- Let
*P*denote the center of the circle. Then ∠*ACP*= ∠*ABP*= π/2 and ∠*BAP*= α/2. Therefore*BP*=*a*tan(α/2) and we see that*ABPC*has area*a*^{2}tan(α/2). Since ∠*BPC*= π - α, we find that the area of the sector*BPC*is (π/2 - α/2)*a*^{2}tan^{2}(α/2). Therefore the area of the curvilinear triangle is*a*^{2}(1 + α/2 - π/2)tan^{2}(α/2). - If we differentiate both sides with respect to
*x*, we obtain 3*f*(*x*)^{2}*f'*(*x*) =*f*(*x*)^{2}. Therefore*f*(*x*) = 0 or*f'*(*x*) = 1/3. In the latter case,*f*(*x*) =*x*/3 +*C*where*C*is a constant. However*f*(0)^{3}= 0 and we see that*C*= 0. We conclude that*f*(*x*) = 0 and*f*(*x*) =*x*/3 are the functions required. - We are given that
α satisfies
(1 +
*x*)*x*^{n+1}= 1, and we want to show that α satisfies (1 +*x*)*x*^{n+2}=*x*. This is clear, by multiplying the first equation by*x*. - Set
*f*(*x*) =*x*^{n}/(*x*+ 1)^{n+1}, the left hand side of the inequality. Then*f'*(*x*) =*x*^{n-1}(*n*-*x*)/(*x*+ 1)^{n+2}.*x*> 0, that*f*(*x*) has its maximum value when*x*=*n*and we deduce that*f*(*x*)≤*n*^{n}/(*n*+ 1)^{n+1}for all*x*> 0. - Clearly there exists
*c*such that*f*(*x*) -*c*has a root of multiplicity 1, e.g.*x*=*c*= 0. Suppose*f*(*x*) -*c*has a multiple root*r*. Then*r*will also be a root of (*f*(*x*) -*c*)' = 5*x*^{4}-15*x*^{2}+ 4. Also if*r*is a triple root of*f*(*x*) -*c*, then it will be a double root of this polynomial. But the roots of 5*x*^{4}-15*x*^{2}+ 4 are ±((15±√145)/10)^{1/2}, and we conclude that*f*(*x*) -*c*can have double roots, but neither triple nor quadruple roots. - Expand (1 - 1)
^{n}by the binomial theorem and divide by*n*!. We obtain for*n*> 01/(0!Clearly the result is true for*n*!) - 1/(1!(*n*- 1)!) + 1/(2!(*n*- 2)!) - ... + (- 1)^{n}/(*n*!0!) = 0.*n*= 0. We can now proceed by induction; we assume that the result is true for positive integers <*n*and plug into the above formula. We find that*a*_{0}/*n*! +*a*_{1}/(*n*- 1)! +*a*_{2}/(*n*- 2)! + ... +*a*_{n-1}/1! + (- 1)^{n}/(*n*!0!) = 0 - Suppose
2/3 <
*a*_{n},*b*_{n}< 7/6. Then 2/3 <*a*_{n+2},*b*_{n+2}< 7/6. Now if*c*= 1.26, then 2/3 <*a*_{3},*b*_{3}< 1, so if*x*_{n}=*a*_{2n+1}or*b*_{2n+1}, then*x*_{n+1}=*x*_{n}/4 + 1/2 for all*n*≥1. This has the general solution of the form*x*_{n}=*C*(1/4)^{n}+ 2/3. We deduce that as*n*-> ∞,*a*_{2n+1},*b*_{2n+1}decrease monotonically with limit 2/3, and*a*_{2n},*b*_{2n}decrease monotonically with limit 4/3.On the other hand suppose

*a*_{n}> 3/2 and*b*_{n}< 1/2. Then*a*_{n+1}> 3/2 and*b*_{n+1}< 1/2. Now if*c*= 1.24, then*a*_{3}> 3/2 and*b*_{3}< 1/2. We deduce that*a*_{n+1}=*a*_{n}/2 + 1 and*b*_{n+1}=*b*_{n}/2. This has general solution*a*_{n}=*C*(1/2)^{n}+ 2,*b*_{n}=*D*(1/2)^{n}. We conclude that as*n*-> ∞,*a*_{n}increases monotonically to 2 and*b*_{n}decreases monotonically to 0. - Let
*A*be a base campsite and let*h*be a hike starting and finishing at*A*which covers each segment exactly once. Let*B*be the first campsite which*h*visits twice (i.e.*B*is the earliest campsite that*h*reaches a second time). This could be*A*after all segments have been covered, and then we are finished (just choose C = {*h*}). Otherwise let*h*_{1}be the hike which is the part of*h*which starts with the first visit to*B*and ends with the second visit to*B*(so*B*is the base campsite for*h*_{1}). Let*h'*be the hike obtained from*h*by omitting*h*_{1}(so*h'*doesn't visit all segments). Now do the same with*h'*; let*C*be the first campsite on*h'*(starting from*A*) that is visited twice and let*h*_{2}be the hike which is the part of*h'*that starts with the first visit to*C*and ends with the second visit to*C*. Then C can be chosen to be the collection of hikes {*h*_{1},*h*_{2},...} to do what is required.

Peter Linnell 2009-06-24