Solutions by H Reuvers

11663 The unit interval is broken at two randomly chosen points along its length. Show that the probability that the lengths of the resulting three intervals are the heights of a triangle is equal to 12√5 log((3+√5)/2)/25 - 4/5.

Solution:

Denote the randomly chosen numbers by u and v with 0 < u < v < 1, and suppose there exists a triangle with heights ha, hb, hc equal to u, v-u, 1-v respectively.
Let a,b,c be the lengths of the corresponding sides of the triangle.
Then triangle geometry learns that (ha, hb, hc) is a scalar multiple of (1, a/b, a/c), and hence (u, v-u, 1-v) is a scalar multiple of (1, a/b, a/c).
Now the only restriction on the existence of the triangle is that each of the lengths a, b, c of the triangle be smaller than the sum of the other two.
So we find that the probability that we have to calculate is equal to the area of the region inside {0 < u < v < 1} where (1-v)/u + (1-v)/(v-u) > 1 and u/(v-u) + u/(1-v) > 1 and (v-u)/u + (v-u)/(1-v) > 1, divided by the area of {0 < u < v < 1}.
With the help of a graph plotter we see that the outcome is 2*(∫01 (3v-2+√(5v2-8v+4))/2 - (2-v-√(5v2-8v+4))/2) dv - ∫4/51 (v/2+√(5v2/4-v))-(v/2-√(5v2/4-v)) dv).
So this should be equal to 12√5 log((3+√5)/2)/25 - 4/5.

11672 A random walk starts at the origin and moves up-right or down-right with equal probability. What is the expected value of the first time that the walk is k steps below its then-current all time high?

Partial solution:

I ran the following Pascal program. The output seems to be k*(k+1).

program p11672;
var k,x,max,t:integer; n,som:real;
begin
while true do
begin
som:=0; n:=0;
while (n<1000000) do
begin
n:=n+1;
x:=0; max:=0; t:=0;
repeat
t:=t+1;
if random>0.5 then x:=x+1 else x:=x-1;
if x>max then max:=x
until (x=(max-k));
som:=som+t;
writeln(som/n:13:4)
end
end
end.

11674 Let a be a negative real number and b a positive real number. Let S be the set of continuous functions f from [0,1] to [a,b] with ∫01 f(x) dx = 0.
Let g be a strictly increasing function from [0,1] to the set of all reals.
(a) Find the supremum of    ∫01 f(x).g(x) dx for f ∈ S.
(b) Prove that this supremum is not attained.

Solution:

Since g is strictly increasing, we find the supremum by letting f approach a function h (with ∫01 h(x) dx = 0) that is maximal for the greater values of x and minimal for the lesser.
So let h(x) = a for x smaller than b/(b-a) and h(x) = b for x greater than b/(b-a).
Then the supremum is ∫01 h(x).g(x) dx (, which can also be expressed in a,b and g).
This supremum is not attained because h is not continuous.

11682 Compute Σn=0n=∞ (-1)nk=1k=∞ (-1)k-1/(n+k))2.

Partial solution:

Using power series and their derivatives, we find that this sum equals Σn=0n=∞ (-1)n (ln(2) - Σk=1k=n (-1)k-1/k)2.
Then, using a Pascal computer program, we find that this second sum equals 0.4112335167...
Finally, using Google on the internet, we find that this outcome is also the outcome for π2/24 = (1/2)*Σn=1n=∞ (-1)n-1/n2.
But we still have to prove that the sum in the problem equals π2/24.

11683 Given a triangle ABC, let FC be the foot of the altitude from the incenter I to AB. Define FA and FB similarly.
Let CA be the circle with center A that passes through FB and FC , and define CB and CC similarly.
The Gergonne point G of a triangle is the point at which segments AFA , BFB and CFC meet.
Determine, up to similarity, all isosceles triangles such that G lies on one of the circles CA , CB or CC.

Solution:

Let triangle ABC be isosceles with top C.
Then C, I, G and FC are collinear, and clearly G can't lie on CA or CB.
If the angle at C is smaller than 60 degrees, then G lies between I and FC, so clearly G can't lie on CC.
If the angle at C is equal to 60 degrees, the G = I, so clearly G can't lie on CC, either.
So let's suppose henceforward that the angle at C is greater than 60 degrees.
Then, without loss of generality, we may give coordinates C(0,0), A(-cos(t),-sin(t)), B(cos(t),-sin(t)) with t between 0 and π/3.
After a bit of calculation, we find:

1) I( 0 , -sin(t) + tan(t/2) cos(t)), and
2) FA( 2s2 - 4s4 , -4s3 √(1-s2) with s = sin(t/2) between 0 and 1/2, and
3) G( 0 , (-8s3 √(1-s2)) / (1+2s2)).

Since G ∈ CC iff the distance between G and C is equal to the distance between C and FA, we find, after another bit of calculation,
G ∈ CC iff 20s4 - 12s2 + 1 = 0.
Since s is smaller than 1/2 this yields s = sin(t/2) = √(10)/10.
Hence sin(t) = 3/5 and B(4/5 , -3/5).
So the triangle is similar to the triangle with two sides of length 5 and one of length 8.

11684 For complex a en z, let φa(z) := (a-z)/(1-az) and ρ(a,z) := |(a-z)/(1-az)|, where a denotes the complex conjugate of a.
(a) Show that, whenever a and b are reals in (-1,1), max|z| at most 1a(z)-φb(z)| = 2*ρ(a,b), and max|z| at most 1a(z)+φb(z)| = 2.
(b) For complex α, β with |α| = |β| = 1, let m(z) := |α φa(z) - β φb(z)|.
Determine the maximum and minimum of m(z) over z with |z| = 1.

Partial solution:

(a) We already noticed that φa(z) = -z for a=0, and φa(z) = a for both a=1 and a=-1.
Furthermore, we think that |φa(z)| ≤ 1 for a ∈ (-1,1). Straightforward calculation shows this amounts to r2(1-a2) ≤ (1-a2) for r ∈ (0,1).
Now we investigate the first assertion in (a) with a Pascal program.
From the output we conclude that the assertion seems to be true, and (if a is smaller than b) the maximum attained for some z for which nearly holds Re(φa(z)) = -Re(φb(z)) = -ρ(a,b) and Im(φa(z)) = Im(φb(z)).
So let z = x + i y = r cos(s) + i r sin(s), for real x and y, s∈[0,2π) and r∈(0,1). Then |φa(z)-φb(z)| = |a-b| √(((1-r2)2+r4sin2(2s))/((1-2ar cos(s)+a2r2)(1-2br cos(s)+b2r2))).
From Im(φa(z)) = Im(φb(z)) we find x = (1+r2)(a+b)/(1+ab), and then using Re(φa(z)) = -Re(φb(z)) we may express x and y in a and b. However, this yields r2=1/(ab), which is impossible since r≤1.
Next, we try to find stationary points, but the calculations are too complicated.
We proceed to include in the output of the program the value of r for which the maximum is attained. This value is 1 for all pairs (a,b). So we have to find the maximum and the corresponding value of s of sin2(2s)/((1-2a cos(s)+a2)(1-2b cos(s)+b2)).
With u:=cos(s), this amounts to finding the zeroes of the quartic polynomial p(u) := 4ab u4 -3(a+b)(1+ab) u3 +2(1+a2)(1+b2) u2 + (a+b)(1+ab) u - (1+a2)(1+b2).
We could use Ferrari's formula to do this (see http://tieba.baidu.com/f?kz=531280692), but there must exist a quicker approach.

As for the second assertion in (a): slightly adapting the program, we see that for each (a,b) the maximum is 2, and this maximum attained for z=1 (φa(1) = φb(1) = -1) and for z=-1 (φa(-1) = φb(-1) = 1).
We can complete the proof with the help of the triangle inequality.

(b) Let α = cos(u) + i sin(u), β = cos(v) + i sin(v), a = a1 + i a2, b = b1 + i b2, where all four reals u,v,a1,a2 are given. Let z = cos(s) + i sin(s), where the real s is variable.
Let A1 := 2a1 + (a22 - a12)cos(s) - 2 a1a2 sin(s) - cos(s),
Let A2 := 2a2 + (a12 - a22)sin(s) - 2 a1a2 cos(s) - sin(s),
Let A := 1 + a12 + a22 - 2 a1cos(s) - 2 a2sin(s).
Let B1, B2, B be defined similarly with b1,b2 instead of a1,a2.
Then |α φa(z) - β φb(z)|2 = ((cos(u) A1B - sin(u) A2B - cos(v) A B1 + sin(v) A B2)2 + (cos(u) A2 B - sin(u) A1 B - cos(v) A B2 - sin(v) A B1)2)/(A2B2).
This function of s is maximal or minimal for the values of s for which the derivative is 0. However, the calculations seem too complicated.

11687 Let T be a solid torus in R3 with center at the origin, tube radius 1 and spine radius r with r greater than 1. Let P be a random nearby plane.
Find the conditional probability, given that P meets T, that the intersection is simply connected. For which value of r is this probability maximal?
(The plane is chosen by first picking a distance A from the origin uniformly between 0 and 1+r, and then picking a normal vector independently and uniformly on the unit sphere.)

Solution:

Consider the torus x2 + y2 = (r +/- √(1-z2))2 and the plane x cos(α) + z sin(α) = A with α between 0 and π/2.
They intersect iff either A is smaller than 1, or α is at most arccos((A-1)/r). (For this last value of α the plane is tangent to the circle (r+cos(t),0,sint(t)) in a point away from the origin.)
This accounts for the denominator of the fraction that represents the conditional probability:
01 π/2 dA + ∫11+r arccos((A-1)/r) dA = π/2 + r.

As to the numerator:
We first notice that the plane is tangent to the circle (r+cos(t),0,sint(t)) in a point close to the origin if A is at most r-1 and α = arccos((1+A)/r).
Furthermore, the plane is tangent to the circle (-r+cos(t),0,sint(t)) in a point close to the origin if A is at most 1 and α = arccos((1-A)/r).
Now the intersection is simply connected iff the plane does intersect the circle (r+cos(t),0,sint(t)) but doesn't intersect the circle (-r+cos(t),0,sint(t)).
We distinguish:

1) r ≥ 2
In this case r-1 is greater than 1 and the numerator is
01 arccos((1-A)/r) - arccos((1+A)/r) dA + ∫1r-1 arccos((A-1)/r) - arccos((1+A)/r) dA + ∫r-1r+1 arccos((A-1)/r) dA = ∫01 arccos((1-A)/r) dA - ∫0r-1 arccos((1+A)/r) dA + ∫1r+1 arccos((A-1)/r) dA

2) r ≤ 2
In this case r-1 is between 0 and 1 and the numerator is
0r-1 arccos((1-A)/r) - arccos((1+A)/r) dA + ∫r-11 arccos((1-A)/r) dA + ∫1r+1 arccos((A-1)/r) dA = ∫01 arccos((1-A)/r) dA - ∫0r-1 arccos((1+A)/r) dA + ∫1r+1 arccos((A-1)/r) dA

So in all cases the conditional probability is (∫01 arccos((1-A)/r) dA - ∫0r-1 arccos((1+A)/r) dA + ∫1r+1 arccos((A-1)/r) dA) / (π/2 + r) = (2r + 2arccos(1/r) - 2√(r2-1)) / (π/2 + r).

With a pascal program I found an approximation for the maximum value of the conditional probability: 0.8107767119 for r = 1.243761.
We also find this value by equating the derivative to zero, which yields sin((π/2)*√(1-(1/r)2)) = 1/r.
Can someone find the 'exact' value of r?

11693 Let T be an equilateral triangle inscribed in the d-dimensional unit cube with d ≥ 2.
As a function of d, what is the maximum possible side length of T?

Solution:

For one of the vertices of the triangle we take the origin A = (0,0,0,...,0).
For the other two points B and C, we try first (x,1,1,...,1) and (1,x,1,...,1) (one x and d-1 1). We choose x ∈ [0,1] so that AB = AC = BC.
If, for larger d, this isn't possible, we try (x,1,x,1,1,...,1) and (1,x,1,x,1,...,1) (two x and d-2 1).
If, for d that's larger yet, this isn't possible, either, we try (x,1,x,1,x,1,1,...,1) and (1,x,1,x,1,x,1,...,1) (three x and d-3 1).
Etc.
We get the following table of values of d,x,a (where a is the side length asked for):

d = 2, x = 2 - √3, a = √6 - √2.
d = 3, x = 0, a = √2.
d = 4, x = 2 - √3, a = 2√3 - 2.
d = 5, x = 2 - (√14)/2, a = √14 - 2.
d = 6, x = 0, a = 2.

We see that, whenever x=0, for the next d we need to try one more x.

Using k instances of x, we get x = 2 - √((d+k)/k), a = √(2d+2k) - √(2k).
So this outcome holds for d that's greater than 3(k-1) and at most 3k.

11695 Provide an algorithm that takes as input a positive integer n and a nonzero constant k and returns polynomials F and G in variables u and v such that when xn is substituted for u and x+k/x for v, then F/G simplifies to x.

Solution:

If n=1 we try for F and G polynomials of degree 1 and find (for arbitrary real numbers λ and μ): F = λk+μu, G = μ-λu+λv.
If n=2 we try for F and G again polynomials of degree 1 and find: F = λk+λu, G = λv.
If n=3 we try for F and G polynomials of degree 2 and find: F = λk2+μu+μkv+λuv, G = -μk+λu+λkv+μv2.
If n=4 we try for F and G again polynomials of degree 2 and find: F = λk2v+λuv, G = -λk2+λu+λkv2.

Next we investigate the general case. We claim that for any n we can find a linear space of solutions, trying polynomials of some degree m that is large enough..

Suppose F and G are polynomials of degree m with coefficients aij and bij respectively.
Then xmF(xn,x+k/x) = Σ(p:=0 to nm+m) cpxp with cp := Σ(i:=0 to m, j:=0 to m-i, h:=0 to j, ni+m+j-2h=p) aij.(j over h).kh.
And xmG(xn,x+k/x) = Σ(p:=0 to nm+m) dpxp with dp := Σ(i:=0 to m, j:=0 to m-i, h:=0 to j, ni+m+j-2h=p) bij.(j over h).kh.
Now we demand c0 = dnm+m = 0 and cp = dp-1 for p:=1 to nm+m, and not all cp equal to zero.
For m=3 and n=5 this algorithm yields solutions F = -μk3-λku+λk3v+μuv+μk2v2+λuv2, G = -λk3+μu-2μk2v+λuv+λk2v2+μkv3.
This confirms our claim, but how can we prove it?

In the general case, we have nm+m+2 linear equations with (m+1)(m+2) variables aij and bij.
Clearly, given any n, we can make m so large that the rank of the matrix of the linear equations is less than (m+1)(m+2). Then the dimension of the space of solutions is at least 1.

11700 Let n be an integer greater than 1. Let a,b,c be complex numbers with a+b+c = an+bn+cn = 0. Prove that the absolute values of a,b,c cannot be distinct.

Partial solution:

Without loss of generality we assume |a| ≤ |b| ≤ |c|.
Furthermore we exclude a=0, since in that case |b| = |c|.
Let B:=b/a and C:=c/a.
(Note that B=(-1+i√3)/2, C=(-1-√3)/2 satisfy the equations for n=2.)

Suppose B+C = Bn+Cn = -1 and 1 < |B| < |C|.
We find (-1-C)n = -1 - Cn and the real part of C is not -1/2.
Now we may derive a contradiction as follows:

Consider the following actions we may exert on complex numbers, viewed upon as vectors in the complex plane:
1) multiplying with -1 ('point reflection');
3) multiplying the argument (modulo 2π) by factor n, while maintaining the modus ('rotation');
4) taking the n-th power of the modus, while maintaining the argument ('point multiplication').
It seems impossible that exerting the actions in the order 1) 2) 3) 4), starting from C, should yield the same final result as exerting them in the order 3) 4) 1) 2).

postscript: O.P.Lossers discovered there does exist a C (with |C| > 1) that satisfies (-1-C)6 = -1 - C6 and whose real part is not -1/2. So the assertion we have to prove is false.
Indeed, substituting C = -1/2 + iz yields for n=6: 64z6 + 240z4 - 60z2 + 33 = 0. There does exist a z that satisfies this equation and for which z2 is about -4. Then the real part of C is about 3/2 or -5/2.

11703 For λ > 0, let Γ(λ) = {z ≥ λ √(x2+y2)} and let C(λ) be the half-cone boundary of Γ(λ).
Prove that every point in the interior of Γ(λ) is the focus of at least one ellipse in C(λ) and find the largest μ such that every ellipse in C(λ) has at least one focus in Γ(μ).

Solution:

Any ellipse in C(λ) with a focus in {y=0} is the intersection of C(λ) with a plane P(a,b)={z=ax+b} with b > 0 and -λ < a < λ.
Let u = x √(1+a2) and v = u - ab √(1+a2)/(λ2-a2).
Then the intersection has an equation of the form v2/A2 + y2/B2 = 1 and focuses (±v,0) with v = ab √(λ2+1)/(λ2-a2), which correspond to the points (x,0,ax+b) with x = (ab/(λ2-a2))(1 ± √(1+λ2)/√(1+a2)).
When we vary a and b through b > 0 and -λ < a < λ}, these focuses vary through {y=0, z > λ|x| > 0}:
(x,0,z) is focus of the ellipse C(λ) ∩ P(a,b) with a = (x√((z22x2)(1+λ2))-xz) / (z2-x2(1+λ2)), b = (z(z22x2) - x2√((z22x2)(1+λ2))) / (z2-x2(1+λ2)).
But the first part of the assertion was already clear enough from the geometry.

As for the second part, μ is the infimum over {-λ < a < λ, b > 0} of |(ax+b)/x| with x = (ab/(λ2-a2))(1 - √(1+λ2)/√(1+a2)), that is:
μ is the infimum over {-λ < a < λ} of |(λ2√(1+a2) - a2√(1+λ2))/(a√(1+a2)-a√(1+λ2))|.
This infimum is λ + 2/λ for a = ± λ. (There is no minimum over {-λ < a < λ}.)

11704 Let S2n denote the symmetric group of all permutations of {1,2,...,2n}, and let T2n denote the set of all fixed-points-free involutions in S2n.
Choose u and v belonging to T2n at random and independently. What is the probability that 1 and 2 will be in the same cycle of the permutation uv (written as a product of disjoint cycles)?

Solution:

The number of elements of T2n is (2n-1)(2n-3)..3.1, so the required probability is the number of possible pairs (u,v) divided by (2n-1)2(2n-3)2..3212.
Straightforward checking learns: for n=1 we get probability 0, for n=2 we get probability 2/9 and for n=3 we get probability 60/225 = 4/15.

Now we look for a general procedure.
The product uv maps 1 to 2 iff v contains cycle (1a) and u cycle (a2) where a does not belong to {1,2}. The number of possible pairs (u,v) is (2n-2)(2n-3)2(2n-5)2....
The product uv maps 1 to b and b to 2 where b does not belong to {1,2} iff v contains cycle (1a) and u cycle (ab) and v cycle (bc) and u cycle (c2) where a and b and c are distinct and do not belong to {1,2}. The number of possible pairs (u,v) is (2n-2)(2n-3)(2n-4)(2n-5)2(2n-7)2....
The product uv maps 1 to b and b to d and d to 2 where b and d are distinct and do not belong to {1,2} iff v contains cycle (1a) and u cycle (ab) and v cycle (bc) and u cycle (cd) and v cycle (de) and u cycle (e2) where a and b and c and d and e are distinct and do not belong to {1,2}. The number of possible pairs (u,v) is (2n-2)(2n-3)(2n-4)(2n-5)(2n-6)(2n-7)2(2n-9)2....
Etc.

So the probability asked for is (2n-2)/(2n-1)2 + (2n-2)(2n-3)(2n-4)/((2n-1)2(2n-3)2) + (2n-2)(2n-3)(2n-4)(2n-5)(2n-6)/((2n-1)2(2n-3)2(2n-5)2) + ... .
(The number of nonzero terms is finite for every n.)

11706 Let ABC and DEF be triangles in a plane.
(a) Provide a compass and straightedge construction, which may use ABC and DEF, of a trangle A'B'C' that is similar to ABC and circumscribes DEF.
(b) Among all triangles A'B'C' of the sort described in part (a), determine which one has the greatest area and which one has the greatest perimeter.

Partial solution:

(a) After some permutation of the names of the vertices, we may assume α, δ, γ are acute.
We can take some ρ with max(α,δ) < ρ < min(π,α+β+δ).
Construct a line l through D that makes angle π-ρ with base DE and a line m through E that makes angle ρ-α with DE, that is: if A':=l.m, beneath triangle DEF, then ∠ A'DE = π-ρ and ∠ A'ED = ρ-α, and hence the angle ∠ EA'D at A' is α.
Next, construct a line n through top F that makes angle α+β+δ-ρ = δ+π-ρ-γ with DF, that is: if C':=l.n, outside triangle DEF, then ∠ DFC' = δ+π-ρ-γ and ∠ FDC' = ρ-δ, and hence the angle ∠ FC'D at C' is equal to γ.
Then at B':=m.n the angle ∠ EB'D is equal to β.

(b) Let h := a'/sin(α) = b'/sin(β) = c'/sin(γ). Then the area of A'B'C' is h2sin(α)sin(β)sin(γ) and its perimeter is h(sin(α)+sin(β)+sin(γ)), so they both are maximal when h is maximal.
Let D=(0,0), E=(1,0) and F=r(cos(δ),sin(δ)).
Let A'B'C' be the triangle formed by the lines through D,E,F with slopes tan(ρ),tan(σ),tan(τ) respectively, where, after some suitable permutation of the names of the vertices of ABC and A'B'C', D is on A'C', E on A'B', F on B'C', and
max(α,δ) < ρ < min(π,α+β+δ), σ = ρ-α, τ = ρ-α-β+π.
Then B'= r(cos(δ),sin(δ)) + (r sin(δ-ρ)/sin(α+β))(cos(ρ-α-β),sin(ρ-α-β)), C'= r(cos(δ),sin(δ)) + ((r sin(δ+α-ρ) + sin(ρ-α))/sin(β))(cos(ρ-α-β),sin(ρ-α-β)), so
h = a'/sin(α), where a'= B'C'= |(r sin(δ-ρ)/sin(α+β)) - (r sin(δ+α-ρ) + sin(ρ-α))/sin(β))|.
To find the value of ρ for which h is maximal, we demand that the derivative be equal to zero and find

ρ must satisfy equation (*): cos(ρ-α)sin(α+β) + r sin(β)cos(δ-ρ) - r cos(δ+α-ρ)sin(α+β) = 0.

We see that the first term vanishes for ρ=α+π/2, and the sum of the latter two for ρ=δ+α+β-π/2, so if β+δ=π then ρ=α+π/2 is a solution.
Moreover, if α=γ and r=1 and δ=π/3, then the solution turns out to be ρ=2π/3, in accordance with what we should expect.
But we don't see an easy general solution ρ for equation (*).

11707 For N at least 1, consider the following random walk on the N+1-cycle with vertices labeled 0,1,...,N.
The walk begins at vertex 0 and continues until every vertex has been visited and the walk returns to vertex 0.
Prove that the expected number of visits to any vertex other than 0 is (2N+1)/3.

Solution:

I ran a pascal program. It seems to confirm the assertion (even extended to including vertex 0, if we don't count the starting position as a visit).

It is straightforward to show that the expected number of visits to any vertex is 1 if N=1 and 5/3 if N=2.
In general, we have to prove we expect (N+1)(2N+1)/3 steps to go around once, starting from 0. (Clearly we expect an equal number of visits to each position.)
To prove this with induction, we have to explain why we expect 5/3 + 4N/3 more steps if we insert 1 more position in the cycle, at place N+1.
This is because, going back to each of the N positions at places 1,2,..,N, we step on the average 4/3 times on it, where 4/3 = 1/(1-(1/4)) = 1 + 1/4 + (1/4)2 + (1/4)3 + ... and 1/4 = (1/2)*(1/2) (to and fro).

11712 In the game of Bulgarian solitaire, we begin with a pile of n identical coins. A move takes one coin from each existing pile to form a new pile with the coins taken.
How many moves are needed to reach a cycle? (That is: to reach a composition of piles that will eventually come back.)

Solution:

We have n --> n-1 1 --> n-2 2 --> n-3 1 2 --> n-4 1 3 --> n-5 2 3 --> n-6 1 2 3 --> n-7 1 2 4 --> n-8 1 3 4 --> n-9 2 3 4 --> n-10 1 2 3 4 --> n-11 1 2 3 5 --> n-12 1 2 4 5 --> n-13 1 3 4 5 --> n-14 2 3 4 5 --> n-15 1 2 3 4 5 --> n-16 1 2 3 4 6 etc,
and we see:

1) After n steps we reach the cycle for the second time.
2) If n = 1 + 2 + 3 + .. + m = m(m+1)/2 then we reach the cycle for the first time after n-m steps.
3) If n = 1 + 2 + .. + (m-k) + (m-(k-2)) + m + (m+1) = m(m+1)/2 + k with 1 ≤ k ≤ m+1, then we reach the cycle for the first time after n-(m+1) steps.

The answer is in observation 3).

11714 Let ABCD be a cyclic quadrilateral (the four vertices lie on a circle). Let e=|AC| and f=|BD|. Let ra be the inradius of BCD, and define rb , rc and rd similarly. Prove that e ra rc = f rb rd .

Solution:

Give coordinates A(cos(α),sin(α)), B(cos(β),sin(β)), C(cos(γ),sin(γ)), D(cos(δ),sin(δ)) (with α=0).
Use that the inradius of a triangle RST is equal to : r sin(σ/2) sin(τ/2)/sin((σ + τ)/2). We find:

ra = f sin((γ-β)/4) sin((δ-γ)/4)/sin((δ-β)/4),
rb = e sin((α-δ+2π)/4) sin((δ-γ)/4)/sin((α-γ+2π)/4),
rc = f sin((α-δ+2π)/4) sin((β-α)/4)/sin((β-δ+2π)/4),
rd = e sin((β-α)/4) sin((γ-β)/4)/sin((γ-α)/4).
Furthermore, straightforwardly from the definitions of e and f, we find : e = 2 sin((γ-α)/2), f = 2 sin((δ-β)/2).

Then we find by straightforward verification that e ra rc = f rb rd.

11717 Given a circle c and line segment AB tangent to c at a point E that lies strictly between A and B, provide a compass and straightedge construction of the circle through A and B to which c is internally tangent.

Solution:

Without loss of generality we may assume the points have coordinates E(0,0), A(-a,0), B(b,0) for some positive a and b, and the circle c has equation x2 + (y+1)2 = 1.
We have to construct a circle c' with equation (x-s)2 + (y-t)2 = r2 that goes through A and B and is tangent to c.
Since c' goes through A and B we find s = (b-a)/2 and r2 = ((a+b)/2)2 + t2.
Since c and c' are tangent, the intersection points must coincide. We find a quadratic equation whose discriminant must be 0.
This leads to 16(1+t)3ab + 4(1+t)2((a-b)2(1-ab)2-a2b2) + 4(1+t)ab(a-b)2 -a2b2(a-b)2 = 0.
Next we'll try 4(1+t) = ab + λ(a-b) and find λ = -(a-b)/ab.

postscript: O.P.Lossers makes use of the point P where the line AEB meets the line that at T:=c.c' is tangent to both c and c'. It is well known that PA.PB = PT2 = PE2.