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 h_{a}, h_{b}, h_{c} 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 (h_{a}, h_{b}, h_{c}) 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*(∫_{0}^{1} (3v-2+√(5v^{2}-8v+4))/2 - (2-v-√(5v^{2}-8v+4))/2) dv -
∫_{4/5}^{1} (v/2+√(5v^{2}/4-v))-(v/2-√(5v^{2}/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

writeln('k?'); readln(k);

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 ∫ _{0}^{1} 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 ∫_{0}^{1} 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 ∫_{0}^{1} 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 ∫_{0}^{1} 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 * Σ_{n=0}^{n=∞} (-1)^{n} (Σ_{k=1}^{k=∞} (-1)^{k-1}/(n+k))^{2}.

__Partial solution:__

Using power series and their derivatives, we find that this sum equals
Σ_{n=0}^{n=∞} (-1)^{n} (ln(2) - Σ_{k=1}^{k=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 π^{2}/24 = (1/2)*Σ_{n=1}^{n=∞} (-1)^{n-1}/n^{2}.

But we still have to prove that the sum in the problem equals π^{2}/24.

11683 *Given a triangle ABC, let F _{C} be the foot of the altitude from the incenter I to AB. Define F_{A} and F_{B} similarly.
*

Let C_{A} be the circle with center A that passes through F_{B} and F_{C} , and define C_{B} and C_{C} similarly.

The Gergonne point G of a triangle is the point at which segments AF_{A} , BF_{B} and CF_{C} meet.

Determine, up to similarity, all isosceles triangles such that G lies on one of the circles C_{A} , C_{B} or C_{C}.

__Solution:__

Let triangle ABC be isosceles with top C.

Then C, I, G and F_{C} are collinear, and clearly G can't lie on C_{A} or C_{B}.

If the angle at C is smaller than 60 degrees, then G lies between I and F_{C}, so clearly G can't lie on C_{C}.

If the angle at C is equal to 60 degrees, the G = I, so clearly G can't lie on C_{C}, 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) F_{A}( 2s^{2} - 4s^{4} , -4s^{3} √(1-s^{2}) with s = sin(t/2) between 0 and 1/2, and

3) G( 0 , (-8s^{3} √(1-s^{2})) / (1+2s^{2})).

Since G ∈ C_{C} iff the distance between G and C is equal to the distance between C and F_{A}, we find, after another bit of calculation,

G ∈ C_{C} iff 20s^{4} - 12s^{2} + 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 φ _{a}(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 1} |φ_{a}(z)-φ_{b}(z)| = 2*ρ(a,b), and
max_{|z| at most 1} |φ_{a}(z)+φ_{b}(z)| = 2.
(b) For complex α, β with |α| = |β| = 1, let m(z) := |α φ_{a}(z) - β φ_{b}(z)|.
Determine the maximum and minimum of m(z) over z with |z| = 1.*

__Partial solution:__

(a) We already noticed that φ_{a}(z) = -z for a=0, and φ_{a}(z) = a for both a=1 and a=-1.

Furthermore, we think that |φ_{a}(z)| ≤ 1 for a ∈ (-1,1). Straightforward calculation shows this amounts to r^{2}(1-a^{2}) ≤ (1-a^{2}) 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(φ_{a}(z)) = -Re(φ_{b}(z)) = -ρ(a,b) and Im(φ_{a}(z)) = Im(φ_{b}(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 |φ_{a}(z)-φ_{b}(z)| =
|a-b| √(((1-r^{2})^{2}+r^{4}sin^{2}(2s))/((1-2ar cos(s)+a^{2}r^{2})(1-2br cos(s)+b^{2}r^{2}))).

From Im(φ_{a}(z)) = Im(φ_{b}(z)) we find x = (1+r^{2})(a+b)/(1+ab), and then using Re(φ_{a}(z)) = -Re(φ_{b}(z))
we may express x and y in a and b. However, this yields r^{2}=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
sin^{2}(2s)/((1-2a cos(s)+a^{2})(1-2b cos(s)+b^{2})).

With u:=cos(s), this amounts to finding the zeroes of the quartic polynomial p(u) :=
4ab u^{4} -3(a+b)(1+ab) u^{3} +2(1+a^{2})(1+b^{2}) u^{2} + (a+b)(1+ab) u - (1+a^{2})(1+b^{2}).

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 (φ_{a}(1) = φ_{b}(1) = -1) and for z=-1 (φ_{a}(-1) = φ_{b}(-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 = a_{1} + i a_{2}, b = b_{1} + i b_{2}, where all four reals u,v,a_{1},a_{2}
are given. Let z = cos(s) + i sin(s), where the real s is variable.

Let A_{1} := 2a_{1} + (a_{2}^{2} - a_{1}^{2})cos(s) - 2 a_{1}a_{2} sin(s) - cos(s),

Let A_{2} := 2a_{2} + (a_{1}^{2} - a_{2}^{2})sin(s) - 2 a_{1}a_{2} cos(s) - sin(s),

Let A := 1 + a_{1}^{2} + a_{2}^{2} - 2 a_{1}cos(s) - 2 a_{2}sin(s).

Let B_{1}, B_{2}, B be defined similarly with b_{1},b_{2} instead of a_{1},a_{2}.

Then |α φ_{a}(z) - β φ_{b}(z)|^{2} = ((cos(u) A_{1}B - sin(u) A_{2}B - cos(v) A B_{1} + sin(v) A B_{2})^{2} +
(cos(u) A_{2} B - sin(u) A_{1} B - cos(v) A B_{2} - sin(v) A B_{1})^{2})/(A^{2}B^{2}).

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 R ^{3} 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 x^{2} + y^{2} = (r +/- √(1-z^{2}))^{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:

∫_{0}^{1} π/2 dA + ∫_{1}^{1+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

∫_{0}^{1} arccos((1-A)/r) - arccos((1+A)/r) dA + ∫_{1}^{r-1} arccos((A-1)/r) - arccos((1+A)/r) dA + ∫_{r-1}^{r+1} arccos((A-1)/r) dA =
∫_{0}^{1} arccos((1-A)/r) dA - ∫_{0}^{r-1} arccos((1+A)/r) dA + ∫_{1}^{r+1} arccos((A-1)/r) dA

2) r ≤ 2

In this case r-1 is between 0 and 1 and the numerator is

∫_{0}^{r-1} arccos((1-A)/r) - arccos((1+A)/r) dA + ∫_{r-1}^{1} arccos((1-A)/r) dA + ∫_{1}^{r+1} arccos((A-1)/r) dA =
∫_{0}^{1} arccos((1-A)/r) dA - ∫_{0}^{r-1} arccos((1+A)/r) dA + ∫_{1}^{r+1} arccos((A-1)/r) dA

So in all cases the conditional probability is
(∫_{0}^{1} arccos((1-A)/r) dA - ∫_{0}^{r-1} arccos((1+A)/r) dA + ∫_{1}^{r+1} arccos((A-1)/r) dA) / (π/2 + r) =
(2r + 2arccos(1/r) - 2√(r^{2}-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 x ^{n} 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 = λk^{2}+μu+μkv+λuv, G = -μk+λu+λkv+μv^{2}.

If n=4 we try for F and G again polynomials of degree 2 and find: F = λk^{2}v+λuv, G = -λk^{2}+λu+λkv^{2}.

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 a_{ij} and b_{ij} respectively.

Then x^{m}F(x^{n},x+k/x) = Σ(p:=0 to nm+m) c_{p}x^{p} with c_{p} :=
Σ(i:=0 to m, j:=0 to m-i, h:=0 to j, ni+m+j-2h=p) a_{ij}.(j over h).k^{h}.

And x^{m}G(x^{n},x+k/x) = Σ(p:=0 to nm+m) d_{p}x^{p} with d_{p} :=
Σ(i:=0 to m, j:=0 to m-i, h:=0 to j, ni+m+j-2h=p) b_{ij}.(j over h).k^{h}.

Now we demand c_{0} = d_{nm+m} = 0 and c_{p} = d_{p-1} for p:=1 to nm+m, and not all c_{p} equal to zero.

For m=3 and n=5 this algorithm yields solutions F = -μk^{3}-λku+λk^{3}v+μuv+μk^{2}v^{2}+λuv^{2},
G = -λk^{3}+μu-2μk^{2}v+λuv+λk^{2}v^{2}+μkv^{3}.

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 a_{ij} and b_{ij}.

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 = a ^{n}+b^{n}+c^{n} = 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 = B^{n}+C^{n} = -1 and 1 < |B| < |C|.

We find (-1-C)^{n} = -1 - C^{n} 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');

2) adding -1 ('translation');

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 - C^{6} 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: 64z^{6} + 240z^{4} - 60z^{2} + 33 = 0. There does exist a z that satisfies this equation and for which z^{2} 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 √(x ^{2}+y^{2})} 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+a^{2}) and v = u - ab √(1+a^{2})/(λ^{2}-a^{2}).

Then the intersection has an equation of the form v^{2}/A^{2} + y^{2}/B^{2} = 1 and focuses (±v,0)
with v = ab √(λ^{2}+1)/(λ^{2}-a^{2}), which correspond to the points (x,0,ax+b) with
x = (ab/(λ^{2}-a^{2}))(1 ± √(1+λ^{2})/√(1+a^{2})).

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√((z^{2}-λ^{2}x^{2})(1+λ^{2}))-xz) / (z^{2}-x^{2}(1+λ^{2})),
b = (z(z^{2}-λ^{2}x^{2}) - x^{2}√((z^{2}-λ^{2}x^{2})(1+λ^{2})))
/ (z^{2}-x^{2}(1+λ^{2})).

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/(λ^{2}-a^{2}))(1 - √(1+λ^{2})/√(1+a^{2})), that is:

μ is the infimum over {-λ < a < λ} of
|(λ^{2}√(1+a^{2}) - a^{2}√(1+λ^{2}))/(a√(1+a^{2})-a√(1+λ^{2}))|.

This infimum is λ + 2/λ for a = ± λ. (There is no minimum over {-λ < a < λ}.)

11704 *Let S _{2n} denote the symmetric group of all permutations of {1,2,...,2n}, and let T_{2n} denote the set of all fixed-points-free involutions in S_{2n}.
*

Choose u and v belonging to T_{2n} 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:__

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 h^{2}sin(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 r _{a} be the inradius of BCD, and define r_{b} , r_{c} and r_{d}
similarly. Prove that e r_{a} r_{c} = f r_{b} r_{d} .*

__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:

r_{a} = f sin((γ-β)/4) sin((δ-γ)/4)/sin((δ-β)/4),

r_{b} = e sin((α-δ+2π)/4) sin((δ-γ)/4)/sin((α-γ+2π)/4),

r_{c} = f sin((α-δ+2π)/4) sin((β-α)/4)/sin((β-δ+2π)/4),

r_{d} = 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 r_{a} r_{c} = f r_{b} r_{d}.

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 x^{2} + (y+1)^{2} = 1.

We have to construct a circle c' with equation (x-s)^{2} + (y-t)^{2} = r^{2} 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 r^{2} = ((a+b)/2)^{2} + t^{2}.

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)^{3}ab + 4(1+t)^{2}((a-b)^{2}(1-ab)^{2}-a^{2}b^{2}) + 4(1+t)ab(a-b)^{2} -a^{2}b^{2}(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 = PT^{2} = PE^{2}.

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