Solutions of old problems by H Reuvers

The American Mathematical Monthly shows the problems of the issues from the years 1894 through 1922 free of charge, so I tried some of them.

2973 A straight brass bar 800 feet long expands 2/3 foot. The ends are fixed, so that it is distorted. If the new shape is that of an arc of a circle of which the original bar is the chord, how far above the center of the bar in its original position will the center be in its new position?

Solution:

Denote the radius and the angle of the arc by r and α. Then we have αr = 800 + 2/3 and sin(α/2) = 400/r.
The distance asked for is d := r - r cos(α/2) = r - √(r2 - 160000).

Let x := 1/(3r). Then we have α/2 = 1200*x + x, and sin(α/2) = 1200*x, where sin(α/2) is approximately α/2 - (α/2)3/6.
So in third degree approximation we have 1200*x + x - (1200*x + x)3/6 = 1200*x.
Hence we find an approximation of x and then an approximation of r (almost 5564 feet) and then an approximation of d, which is about 14 feet and 1/7 foot.

2974 A man standing on a straight railway track watches a train starting from a station half a mile distant and notices that it approaches at such a speed that each puff of the exhaust is heard at the same instant that the next succeeding puff is seen.
How long will it be before the train reaches him, the drive wheels of the locomotive being sixteen feet in circumference?

Solution:

We reckon a mile is 1609 meters, a foot is 0.348 meter, the speed of light is c=299793000 meters/second and the speed of sound is a=343.2 meters/second (supposing the air is dry and the temperature is 20 degrees celsius).
Furthermore, we suppose that the train keeps approaching with the velocity it starts with.

If the two succeeding puffs occur at instants t0 and t1, the instant of the two perceptions is t0 + (1609/2)/a = t1 + (1609/2)/c, so the time t between two puffs is t = t1 - t0 = (1609/2)(1/a - 1/c) seconds.
In time t the driving wheel performs one rotation, so in this time the train travels 16*0.3048 meters and in one second 16*0.3048/t meters.
So the time needed for the train to get to the man is (1609/2)/(16*0.3048/t) = (1609/2)2(1/a - 1/c)/(16*0.3048) seconds = 386.6957... seconds.

If we neglect the time for the light needed to reach the man and we use the formula a = 331 √(1+T/273) where T is the temperature in degrees celsius, we get traveling time 400.949/√(1+T/273) instead.
For T=20 we get about 387 seconds, for T=0 about 401 seconds. So the temperature is important.

2976 The base of a variable triangle is fixed, the opposite vertex describing a straight line. Find the locus of the symmedian point, the locus of the center of the nine-point-circle, and the envelope of the Euler line.

Sketch of a partial solution:

Give coordinates so that A(0,0), B(1,0) and C((p,q)+λ(s,t)) for given p,q,s,t.

The symmedian point is the point where the sum of the squares of the distances to the sides is minimal.
So let f:= y2 + (-x(q+λt)+y(p+λs))2/((p+λs)2+(q+λt)2) + (-x(q+λt)+y(p+λs-1)+(q+λt))2/((p+λs-1)2+(q+λt)2).
Then f 'x = 0 is equivalent to (-x(q+λt)+y(p+λs))((p+λs-1)2+(q+λt)2) + (-x(q+λt)+y(p+λs-1)+(q+λt))((p+λs)2+(q+λt)2) = 0.
Hence we can express λ in x and y, using the formula of Cardano.
Furthermore we have f 'y=0. When we calculate f 'y and substitute for λ the expression we find with Cardano, we get an equation for the locus of the symmedian point.

By straightforward calculation, we find the orthocenter O1(p+λs,(p+λs)(q+λt)/(p+λs-1)) and the circumcenter O2(1/2,((p+λs)2+(q+λt)2-(p+λs))/(2(q+λt))).
The center of the nine-point-circle is the midpoint of the segment O1O2, so M: (x,y) = (1/2)(p+λs,(p+λs)(q+λt)/(p+λs-1)) + (1/2)(1/2,((p+λs)2+(q+λt)2-(p+λs))/(2(q+λt))).
When we eliminate λ, substituting λ = (4x-2p-1)/(2s) in y = (1/2)(p+λs)(q+λt)/(p+λs-1)) + (1/2)((p+λs)2+(q+λt)2-(p+λs))/(2(q+λt)), we find an equation for the locus of the center of the nine-point-circle.

The Euler line is the straight line through O1 and O2.
It is straightforward to find an equation X.N(λ) = c(λ) of this Euler line, where X = (x,y) and N(λ) is a unit normal vector and c(λ) a constant.
Thereafter, we can calculate an equation of the envelope: eliminating λ from X.N(λ) = c(λ) and X.N'(λ) = c'(λ), where the accent denotes the derivative.

2977 A point moves in such a way that its polars with respect to two given conics intersect at right angles. Prove that the locus of the intersection is a rational quartic curve, and find its double points.

Partial solution:

We give an example: consider the conics C1: y - 3 = 6x2, C2: x2 + 2y2 = 3.
Let the point be P(p,q). The polars are p1: y = 12 px +6 - q, p2: px + 2qy = 3. These polars intersect at right angles iff q = 6p2.
The points of intersection are given by (x,y) = ((72p4 - 72p2 +3)/(144p3 + p) , (-6p2 + 42)/(144p2 + 1)).
This is a rational quartic curve. It turns out this one has no double points.

2978 Given the equation of the general cubic f(x,y) = a x3 + 3b x2y + 3c xy2 + d y3 + a1 x2 + 2b1 xy + c1 y2 + a2 x + b2 y + c3 = 0, show that the three asymptotes of the curve will be concurrent if det(a b c | b c d | a1 b1 c1 ) = 0.

Sketch of a solution:

We calculate f(x,mx+n). To find the best values of m and n, we demand that the coefficients of x3 and x2 be zero.
This amounts to a + 3bm + 3c m2 + d m3 = 0 = 3bn + 6cmn + 3d m2n + a1 + 2b1m + c1 m2.
From the first of these two equations we find three values mi of m, for which holds -a = d m1m2m3, 3b = d (m1m2 + m1m3 + m2m3), -3c = d (m1 + m2 + m3).
Then from the second we find three corresponding values ni of n: ni = -(a1 + 2b1mi + c1 mi2)/(3b + 6cmi + 3d mi2) = ... (use 3b = d (m1m2 + m1m3 + m2m3), etc) .
This way we find the three asymptotes y = mix + ni.
Now these three asymptotes are concurrent iff (m2 - m1)(n3 - n1) = (m3 - m1)(n2 - n1).
After substituting in this last equation the values we found for ni and in det(a b c | b c d | a1 b1 c1) = 0 the expressions we found for a,b,c, we must find identical relations between the mi (and a1,b1,c1).

2979 If a tangent be drawn from a variable point of an ellipse of length equal to n times the conjugate semidiameter at the point, the locus of its extremity will be a concentric elipse with semi-axes equal to √(n2+1) times the semi-axes of the original ellipse.

Solution:

Let the ellipse have equation x2/a2 + y2/b2 = 1 with a > b. The variable point is P: (x,y) = (a cos(t), b sin(t)) with 0 ≤ t < 2π.
The corresponding semidiameter is OP and its conjugate OQ with Q: (x,y) = (a cos(t+π/2), b sin(t+π/2)) = (-a sin(t),b cos(t)) and length √(a2sin2(t)+b2cos2(t)).
Now the normal vector at P to the ellipse has the same direction as (cos(t)/a, sin(t)/b) and (b cos(t), a sin(t)), so one of both families of tangents with the prescribed length has extremities

E: (x,y) = (a cos(t), b sin(t)) + n √(a2sin2(t)+b2cos2(t)) (-a sin(t), b cos(t))/√(a2sin2(t)+b2cos2(t)).
or E: (x,y) = (a cos(t) - na sin(t), b sin(t) + nb cos(t)) = (a √(n2+1) cos(t+s), b √(n2+1) sin(t+s)), where cos(s) = 1/√(n2+1), sin(s) = n/√(n2+1).
Hence the locus of E has equation x2/(a2(n2+1)) + y2/(b2(n2+1)) = 1.

2980 Locate a point such that the sum of its distances from the vertices of a given polygon shall be a minimum.

Sketch of a numerical solution:

Let the vertices be end points of vectors a1, a2, ... , an. The point X we look for is end point of vector x = α1a1 + α2a2 + ... + αnan for some nonnegative reals αi that sum to 1.
So we may vary α1 from 0 to 1 with steps of, say, 0.01, and likewise α2 from 0 to 1 - α1, etc, each time calculating the sum of the distances from X to the vertices, and determine both the minimum of the sum and the position of the corresponding X.

2981 Find the envelope of the (variable) circle on which two diametrically opposite points divide in a given ratio the focal radii of a variable point on an ellipse or hyperbola.

Solution:

We consider the ellipse (x/a)2 + (y/b)2 = 1 with a > b. (The case of the hyperbola is similar.)
The foci are (±c,0) with c = √(a2-b2). The variable point is (a cos(t), b sin(t)) with 0 ≤ t < 2π.
Let u < 1 be the given ratio.
Then the diametrically opposite points are (±c,0) + u((a cos(t), b sin(t)) - (±c,0)).
The corresponding circle has center u((a cos(t), b sin(t)) and radius c(1-u), so equation (x - ua cos(t))2 + (y - ub sin(t))2 = (c(1-u))2.
Now if we intersect the circles with parameters s and t and let s approach t, we find the two limits of the two intersection points also lie on the straight line ax sin(t) - by cos(t) = uc2sin(t)cos(t).
Hence we find the parametrization of the envelope: (x,y) = (ua cos(t) ± bc(1-u)cos(t)/√(b2cos2(t)+a2sin2(t)), ub sin(t) ± ac(1-u)sin(t)/√(b2cos2(t)+a2sin2(t))).

2982 A triangular yard has a post at each corner. Given the lengths of the sides of the yards and the posts, find the length and the position of the foot of a ladder that will just reach the top of each post without changing this position.

Sketch of a solution:

Let the triangle ABC have side lengths a,b,c and angles as usual, so cos(α) = (b2 + c2 - a2)/(2bc), etc.
Let the post tops have coordinates (0,0,pA), (c,0,pB), (b cos(α), b sin(α), pC), and the foot of the ladder (r cos(δ), r sin(δ), 0).
The conditions on the ladder amount to the following three equations from which r, δ and m have to be found in terms of a,b,c,pA,pB,pC:

(1) r2 + pA2 = m2
(2) r2 + pB2 = m2 - c2 + 2rc cos(δ)
(3) r2 + pC2 = m2 - b2 + 2rb cos(α-δ)

Since cos(α-δ) = cos(α)cos(δ) + sin(α)sin(δ) = ((b2 + c2 - a2)/(2bc))cos(δ) + √(1 - ((b2 + c2 - a2)/(2bc))2) √(1-cos(δ)2),
we can first from (3) express cos(δ) in m and r and the given lengths, then from (2) express r in m and the given lengths and finally from (1) express m in the given lengths.

2983 (Calculate the indicated sums.)

Sketch of a partial solution for (1)-(4):

(1)+(2): four variants, for instance
(2m+1 over 1) + (2m+1 over 3) + (2m+1 over 5) + ... + (2m+1 over 2m-1) + (2m+1 over 2m+1) = (2m over 0) + (2m over 1) + (2m over 2) + (2m over 3) + (2m over 4) + (2m over 5) + ... + (2m over 2m-2) + (2m over 2m-1) + (2m over 2m) = 22m.

(3)+(4): eight variants, for instance (to illustrate the general way of working out the sum)
(7 over 1) - (7 over 3) + (7 over 5) - (7 over 7) = (6 over 0) + (6 over 1) - (6 over 2) - (6 over 3) + (6 over 4) + (6 over 5) - (6 over 6) = (5 over 0) + (5 over 0) + (5 over 1) - (5 over 1) - (5 over 2) - (5 over 2) - (5 over 3) + (5 over 3) + (5 over 4) + (5 over 4) + (5 over 5) - (5 over 5) = 2((5 over 0) - (5 over 2) + (5 over 4)) = 2((4 over 0) - (4 over 1) - (4 over 2) + (4 over 3) + (4 over 4)) = 2((3 over 0) - (3 over 0) - (3 over 1) - (3 over 1) - (3 over 2) + (3 over 2) + (3 over 3) + (3 over 3)) = 4(- (3 over 1) + (3 over 3)) = 4(- (2 over 0) - (2 over 1) + (2 over 2)) = 4(- (1 over 0) - (1 over 0) - (1 over 1) + (1 over 1)) = 8(- (1 over 0)) = ...

2984 Find the number of numbers of n digits that can be written with n consecutive digits, allowing all possible repetitions, such that the sum of the digits in each number is a multiple of n.

Partial solution:

The base of our calculation is the fact that a number with n equal digits satisfies the requirements.
Then we can easily count the other numbers that satisfy the requirements.
For n=8,9,10 we are still counting.

2985 Find the number of combinations of n digits each that can be made with the first n consecutive digits, allowing repetitions, and such that the sum of the digits in each combination is a multiple of n.

Solution:

Counting 0 as the first digit and 9 as the last one, we find the following list of answers (n,answer(n)):

(0,0), (1,1), (2,2), (3,4), (4,10), (5,26), (6,80), (7,246), (8,810), (9,2704), (10,9252).

For instance, for n=5 we count by using the following procedure:
a:=0; for d1:=0 to 4 do for d2:=0 to d1 do for d3:=0 to d2 do for d4:=0 to d3 do for d5:=0 to d4 do if ((d1+d2+d3+d4+d5) mod 5)=0 then a:=a+1; write(a);

2986 A triangle is inscribed in a circle. The arcs in which it divides the circumference are bisected at points forming the vertices of a second triangle. A third triangle is derived in the same way from the second, and so on.
Prove that each set of alternate triangles approaches a limiting position.

Solution:

Let the vertices of the initial triangle be (cos(ρ),sin(ρ)), (cos(σ),sin(σ)) and (cos(τ),sin(τ)), with 0 ≤ ρ < σ < τ < 2π.
I calculated the vertices (cos(s),sin(s)) of the second triangle, the third, etc, until it became clear that the limiting position is the triangle with vertices where modulo 2π we have s = (ρ+σ+τ)/3, s = (ρ+σ+τ)/3 + 2π/3, and s = (ρ+σ+τ)/3 + 4π/3, repectively.
This is a consequence of the fact we continue to take the averages of the next two points, going around again and again.

2987 A flexible chain of length l and uniform weight is fastened at one end to the ridge of a roof with pitch p and slant height L. If the eaves of the roof are at height h from the ground and the coefficient of friction between the chain and the roof is μ, how long will it take, after releasing the chain, for the highest end to reach the ground?

Solution:

Let the constant c be defined by c := (1-μ)p/√(1-p2) and let g be the acceleration of gravity.
For the first L meters, the top end of the chain will move with acceleration cg and, after t seconds, velocity cgt, so the time t=t1 needed to move along the slant satisfies cgt2/2 = L, so t1 = √(2L/(cg)).
For the last h meters, the top end falls with acceleration g, initial velocity v0 = cgt1, and, after t seconds, velocity v0 + gt, so the time t=t2 needed to fall satisfies h = v0t + gt2/2.
This way we calculate the total time t1 + t2, which doesn't depend on the length l of the chain.

2988 Prove geometrically that if in an ellipse the tangent at P cuts the directrices in Z and Z', and the remaining tangents from Z and Z' to the ellipse meet at T, then PT is normal to the ellipse at T.

Sketch of a partial solution:

We give a sketch of an analytical solution.

Let's give coordinates such that the ellipse has equation x2/a2 + y2/b2 = 1 with a > b, and the point P coordinates (a cos(t), b sin(t)) for some real t.
The directrices have equations x = ± a2/√(a2-b2).
Calculate the coordinates of Z and Z' and determine equations for the remaining tangents to find the coordinates of T. Show that PT has the same direction as the normal at P to the ellipse, that is (cos(t)/a, sin(t)/b).

2989 How should the following questions be answered, assuming that the place referred to is in latitude 34 degrees 8 minutes?
A building twelve feet high has been erected 49 inches south of our lot line. We desire to erect a wall on our line six inches in thickness.
(a) How high can we build the wall and have it wholly within the shadow cast by the building?
(b) How high can we build the wall and have it within the shadow cast by the building during the winter months?

Sketch of a solution:

Let au be the distance between the earth and the sun, r the radius of the earth, b the height of the building, and w the heigth of the wall, all expressed in meters.
Let ε := (180d)/(πr) where d is the number of meters corresponding to 49 inches. Let δ := (180e)/(πr) where e is the number of meters corresponding to 3 inches.

Give coordinates such that the center of the earth is in (0,0) and the sun in (0,au) (after a suitable rotation, which is distinct in (a) and (b)) .
(a) The northern top end of the building has coordinates (r+b)(sin(s-ε),-cos(s-ε)), where s = 34o8' - 23o27'.
Calculate w such that the line through the sun and the northern top end of the building goes through (r+w)(sin(s+δ),-cos(s+δ)).
(b) The same as (a) except that s = 34o8'.

(Because the nearest overall position of the sun is above the tropic of cancer, but the nearest in winter above the equator.)

2990 If a circle be bitangent to a conic, its center lies on one of the axes of the curve.

Sketch of a solution:

We'll prove it for the case that the conic is an ellipse. In the other two cases, we'll have to make only slight changes to finish the proofs.

Let the ellipse have equation (x/a)2 + (y/b)2 = 1 (with a > b), and let the circle have parametrization (x,y) = (s + r cos(u), t + r sin(u)).
The points of intersection satisfy equation
(*): b2(s + r cos(u))2 + a2(t + r sin(u))2 = a2b2.

Now suppose the circle is bitangent to the ellipse. We have to prove that st = 0.

Let cos(u) = W, so sin(u) = ± √(1-W2). From equation (*) we derive a polynomial equation in W of degree 4 which for some p,q must equal (W-p)2(W-q)2=0.
Hence we find:

(p + q) = 2rsb2/(r2a2-r2b2)
(p+q)2 + 2pq = (4r2s2b4 + 4r2t2a4 + 2(r2b2-r2a2)(r2a2+s2b2+t2a2-a2b2)/(r2a2-r2b2)2
pq(p+q) = (2rsb2(r2a2+s2b2+t2a2-a2b2))/(r2a2-r2b2)2
p2q2 = ((r2a2+s2b2+t2a2-a2b2)2 - 4a4r2t2)/ (r2a2-r2b2)2.

If s is not equal to 0, we may use the first of these 4 equations to derive from the latter three equations three expressions for pq. Then we see that t must be equal to 0.

2991 Sum the infinite series 1 + 3x2/2! + 4x4/4! + 6x6/6! + ... , where the numerators of the coefficients form a series of numbers whose third differences are all equal to 2.

Solution:

We find Σk=0 (k3/3 - 3k2/2 + 19k/6 + 1)x2k/(2k)! = (x3/24 + 5x/4)sinh(x) + (-x2/4 + 1)cosh(x).

2949 (revised) Find the lateral area of the cone with vertex at (0,0,h) and whose base is the epicycloid: 2x = a(3 cos(θ) - cos(3θ), 2y = a(3 sin(θ) - sin(3θ).

Solution:

After some goniometric calculations, we may rewrite the base as (x,y) = a(3 cos(θ) - 2 cos3(θ), 2 sin3(θ)).
So the cone has parametrization (x,y,z) = (0,0,h) + λ (3a cos(θ) - 2a cos3(θ), 2a sin3(θ), -h), 0 ≤ λ ≤ 1, 0 ≤ θ ≤ 2π.
We make a sketch of the base and see that the area asked for is equal to:

4 ∫0 ≤ λ ≤ 1 0 ≤ θ ≤ π/2 ||xλ x xθ|| dθ dλ = 4 ∫0 ≤ λ ≤ 1 0 ≤ θ ≤ π/2 3a λ sin(θ) √(h2 +4a2sin2(θ)) dθ dλ.

Now we do some further calculation and consult a table of integrals and find: 3 ∫0 ≤v≤2a √(h2+v2) dv = (3h2/2) (p√(1+p2) + ln(p + √(1+p2)/2)) with p=2a/h.

2992 A semi-circle rotates at a uniform velocity about its diameter and slides along the line of that diameter while making one revolution about it. Find the equation of the surface thus generated.

Solution:

We find (x,y,z) = (cos(θ) + φ/π, sin(θ)sin(φ), sin(θ)cos(φ)), -π ≤ φ ≤ π, and hence (x ± arccos(z/√(y2+z2)))2 + y2 + z2 = 1.

2993 Let ABC be any triangle, and O the center of its circum circle. Bisect the arcs AB, BC and CA at F, D and E. With F,D and E as centers draw arcs passing in each instant through the adjacent corners of the triangle. Prove that these arcs intersect at the in-center of triangle ABC.

Provisional solution:

Give coordinates O(0,0), F(0,-1), A(-cos(ρ),sin(ρ)), B(cos(ρ),sin(ρ)), C(cos(σ),sin(σ)), D(cos((ρ+σ)/2),sin((ρ+σ)/2)), E(cos((σ+π-ρ)/2),sin((σ+π-ρ)/2)).
The in-center I is the point of intersecton of CF and AD.
We find I:(-1,0) + μ(cos(σ),sin(σ)+1) with μ = (-sin((3ρ+σ)/2) - cos(ρ) - cos((ρ+σ)/2))/(sin((ρ-σ)/2) - sin(ρ+σ) - cos((ρ+σ)/2) - cos(ρ)).
After some calculation it turns out that the distance from F to I is equal to the distance of F to A and B, that is √(2+2sin(ρ)).
From the symmetry we conclude that I lies on the proposed arcs with centers D and E as well.

2994 Can the following construction be made without the use of a regulus? Construct a line which meets four given skew lines.

Partial solution:

From the following observation we see it must be possible to do so with the help of a regulus:

Let the four lines have parametrizations (x(λ),y(λ),z(λ)), (x(μ),y(μ),z(μ)), (x(ρ),y(ρ),z(ρ)), (x(σ),y(σ),z(σ)), where all twelve coordinate functions are linear.
Then in general we can find the parameter values of the points on the line we seek by solving four independent equations with four unknown values λ,μ,ρ,σ :
(x(λ)-x(μ))/(x(λ)-x(ρ)) = (y(λ)-y(μ))/(y(λ)-y(ρ)) = (z(λ)-z(μ))/(z(λ)-z(ρ)),
(x(λ)-x(μ))/(x(λ)-x(σ)) = (y(λ)-y(μ))/(y(λ)-y(σ)) = (z(λ)-z(μ))/(z(λ)-z(σ)).

(It is well known there are in general two lines that meet four given skew lines.)

2998 A cube has removed from it a right pyramid, whose base is a face of the cube and whose altitude is the altitude of the cube.
How far from the base of the cube must a plane be passed parallel to the removed face so as to divide the remaining volume of the cube into two equal parts?

Solution:

Let the cube be {0≤x≤1, 0≤y≤1, 0≤z≤1}, and the removed pyramid {0≤x≤1, 0≤y≤1-x, 0≤z≤x} ∪ {0≤x≤1, 1-x≤y≤1, 0≤z≤1-y}.
Let the dividing plane be z=a (a is some real between 0.5 and 1).

Because of the symmetry, we can restrict ourselve to the half cube where y≤1-x.
Hence we only demand

0a01-xxa dz dy dx + ∫a101-xax dz dy dx = ∫0a01-xa1 dz dy dx + ∫a101-xx1 dz dy dx .

After a bit of calculation we find 3a3 - 9a2 + 3a +1 = 0 and a = 0.6441329173543893..

2999 Given n positive numbers a1, ... , an , show that (Σ ai)*(Σ aj-1) > n2 unless ai = aj for all i and j.
Show by passing to limits that if φ(x) is positive and continuous, then ( ∫ab φ(x) dx)*( ∫ab φ(x)-1 dx) takes its minimum if φ(x) is constant over [a,b].

Solution:

The inequality holds since for distinct positive s and t holds s/t + t/s > 2.
The second assertion then follows by considering Riemann sums for the integrals, with φ(xi) = ai and δx = (b-a)/n.

3000 Construct a circle whose area is n times the area of a given circle.

Solution:

Let the given circle have radius 1.
We can construct the radius √(n) of the new circle as the length of the hypotenusa of a rectangular triangle with sides of length 1 and √(n-1).
So we can construct √(n) by constructing rectangular triangles with sides of length 1, √(k), √(k+1) for k = 1,2,..,n-1 successively.

3001 In the plane of a given circle, a second circle with a given radius is drawn so that the radical axis of the two circles passes through a given fixed point. Find the locus of the center of the second circle.

Solution:

Let the given circle be C: x2 + y2 = 1, the given point P:(p,q), and the given radius R > 0.
Let the second circle be S: (x-a)2 + (y-b)2 = R2.
Since P has the same power respective to C and S, we have (p-a)2 + (q-b)2 - R2 = p2 + q2 - 1,
so the locus of the center (a,b) is a circle with center (p,q) and radius √(R2 + p2 + q2 - 1), provided R2 + p2 + q2 ≥ 1.

3002 Let a and b be the lengths of the diagonals of a quadrilateral, and A the angle between them. Show that the greatest area of a rectangle that circumscribes the quadrilateral is ab(1 + sin(A))/2.

Solution:

This seems wrong, since in the image below A = φ + ψ 3004 Is there a plane curve such that two tangents whose lengths are in the constant ratio n:1 (n not equal to 1) may be drawn to it from any point in its plane? If so, discuss its properties.

Solution:

For n=1 there is the circle, if we replace "any point" with "any exterior point".
Of course, the proposer meant "any point outside the curve". I take it he meant "differentiable curve".
Without loss of generality we may suppose n > 1.
If there is such a curve, it must be a spiral.
But the answer is no: if we draw two tangents to the spiral from some point P, and one of the two contact points on the curve is n times as far away as the other, we may move P in the direction of the distant contact point, and the ratio of the distances to the two contact points (one of them new) will become less than n.

3005 (from genetics)
Given the recurrence formulae x(n+1) = x(n)(x(n)+y(n)), y(n+1) = y(n)(2x(n)+y(n)) with initial values x(1)=y(1)=1.
Find good approximations to x(n)/(2x(n)+y(n)) for n ranging from 20 to 30.

Solution:

We can easily find the values of x(n) and y(n) for small n, but for greater n these values soon become too great, even with the help of an electronic calculator.
However, we observe x(n+1)/(2x(n+1)+y(n+1)) = 1/(2+y(n+1)/x(n+1)).
With t(n) := y(n)/x(n) we find t(n+1) = t(n)(1 + 1/(1+t(n))), and the values of t(n) can be calculated for greater n.
The electronic calculator gives the following values of x(n)/(2x(n)+y(n)) = 1/(2+t(n)) for n = 20 through 30:
0.05306, 0.05053, 0.04823, 0.04611, 0.04418, 0.04239, 0.04074, 0.03921, 0.03779, 0.03646, 0.03523.

3007 Given two opaque spheres that don't intersect, locate the points on either from which the maximum surface of the other can be seen.

Solution:

Let M and N be the centers of the spheres, and AB a common tangent (see the figure).
The points on the upper sphere from where the maximum surface of the lower sphere can be seen are these points A on a 'circle of latitude' perpendicular to MN: from there you can see the part of the lower sphere above and including the circle through X and B perpendicular to AN.
(Likewise, the points on the lower sphere from where the maximum surface of the upper sphere can be seen are these points B on a 'circle of latitude' perpendicular to MN: from there you can see the part of the upper sphere beneath and including the circle through Y and A perpendicular to BM.) 3008 The altitude of a right circular cone is a, the radius of its base is b, and its slant height is c. A string is wrapped n times about the cone, starting at the vertex and ending at the base, in such a manner that for any complete circuit the vertical rise (the cone being supposed to rest on its base) is the same. A bird at the vertex takes the end of the string in its beak and flies around the cone, unwinding the string, keeping it taut and always tangent to the curve of the string as it lies around the cone. Find an expression for the distance that the bird has flown when the string is completely unwound (a) if it starts at the vertex (b) if it starts at the base.

Numerical solution:

Let the curve on the cone have parametrization x(t) and the curve of the bird y(t), 0 ≤ t ≤ 1. The core of a computer program to calculate this distance (in case (a)) is as follows:

dt:=0.000001; t:=1; xold:=(0,0,a); yold:=(0,0,a); arclength:=0; distance:=0;
while t ≥ 0 do
begin
t:=t-dt; xnew:=(b(1-t)cos(2πnt), b(1-t)sin(2πnt), at);
arclength:=arclength + ||xnew-xold||;
ynew:=xnew - arclength*(xnew-xold)/||xnew-xold||;
distance:=distance + ||ynew-yold||;
xold:=xnew; yold:=ynew
end

3009 Rectangular pieces of cardboard of the same dimensions are piled so that they overhang to the greatest extent possible; what curve do the edges touch?

Solution: It's mostly a kind of self-intersecting spiral, but it's a circle if 1/b = sin(2*π/n) for some divisor n≥5 of 360.

In hindsight I see other solvers make the k-th card from above overhang the pile under it with 100/(k+1) percent, so that the edges of the cards touch a logarithmic curve.

3012 ... Show how to construct a polygon with its vertices concyclic and with its sides in order equal to the sides of a given polygon.

Numerical solution:

We have to find the radius R of the circle wherein a polygon with the given sides can be inscribed.
If in this circle a side with length L is a chord, it's subtended by an angle α at the center of the circle equal to 2*arcsin(L/(2R)).
With trial and error and the help of a computer program, we can get close to the right value of R for which the sum of the angles 2*arcsin(L/(2R)) is equal to 2*π.

3014 If 1,p,d are the radius and sides of the regular inscribed pentagon and decagon, and if a triangle be formed from these three lengths, determine the angle opposite p without the use of trigonometric tables.

Solution:

Applying the cosine rule in the triangle with sides 1,1,d, we get d2 = 2 - 2 cos(36°) = 4 sin2(18°). Likewise, p2 = 2 - 2 cos(72°) = 4 sin2(36°).
Now applying the cosine rule in the triangle with sides 1,d,p, we get p2 = 1 + d2 - 2d cos(x°), where x is the angle we have to determine. Hence we find cos(x°) = (1 -12 sin2(18°) + 16 sin4(18°))/(4 sin(18°)).
But it's well known that sin(18°) = (√5 - 1)/4, which yields cos(x°) = 0 and then x = 90°.

3021 Prove that the two lines joining the points of intersection of two orthogonal circles to any point of one of them meet the other circle in two diametrically opposite points.

Provisional solution: See the picture. We may calculate the coordinates of X2 and Y2, and check that X2 and Y2 and M are collinear.

3029 To locate two points D and E on the sides AB and BC of a triangle ABC such that AD:DE:EC shall be equal to p:q:r, where p,q,r are given line segments.

Solution:

Let A:(0,0), B:(1,0), C:(R*cos(α),R*sin(α)).
We have to find m,D,E, such that AD = mp/(p+q+r), EC = mr/(p+q+r), and DE = mq/(p+q+r).
Take D:(mp/(p+q+r),0) and E:(R*cos(α),R*sin(α)) + (mr/(p+q+r))*((1-R*cos(α),-R*sin(α))/√(R2-2R*cos(α)+1).
Demanding DE = mq/(p+q+r), we find a quadratic equation in m and hence m.

3034 If every root of the equation f '(x) = 0 be subtracted from every root of the equation f(x) = 0, find the sum of the reciprocals of the differences.

Solution:

If f(x) is a polynomial, say f(x) = f = (x-a1)(x-a2) ... (x-an), where the zeros ai need not be distinct, then log(f) = Σ log(x-ai), so f '/f = Σ 1/(x-ai), and for each zero bj of f ' we find Σ 1/(bj-ai) = 0.
I think the proposer meant that f(x) may be approached by polynomials as closely as we want.
Then I think the general answer is 0, too.

3046 What is the probability that there will be at least r consecutive heads out of n tosses of a coin?

Solution:

If r is larger than n, the probability is 0.
Otherwise, the following events are disjoint and their probabilities sum up to the answer.
Event number k (0 ≤ k ≤ n-r) is: the first k tosses are tails, then follow r heads.
So the answer is the sum of 2-k-r, which is 21-r - 2-n.