- We change to polar coordinates. Thus
*x*=*r*cosq,*y*=*r*sinq, and*dA*=*r**drd*q. The circle (*x*- 1)^{2}+*y*^{2}= 1 becomes*r*^{2}- 2*r*cosq = 0, which simplifies to*r*= 2cosq. Also as one moves from (2, 0) to (0, 0) on the semicircle*C*(see diagram below), q moves from 0 to p/2. Therefore*x*^{3}/(*x*^{2}+*y*^{2}) d*A*= ( *r*^{3}cos^{3}q/*r*^{2})*rdrd*q =*r*^{2}cos^{3}q d*rd*q= (8/3)cos ^{6}q dq = (1/3)(1 + cos 2q)^{3}dq= (1/3)(1 + 3cos ^{2}2q) dq = 5p/12.

- Since
*r*_{1}*r*_{2}= 2, the roots*r*_{1},*r*_{2}will satisfy a quadratic equation of the form*x*^{2}+*px*+ 2 = 0, where*p***C**. Therefore we may factor*x*^{4}-*x*^{3}+*ax*^{2}- 8*x*- 8 = (*x*^{2}+*px*+ 2)(*x*^{2}+*qx*- 4)*q***C**. Equating the coefficients of*x*^{3}and*x*, we obtain*p*+*q*= - 1 and 2*q*- 4*p*= - 8. Therefore*p*= 1 and*q*= - 2. We conclude that*a*= - 4 and*r*_{1},*r*_{2}are the roots of*x*^{2}+*x*+ 2, so*r*_{1}and*r*_{2}are (- 1±*i*)/2. - The number of different combinations of possible flavors is the same
as the coefficient of
*x*^{100}in(1 +This is the coefficient of*x*+*x*^{2}+ ...)^{4}*x*^{100}in (1 -*x*)^{-4}, that is 103!/(3!100!) = 176851. - We can represent the possible itineraries with a matrix. Thus we let
*A*=( 1 3 1 1 ) and let

*a*_{ij}indicate the (*i*,*j*)th entry of*A*. Then for a one day period,*a*_{11}is the number of itineraries from New York to New York,*a*_{12}is the number of itineraries from New York to Los Angeles,*a*_{21}is the number of itineraries from Los Angeles to New York, and*a*_{22}is the number of itineraries from Los Angeles to Los Angeles. The number of itineraries for an*n*day period will be given by*A*^{n}; in particular the (1, 1) entry of*A*^{100}will be the number of itineraries starting and finishing at New York for a 100 day period.To calculate

*A*^{100}, we diagonalize it. Then the eigenvalues of*A*are 1± and the corresponding eigenvectors (vectors**u**satisfying*A***u**= l**u**where l = 1±) are (±, 1). Therefore if*P*=( - 1 1 ) then

*P*^{-1}*AP*=( 1 + 0 0 1 - ) Thus

*A*^{100}=*P*( (1 + ) ^{100}0 0 (1 - ) ^{100}) *P*^{-1}We conclude that the (1, 1) entry of

*A*^{100}is ((1 + )^{100}+ (1 - )^{100})/2, which is the number of itineraries required. - For each city
*x*in S, let*G*_{x}S denote all the cities which you can travel from*x*(this includes*x*). Clearly*G*_{x}is well served and |*G*_{x}|__>__3 (where |*G*_{x}| is the number of cities in*G*_{x}). Choose*x*so that |*G*_{x}| is minimal. We need to show that if*y*,*z**G*_{x}, then one can travel from*y*to*z*stopping only at cities in*G*_{x}; clearly we need only prove this in the case*z*=*x*. So suppose by way of contradiction*y**G*_{x}and we cannot travel from*y*to*x*stopping only at cities in*G*_{x}. Since*G*_{y}*G*_{x}and*x**G*_{y}, we have |*G*_{y}| < |*G*_{x}|, contradicting the minimality of |*G*_{x}| and the result follows. - Let
*O*denote the center of the circle with radius 2cm., let*C*denote the center of the disk with radius 1cm., and let*H*denote the hole in the center of the disk. Choose axes so that the origin is at*O*, and then let the initial position have*C*and*H*on the positive*x*-axis with*H*furthest from*O*. The diagram below is in general position (i.e. after the disk has been moved round the inside of the circle). Let*P*be the point of contact of the circle and the disk, (so*OCP*will be a straight line), let*Q*be where*CH*meets the circumference of the disk (on the*x*-axis, though we need to prove that), and let*R*be where the circle meets the positive*x*-axis. Since the arc lengths*PQ*and*PR*are equal and the circle has twice the radius of that of the disk, we see that*PCQ*= 2*POR*and it follows that*Q*does indeed lie on the*x*-axis.Let (

*a*,*b*) be the coordinates of*C*. Then*a*^{2}+*b*^{2}= 1 because the disk has radius 1, and the coordinates of*H*are (3*a*/2,*b*/2). It follows that the curve*H*traces out is the ellipse 4*x*^{2}+ 36*y*^{2}= 9. We now use the formula that the area of an ellipse with axes of length 2*p*and 2*q*is p*pq*. Here*p*= 3/2,*q*= 1/2, and we deduce that the area enclosed is 3p/4.

- Let
*x*= {*x*_{0},*x*_{1},...,*x*_{n}} J. Then

*Tx*= *LA*({*x*_{0},*x*_{0}+*x*_{1},*x*_{0}+*x*_{1}+*x*_{2},...})= *L*({1 +*x*_{0}, 1 +*x*_{0}+*x*_{1}, 1 +*x*_{0}+*x*_{1}+*x*_{2},...})= {1, 1 + *x*_{0}, 1 +*x*_{0}+*x*_{1}, 1 +*x*_{0}+*x*_{1}+*x*_{2},...}.

Therefore*T*^{2}*y*=*T*({1, 2, 3,...}) = {1, 1 + 1, 1 + 1 + 2, 1 + 1 + 2 + 3,...}. We deduce that*T*^{2}*y*= {1, 2, 4, 7, 11, 16, 22, 29,...} and in general (*T*^{2}*y*)_{n}=*n*(*n*+ 1)/2 + 1.Suppose

*z*= lim_{i - > }*T*^{i}*y*exists. Then*Tz*=*z*, so 1 =*z*_{0}, 1 +*z*_{0}=*z*_{1}, 1 +*z*_{0}+*z*_{1}=*z*_{2}, 1 +*z*_{0}+*z*_{1}+*z*_{2}=*z*_{3}, etc. We now see that*z*_{n}= 2^{n}. To verify this, we use induction on*n*, the case*n*= 0 already having been established. Assume true for*n*; then*z*_{n + 1}= 1 +*z*_{0}+*z*_{1}+ ... +*z*_{n}= 1 + 1 + 2 + ... + 2^{n}= 2^{n + 1},