-60) = abs(an-60)/2, and likewise abs(bn+1-60) = abs(bn-60)/2 and abs(cn+1-60) = abs(cn-60)/2.
This proves that an tends to 60, and likewise bn, cn.
(b) In this case we get exactly the same recurrence relations, so the results are the same as well.

10996 Prove that there exists a point P in the interior of triangle ABC such that angle(PCB) = angle(PAB) and angle(PBC) = angle(PAC) if and only if ABC is an acute triangle.

Solution:

Give coordinates A(0,0) and B(1,0). Let alpha, beta, gamma be the angles at A,B,C.
Suppose P is inside the triangle and angle(PAC) = angle(PBC) = delta.
We can calculate the coordinates of P and C in terms of alpha, beta, gamma, delta.
Now if angle(PCB) = angle(PAB) = alpha - delta, we can calculate delta, using the cosine rule in triangle PBC.
We find that the angles PAC, PAB, PBA, PBC, PCB, PCA must be 90-gamma, 90-beta, 90-alpha, 90-gamma, 90-beta, 90-alpha.
If the triangle is acute, we can construct P. If not, there clearly is no solution.

10999 Give a simple example of a uniformly bounded sequence of infinitely differentiable functions from the reals into the reals that converges pointwise to a nowhere continuous limit function.

Solution:

I learned from the web that the function f with f(x)=0 if x has a finite decimal representation and f(x)=1 otherwise is nowhere continuous. This makes me think of the following:
Let fk(x):=sin(10kx*pi/2). Then if x has a finite decimal representation, fk(x) has limit f(x)=0 and fk(x+10-k) has limit 1.
But fk(x) has no limit if x has no finite decimal representation.
Again, g with g(x)=0 for rational x and g(x)=1 for irrational x is nowhere continuous.
Let gk(x):=sin((k!)*x*pi/2). Then gk(x) has limit g(x)=0 for rational x, and gk(x+1/(k!)) has limit 1.
But gk(x) has no limit if x is irrational.
Hold on, we can take fk(x) := sin(sin(sin...(sin(10kx*pi/2)))) with k times sin. Then fk(x) tends (for all x) to the solution of sin(t)=t, but that is only t=0.
With cos instead of sin we would for all x get the solution of cos(t)=t as a limit.
But take F(t)=sin(t*pi/2). The equation F(t)=t has solutions t=0,1,-1. Take fk(x):=F(F(F(...(10kx)))) with k times F.
If x has a finite decimal representation, then fk(x) tends to 0. If not, since abs(F') is larger than 1 in a neighbourhood of 0 and smaller than 1 in a neighbourhood of 1 or -1, fk(x) tends for many x to 1 or -1 (successive substitution, look at the graph of F for t between -1 and 1; one can take 10k times F instead of k times).
My son Erik comments that if the decimals of x are 0's and 1's, each an infinite number, and after each 1 there follows a sufficiently fast increasing number of 0's, fk(x) does not converge.
Or take G(t)=sin(t*pi/2) and gk(x):=G(G(G(...((k!)*x)))) with, say, k! times G.
If x is rational, then gk(x) tends to 0. If not, then gk(x) tends to 1 or -1, except when k!*x is too close to integer again and again.
A simpler approach: let hk(s)=1 iff s has a decimal representation with from the k+1-th place after the decimal point only 0, and draw a smooth graph with top (s,1) which descends to the x-axis where it arrives at (s+10-2k,0), coincides with the x-axis until (s+10-k-10-2k,0), and ascends to the next top (s+10-k,1). But, like in the preceding examples, we can construct a bad x where hk(x) does not converge. This difficulty is structural.
I think that the proposer of the problem didn't think of anything better than this last example. We simply throw all the bad x out of the domains. Since the set of bad x has measure 0, for almost all x with infinite decimal representation hk(x) tends to 0. And hk(x) tends to 1 if x has finite decimal representation.

11003 Given a triangle ABC, reflect the circumcenter O through each of the sides to give new points A', B', C'. Show that the lines AA', BB', CC' are concurrent at the midpoint F of the segment joining the circumcenter O and the orthocenter H of ABC. (This point F is the center of the nine-point circle of ABC.) Show also that the triangle A'B'C' is congruent to ABC and rotated with respect to it 180 degrees around F.

Solution:

Give coordinates O(0,0), A(-cos(t),sin(t), B(cos(t),sin(t)), C(cos(s), sin(s)).
Then we find A'(cos(s)+cos(t),sin(s)+sin(t)), B'(cos(s)-cos(t), sin(s)+sin(t)), C'(0,2*sin(t)).
Write x for the vector from O to X.
Since a+a' = b+b' = c+c', we see that the lines AA', BB', CC' are concurrent and that reflection in the point F of concurrence maps A,B,C to A',B',C' resp.
Let h:= 2*f = (cos(s),sin(s)+2*sin(t)) . We can easily check that H is the orthocenter of triangle ABC.

11004 Prove that, in hyperbolic geometry, the diagonal of the Saccheri quadrilateral can be less than, congruent to, or greater than, the summit.

Solution:

We use the model where the hyperbolic points are (p,q) with p2+q2 smaller than 1.
The hyperbolic distance between (p1,q1) and (p2,q2) is arccosh(abs(1-p1q1-p2q2)/ (sqrt(1-p12-p22)* sqrt(1-q12-q22)).
The lines orthogonal to a given line are those that go through the pole of the given line with respect to the unit circle.
I searched systematically for solutions to the problem, using a computer program.
I found Saccheri quadrilaterals with vertices A(-a,-b), B(a,-b), C(x,-y), D(-x,-y), where, for example,
if a=0.4, b=0.5, x=0.3, y=0.875 then CD is smaller than BD,
if a=0.5, b=0.5, x=0.4, y=0.8 then CD is greater than BD.
By continuity, there must be a Saccheri quadrilateral in between where CD is as long as BD.

11006 Let ABC be an acute triangle, T the midpoint of arc BC of the circle circumscribing ABC. Let G and K be the projections of A and T respectively on BC, let H and L be the projections of B and C on AT, and let E be the midpoint of AB. Prove that:
a) KH is parallel to AC, GL to BT, GH to TC and LK to AB.
b) G,H,K,L are non-cocyclic.
c) The center of the circle through G,H,K lies on the Euler circle of ABC.

Sketch of solution:

Without loss of generality, we may assume that s,t exist between -pi/2 and pi/2 with s smaller than t such that
A(cos(s),sin(s)), B(cos(t),sin(t)), T(0,1), C(-cos(t),sin(t)), K(0,sin(t)), G(cos(s),sin(t)) and
H and L on AT:(0,1)+m(cos(s),sin(s)-1) with
mH=(1-sin(s)-sin(t)+cos(s-t))/(2-2sin(s)), mL=(1-sin(s)-sin(t)-cos(s+t))/(2-2sin(t)).
Now it is straightforward to check the truth of a),b),c) by calculation. Note that the Euler circle passes through E,G,K.

11010 (revised) Let C1C2...Cn be a regular n-gon, and let Cn+1=C1. Let O be the inscribed circle, with radius 1.
Let Tk be the point at which O is tangent to CkCk+1.
Let Mk be the midpoint of chord TkTk+1.
Let X and Y be arbitrary points on O.
Let Bk be the second point at which the line XCk meets O, and let pk = d(X,Bk)*d(X,Ck).
Let Nk be the second point at which the line YMk meets O, and let qk = d(Y,Mk)*d(Y,Nk).
Calculate SUM(1 to n: pk) - SUM(1 to n: qk).

Solution:

Without loss of generality, we can assume the following coordinates:
X:(cos(x),sin(x)), Y:(cos(y),sin(y)),
Ck: (1/cos(pi/n))*(cos(2k*pi/n),sin(2k*pi/n)), Mk-1: cos(pi/n)*(cos(2k*pi/n),sin(2k*pi/n)).
Using the power of a point with regard to the unit circle, we find
pk = d(Ck,X)*d(Ck,Bk) - d(Ck,X)2 = 1/cos2(pi/n) - 1 - d(Ck,X)2 = tan2(pi/n) - d(Ck,X)2,
and likewise qk = sin2(pi/n) - d(Mk,Y)2.
Now d(Ck,X)2 = (cos(x)-cos(2k*pi/n)/cos(pi/n))2 + (sin(x)-sin(2k*pi/n)/cos(pi/n))2 =
2 + tan2(pi/n) - 2cos(x-2k*pi/n)/cos(pi/n),
and likewise d(Mk,Y)2 = 2 - sin2(pi/n) - 2cos(y-2k*pi/n)*cos(pi/n).
Since SUM(1 to n: cos(x-2k*pi/n)) = 0, and likewise with y instead of x, we find:
SUM(1 to n: pk-qk) = 2n*sin2(pi/n).

11012 For natural n, find the minimum value of
(x13+x23+...+xn3)/(x1+x2+...+xn), where x1, x2, ... , xn are distinct positive integers
.

Conjecture:

I think that the minimum appears when xk=k, since the numerator increases much faster than the denominator. We still have to prove this. Then the outcome is (n(n+1)/2)2/(n(n+1)/2) = n(n+1)/2.

11015 Given a triangle with sides a,b,c, angles alpha,beta,gamma, radius r of inscribed circle and R of circumscribed circle, area S and perimeter 2p. Prove:
a) 1/sin(alpha/2)+1/sin(beta/2)+1/sin(gamma/2) is at least sqrt(a2+b2 +c2+4S*sqrt(3))/(4r)
b) 1/sin(alpha/2)+1/sin(beta/2)+1/sin(gamma/2) is at least 1/2 - r/(4R) + p*sqrt(3)/(4R).

Sketch of a solution:

Using 2S = r(a+b+c) and R = abc/(4S), we can express both right hand sides in a/r,b/r,c/r.
We get sqrt((a/r)2+(b/r)2+(c/r)2+2(a/r+b/r+c/r)*sqrt(3))/4 and
1/2 - (a/r+b/r+c/r)/(2*(a/r)*(b/r)*(c/r)) + (a/r+b/r+c/r)2*sqrt(3)/(4*(a/r)*(b/r)*(c/r)) respectively.
Here a/r = cotg(beta/2)+cotg(gamma/2), b/r = cotg(alpha/2)+cotg(gamma/2), c/r = cotg(alpha/2)+cotg(beta/2).
Hence the inequalities reduce to goniometric inequalities with arguments alpha/2,beta/2,gamma/2.
Using gamma=pi-alpha-beta, these can both be formulated in the form
p(cos(alpha/2),sin(alpha/2),cos(beta/2),sin(beta/2)) is at least 0,
where p(X,Y,U,V) is a polynomial with real coefficients.
Using Maple, we find that the inequalities hold.

remark: a solution following this outline cannot be heuristic, but the problem does not challenge to find heuristic solutions.

11017 a) Let a,b,c be the sides of a non-equilateral triangle. Prove that the Brocard axis goes through a vertex if and only if a.b,c are in geometric progression.
b) Show that, if alpha is the greatest angle, omega is smaller than minimum(pi/6,(beta+gamma)/3))

Solution:

a) Denote the angles of the triangle by alpha, beta, gamma, and the Brocard angle by omega.
We find in the literature: cotan(omega) = cotan(alpha)+cotan(beta)+cotan(gamma).
The Brocard axis goes through a vertex if and only if omega is half the angle at that vertex.
If, for instance, omega=beta/2, we find: cotan(beta/2) = cotan(alpha)+cotan(beta)+cotan(gamma).
Using the goniometric formulas, we find 1/sin(beta) = cos(alpha)/sin(alpha) + cos(gamma)/sin(gamma).
Using the sine rules, we find 1/b = cos(alpha)/a + cos(gamma)/c.
Using the cosine rules to express cos(alpha) and cos(gamma) in a,b,c, we find b2 = ac.

b) It is straightforward to show (using alpha+beta+gamma=pi) that cotan(alpha)+cotan(beta)+cotan(gamma) is at least sqrt(3) = cotan(pi/6), with equality if and only if alpha, beta and gamma are equal to pi/3. It is also straightforward to show that cotan(alpha)+cotan(beta)+cotan(gamma) is at least cotan((beta+gamma)/3)).

11022 Let T1 and T2 be triangles with circumradius R1 and R2 respectively, and inradius r1 and r2 respectively. Denote the sidelengths as a1, b1, c1 and a2, b2, c2.
Prove that 8*R1*R2 + 4*r1*r2 >= a1*a2 + b1*b2 + c1*c2 >= 36*r1*r2.
When does equality hold?

Solution:

We use, for each of both triangles: r=(2*O)/(a+b+c), R=(abc)/(4O), where O is the area of the triangle. Then we find that the inequalities that we have to prove can be written as follows:
a1*b1*c1*a2*b2*c2/(2*O1*O2) + 16*O1*O2/((a1+b1+c1)*(a2+b2+c2)) >= a1*a2 + b1*b2 + c1*c2 >=
144*O1*O2/((a1+b1+c1)*(a2+b2+c2)).
From O = a*b*sin(gamma)/2 = b*c*sin(alpha)/2 = a*c*sin(beta)/2, we deduce
O=(1/2)*(abc)2/3*(sin(alpha)*sin(beta)*sin(gamma))1/3 <= (1/4)*(abc)2/3*sqrt(3).
Now the second inequality follows from the arithmetic-geometric mean inequality, applied three times.
We have equality if and only if a1=b1=c1 and a2=b2=c2, so if both triangles are equilateral.
As to the first inequality: with a/sin(alpha) = b/sin(beta) = c/sin(gamma) = h, we get
O = (1/2)*h2*sin(alpha)*sin(beta)*sin(gamma).
Then, the first inequality can be written as follows:
2*h1*h2 + 4*h1*h2*sin(alpha1)*sin(alpha2)*sin(beta1)*sin(beta2)* sin(gamma1)*sin(gamma2)/((sin(alpha1)+sin(beta1)+sin(gamma1))* (sin(alpha2)+sin(beta2)+sin(gamma2)))
is at least
h1*h2*((sin(alpha1)*sin(alpha2)+sin(beta1)* sin(beta2)+sin(gamma1)*sin(gamma2)),
so also as follows:
2*(sin(alpha1)+sin(beta1)+sin(gamma1))* (sin(alpha2)+sin(beta2)+sin(gamma2)) + 4*sin(alpha1)*sin(alpha2)*sin(beta1)*sin(beta2)* sin(gamma1)*sin(gamma2)
is at least
(sin(alpha1)+sin(beta1)+sin(gamma1))* (sin(alpha2)+sin(beta2)+sin(gamma2))*((sin(alpha1)*sin(alpha2)+sin(beta1)* sin(beta2)+sin(gamma1)*sin(gamma2)).
It is straightforward to prove this last inequality, using gamma=pi-alpha-beta, or to check it with Maple. Again, we have equality if and only if both triangles are equilateral.

As to the heuristics of this problem, the inequalities could have been found as follows:
1) You seek inequalities with the circumradius of both circles in the greatest term, the sides "inner product" a1*a2+b1*b2+c1*c2 in the middle, and the inradius in the least term.
2) You guess that equality must be reached if the triangles are equilateral.
3) For equilateral triangles we find R1*R2=b1*b2/3, the "inner product" is 3*b1*b2, and r1*r2=b1*b2/12.
4) So you guess 9*R1*R2 >= a1*a2+b1*b2+c1*c2 >= 36*r1*r2.
5) You try to improve the inequalities, using that R is at least 2*r.