Let L be the side of the square, 50m, and let D be the distance the dog travels.

Let v1 be the soldiers' marching speed and v2 be the speed of the dog. Then v1 = L / (1 time unit) and v2 = v1*D/L.

Let t1, t2, t3, t4 be the time the dog takes to traverse each side of the square, in order. Find t1 through t4 in terms of L and D and solve t1+t2+t3+t4 = 1 time unit.

While the dog runs along the back edge of the square in time t1, the soldiers advance a distance d=t1*v1, so the dog has to cover a distance sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2. Solving for t1 gives t1=L/sqrt(v2^2 - v1^2).

The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1).

In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by using v1=L/(1 time unit), obtaining

2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1

which can be turned into
D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0

which has a root D = 4.18113L = 209.056m.

e^(pi). Put x = pi/e - 1 in the inequality e^x > 1+x (x>0).

1) Assuming f finite everywhere, (*) ==> x<>y ==> f(x)<>f(y)

2) Exchanging x and y in (*) we see that f(-x) = -f(x).

3) a <> 0 ==> f((a-a)/(a+a)) = (f(a)-f(a))/(f(a)+f(a)) ==> f(0) = 0.

4) a <> 0 ==> f((a+0)/(a-0)) = f(a)/f(a) ==> f(1) = 1.

5) x<>y, y<>0 ==> f(x/y) = f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) = f(x)/f(y)
==> f(xy) = f(x)f(y) by replacing x with xy and by noting that f(x*1) = f(x)*1 and f(x*0) = f(x)*f(0).

6) f(x*x) = f(x)*f(x) ==> f(x) > 0 if x>0.

7) Let a=1+\/2, b=1-\/2; a,b satisfy (x+1)/(x-1) = x ==>

f(x) = (f(x)+1)/(f(x)-1) ==> f(a)=a, f(b)=b. f(1/\/2) = f((a+b)/(a-b)) = (a+b)/(a-b) = 1/\/2 ==> f(2) = 2.

8) By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1) we get that f(n)=n for all integer n. #5 now implies that f fixes the rationals.

9) If x>y>0 (*) ==> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6. Thus f is order-preserving.
Since f fixes the rationals *and* f is order-preserving, f must be the identity function.

This was E2176 in _The American Mathematical Monthly_ (the proposer was R. S. Luthar).

By induction f(mx) = m(f(x)+C)-C.

Let x=1/n, m=n and find that f(1/n) = (1/n)(f(1)+C)-C.

Now let x=1/n and find that f(m/n) = (m/n)(f(1)+C)-C.

f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x)

(since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C.

Since f is monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem).

First, note that if f(0) is 0, then by substituting u=ax in the integral of f(x)/x, our integral is the difference of two equal integrals and so is 0 (the integrals are finite because f is 0 at 0 and differentiable there. Note I make no requirement of continuity).

Second, note that if f is the characteristic function of the interval [0, 1]--- i.e.

f (x) = 1, 0<=x<=1
f (x) = 0, otherwise

then a little arithmetic reduces our integral to that of 1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar), which is ln(a) = f(0)ln(a) as required. Call this function g.

Finally, note that the operator which takes the function f to the value of our integral is linear, and that every function meeting the hypotheses (incidentally, I should have said `differentiable from the right', or else replaced the characteristic function of [0,1] above by that of (-infinity, 1]; but it really doesn't matter) is a linear combination of one which is 0 at 0 and g, to wit

f(x) = f(0)g(x) + (f(x) - g(x)f(0)).

Yes. Center the stamp at (0,0), (1,0), and (0,pi).

Suppose there is a point (x,y) which is not covered. Then there are rational numbers a,b,c satisfying the following equations:

(1) x^2 + y^2 = a
(2) (x-1)^2 + y^2 = b
(3) x^2 + (y-pi)^2 = c

Subtract (2) from (1) and solve for x. Thus x is rational. From equation (2), y is algebraic. But equation (3) implies that y is transcendental, contradiction.

Let the speed of the car be v, the speed of a walking person be u, and the distance between cities by L. We want to minimize T, the time t at which all persons are at displacement x=L (city B), when they all start at displacement x=0 (city A) at time t=0.

I'll assume that the solution has everyone starting out from city A at the same time t=0 arriving in city B at the same time t=T so nobody is standing around idly waiting. Let's plot everyone's movements on a graph showing coordinates (t,x). Then at time t just after 0, (N-2) walkers are on the line L0 through (0,0) with slope dx/dt = u, and 2 in the car are on a line through (0,0) with slope v, and at t just before T, (N-2) walkers are on the line L1 through (T,L) with slope u, and 2 in the car are on a line through (T,L) with slope v. Obviously L1 lies "above" L0 (greater x coordinate given the same t coordinate). In between t=0 and t=T, the car zigzags between L0 and L1 along lines of slope v and -v, picking up people from L0 and dropping them off at L1. I will not prove that this is the optimal strategy; in fact you can make an infinite number of variations on it which all come up with the same elapsed time.

Now examine the graph again. Say the car travels distance r between picking someone up and dropping that person off, and distance s back to pick up the next person. The car makes (N-1) trips forward and (N-2) trips back to pick up and ferry everyone, so its total travel is

vT = (N - 1)r + (N - 2)s = (N - 2)(r + s) + r

Moreover the car makes (r-s) net displacement on each round trip, plus r displacement on the extra forward trip, so

L = (N - 2)(r - s) + r

Note that a person walks distance (r-s) in the time it takes the car to go (r+s), so

r - s = (u/v)(r + s)

A little algebraic manipulation of this equation shows us that

r - s = r * 2u/(v + u)
r + s = r * 2v/(v + u)

Plug this into the equation for L, and we get the first important piece of information, how far the car should drive before dropping off the passenger (once you know this, you tell it to the driver and this guarantees the people get to B in minimum time):

L = r + (N - 2) r * 2u/(v + u)

Note that a person walks distance (r-s) in the time it takes the car to go (r+s), so

r - s = (u/v)(r + s)

A little algebraic manipulation of this equation shows us that

r - s = r * 2u/(v + u)
r + s = r * 2v/(v + u)

Plug this into the equation for L, and we get the first important piece of information, how far the car should drive before dropping off the passenger (once you know this, you tell it to the driver and this guarantees the people get to B in minimum time):

L = r + (N - 2) r * 2u/(v + u)
L = r * (v + u + (N - 2)*2u)/(v + u)
r = (v + u)*L/(2uN + v - 3u)

We can also find out what the elapsed time T will be:

vT = (N - 2)r*2v/(v + u) + r
vT = r * ((N - 2)*2v + v + u)/(v + u)
T = (1/v) * (2vN - 3v + u)*r/(v+u)

Therefore T = (2vN - (3v - u))*(L/v)/( 2uN + v - 3u)

-- David Karr (karr@cs.cornell.edu)

Assumptions:

1. x(0) = 0; x(T) = X

2. v(0) = 0; v(T) = V

3. d(a)/dt <= 0

Solution:

a(t) = constant = A = V^2/2X which implies T = 2X/V.

Proof:

Consider assumptions as they apply to f(t) = A * t - v(t):

1. integral from 0 to T of f = 0

2. f(0) = f(T) = 0

3. d^2(f)/dt^2 <= 0

From the mean value theorem, f(t) = 0. QED.

Period 2. Clearly, the sum of periodic functions of periods 2 and three is 6. So take the function which is the sum of that function of period six and the negative of the function of period three and you have a function of period 2.

This proves that a period-2 solution exists, but not that it is minimal. Since we're talking about integers, the only lower possibility is 1. But the sum or difference of a period-1 and a period-3 function must have period 3, not 6, therefore 1 is indeed impossible.

Let w = speed of bug and N = ratio of car speed/bug speed = 100. Paint N+1 equally spaced stripes on the rubberband. When the bug is standing on one stripe, the next stripe is moving away from him at a speed slightly < w (relative to him). Since he is walking at w, clearly the bug can reach the next stripe. But once he reaches that stripe, the next one is only receeding at < w. So he walks on down to the car, one stripe at a time.

The bug starts gaining on the car when he is at the next to last stripe.

Throw 0 into the sequence; there are now n numbers, so some pair must have fractional parts within 1/n of each other; their difference is then within 1/n of an integer.

Answer:11:22:55.077 am.

Method:
Let b = the depth of the snow at noon, a = the rate of increase in the depth. Then the depth at time t (where noon is t=0) is at+b, the snowfall started at t_0=-b/a, and the snowplow's rate of progress is ds/dt = k/(at+b).

If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for x and t_0 = -(1 hour)/x.

The exact answer is 11:(90-30 Sqrt[5]).

_American Mathematics Monthly_, April 1937, page 245 E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa.

The solution appears, appropriately, in the December 1937 issue, pp. 666-667. Also solved by William Douglas, C. E. Springer, E. P. Starke, W. J. Taylor, and the proposer.

See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff.

ANSWER: e^(1/e)

Let N be the number in question and R the result of the process. Then R can be defined recursively by the equation:

(1) R = N^R

Taking the logarithm of both sides of (1):

(2) ln(R) = R * ln(N)

Dividing (2) by R and rearranging:

(3) ln(N) = ln(R) / R

Exponentiating (3):
(4) N = R^(1/R)

We wish to find the maximum value of N with respect to R. Find the derivative of N with respect to R and set it equal to zero:

(5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0

For finite values of R, (5) is satisfied by R = e. This is a maximum of N if the second derivative of N at R = e is less than zero.

(6) d2(N)/d2(R) | R=e = (2 * ln(R) - 3) / R^3 | R=e = -1 / e^3 < 0

The solution therefore is (4) at R = e:
(7) Nmax = e^(1/e)