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.


7 miscellaneous, reverend Gridley (From the second issue of 1894)
Making no allowance for the curvature of the earth, and supposing the sun to rise in the east and set in the west, what would be the course of a man who should walk constantly toward the sun from morning until night? How far and in what direction from the starting point would he be, walking three miles per hour, at the end of three days?

Partial solution:

In fact it's easier to allow for the earth's spherical curvature, if only we let the man walk on the equator on a day that runs from 6 o'clock in the morning to 6 o'clock in the evening, starting at noon of the first day.
It turns out the man walks 0.0432 degrees per hour and the sun turns 15 degrees per hour.
The man and the sun first meet after t1 hours at position x1 degrees where x1 = t1*0.0432 = 90 - 15*t1, so t1 = 5.98277 hours, x1 = 0.25845565 degrees.
The rest of the first day, during t2 = 0.01723 hours, the man follows the sun, walking over 0.01723*0.0432 = 0.000744 degrees to position x2 = 0.25845565 - 0.000744 = 0.25771165 degrees.
Next morning, the man and the sun meet after t3 hours at position x3 degrees where x3 = x2 + 0.0432*t3 = 90 - 15*t3, so t3 = 5.965638 hours, and x3 = 0.515427 degrees. The rest of the second day, during t4 = 12 - 5.965638 = 6.034362 hours, the man follows the sun, walking over ... (etc.).


14 geometry (From the very first issue of 1894) Through a given point to draw four circles tangent to two given circles.

Sketch of a solution:

Let the given point be (0,0). Let the center and radius of one the given circles be (a,b) and r, and those of the other (d,e) and s.
We seek a circle with center (u,v) and radius t = √(u2+v2) which is tangent to both given circles.
In any case this amounts to √((u-a)2+(v-b)2) = ± (r ± t) and √((u-d)2+(v-e)2) = ± (r ± s) for certain values of ±.
Using (u,v) = p(cos(t),sin(t)) we find:

2ap cos(t) + 2bp sin(t) = a2+b2-r2±2rp and 2dp cos(t) + 2ep sin(t) = d2+e2-s2±2sp.
Solving for cos(t) and sin(t), and using sin2(t) + cos2(t) = 1, we find in the end four values of (p,cos(t),sin(t)) which may be constructed with compass and straightedge.

This reasoning may be extended to: 13 Through a given point to pass four spherical surfaces tangent to two given spheres.


15 calculus From a given quantity of material a cylindrical cup with circular bottom and open top is to be made, the cup to contain the greatest amount. What must be its dimensions?

Solution:

Denote by g the volume of the material, by d the thickness of cylinder and bottom, by r the radius of the circular bottom inside the cup, and by h the height of the cylinder inside the cup.
Then we have the restriction g = πr2d + 2πrhd, or r2 + 2rh = a for a:=g/(πd), so h = (a - r2)/(2r).
Now we must maximize the contents of the cup, that is hπr2 = (π/2)(ar - r3).
With (d/dr)(ar - r3) = 0 we find r = √(a/3) and then, using the restriction, h = r.


17 geometry Draw a circle bisecting the circumferences of three given circles.

Correction:

This isn't always possible.

Give coordinates in such a way that the given circles have equations x2 + y2 = 1, (x-a)2 + (y-b)2 = r2, (x-c)2 + (y-d)2 = s2
and so that the circle we seek passes through (-1,0) and (1,0) and has equation x2 + (y-q)2 = 1 + q2.

There must exist some φ such that both (a + r cos(φ),b + r sin(φ)) and (a -r cos(φ),b - r sin(φ)) lie on x2 + (y-q)2 = 1 + q2,
and some ψ such that both (c + s cos(ψ),d + s sin(ψ)) and (c - s cos(ψ),d - s sin(ψ)) lie on x2 + (y-q)2 = 1 + q2.
It follows that q = (a2 + b2 + r2 - 1)/(2b) = (c2 + d2 + s2 - 1)/(2d),
and the latter equality is imposing a restriction on the given circles.


19 geometry If any point be taken on the circumference of a circle, and lines be drawn from it to the angles of an inscribed equilateral triangle, prove that the middle line so drawn is equal to the sum of the other two.

Solution:

Give coordinates in such a way that the arbitrary point is P(cos(t),sin(t)) for some t between 0 and π/2, and the angles are at A(1,0), B(-1/2,√3/2), C(-1/2,-√3/2).
It is straightforward to check that PB = PA + PC.


90 average and probability (january 1900)
During a heavy rain storm a circular pond is formed in a circular field. If a man undertakes to cross the field in the dark, what is the chance that he will walk into the pond?

Criticism and partial solution:

Give coordinates in such a way that the circular field has center (0,0) and radius 1, and the man starts from (-1,0).
We take for granted the man walks along (-1,0) + λ(cos(θ),sin(θ)), λ > 0, where every direction θ between -π/2 and π/2 is equally probable.
Now the circular pond may have equation (x - ρcos(α))2 + (y - ρsin(α))2 = R2 for any α ∈ [0,2π) and any ρ < 1 and any R between 0 and 1-ρ.
We assume every α is equally probable, so the probability of α lying in an infinitesimal range δα is δα/(2π).
We also assume that the distance of the center of the pond from the center of the field is lying between ρ and ρ + δρ for some positive infinitesimal δρ with a chance equal to 2πρ δρ/π.
But we can't make any reasonable assumption about the radius R of the pond, given ρ, without further info.
We have to know the probability density function fρ(R) with ∫01-ρ fρ(R) δR = 1.

We now calculate the probability of the man walking into the pond if α, ρ and R are given.
This probability is P(α,ρ,R) = (θ2 - θ1)/π, with θi corresponding to the pair of walks that are tangent to the pond.

I find cos(2θi ) = (BC ± A √(A2 + B2 - C2))/(A2 + B2), where
A = ρsin(α) + ρ2sin(α)cos(α),
B = 1/2 + ρcos(α) + ρ2cos(2α)/2,
C = ρ2/2 + ρcos(α) + 1/2 - R2.

The chance asked for is ∫00101-ρ P(α,ρ,R) fρ(R) 2πρ δR δρ δα/(2π2).


171 number theory and diophantine equations (march 1910)
Solve completely: 2x2 - 1 = y, 2y2 - 1 = z, 2z2 - 1 = w, 2w2 - 1 = x.

Solution:

We find:
y = 2x2 - 1,
z = 8x4 - 8x2 + 1,
w = 128x8 - 256x6 + 160x4 - 32x2 + 1,
x - 1 = (215x16 - 217x14) + (13*214x12 - 11*214x10) + (165*29x8 - 21*210x6) + (21*27x4 - 27x2).
Now we see: if |x| ≤ 1 then x=1 and if |x| ≥ 2 there is no solution.
So the only solution is x=y=z=w=1.


208 average and probability (december 1911)
Find the chance that the distance between two points in a square shall not exceed a side of the square.

Numerical solution:

If I take two random points an indeterminate number of times, the outcome seems to fluctuate a bit below 0.975.

The corresponding pascal program is the following:
program chance;
var a,b,x1,x2,y1,y2:real;
begin
  a:=0; b:=0;
  while true do
    begin
      x1:=random; x2:=random;
      y1:=random; y2:=random;
      if (((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)) <= 1) then a:=a+1 else b:=b+1;
      writeln(a/(a+b):13:11)
    end
end.

If I take 100*100*100*100 lattice points, the outcome is 0.9750...

The corresponding pascal program is the following:
program integraal;
var j,k,m,n:integer; p,q:real;
begin
  p:=0; q:=100*100*100*100;
  for j:=0 to 99 do for k:=0 to 99 do
  for m:=0 to 99 do for n:=0 to 99 do
    if (((j-k)*(j-k)+(m-n)*(m-n)) <= 10000) then p:=p+1;
  writeln(p/q:10:8); readln
end.

But if I take 200 and then 300 and then 400 and then 500 instead of 100, it becomes clear the right answer is 0.9749 in four decimals and the fifth decimal is probably 3.


295 calculus (june 1910)
A hawk can fly v feet/second, a hare can run w feet/second. The hawk, when A feet vertically above the hare, gives chase and catches the hare when the hare has run B feet. Find the length of the curve of pursuit.

Solution and criticism:

We suppose both go always at maximal speed.
If the hare chooses a path with parametrization x(t), and he runs his B feet during a time of T seconds, we have B = ∫0T ||x'(t)|| dt = Tw.
If the curve of pursuit of the hawk has parametrization y(t), and he flies L feet during the T seconds, we have L = ∫0T ||y'(t)|| dt = Tv = Bv/w.
But, in fact, B can't be known beforehand, and neither can x(t).
If we knew x(t), we could find y(t), and hence T and B, by solving y' = w(x-y)/||x-y|| with x(0)=(0,0,0) and y(0)=(0,0,A).


362 number theory (october 1911)
Show that the number of solutions in non-negative integers of the equation x + 2y + 3z = 6n is equal to 3n2 + 3n + 1.

Solution:

We proceed with induction.
The assertion is clearly true for n=0.
Suppose the assertion is true for n=m, i.e. suppose the number of solutions in non-negative integers of the equation x + 2y + 3z = 6m is 3m2 + 3m + 1.
Then any of these solutions, say (x,y,z) = (a,b,c), gives a solution (a+1,b+1,c+1) of x + 2y + 3z = 6(m+1).
On the other hand, each solution (A,B,C) of x + 2y + 3z = 6(m+1) comes from a solution (A-1,B-1,C-1) of x + 2y + 3z = 6m, unless ABC = 0.
Furthermore, the number of solutions of x + 2y = 6m+6 is 3m + 4, the number of solutions of x + 3y = 6m+6 is 2m + 3, and the number of solutions of 2y + 3z = 6m+6 is m + 2,
and among these solutions with ABC = 0, three of them are counted twice.
So the total number of solutions of x + 2y + 3z = 6m+6 is (3m2 + 3m + 1) + (3m + 4) + (2m + 3) + (m + 2) - 3 = 3(m+1)2 + 3(m+1) + 1.


382 geometry (february 1911)
Between the side of a given rhombus and its adjacent side produced, to insert a straight line of a given length and directed to the opposite corner. (Euclidean constructions are particularly desired.)

Numerical solution:

Let the given length be L, so L is a given positive real.
Take coordinates so that the vertices of the rhombus are A(-1,0), B(0,t), C(1,0) and D(0,-t) for some given positive real t.
Consider the side AD and take for positive real p a variable point P(-1-p,-pt) on BA produced. We have to find p so that the line through P and C intersects AD in Q with PQ=L.
We find p4(1+t2) + 4p3 + 4p2 = L2(p+1)2.
Now if we don't want to use the formulas of Cardano cs, we may find a rational approximation for p that's as accurate as we want.


389 geometry (april 1911)
On the same side of a given base, triangles are erected such that the bisectors of their vertex angles all pass through a given point. Find the locus of the vertices.

Solution:

Let the given base be AB with A(-1,0) and B(1,0). Let the given point be P(a,b) for some negative b.
Let the vertex be V(s,t) for some positive t.

For the vertex angle φ we find cos(φ) = (s2+t2-1)/(√((s+1)2+t2)√((s-1)2+t2)).

Now we demand that the angle AVP be φ/2, so cos(φ/2) = (s2+t2+s-as-bt-a)/(√((s+1)2+t2)√((s-a)2+(t-b)2)).

The required locus is found by cos(φ) = 2cos2(φ/2) - 1, which yields

(s2+t2-1)/(√((s+1)2+t2)√((s-1)2+t2)) = 2((s2+t2+s-as-bt-a)/(√((s+1)2+t2)√((s-a)2+(t-b)2)))2 - 1.

We can plot this for any given a,b, using https://www.desmos.com/calculator.


399 geometry (january 1912)
A race track is to be composed of two tangents and the arc of a circle which is concave towards the point of intersection of the two tangents, each tangent and the arc of the circle being 1 mile.
What is the radius of the circle?

Solution:

If the tangents touch the circle in r(-cos(t),sint(t)) and r(cos(t),sin(t)), we find r = tan(t) = 1/(π-2t).
Hence t is about 0.405235 radians and r is about 0.429 mile.
I don't think a much more exact answer is possible.


2958 Over a frictionless pulley a weightless cord sustains at one end a mass M, while the other end is wound on an axle of a wheel of mass M and moment of inertia N.
At the zero of time the wheel revolves with angular velocity ω and tends to wind up the cord.
Describe the motion, neglecting friction.

Solution:

For the angular velocity at time t we have ω(t) = dφ(t)/dt = dr(t)/dt, where the angle φ(t) and the loose end of length r(t) of the cord satisfy φ(t) - φ(0) = r(t) - r(0) (mod 2π).
Since the moment of inertia comes from gravity, we have N = Mg = (dω(t)/dt) M r(t)2 = (d2r(t)/dt2 M r(t)2 as long as ω(t) ≤ 0, so
r(t)2 (d2r(t)/dt2 = g, with r(0) = r0 and r'(0) = ω.
To solve this differential equation we try a power series r(t) = r0 + ωt + r2t2 + r3t3 + r4t4 + ...
We find r2 = g/(2r02), r3 = -gω/(3r03), r4 = (g2-3gr0ω2)/(12r05), etc.


2959 Solve the functional equation g(x)2 = -2x + g(x2).

Solution:

If g is a real solution then 1 and -1 are not in its domain, because there's no real c with c2 = -2 + c.
Furthermore, g(0) must be either 0 or 1.

Whenever A is a real that doesn't belong to {0,1,-1}, and c is any real, there exists infinitely many solutions with g(A) = c and g(A2) = c2 + 2A.
For example, if g(1/2) = c then g(1/4) = c2 + 1 and g(1/16) = (c2 + 1)2 + 1/2, etc.
This solution may be combined with, for instance, g(1/3) = c' and g(1/9) = c'2 + 2/3, etc, for any real c'. However, g(-1/2) must be equal to ± g(1/2).

If the domain of g consists of all reals not equal to 1 or -1, g can't be continuous, because the sequence g(1/2), g(1/4), (g(1/16) ... can't tend to g(0).


2965 If a quadrilateral inscribed in a square has the diagonals a and b, and the area A, show that the area of the square is (a2b2-4A2)/(a2+b2-4A).

Sketch of a solution:

Give coordinates so that the square has vertices (0,0), (r,0), (r,r) and (0,r).
Let the quadrilateral have vertices (0,s) and (t,0).
Then the other two vertices must be ((s(r-s)/t,r) and (r,t(r-t)/s) and we must have r = (t3+t2s+ts2+s3)/(t2+s2).
So for the diagonals we find a = √(t4+s4-2s3r-2t2sr+2s2t2+s2r2+r2t2)/t and b = √(t4+s4-2t3r-2s2tr+2s2t2+s2r2+r2t2)/s,
and for the area we find A = √(s2+t2)√(r2s2+t2s2-2rs2t+t2r2+t4-2t3r)/s.

Now we still have to work out that r2 equals (a2b2-4A2)/(a2+b2-4A), but perhaps 4A is an error and should be should be 4A2.
We leave this to someone more diligent.


2967 A plane revolves about one of two non-coplanar lines as an axis. Find the locus, in the plane, of the intersection of the plane and the other line.

Solution:

By the principle of relativity, we may keep the plane fixed and let the rest of space merrily go around.
Let the plane have parametrization (x,y,0) and the axis λ(sin(α),0,cos(α)) and the other line (a,b,0) + μ(p,q,1).
During the revolution of the axis about the normal of the plane, the angle between the direction vector of the axis and the normal vector (0,0,1) of the plane remains α, so after φ degrees of revolution the revolved axis has parametrization λ(sin(α)cos(φ),sin(φ),cos(α)).
Then the other line in its revolved position has parametrization ((a+μp)cos(φ),(b+μq)sin(φ),μ) and its point of intersection with the plane is (a cos(φ),b sin(φ),0).
So the locus of this point of intersection is an ellipse.


2970 For which positive integers k,m,n is k times cosm(2π/n) a quadratic integer?

Partial solution:

For n=1, any m, and k a quadratic integer;
for n=2, even m, and k a quadratic integer;
for n=3, even m, and k equal to (2m times a quadratic integer).
for n=4, and arbitrary m and k;
for n=6, any m, and k equal to (2m times a quadratic integer);
for n=8, even m, and k equal to (2m/2 times a quadratic integer).

For each n, we can find (infinitely many) k iff cosm(2π/n) is a (positive) rational number u/v: take k := (v times (any j such that j times u is a quadratic integer)).
I guess the only other possible n is n=12: then for even m we have u/v = (3/4)m/2.


2971 Prove in an analytical way that the two tangents drawn to an ellipse from any external point subtend equal angles at a focus.

Sketch of a solution:

Give coordinates such that the ellipse E has equation x2/a2 + y2/b2 = 1 (a ≥ b) with focuses F(±c,0) where c = √(a2-b2).
Let S and T be any two points on E, say S(a cos(s), b sin(s)), T(a cos(t), b sin(t)) for s,t ∈ [0,2π).
The tangents at S and T intersect at P(a(sin(s)-sin(t))/sin(s-t),b(cos(t)-cos(s))/sin(s-t)).
We have to prove that the angles ∠PFS and ∠PFT are equal, which could be done straightforwardly by using the cosine rule.


2972 We learn about the following proposed "approximate compass and straightedge tripartitions of 'any' angle":

(a) Let β be the angle ABC, where AB=CB. Find P within this angle so that the line through P and C is parallel to the line through A and B and PC=AC.
Then the angle φ := ABP is approximately equal to β/3.

(b) In the situation of (a), find Q on CP so that CQ=2*AB and find R between P and Q so that PR:RQ = 5:4.
Then the angle φ := ABR is approximately equal to β/3.

Study these approximations and discuss how good they are.

(c) Propose better approximations yourself.

Solution:

(a) After applying the cosine rule in triangles ABC, ACP, BCP and ABP, respectively, and using some goniometrical formulas, we find :

cos(φ) = (1 + 4(4t2-1)3(1-t2)√(1-t2) - 2(4t2-1)2(1-t2) + 2(4t2-1)√(1-t2)(32t6-48t4+18t2-1)) / √(1 + 4(4t2-1)2(1-t2) + 4(4t2-1)√(1-t2)(32t6-48t4+18t2-1)) , with t = cos(β/6).

Now we calculate φ for each integer β in the range from 1 to 179 degrees with the help of a device we call 'computer'.
We see the estimate φ is too large for small β and too small for large β.
It turns out φ is only between 80 and 120 percent of β/3 for β between 87 and 128 degrees. And β/3 = φ only for β=108 degrees.

(b) In a way analogous to (a), we find:

cos(φ) = (19 + 8u - 18u2) / √(361 + 304u -296u2 -288u3), with u = sin(β/2).
(After my experience with (a), I don't think it's useful to express u in t = cos(β/6).)

Of course, since φ is smaller than in (a), we expect this estimate to be better than in (a) for small angles, and worse for large angles.
It turns out φ is between 80 and 120 percent of β/3 for β between 46 and 121 degrees. And the best fit is φ=1.002*β/3 for β=90 degrees.
So this one is a reasonable estimate for a much larger range of β.

(c) In order to find better approximations, we may construct a line segment of length z:= p*AC + q*AB for rational p and q.
Let T be the point within angle ABC so that CT is parallel to AB and T is at distance z from C.
Let φ be the angle ABT.

Again, by using the cosine rule four times, we find:
cos(φ) = (1+2pu +q-2u2) / √(1+(2pu+q)(2pu+q+2-4u2)), with u = sin(β/2).

Now for all p and q such that 90p and 90q are integers belonging to {0,1,..,360}, we calculate the number of integer β between 0 and 180 degrees such that φ is between 80 and 120 percent of β/3.
We find this number is maximal for p=0 and q=135/90=3/2, and then the range of β for which the approximation is reasonable is {1,2,3...,147}.

When proposers of problems wouldn't lead us astray, we would simply divide AC in three equal parts:
Take S on AC so that AS=AC/3. Let φ:= angle ABS.
Then cos(φ) = (3-2u2)/(sqrt(9-8u2), and φ is between 80 and 120 percent of β/3 for all β ≤ 111 degrees (and, in fact, φ is always smaller than β/3).
So this one is better than the constructions proposed in (a) and (b), if we only look at the range of β for which the estimate is reasonable.

If we take S on AC so that AS=a*AC/3 for some rational a between 1 and 1.3 for which 100a is integer, we get cos(φ) = (3-2*a*u2)/(sqrt(9-12a*u2+4*a2u2).
It turns out the range of reasonable β is maximal for a=1.19. Then this range is {1,2,..,146}.
But I think B and the best points S and T must be collinear, which holds for p=0, q=3/2, a=1.2 (with AS = 2AC/5).
The difference beween 1.19 and 1.2 clearly stems from rounding errors.

We conclude the best (and easiest to perform) approximate straightedge and compass angle tripartition that we found is as follows:
1) Construct through C a line parallel to AB.
2) Find on this line the point T within angle ABC so that TC = 3AB/2.
3) The angle φ := ABT is a reasonable (between 80 and 120 percent) estimate of β/3 if β is at most 147 degrees.
(We also find φ is between 90 and 110 percent of β/3 for β between 99 and 139 degrees, and the estimate is best for β=124 degrees.)


Further solutions: click here .


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