# Morning Session ## Problem 1 A solid is bounded by two bases in the horizontal planes $z=h / 2$ and $z$ $=-h / 2$, and by such a surface that the area of every section in a horizontal plane is given by a formula of the sort$\text { Area }=a_0 z^3+a_1 z^2+a_2 z+a_3$(where as special cases some of the coefficients may be 0). Show that the volume is given by the formula $V=\frac{1}{6} h\left[B_1+B_2+4 M\right],$where $B_1$ and $B_2$ are the areas of the bases, and $M$ is the area of the middle horizontal section. Show that the formulas for the volume of a cone and of a sphere can be included in this formula when $a_0=0$. > [!sol]- Click for Solution! > Let $p(z)=az^3+bz^2+cz+d$ be the [[polynomial#Degree 3 - Cubic|cubic]] such that $p(k)$ is the [[area]] of the [[cross-section]] at $z=k$. > $B,M,T$ are the following: > $\begin{aligned} > B&=p\left(\frac{-h}{2}\right)&=&a\left(\frac{-h}{2}\right)^3+b\left(\frac{-h}{2}\right)^2+c\left(\frac{-h}{2}\right)+d&&=-\frac{ah^3}{8}+\frac{bh^2}{4}-\frac{ch}{2}+d \\ > M&=p\left(0\right)&=&a\left(0\right)^3+b\left(0\right)^2+c\left(0\right)+d&&=d\\ > T&=p\left(\frac{h}{2}\right)&=&a\left(\frac{h}{2}\right)^3+b\left(\frac{h}{2}\right)^2+c\left(\frac{h}{2}\right)+d&&=\frac{ah^3}{8}+\frac{bh^2}{4}+\frac{ch}{2}+d > \end{aligned}$ > Build $\frac{h(B + 4M + T)}{6}$ starting with $B+T$ > $\begin{aligned} > B+T&=&\left(-\frac{ah^3}{8}+\frac{bh^2}{4}-\frac{ch}{2}+d \right)+\left(\frac{ah^3}{8}+\frac{bh^2}{4}+\frac{ch}{2}+d\right)&=&\frac{bh^2}{2}+2d\\ > B+T+4M&=& &=&\frac{bh^2}{2}+6d\\ > \frac{h(B+4M+T)}{6}&=& &=& \boxed{\frac{bh^3}{12}+dh} > \end{aligned}$ > Calculating [[volume]] of [[solid]] > $\begin{aligned} > \int_{-\frac{h}{2}}^{\frac{h}{2}}{\left(az^3+bz^2+cz+d\right)dz} = \left.\frac{a z^{4}}{4}+\frac{b z^{3}}{3}+\frac{c z^{2}}{2}+dz\right|_{-\frac{h}{2}}^{\frac{h}{2}} = \\ > \frac{a \left(\frac{h}{2}\right)^{4}}{4}+\frac{b \left(\frac{h}{2}\right)^{3}}{3}+\frac{c \left(\frac{h}{2}\right)^{2}}{2}+d\left(\frac{h}{2}\right) - \left(\frac{a \left(-\frac{h}{2}\right)^{4}}{4}+\frac{b \left(-\frac{h}{2}\right)^{3}}{3}+\frac{c \left(-\frac{h}{2}\right)^{2}}{2}+d\left(-\frac{h}{2}\right)\right)=\\ > \frac{ah^4}{64}+\frac{bh^3}{24}+\frac{ch^2}{8}+\frac{dh}{2}-\left(\frac{ah^4}{64}+\frac{bh^3}{24}+\frac{ch^2}{8}+\frac{dh}{2}\right)=\\ > \boxed{\frac{bh^3}{12}+dh} > \end{aligned}$ > The claim from the problem is verified. > - Deriving > - Volume of Cone > Since the cross-sections of a cone with base on $z=\frac{-b}{2}$ and vertex on $z=\frac{b}{2}$ would have area $p(z)=\pi \left(\frac{1}{2}-\frac{z}{h}\right)^2r^2$, a quadratic of $z$, the formula from the problem applies. > $B=\pi r^2$ > $M=\pi \left(\frac{r}{2}\right)^2= \frac{\pi r^2}{4}$ > $T=0$ > $B+4M+T=2\pi r^2\implies \frac{h(B + 4M + T)}{6}=\frac{\pi r^2h}{3} \checkmark$ > - Volume of Sphere > Since, the cross-sections of a sphere centered at the origin with a diameter of $h=2r$ would have an area of $p(z)=\pi\left(r^{2}-z^{2}\right)$, a quadratic of $z$, the formula from the problem applies. > $B=0$ > $M=\pi r^2$ > $T=0$ > $\frac{h(B + 4M + T)}{6}=\frac{(2r)(4\pi r^2)}{6}=\frac{4}{3}\pi r^3 \checkmark$ ## Problem 2 A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume? > [!sol]- Click for Solution! Let $r$ be the [[radius]] of the [[cylinder]], and $h$ its [[altitude]]. The given condition is > $ > S=2 \pi r h+2\left(\pi r \sqrt{h^2+r^2}\right)=\text { constant, } \tag{1} > $ > and the volume of the buoy is > $ > V=\pi r^2 h+\frac{2 \pi r^2 h}{3}=\frac{5 \pi r^2 h}{3} .\tag{2} > $ > The required problem is to find the [[maximum]] value of $V$ subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for $h$ and express $V$ as a function of $r$. We have > $ > (S-2 \pi r h)^2=4 \pi^2 r^2\left(h^2+r^2\right), > $ > whence > $ > h=\frac{S^2-4 \pi^2 r^4}{4 \pi r S}, \tag{3} > $ > and the expression for $V$ becomes > $V=\frac{5 r}{12 S}\left(S^2-4 \pi^2 r^4\right). \tag{4}$ > Since $r$ and $V$ must be positive, the domain of interest is given by > $0<r<\sqrt[4]{S^2 / 4 \pi^2}.$ > We compute the derivative and equate it to zero to get > $ > \frac{d V}{d r}=\frac{5 S}{12}-\frac{100 \pi^2 r^4}{12 S}=0 . > $ > The only critical value is > $ > r_0=\sqrt[4]{\frac{S^2}{20 \pi^2}} . > $ > Since $V \rightarrow 0$ as $r \rightarrow 0$ or as $r \rightarrow \sqrt[4]{S^2 / 4 \pi^2}$, and is positive in between, the critical value $r_0$ yields a maximum for $V$. > The corresponding value of $h$ is found from (3) to be $h_0=\frac{2}{5} \sqrt{5} r_0$. The shape of the buoy is completely determined by the ratio > $ > \boxed{\frac{h_0}{r_0}=\frac{2}{5} \sqrt{5}} > $ ## Problem 3 If a particle moves in the plane, we may express its coordinates $x$ and $y$ as functions of the time $t$. If $x=t^3-t$ and $y=t^4+t$, show that the curve has a point of inflection at $t=0$ and that the speed of the moving particle has a local maximum at $t=0$. > [!sol]- Click for First Solution! > If the velocity vector at time $t$ is of length $v$ and has direction $\theta$, then $\dot{x}=v \cos \theta, \dot{y}=v \sin \theta$, and $\ddot{x}=\dot{v} \cos \theta-v \dot{\theta} \sin \theta$, $\ddot{y}=\dot{v} \sin \theta+v \dot{\theta} \cos \theta$. Thus $\dot{x} \ddot{y}-\dot{y} \ddot{x}=v^2 \dot{\theta}$ and $v^2=\dot{x}^2+\dot{y}^2$. From the given parametric data, > $ > \begin{aligned} > v^2 \dot{\theta} & =\left(3 t^2-1\right) 12 t^2-\left(4 t^3+1\right) 6 t=6 t\left(2 t^3-2 t-1\right) \\ > v^2 & =16 t^6+9 t^4+8 t^3-6 t^2+2 . > \end{aligned} > $ > From the $v^2 \dot{\theta}$ relation, $\dot{\theta}$ changes sign as $t$ passes through 0 . To rule out the possibility of a cusp point, one notes that $v^2 \neq 0$ at $t=0$. Hence the curve has an inflection point at $t=0$. Also > $ > \begin{aligned} > \frac{d\left(v^2\right)}{d t} & =96 t^5+36 t^4+24 t^2-12 t \\ > \left.\frac{d\left(v^2\right)}{d t}\right]_{t=0} & =0 > \end{aligned} > $ > and > $ > \left.\frac{d^2\left(v^2\right)}{d t^2}\right]_{t=0}=-12 > $ > So $v^2$ has a (local) maximum at $t=0$. Hence the speed $v$ also has a local maximum. > [!sol]- Click for Second Solution! > Since $d x / d t$ does not vanish for $t=0$, we can solve for $t$ in terms of $x$ near $t=0$ and write $y$ as a function of $x$. Then we have > $ > \frac{d y}{d x}=\frac{d y}{d t} \mid \frac{d x}{d t}=\frac{4 t^3+1}{3 t^2-1}=-1-3 t^2-\cdots . > $ > Therefore, $d y / d x$ has a (local) maximum at $t=0$, so the curve has an inflection point at $t=0$. > The magnitude $v$ of the velocity is given by > $ > v^2=\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2=\left(3 t^2-1\right)^2+\left(4 t^3+1\right)^2=2-6 t^2+\cdots . > $ > Hence $v$ has a (local) maximum at $t=0$. ## Problem 4 A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle $\theta$ along a horizontal line through the axis of the cylinder. If $\theta$ is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal. ![[putnam-cylinder-wedge.svg|-noinv-dmo]] > [!sol]- Click for First Solution! > Suppose $0 \leq \alpha_1<\alpha_2<\alpha_2+\theta<\pi / 2$; then the wedgeshaped solid between the planes at angles $\alpha_1$ and $\alpha_1+\theta$ is smaller than the wedge between $\alpha_2$ and $\alpha_2+\theta$, because a simple rotation of the former through an angle $\alpha_2-\alpha_1$ makes it a proper subset of the latter. > ![[1938-04-16 Putnam P4Sol1.svg]] > Consider now any asymmetrical wedge of angle $\theta$ with cross-section $A O B$. If $A$ and $B$ are on the same side of the horizontal through $O$, then the above argument shows that the wedge does not have minimal volume. > ![[1938-04-16 Putnam P4Sol1-cont.svg]] > Suppose then that $A$ is below the horizontal, and $B$ above it. By symmetry we can assume that $A O B$ lies below the symmetrical wedge $S O T$ of angle $\theta$, as shown. The wedge $A O S$ is congruent by symmetry with the wedge $A^{\prime} O T$, which is, in turn, larger than wedge $B O T$ (as shown above). Hence > $ > \begin{aligned} > \text { wedge } A O B& = \text { wedge } A O S+\text { wedge } S O B \\ > & >\text { wedge } S O B+\text { wedge } B O T \\ > & = \text { wedge } S O T > \end{aligned} > $ > Thus the symmetrical wedge is a strict minimum. > [!sol]- Click for Second Solution! Let $a$ be the radius of the cylindrical tree, and let $\alpha$ and $\beta$ be the angles between the planes of the cut and the horizontal; > $ > \beta=\alpha+\theta . > $ > The volume of the wedge is > $ > \begin{aligned} > V & =\int_0^a 2 x(\tan \beta-\tan \alpha) \sqrt{a^2-x^2} d x \\ > & =A(\tan \beta-\tan \alpha) > \end{aligned} > $ > ![[1938-04-16 Putnam P4Sol2.svg]] > (It is easy, but unnecessary, to evaluate the integral; in fact, $A=2 a^3 / 3$.) We seek to minimize $V$ by choice of $\alpha$. This is equivalent to minimizing > $ > W=\tan (\alpha+\theta)-\tan \alpha > $ > for $-\pi / 2<\alpha<\pi / 2-\theta$, since $-\pi / 2<\alpha$ and $\beta<\pi / 2$. The critical points are found by solving > $ > \frac{d W}{d \alpha}=\sec ^2(\alpha+\theta)-\sec ^2 \alpha=0 > $ > Since both $\sec (\alpha+\theta)$ and $\sec \alpha$ are positive through the interval in question, $\sec (\alpha+\theta)=\sec \alpha$, whence $\alpha+\theta=\pm \alpha$. Since $\theta$ is not zero, the only critical point is given by $\alpha=-\theta / 2$. It is easily seen to correspond to a minimum. When $\alpha=-\theta / 2, \beta=\theta / 2$ and the horizontal plane bisects the wedge. ## Problem 5 Evaluate the following limits: > [!i] $\lim _{n \rightarrow \infty} \frac{n^2}{e^n}$ > [!ii] $\lim _{x \rightarrow 0} \frac{1}{x} \int_0^x(1+\sin 2 t)^{1 / t} d t$ > [!sol]- Click for Solution! > > [!i] It follows from L'Hospital's rule that > > $ > > \lim _{x \rightarrow \infty} \frac{x^2}{e^x}=\lim _{x \rightarrow \infty} \frac{2 x}{e^x}=\lim _{x \rightarrow \infty} \frac{2}{e^x}=0, > > $ > > whence the desired limit is zero. > > Alternatively, one could use the fact that for $x>0$, > > $ > > e^x=\sum_{n=0}^{\infty} \frac{x^n}{n !}>\frac{x^3}{6}, \quad \text { whence } 0<\frac{x^2}{e^x}<\frac{6}{x} > > $ > > and, as $x \rightarrow \infty, \lim _{x \rightarrow \infty} 6 / x=0$, whence the desired limit is zero. > > > [!ii] By L'Hospital's rule, > > $ > > \lim _{x \rightarrow 0} \frac{1}{x} \int_0^x(1+\sin 2 t)^{1 / t} d t=\lim _{x \rightarrow 0}(1+\sin 2 x)^{1 / x} > > $ > > provided the latter limit exists. Let $\phi(x)=(1+\sin 2 x)^{1 / x}$. Then > > $ > > \log \phi(x)=\frac{\log (1+\sin 2 x)}{x} . > > $ > > Again using L'Hospital's rule, > > $ > > \lim _{x \rightarrow 0} \log \phi(x)=\lim _{x \rightarrow 0} \frac{2 \cos 2 x}{1+\sin 2 x}=2 . > > $ > > Since the exponential function is continuous, > > $ > > \lim _{x \rightarrow 0} \phi(x)=e^2 . > > $ > > Therefore, > > $ > > \lim _{x \rightarrow 0} \frac{1}{x} \int_0^x(1+\sin 2 t)^{1 / \tau} d t=e^2 > > $ ## Problem 6 A swimmer stands at one corner of a square swimming pool and wishes to reach the diagonally opposite corner. If $w$ is his walking speed and $s$ is his swimming speed $(s<w)$, find his path for shortest time. (Consider two cases: `(i)` $w / s<\sqrt{2}$, and `(ii)` $w / s>\sqrt{2}$.) > [!sol]- Click for Solution! > Let the square pool be denoted by $A B C D$, with the swimmer initially at $A$ and desirous of reaching $C$. The path of least time can evidently be described as follows. The swimmer walks from $A$ to $E$ (a point on side $A B$ ), swims from $E$ to $F$ where $F$ is on $B C$, and then walks from $F$ to $C$. Note that a path like $A G H C$ is time equivalent to a path of the type described with $F=C$. > ![[1938-04-16 Putnam P6Sol.svg]] > Let $\overline{A E}=x, \overline{E F}=y, \overline{F C}=z$. Then the time $T$ is given by $T=$ $(x+z) / w+(y / s)$. If the sum $x+z$ is fixed, then the sum $y \sin \alpha+$ $y \cos \alpha$ is also fixed, and $y$ is minimal when ( $\sin \alpha+\cos \alpha$ ) is maximal. This maximum is attained for $\alpha=45^{\circ}$. > > Thus for a minimal time path, $x=z$ and $y=\sqrt{2}(l-x)$, where $l$ is the length of a side of the pool. Accordingly, we have to minimize $T=$ $(2 x / w)+\sqrt{2}(l-x) / s$ for $0 \leq x \leq l$. > > But $T$ is a linear function of $x$, so its maximum occurs at an endpoint of the interval. If $x=0, T=\sqrt{2 l} / s$, and if $x=l, T=2 l / w$. > > If $\sqrt{2} l / s<2 l / w$ then $w / s<\sqrt{2}$, and conversely. Hence, if $w / s<\sqrt{2}$ the minimal path is unique and the swimmer should swim diagonally across the pool from $A$ to $C$. If $w / s>\sqrt{2}$, he should walk from $A$ to $B$ to $C$. Finally, if $w / s=\sqrt{2}, T$ is independent of $x$ and there are infinitely many minimizing paths, in fact any path $A E F C$ for which $\alpha=45^{\circ}$. ## Problem 7 Take either `(i)` or `(ii)`. > [!i] Show that the gravitational attraction exerted by a thin homogeneous spherical shell at an external point is the same as if the material of the shell were concentrated at its center. > [!ii] Determine all the straight lines which lie upon the surface $z=x y,$ and draw a figure to illustrate your result. # Afternoon Session ## Problem 8 Take either `(i)` or `(ii)`. > [!i] Let $A_{i k}$ be the cofactor of $a_{i k}$ in the determinant $d=\left|\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\a_{21} & a_{22} & a_{23} & a_{24} \\a_{31} & a_{32} & a_{33} & a_{34} \\a_{41} & a_{42} & a_{43} & a_{44}\end{array}\right|$Let $D$ be the corresponding determinant with $a_{i k}$ replaced by $A_{i k}$. Prove $D=$ $d^3$. > [!ii] Let $P(y)=A y^2+B y+C$ be a quadratic polynomial in $y$. If the roots of the quadratic equation $P(y)-y=0$ are $a$ and $b(a \neq b)$, show that $a$ and $b$ are roots of the biquadratic equation $P[P(y)]-y=0$. Hence write down a quadratic equation which will give the other two roots, $c$ and $d$, of the biquadratic. Apply this result to solving the following biquadratic equation:$\left(y^2-3 y+2\right)^2-3\left(y^2-3 y+2\right)+2-y=0$ ## Problem 9 Find all the solutions of the equation $y y^{\prime \prime}-2\left(y^{\prime}\right)^2=0$ which pass through the point $x=1, y=1$. ## Problem 10 A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc. ## Problem 11 Given the parabola $y^2=2 m x$, what is the length of the shortest chord that is normal to the curve at one end? ## Problem 12 From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve. ## Problem 13 Find the shortest distance between the plane $A x+B y+C z+1=0$ and the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$. For brevity, let $h=\frac{1}{ \sqrt{A^2+B^2+C^2}}\text{\quad and\quad}m=\sqrt{a^2 A^2+b^2 B^2+c^2 C^2}.$ State algebraically the condition that the plane shall lie outside the ellipsoid.