Solutions by H Reuvers, part 3

11023 Find all pairs (x,y) of integers such that
x2 + 3xy + 4006(x+y) + 20032 = 0.

Solution:

We factorize and get x(x+3y) = -2003(2x+2y+2003). Notice that 2003 is prime.
We have two cases:
1) x=k*2003, and 2) x+3y=k*2003 (for integer k).

case 1)
Then k(k*2003+3y) = -(2k*2003+2y+2003), so y(3k+2) = -(k+1)22003.
Now we have two subcases:
1a) y=n*2003, and 1b) 3k+2=n*2003 (for integer n)

case 1a)
Then n(3k+2) = -(k+1)2.
We get k2 + k(2+3n) + 1+2n = 0. The discriminant must be an integer square, say
9n2 + 4n = u2, and (3n+u)(3n-u) = -4n.
So abs(3n+u)*abs(3n-u) = 4*abs(n).
One of the integer factors at the left hand side must be at least 3*abs(n), so n=0 and hence k=-1.
So we get in this subcase only one solution: (x,y)=(-2003,0).

case 1b)
Now 3k+2=n*2003 and yn = -(k+1)2.
We deduce: n=1+3p for integer p, so k = 667+2003*p and y(1+3p) = -(668+2003*p)2.
Now 668+2003*p = 667(1+3p) + (1+2p), so (1+2p)2 must be divisible by 1+3p, say (1+2p)2 = m(1+3p) for integer m (m not 0).
We get 4p2 + p(4-3m) + 1-m = 0. The discriminant must be an integer square, say
9m2 - 8m = v2, and (3m-v)(3m+v) = 8m.
So abs(3m-v)*abs(3m+v) = 8*abs(m).
Since m is not 0, one of the integer factors at the left hand side must be at least 3*abs(m) and the other at least 1, so we have
(3m-v=+1 and 3m+v=+8m) or (3m-v=+2 and 3m+v=+4m) and 6 more possibilities, depending on the choice of the signs.
There is only one integer solution: m=1 and hence p=0, n=1, k=667.
So we get in this subcase only one solution: (x,y)=(667*2003,-6682).

case 2)
In this case we have (k*2003-3y)*k = -(2k*2003-4y+2003), so y(3k+4) = (k+1)22003.
Again, we have two subcases:
2a) y=n*2003, and 2b) 3k+4=n*2003 (for integer n).

case 2a)
Then n(3k+4) = (k+1)2.
We get k2 + k(2-3n) + 1-4n = 0, with discriminant 9n2+4n, the same as in case 1a). As above in case 1a) we find n=0.
Then k=-1, y=0, x=-2003. So we have the same solution as in case 1a).

case 2b)
Now 3k+4=n*2003 and yn = (k+1)2.
We deduce: n=2+3p for integer p, so k = 1334+2003*p and y(2+3p) = (1335+2003*p)2.
Now 1335+2003*p = 667*(2+3p) + (1+2p), so (1+2p)2 must be divisible by 2+3p, say (1+2p)2 = m(2+3p) for integer m (m not 0).
We get 4p2 + p(4-3m) + 1-2m = 0. The discriminant must be an integer square, say 9m2+8m = v2, and (3m-v)(3m+v) =-8m.
So abs(3m-v)*abs(3m+v) = 8*abs(m).
Since m is not 0, one of the integer factors at the left hand side must be at least 3*abs(m) and the other at least 1, so we have
(3m-v=+1 and 3m+v=-8m) or (3m-v=+2 and 3m+v=-4m) and 6 more possibilities, depending on the choice of the signs.
There is only one integer solution: m=-1, and hence p=-1, n=-1, k=-669.
So in this subcase we get as third and last solution: (x,y)=(-1335,-6682).

11027 Call a triangle T good if it has area one and can be inscribed in a circle of radius one.
Let d(P,T) denote the sum of the distances from a point P to the vertices of T. Let d(T) be the minimum of d(P,T) over all P that are inside or on the boundary of T.
Let D(T), s(T) and S(T) be the analogous minimums for the sum of the squared distances to the vertices, the sum of the distances to the sides, and the sum of the squared distances to the sides.
(a) Prove that every good triangle minimizing d(T), D(T), or s(T), is a right isosceles triangle.
(b) Prove that every good triangle minimizing S(T) is an isosceles triangle, but not a right triangle.

Solution:

I myself proposed this problem to the problem section of the American Mathematical Monthly.
My solution can be found here.

11030 Prove that for each d<-1 there are exactly two solutions to the functional equation
f(x+y) - f(x)f(y) = d*sin(x)*sin(y).

Solution:

Suppose f(pi/2)=a and f(pi)=b.
Then for all x we have f(x+pi/2) - a*f(x) = d*sin(x), f(x+pi) - b*f(x) = 0.
With x=pi/2 in the first identity we get b-a2=d; with x=0: a - a*f(0) = 0; with x=0 in the second we get b - b*f(0) = 0.
Since d<-1, a and b cannot be both 0, so f(0)=1.
From the original equation, substituting x=pi/2 and y=x+pi/2, we get
f(x+pi) - a* f(x+pi/2) = d*sin(x+pi/2) = d*cos(x).
So, using the two identities in the first lines of the proof, we find
b*f(x) - a*(a*f(x) + d*sin(x)) = d*cos(x).
Using b-a2=d and d<>0, we then get
f(x) = cos(x) + a*sin(x).
From this last identity we deduce f(pi) = b = -1, so d=-1-a2. Hence for each d<-1 there are two values of a.
Indeed, substitution of f(x) = cos(x) + a*sin(x) in the original equation, with d=-1-a2, learns that these functions are solutions of the equation. (Use the goniometric formulas for sin(x+y), cos(x+y).)

Nota bene:
Professor van Lint gave already a presentation of a solution. But as soon as he proposed to set f(pi/2)=a and f(pi)=b, I didn't need more hints and didn't listen any longer.

11032 For each nonnegative integer N, let P be the tridiagonal square matrix of dimension N+3 with entries Pi,j (for i,j in {0,1,..,N+2}) defined by
P0,1 = PN+2,N+1 = 1, Pi,i+1 = i/(N+2), Pi,i-1 = (N+2-i)/(N+2) for i varying from 1 to N+1, and Pi,j = 0 otherwise.
This is a row-stochastic matrix.
The stationary distribution x0, x1, ... , xN+2 is the solution of the system
SUM(i from 0 to N+2 : xiPi,j) = xj (for j in {0,1,..,N+2})
with SUM(j from 0 to N+2 : xj) = 1.
In terms of N, find x and the positions in x of its maximal and minimal entries.

Solution:

The system of linear equations can be reduced to the following system of N+2 equations in N+3 variables:
(read first equations number 0 and number N+2, then 1 and N+1, etc, until you have read all N+2 equations)
(N+2)*x0 = (N+1)*x1
x1 = N*x2
2*x2 = (N-1)*x3
3*x3 = (N-2)*x4
4*x4 = (N-3)*x5
....
(N-2)*xN-2 = 3*xN-1
(N-1)*xN-1 = 2*xN
N*xN = xN+1
(N+1)*xN+1 = (N+2)*xN+2.
So we have for some positive m (if N is large enough):
xN+2 = (N+1)*m = x0
xN+1 = (N+2)*m = x1
xN = ((N+2)/N)*m = x2
xN-1 = (2/(N-1))*((N+2)/N)*m = x3
xN-2 = (3*2)/((N-2)*(N-1))*((N+2)/N)*m = x4
xN-3 = (4*3*2)/((N-3)*(N-2)*(N-1))*((N+2)/N)*m = x5
.... (etc)
The entries xj are maximal for j=1,N+1.
If N is at least 2, they are minimal "in the middle" (j=(N+3)/2 or j=(N+2)/2,(N+4)/2).
(For N=0,1, they are minimal for j=0,N+2.)
Since the sum of xj is 1, m is is equal to:
1 divided by
2*(N+1) + (N+2)*SUM(k from 0 to N: 1/(N over k)).
I don't know whether this expression for m can be given in closed form.

11034 In this problem, integers in a field are to be interpreted as reduced modulo the characteristic of the field.
The adjoint of a matrix A is that matrix A' so that AA' = (det A)I.
Let B = (1 1 3 / -1 3 -3 / 2 2 3).
Find the smallest field K such that there is a matrix A over K with adj(A) = B.
Find one such A. How many are there?
Same questions with C = (1 9 3 / 1 -8 3 / 2 1 6) instead of B.

Solution :

I first assume that the proposer meant to imply that det(A) is not zero. Otherwise there is no unique non-zero adjoint A'.
Hence we must have det(B) = 1 and A/det(A) = B-1.
We try prime characteristics k = 2,3,5,7,11,13,... The first k for which det(B)=1 is k=13.
So K = Z13.
We can take A = B-1 with det(A)=1, so A = (2 3 1 / 10 10 0 / 5 0 4).
But we can also multiply A with a non-zero constant . So there are twelve such A.
If we replace B by C, then there is no such field K, because det(C) = 0 in every field.

If the proposer didn't mean to imply that det(A) is not zero, then K = Z2 already does:
(a b c / d e f / g h i) (1 1 1 / 1 1 1 / 0 0 1) = (0 0 0 / 0 0 0 / 0 0 0) yields
A = (a a 0 / d d 0 / g g 0), so eight possible A with B.
(a b c / d e f / g h i) (1 1 1 / 1 0 1 / 0 1 0) = (0 0 0 / 0 0 0 / 0 0 0) yields
A = (a a a / d d d / g g g), so eight possible A with C as well.

11038 Let ABC be a triangle and let a circle intersect line AB in C1 and C2, AC in B1 and B2, BC in A1 and A2. Suppose that the chords A1B2, B1C2, C1A2 are parallel. Prove that these sides are parallel to one of the sides of Morley's equilateral trisector triangle in ABC.

Solution (outline):

See the picture and text here .

11043 Let f and g be distinct L1-functions with only positive real values on a finite interval I=[a,b], such that
INTEGRAL(x from a to b/ f(x) dx) = INTEGRAL(x from a to b/ g(x) dx) = 1.
Show that for each natural number q there exists a subinterval J of I such that
INTEGRAL(x from a to b/ f(x) dx) = INTEGRAL(x from a to b/ g(x) dx) = 1/q.

Trial:

Draw the graphs of f and g in one figure. There is at least one s in (a,b) such that f-g changes sign in s. With graphical analysis we see:
For each small positive epsilon there is a small positive delta such that
INTEGRAL(x from s-delta to s+epsilon/ f(x)-g(x) dx) = 0.
For each large q we can choose epsilon in such a way that
INTEGRAL(x from s-delta to s+epsilon/ f(x) dx) = 1/q.
By suitable choice of s and epsilon, this can be done for each natural number q (because s+epsilon (or s-delta) may be chosen to be greater (rep. smaller) than another s where f-g changes sign).

11045 Prove that, if x is a sufficiently large positive integer, there exists a finite set S of primes such that the sum of [x/p] with p in S is equal to x. ([x/p] is the integral part of x/p)

Sketch of a solution:

We form an infinite set T of primes such that the sum of 1/p with p in S is equal to 1.
This sum begins as follows: 1/2 + 1/3 + 1/7 + 1/43 + ... (take each time the next smallest p such that 1/p can be added, keeping the partial sum less than 1).
If we take the first k terms of this sum, there is a deficit, for example with k=4 the deficit is 1 - (1/2 + 1/3 + 1/7 + 1/43) = 1/1806.
Then x - ([x/2] + [x/3] + [x/7] + [x/43]) is at most x/1806 + 4.
Now for p greater than x/2 and at most x we have [x/p] = 1.
If the number of primes greater than x/2 and at most x is at least x/1806 + 4, we can form out of these primes together with 2,3,7,43 a set S as asked for, thus supplying a supplement that eliminates the deficit.
According to Hardy and Wright, the number of primes greater than x/2 and at most x is equal to
A(x)*x/log(x) - A(x/2)*(x/2)/log(x/2), with A(x) rapidly tending to 1.
So we can eliminate the deficit in this way for x (not too small and) at most C4, where C4 is about exp(900).
In the same way we can find a set S for larger x, using the first k terms of T, with k>4.
In fact, the deficit can then be eliminated for x at most Ck, and Ck tends to infinity if k tends to infinity.

11046 Let T be a triangle ABC with incircle I. Let the circles K1, K2, K3 be tangent to I and to two sides of T.
Denote the radius of I as r, the radius of Ki as ri (i=1,2,3).
Prove that r = sqrt(r1r2) + sqrt(r1r3) + sqrt(r2r3).

Solution:

Denote the centers of the circles as M, M1, M2, M3, and the angles of T as alpha, beta, gamma.
Since M and M1 are both on the bisectrix through (say) A, we have for some lambda and mu
M1=lambda(cos(alpha/2),sin(alpha/2)), M=mu(cos(alpha/2),sin(alpha/2)).
Since mu=(r/r1)*lamda and distance(M,M1)=r+r1, we find
mu = r*(r+r1)/(r-r1).
Furthermore, r = mu*sin(alpha/2), so
1 = ((r+r1)/(r-r1))*sin(alpha/2).
Hence r1 = r*(1-sin(alpha/2))/(1+sin(alpha/2)) = r*cos2(alpha/2)/(1+sin(alpha/2))2,
and we have similar expressions for r2 and r3.
Then sqrt(r1r2) + sqrt(r1r3) + sqrt(r2r3) =
r*(cos(alpha/2)*cos(beta/2)/((1+sin(alpha/2))*(1+sin(beta/2))) + cos(alpha/2)*cos(gamma/2)/((1+sin(alpha/2))*(1+sin(gamma/2))) + cos(beta/2)*cos(gamma/2)/((1+sin(beta/2))*(1+sin(gamma/2)))).
Since alpha+beta+gamma=pi, the right hand side in the last equation is equal to r.

11048 For two non-antipodal points A and B on a sphere of radius 1, the "spherical segment" AB is the shorter of the arcs into which the points A and B divide the great circle through A and B.
Given three points A,B and C no two of which are antipodal, the "spherical triangle" ABC is the union of the spherical segments AB,AC and BC which are called the "sides" of ABC.
The "diameter" of a closed set on the sphere is the maximum distance on the sphere between two points of the set.
A spherical triangle is "ordinary" if its diameter is the maximum of the lengths of its sides.
Prove that a spherical triangle with sides of length s1,s2,s3 is ordinary if and only if at most one of the three inequalities cos(si)*cos(sj)>cos(sk) holds, where (i,j,k) ranges over the even permutations of (1,2,3).

Solution:

I suppose that the distance between two non-antipodal points is the length of the spherical segment AB. Hence this distance is always smaller than pi.
The relation between the sides and angles in a triangle is given by three formulas like
cos(a) = cos(b)*cos(c) + sin(b)*sin(c)*cos(alpha), where alpha is the angle opposite side a.
So cos(b)*cos(c) > cos(c) is equivalent to alpha > pi/2.
Thus the proposed necessary and sufficient condition for ordinarity is equivalent to "at most one of the angles of the triangle is obtuse".
But a triangle with two angles of 100 degrees and one of, say, 30 degrees, is still ordinary, unless antipodal points are taken together to be one point (as in elliptical geometry). So I think that the proposers mean to imply that antipodal points are taken together, although they don't say that clearly.

11049 Let X,Y,Z be three distinct points in the interior of an equilateral triangle ABC. Let alpha, beta, gamma be positive numbers summing up to pi/3 with the property that the angles XBA and YAB are equal to alpha, YCB and ZBC to beta, ZAC and XCA to gamma. Find the angles of triangle XYZ in terms of alpha, beta and gamma.

Experimental solution:

It seems that we can't find them with the figure alone, and with vectorcalculus it would be too much work. So I decided to try with Cabri. It appears that the angles at X,Y,Z are 3*beta, 3*gamma, 3*alpha respectively.

11050 Let a,b,c be positive integers. Describe the positive rational solutions of
y2+z2 = c2 and c/y = x/a and (c-x)/b = c/z.
For which (a,b,c) is there exactly one such solution (x,y,z)?

Solution for the first part:

Suppose that x,y,z are positive rational solutions.
It is well known that then y = c(t2-s2)/(t2+s2), z = 2cts/(t2+s2) for some natural numbers t,s with t>s and gcd(t,s)=1.
We find x = a(t2+s2)/(t2-s2) and
2ats(t2+s2) + b(t2-s2)(t2+s2) = 2cts(t2-s2).
Hence, since gcd(t,s)=1, we find:
1) 2a= A(t2-s2), b=Bts, 2c= C(t2+s2)
for some natural numbers A,B,C with A+B=C.
2) x,y,z are natural numbers and
2x = A(t2+s2), 2y = C(t2-s2), z = Cts.
Conversely, if a,b,c can be written (resp. uniquely written) as in 1) with gcd(t,s)=1, then x,y,z as in 2) are solutions (resp. unique solutions).

11051 A game show features a weekly contestant. Each week, a sequence of up to four random numbers is determined during play. When play begins, the first number m is chosen at random from a uniform distribution on [0,1] and is not disclosed until the game is over. The next three numbers, x1, x2, and x3, say, are chosen independently and at random from a uniform distribution on [0,1/m].
As each successive number xi is offered, the contestant must accept it or reject it, accepting x3 if x1 and x2 have been rejected. If xi is accepted, then the number m is revealed, the contestant receives m.xi dollar, and the game ends.
Contestants Ann, Beth and Carlos are scheduled to appear on successive weeks. They plan to employ the following strategies:
Ann accepts x1 if and only if x1>1 and x2 if and only if x2>1.
Beth uses positive real numbers s and t chosen in advance. She accepts x1 if and only if x1>s and x2 if and only if x2>t.
Carlos uses three real numbers u,v and w chosen in advance. He accepts x1 if and only if x1>u and x2 if and only if x2>v+wx1.
(a) What are the expected values of the three contestants' strategies, assuming optimal choice of the parameters in the case of Beth and Carlos? What are those optimal choices?
(b)* What is the optimal strategy?

Partial solution:

(a) The expected value of Ann is INTEGRAL(m from 0 to 1: (1-m)(m+1)/2 + m(1-m)(m+1)/2 + m.m/2) dm = 5/8.

To calculate the expected value of Beth, we first think about s and t as follows:
In the beginning we expect m=1/2 and x1=1. If x1<1 then we expect 1/m<2 (so m>1/2) and x2<1. So it seems that s must be 1 and t must be smaller than 1.
Then we get as Beth's expected value
INTEGRAL(m from 0 to 1: (1-m)(m+1)/2 + m(1-mt)(mt+1)/2 + m.mt/2) dm = 7/12 + t/6 - t.t/8.
This is maximal if t=2/3 and the maximum is 23/36 and this is greater than 5/8.
(I calculated a lot, observing all pairs (s,t); I considered four cases (s and t can both be greater than 1 or not); in the end I found s=1 and t=2/3).

As to Carlos, we probably have u=1 (see Beth).
I will pay attention to the case u>1, but I first assume that both u and v+uw are positive and at most 1.
The expected revenue then is
INTEGRAL(m from 0 to 1: (1-mu)(mu+1)/2 + (1/m)*INTEGRAL(x1 from 0 to 1/m : mu(1-m(v+wx1))(m(v+wx1)+1)/2 + mu*m(v+wx1)/2) dx1 ) dm =
(6+3u-2u2)/12 + u(4v+3w-3v2-4vw-2w2)/24.
This is maximal if u=1,v=1/2,w=1/4. The maximum is 7/12 + 11/192 = 123/192 and this is greater than 23/36.
(I did calculate the expected value f(u,v,w) in the case u>1, v+uw<1, and the gradient of f in (u,v,w)=(1,1/2,1/4). This gradient points in the direction (-1,0,0), indicating that f decreases if we make u larger than 1. In our context, I don't believe that there can exist a relative optimum next the global one.)

(b) After the offer of x1, we could first guess m.x1=1/2, reject x1, and accept x2 if and only if x2 > x1.
However, the expected value of this strategy is only (1/2).(3/4)+(1/2).(1/2) = 5/8. This is Ann's revenue, but Beth does better.
We see that Carlos does even better than Beth, but I don't think that Carlos does best possible: we don't have to restrict ourselves to linear decision functions v+w.x1.
Maybe we can find the optimum as a limit, using better decision functions again and again.
If we take u=1 as before and call our decision function f(x1) = g(y) with y=m.x1, then the expected revenue is
7/12 + INTEGRAL(m from 0 to 1: (m/2)*INTEGRAL(y from 0 to 1: mg(1-mg)) dy) dm.
The inner integral is less than 1/4 (we get 1/4 for mg=1/2, but it is not possible to make g(y) equal to 1/(2m) for all y). So we see that the expected value is less than 7/12 + 1/16 = 31/48 = 124/192, so Carlos' 123/192 seems almost optimal.

11053 For each of the conditions (a) and (b), find all real functions such that the stated condition holds for all real x,y.
(a) f(x+f(x)f(y)) = f(x)+xf(y) (b) f(x+f(xy)) = f(x)+xf(y)

Partial solution:

If f is a polynomial, both in (a) and in (b) we readily see by substitution that f cannot have degree greater than 1.
By substitution of f(z)=c+dz, we find:
(a) c=0 and d3=d, so f(z)=z or f(z)=0 or f(z)=-z;
(b) c=0 and d2=d, so f(z)=z or f(z)=0.
So these are the only solutions for polynomials f(z).

11054 Determine the set of all solutions in integers to
19982x2+1997x+1995-1998x1998 = 1998y4+1993y3-1991y1998-2001y.

Solution:

We notice that 1997 is prime. Modulo 1997 we have by Fermat's little theorem
x1997=x and y1997=y, so the equation reduces modulo 1997 to
x2+0-2-x2 = y4-4y3+6y2-4y, or
(y-1)4=-1(mod 1997).
Hence I find with a little Pascal program
(y-1)2=412(mod 1997) or (y-1)2=1585(mod 1997).
But the same Pascal program tells me that 412 and 1585 are no quadratic residues modulo 1997.
So the set of solutions is empty.

11055 Let ABC be an acute triangle with semiperimeter p and with inscribed and circumscribed circles of radius r and R respectively.
a) Show that ABC has a median of length at most p/sqrt(3).
b) Show that ABC has a median of length at most R+r.
c) Show that ABC has an altitude of length at least R+r.

Solution:

Suppose 0 < alpha <= beta <= gamma < 90 degrees.
Let a = h*sin(alpha), b = h*sin(beta), c= h*sin(gamma).
The smallest median has length (h/2)*sqrt(sin2(alpha)+sin2(beta)+2*sin(alpha)*sin(beta)*cos(gamma)).
The largest altitude has length h*sin(beta)*sin(gamma).
Furthermore, p/sqrt(3) = h*(sin(alpha)+sin(beta)+sin(gamma))/(2*sqrt(3)), and
R+r = h/2 + h*(sin(alpha)*sin(beta)*sin(gamma))1/3/(sin(alpha)+sin(beta)+sin(gamma)).
So in each of a),b),c) we have a goniometric inequality that can be checked with Maple or Pascal, using gamma=pi-alpha-beta.

11057 Let x,y,z be positive real numbers. What is the maximum area possible for a rectangle ABCD, given that it has an interior point P such that the lengths of AP, BP and CP are x,y and z respectively?

Solution:

Let h be the distance from P to AB (idea of Benne de Weger). We can express the area in h,x,y,z. By differentiation we find that the maximum area appears if h2 = x2y2/(x2+z2). The area then is xz + y sqrt(x2+z2-y2).
Remark: Wim Nuij had the idea to involve t:=DP in the discussion and to use Ptolemy. We can cut the rectangle into 4 pieces and paste 2 pieces together to make a quadrangle with sides x,y,z,t and such that the area of the rectangle is equal to the product of the diagonals of the quadrangle. According to Ptolemy, this product is equal to xz+yt = xz + y sqrt(x2+z2-y2).

11060 Let [n] de note the set of integers {1,2,..,n}. Let Gn be the union of all closed line segments joining any two elements of [n]x[n] along a horizontal or vertical line, or along a line with slope 1 or -1. Determine the combined total Fn of the number of (nondegenerate) triangles and rectangles for which the set of edges is a subset of Gn. (The vertices of these figures need not be in [n]x[n]).

Solution:

First we determine the numbers of lines:
slope 1: 2n-3, slope -1: 2n-3, vertical: n, horizontal: n.
Next we determine the number of all triangles minus the number of degenerate ones:
slope 1, slope -1, horizontal: (2n-3)2n - (n2-4);
slope 1, slope -1, vertical: (2n-3)2n - (n2-4);
slope 1, horizontal, vertical: (2n-3)n2 - (n2-2);
slope -1, horizontal, vertical: (2n-3)n2 - (n2-2).
Then the number of all rectangles minus the number of degenerate ones:
two slope 1, two slope -1: ((2n-3) over 2)2;
two vertical, two horizontal: (n over 2)2.
All together 17*n4/4 - 33*n3/2 + 157*n2/4 - 66*n + 48.
For n=2 we get 9, and that is correct. For n=3 we get 102.

11061 Find all pairs (x,y) of rational numbers such that
sqrt(2) = sqrt(1/(y-1) + 1/(y+1)) / (1/sqrt(x-1) - 1/sqrt(x+1)).

Partial solution:

We see that x>1 and either y>1 or 0>y>-1.
Let c:=2*x/(x2-1)-2/sqrt(x2-1).
Then c*y2-y-c=0, so y=(1+sqrt(1+4*c2))/(2*c) or y=(1-sqrt(1+4*c2))/(2*c) (the product of these 2 values of y is -1).
Since 1+4*c2 is the square of a rational number, we find c=u*v/(u2-v2) for some natural numbers u,v with u>v and gcd(u,v)=1 (then y=u/v or y=-v/u).
Since x2-1 is the square of a rational number, we find for some natural numbers m,n with m>n and gcd(m,n)=1:
either x = (m2+n2)/(2mn) (case (1))
or x = (m2+n2)/(m2-n2) (case (2)).

In case (1) we get c = 8*m*n3/(m2-n2)2 = u*v/(u2-v2).
Since the second fraction is in lowest terms and the first one too or almost, we find:
either u*v = 8*m*n3 and u2-v2 = (m2-n2)2 (one of m,n is even; subcase (1a))
or u*v = m*n3 and u2-v2 = (m2-n2)2/2 (m and n odd; subcase (1b)).

In case (2) we get c = (m+n)(m-n)3/(2m2n2) = u*v/(u2-v2).
Since both fractions are in lowest terms (or almost), we find:
either u*v = (m+n)(m-n)3, u2-v2 = 2m2n2 (one of m,n is even; subcase (2a))
or 2*u*v = (m+n)(m-n)3, u2-v2 = m2n2 (m,n both odd; subcase (2b)).

In subcase (2a) we find the solution u=3,v=1,m=2,n=1, so x=5/3 and (y=3 or y=-1/3).
I searched for other rational solutions (x,y) of c(x)*y2-y-c(x)=0, using a Pascal program.
There are no other solutions with both numerator and denominator of x and y at most 100.

11064 A curve from P to Q in the plane is wrapped if it completely surrounds the segment PQ, in the sense that every ray with vertex on the segment meets the curve at some point not on the segment. Show that the length of such a curve is at least three times the length of the segment PQ.

Solution:

It is clear that the proposer meant to imply that the curve is continuous, otherwise he should say two times instead of three times.
Let P and Q have rectangular coordinates (-1/2,0) and (1/2,0) respectively.
If we want to cross all rays with a vertex on the segment in points not on the segment, we have to travel around the segment.
If we travel around the segment one time, following a curve that is not too long, we must avoid passing any ray twice and don't cross the segment.
So we go around clockwise or counterclockwise, travelling a distance greater than 2 before we are back in some point (a,0) with a < -1/2.
Then we still have to travel back to Q along a distance greater than 1.