and because \(V\sim U\sim T\)\(\frac{b\left(S\right)}{b\left(U\right)}=\frac{b\left(V\right)}{b\left(U\right)}=\frac{h\left(V\right)}{h\left(U\right)}\)\(b\left(S\right)=\frac{h\left(V\right)b\left(U\right)}{h\left(U\right)}=\frac{\left(h\left(U\right)-h\left(S\right)\right)b\left(U\right)}{h\left(U\right)}\).  Now
\(\frac{A\left(S\right)}{A\left(U\right)}=2\frac{b\left(S\right)h\left(S\right)}{b\left(U\right)h\left(U\right)}=\frac{2h\left(S\right)\left(h\left(U\right)-h\left(S\right)\right)}{h\left(U\right)^2}=-\frac{2}{h\left(U\right)^2}h\left(S\right)_{ }^2+\frac{2h\left(S\right)}{h\left(U\right)}\), which is \(\frac{2}{h\left(U\right)^2}\left[h\left(U\right)h\left(S\right)-h\left(S\right)^2\right]\).  Consider the ratio to be a function of \(h\left(S\right)\), ranging from 0 to \(h\left(U\right)\).  As a y-axis aligned parabola with zeros at 0, \(h\left(U\right)\), the ratio has a maximum of 1/2 at 1/2 \(h\left(U\right)\).
Now that we know that for any choice of \(R\) we can achieve the maximum ratio by taking  \(h\left(S\right)\) equal to half \(h\left(U\right)\).  
Shortcut: let the ratio 
\(f\left(h\left(S\right)\right)=\frac{A\left(S\right)}{A\left(U\right)}\).  Then \(f\) is a quadratic function with domain \(\left[0,h\left(U\right)\right]\), which is zero at the endpoints of its domain.  Therefore, it has a maximum at the center \(\frac{1}{2}h\left(U\right)\) of its domain.
Now, let the ratio in the problem be \(f\left(h\left(R\right)\right)\), a quadratic function with domain \(\left[0,h\left(T\right)\right]\).  This function takes values \(\frac{1}{2}\) and \(0\) at the endpoints of its domain. Renormalize the domain of \(f\) to be the unit interval.  Thus, using the notation for the coefficients:
\(f_2h^2+f_1h+f_0\)
we have \(f_0=\frac{1}{2}\),  \(f_2+f_1+\frac{1}{2}=0\).  Further, we have \(f\left(\frac{1}{3}\right)=\frac{2}{3}\), so that \(\frac{1}{9}f_2+\frac{1}{3}f_1+\frac{1}{2}=\frac{2}{3}\), i.e.  \(f_2+3f_1=\frac{3}{2}\)\(f_1=1,\ f_2=-\frac{3}{2}\).    Then it is easy to see that \(f\) has its vertex (maximum) at \(\frac{1}{3}\).
In the solutions, they have a generalization to the case of  n rectangles stacked on the base of the triangle and inscribed in it.  Some elementary geometry is used to show that the solution is equivalent to minimizing \(\sum_{_{i=1}}^na_i^2\) subject to the constraints that \(a_i>0,\ \sum_{_{i=1}}^na_i=1\).  Geometrically, this is saying to minimize the norm of a vector on a hyperplane in the first octant.  They show this is minimized at \(a_1=\cdots a_n=\frac{1}{n}\) as a special case of the Cauchy-Schwarz Inequality, the Chebyshev inequality, the Power Mean Inequality, and Jensen's Inequality.
A3.  \(e^d-1\), because if we apply the transformations \(d\mapsto\ \frac{d}{2}\), followed by \(x\ \mapsto x^2+2x\) to \(e^d-1\ -\epsilon\) we obtain \(e^d-1\ -\epsilon\left(2e^{\frac{d}{2}}-\epsilon\right)\)...  Which shows that  \(e^d-1\) is a stationary point of these transformations.  They have a more conventional/elementary solution depending on completing the square.  ,
A4. First, observe \(x^{20}\equiv1\text{mod}100\) for any x satisfying \(\mathrm{\gcd\left(x,100\right)=1}\).   (This can be derived from the fact that the Carmichael lambda function of 100 is 20).  Then \(a_4\%\ 100=3^{a_3\%20}\%100\). Further \(a_3^{ }\%20=3^{a_2\%4}\%20=7\).  Thus \(a_4\%\ 100\ =3^7\%^{\ }100=87\).  Thereafter, there are several ways to see that the sequence is constant.
A5. \(\cos mx\ =\ \frac{e^{imx}+e^{-imx}}{2}\).  Thus the \(m\) where \(I_m\ne0\) are just those m for which the constant term of the Laurent polynomial\[\prod_{n=1}^m\left(x^n+x^{-n}\right)\] is nonzero.  All the coefficients are positive, so we just have to record, for each m, the coefficients of the Laurent polynomial which are nonzero:
m=1: -1, 1: \(\pm1\)
m=2: -3,-1,1,3: \(\pm\left\{1,3\right\}\)
m=3: -6,-4,-2,0,2,4,6: \(0,\ \pm\left\{2,4,6\right\}\)
m=4: \(0,\pm\left\{10,8,6,4,2\right\}\)
m=5: \(\pm\left\{1,3,5,7,9,11,13,15\right\}\)
m=6: \(\pm\left\{1,3,\ldots,19,21\right\}\)
m=7: \(0,\pm\left\{2,4,\ldots,26,\ 28\right\}\),
m=8: \(0,\ \pm\left\{2,\ldots,\ 36\right\}\)
3,4,7,8 are the answers.
A6.  
B1. \(x\left(x^2-1\right)^{ }\left(x^2-4\right)^{ }\), for example, minimizes the number of nonzero coefficients at 3.  If the polynomial has only 2 nonzero coefficients, then Descartes rule of signs can be used to show that the number of distinct real roots is less than or equal to 3, but the requirement is that there be 5 distinct integer roots.
B2.  Show by induction that for \(n\ge2\)\(f_n\left(x\right)=x\left(x+n\right)^{n-1}\).  The point is that it's much easier to show that the conditions given uniquely determine that for \(n\ge2\)\(f_n\left(x\right)=x\left(x+n\right)^{n-1}\).  The point is that it's much easier to show that the conditions given uniquely determine the sequence of functions, and then show that the functions of this given form satisfy the conditions, than it is to prove the functional form sequence of functions, and then show that the functions of this given form satisfy the conditions, than it is to prove the functional form by "straightforward induction"."straightforward induction".
B3.  
B4.  \(\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\frac{4\cos\theta\sin\theta}{\pi}\text{d}\theta=\frac{4}{\pi^2}\), where the integrand is the conditional probability of the R lying completely inside C conditioned on the choice of p being angle \(\theta\).
B5.  Break up into the intervals from 1/B to 1 and 1 to B.  Perform some substitutions to get it into the form indicated in the hint.
B6.  \(\Sigma:=\sum_{_{i=1}}^rM_i\). Use the group property to show that \(\Sigma\) satisfies the polynomial \(x^2-rx\).  The minimal polynomial is either (1) \(x^2-rx\), in which case the set of eigenvalues is \(\left\{0,r\right\},\ \)(2) \(x-r\), in which case the eigenvalues are all \(r\) (3) x.  In neither of the first two cases is it possible for the trace (the sum of the eigenvalues) to be 0.

1986 Putnam

A1.  \(x^4+36\le13x^2\) is equivalent to \(-3\le x\le-2\) or \(2\le x\le3\).  Since \(f'\left(x\right)=3x^2-3=3\left(x+1\right)\left(x-1\right)\)\(f\) is increasing on \(\left(-\infty,\ -1\right),\ \left(1,\ \infty\right)\), decreasing on \(\left(-1,1\right)\).  Therefore, the maximum of \(f\) on the domain \(\left[-3,-2\right]\cup\left[2,3\right]\) must occur at one of the right endpoints -2 or 3.  Evaluating \(f\left(-2\right)=-2\)\(f\left(3\right)=18\), so the maximum value of \(f\)on the domain is 18 at 3.
A2. Define \(D=10^{20000}=10^{20k}\) (k is shorthand for \(10^3\)), \(p=10^{100}+3\).  By the Remainder Theorem, there are unique \(q\in\mathbf{Z}\)\(r\in\mathbf{Z}_{\geq0}\)  with \(0\le r<p\) such that \(D=pq+r\).  The question is to find the units digit of q.  Define \(q_0:=10^{20k-100\cdot1}=10^{19,900}\).  For \(i\ge1\) define \(\delta_i=\left(-1\right)^i3^i10^{20k-100\cdot\left(i+1\right)}\).  For \(i\ge0\) define \(q_{i+1}\) by the recurrence relation \(q_{i+1}=q_i+\delta_{i+1}\).  For  \(i\ge0\), define \(r_i:=\left(-1\right)^{i+1}\cdot3^{i+1}\cdot10^{20k-100\cdot\left(i+1\right)}\).  Then we claim that 
\[D=pq_i+r_i\] for all \(i\ge0\).  We prove this by induction.  The base case is as follows:
\(p\cdot q_0+r_0=\left(10^{100}+3\right)10^{19,900}-3\cdot10^{19,900}=10^{100+19,900}=D\)
The induction step is as follows.  Assume that the above relation holds for a given \(i\geq 0\).  Then 
\(p\cdot q_{i+1}+r_{i+1}=\left(10^{100}+3\right)\left(q_i+\delta_{i+1}\right)+\left(-1\right)^{i+2}\cdot3^{i+2}\cdot10^{20k-100\cdot\left(i+2\right)}\).  By the induction hypothesis, \(pq_i=D-r_i\), which we substitute into the right side of the previous equation to obtain \(D-r_i+\left(10^{100}+3\right)\delta_{i+1}+\left(-1\right)^{i+2}3^{i+2}10^{20k-100\left(i+2\right)}\).  Here the first term from applying the distributive law to \(\left(10^{100}+3\right)\delta_{i+1}\)cancels with \(-r_i\) and the second term cancles with the final term of the above, which is to say,  after everything cancels out, \(D\) is left. 
We calculate that \(0<r_{199}=3^{200}=9^{100}<10^{1000}<p\).  Therefore, \(r=r_{199,\ }q=q_{199}\).  By the recurrence relation defining \(q_i\)\(q_i=q_0+\ \sum_{i=1}^i\delta_i\).
Because \(q_0\) and all the \(\delta_i\) for \(0<i\le198\) are divisible by \(10\)\(q_{199}\equiv\delta_{199}\text{mod}10\).  \(\delta_{199}\equiv-3^{199}\text{mod}10\equiv-3^{199\%\phi\left(10\right)}\text{mod}10\), where the latter equality follows from Euler's formua \(a^{\phi\left(n\right)}\equiv1\text{mod}n\) for gcd(a,n)=1.  for gcd()  Since \(\phi\left(10\right)=\left(5-1\right)\cdot\left(2-1\right)=4\cdot1=4\), the answer is \(\delta_{199}\equiv-3^3\text{mod}10\equiv-27\text{mod}10\equiv3\text{mod}10\).
 A3.  Define \(f\left(n\right)=n^2+n+1\).  Define \(\theta_n=\cot^{-1}\left(f\left(n\right)\right)\).  Define \(\varphi_m=\sum_{_{n=0}}^{\infty}\theta_n\).  We are to evaluate \(\lim_{m\rightarrow\infty}\varphi_m\).  Since \(\varphi_m=\theta_m+\varphi_{m-1},\) we have \(\cos\varphi_m=\cos\theta_m\cos\varphi_{m-1}-\sin\theta_m\sin\varphi_{m-1}\) and 
\(\sin\varphi_m=\sin\theta_m\cos\varphi_{m-1}+\cos\theta_m\sin\varphi_{m-1}\), and
Thus, if we define