- It is easily checked that 101 is a prime number (divide 101 by the
primes whose square is less than 101, i.e. the primes
≤7).
Therefore for
1≤
*r*≤100, we may choose a positive integer*q*such that*rq*≡1 mod 101. Since (*I*+*A*+ ... +*A*^{100})(*I*-*A*) =*I*-*A*^{101}, we see that*A*^{101}=*I*, in particular*A*is invertible with inverse*A*^{100}. Suppose 1__<__*n*≤100 and set*r*= 101 -*n*. Then 1≤*r*≤100 and*A*^{n}+ ... +*A*^{100}is invertible if and only if*I*+ ... +*A*^{r-1}is invertible. We can think of*I*+ ... +*A*^{r-1}as (*I*-*A*^{r})/(*I*-*A*), which should have inverse (*I*-*A*)/(*I*-*A*^{r}). However*A*= (*A*^{r})^{q}and so (*I*-*A*)/(*I*-*A*^{r}) =*I*+*A*^{r}+ ... + (*A*^{r})^{q-1}. It is easily checked that(It follows that*I*+ ... +*A*^{r-1})(*I*+*A*^{r}+ ... + (*A*^{r})^{q-1}) =*I*.*A*^{n}+ ... +*A*^{100}is invertible for all positive integers*n*≤100. We conclude that*A*^{n}+ ... +*A*^{100}has determinant ±1 for all positive integers*n*__<__100. - First we will calculate
*f*_{n}(75)mod 16. Note that if*a*,*b*are odd positive integers and*a*≡*b*mod 16, then*a*^{a}≡*b*^{b}mod 16. Also 3^{3}≡11 mod 16 and 11^{11}≡3 mod 16. We now prove by induction on*n*that*f*_{2n-1}(75)≡11 mod 16 for all*n*∈**N**. This is clear for*n*= 1 so suppose*f*_{2n-1}(75)≡11 mod 16 and set*k*=*f*_{2n-1}(75) and*m*=*f*_{2n}(75). Then*f*_{2n}(75)≡*k*^{k}≡11^{11}≡3 mod 16 *f*_{2n+1}(75)≡*m*^{m}≡3^{3}≡11 mod 16

and the induction step is complete. We now prove that*f*_{n}(*a*)≡*f*_{n+2}(*a*)mod 17 for all*a*,*n*∈**N**with*a*prime to 17 and*n*even. In fact we have*f*_{n+1}(*a*)≡*a*^{3}mod 17,*f*_{n+2}(*a*)≡(*a*^{3})^{11}≡*a*mod 17.*f*_{100}(75)≡*f*_{2}(75)mod 17. Therefore*f*_{100}(75)≡7^{11}≡14 mod 17. - First note the
*e*^{2πi/7}satisfies 1 +*x*+ ... +*x*^{6}= 0, so by taking the real part, we obtain ∑_{n=0}^{n=6}cos 2*n*π/7 = 0. Since cos 2π/7 = cos 12π/7, cos 4π/7 = cos 10π/7 = - cos 3π/7 and cos 6π/7 = cos 8π/7 = - cosπ/7, we see that 1 - 2 cosπ/7 + 2 cos 2π/7 - 2 cos 3π/7 = 0.Observe that if 1 - 2 cosθ +2 cos 2θ -2 cos 3θ = 0, then by using cos 2θ = 2 cos

^{2}θ - 1 and cos 3θ = 4 cos^{3}θ -3 cosθ, we find that cosθ satisfies 8*x*^{3}-4*x*^{2}- 4*x*+ 1 = 0. Thus in particular cosπ/7 satisfies this equation. Next note that 1 - 2 cos 3π/7 + 2 cos 6π/7 - 2 cos 9π/7 = 1 - 2 cos 3π/7 - 2 cosπ/7 + 2 cos 2π/7, so cos 3π/7 is also a root of 8*x*^{3}-4*x*^{2}- 4*x*+ 1. Finally since the sum of the roots of this equation is 1/2, we find that -cos 2θ is also a root. Thus the roots of 8*x*^{3}-4*x*^{2}- 4*x*+ 1 are cosπ/7, -cos 2π/7, cos 3π/7. - The equation
4
*A*+ 3*C*= 540^{o}tells us that*A*= 3*B*. Let*D*on*BC*such that ∠*ADC*= 3*B*, and then let*E*on*BD*such that ∠*AED*= 2*B*.

Since triangles

*ABD*and*AED*are similar, we see that*BD*/*AD*=*AD*/*ED*=*AB*/*AE*.*BE*=*AE*because*B*= ∠*BAE*, and*BE*=*BD*-*BE*. We deduce that*BD*^{2}=*AD*^{2}+*AB*·*AD*. Since*BD*=*BC*-*CD*, we conclude that (*BC*-*CD*)^{2}=*AD*^{2}+*AB*·*AD*.Next triangles

*ABC*and*ADC*are similar, consequently*BC*/*AC*=*AB*/*AD*=*AC*/*CD*.*AD*=*AB*·*AC*/*BC*and*CD*=*AC*^{2}/*BC*. We deduce that(Therefore (*a*-*b*^{2}/*a*)^{2}=*c*^{2}*b*^{2}/*a*^{2}+*c*^{2}*b*/*a*.*a*^{2}-*b*^{2})^{2}=*bc*^{2}(*a*+*b*) and the result follows. - Let
*X*denote the center of*A*, let*Y*denote the center of*B*, let*Z*be where*A*and*B*touch (so*X*,*Y*,*Z*are collinear), and let θ = ∠*PXY*. Note that*YQ*makes an angle 2θ downwards with respect to the horizontal, because ∠*QYZ*= 3θ.

Choose (

*x*,*y*)-coordinates such that*X*is the origin and*XP*is on the line*y*= 0. Let (*x*,*y*) denote the coordinates of*Q*. Then we have*x*= 2 cosθ + cos 2θ *y*= 2 sinθ - sin 2θ.

By symmetry the area above the*x*-axis equals the area below the*x*-axis (we don't really need this observation, but it may make things easier to follow). Also θ goes from 2π to 0 as circle*B*goes round circle*A*. Therefore the area enclosed by the locus of*Q*is2∫ _{π}^{0}*y*(*dx*/*d*θ)*d*θ= 2∫ _{π}^{0}(2 sinθ - sin 2θ)(- 2 sinθ -2 sin 2θ)*d*θ= 2∫ _{0}^{π}(4 sin^{2}θ +2 sinθsin 2θ -2 sin^{2}2θ)*d*θ= ∫ _{0}^{π}(4 - 4 cos 2θ +2 cosθ -2 cos 3θ -2 + 2 cos 4θ)*d*θ = 2π.

- Note that if
0 <
*x*,*y*< 1, then 0 < 1 -*y*/2 < 1 and 0 <*x*(1 -*y*/2) < 1, and it follows that (*a*_{n}) is a positive monotone decreasing sequence consisting of numbers strictly less that 1. This sequence must have a limit*z*where 0≤*z*≤1. In particular*a*_{n+2}-*a*_{n+1}=*a*_{n}*a*_{n+1}/2 has limit 0, so lim_{n-> ∞}*a*_{n}*a*_{n+1}= 0. It follows that*z*= 0.Set

*b*_{n}= 1/*a*_{n}. Then*b*_{n+2}=*b*_{n+1}/(1 -*a*_{n}/2) =*b*_{n+1}(1 +*a*_{n}/2 +*O*(*a*_{n}^{2})). Therefore*b*_{n+2}-*b*_{n+1}=*b*_{n+1}(*a*_{n}/2 +*O*(*a*_{n}^{2})). Also*a*_{n}/*a*_{n+1}= (1 -*a*_{n-1}/2)^{-1}= 1 +*O*(*a*_{n})*b*_{n+2}-*b*_{n+1}=*b*_{n}(1 +*O*(*a*_{n}))(*a*_{n}/2 +*O*(*a*_{n}^{2})). Therefore*b*_{n+2}-*b*_{n+1}= 1/2 +*O*(*a*_{n}). Thus given ε > 0, there exists*N*∈**N**such that |*b*_{n+1}-*b*_{n}-1/2| < ε for all*n*>*N*. We deduce that if*k*is a positive integer, then |*b*_{n+k}/*k*-*b*_{n}/*k*- 1/2| < ε. Thus for*k*sufficiently large, |*b*_{n+k}/(*n*+*k*) - 1/2| < 2ε. We conclude that lim_{n-> ∞}*b*_{n}/*n*= 1/2 and hence lim_{n-> ∞}*na*_{n}= 2. - It will be sufficient to prove that
∑
_{n=1}^{∞}*n*^{2}/(1/*a*_{1}^{2}+ ... +1/*a*_{n}^{2}) is convergent. Note we may assume that (*a*_{n}) is monotonic decreasing, because rearranging the terms in series ∑*a*_{n}does not affect its convergence, whereas the terms of the above series become largest when (*a*_{n}) is monotonic decreasing. Next observe that if ∑_{n=1}^{∞}*a*_{n}=*S*, then*a*_{n}≤*S*/*n*for all positive integers*n*. Now consider (2*n*)^{2}/(1/*a*_{1}^{2}+ ... +1/*a*_{2n}^{2}). This is ≤(2*n*)^{2}*S*/(1/*a*_{1}+2/*a*_{2}+ ... +2*n*/*a*_{2n})≤4*n*^{2}*S*/(*n*^{2}/*a*_{n}) = 4*Sa*_{n}. The result follows.

Peter Linnell 2010-11-04