The 1878 Easy Problems paper

Let the angles of the triangle be $A$, $B$, $C$, and let the line $AO$ (which bisects the angle at $A$) meet the circle round $ABC$ again at $M$.

Angle at $B$. The angles $MBC$ and $MAC$ stand on the same arc $MC$, so $\angle MBC = \tfrac12A$. Since $BO$ bisects the angle at $B$, $\angle OBC = \tfrac12B$. So $$\angle MBO = \tfrac12 A + \tfrac12 B.$$

Angle at $O$. The angle $MOB$ is an exterior angle of triangle $ABO$, so it equals the sum of the two opposite interior angles: $$\angle MOB = \angle OAB + \angle OBA = \tfrac12A + \tfrac12B.$$

So triangle $MBO$ has two equal angles, and $MB = MO$. In the same way $MC = MO$. The point $M$ is equally far from $B$, $O$ and $C$, so it is the center of the circle through them, and it lies on $AO$.

Let $n_{AB}$ be the number of families that have left $A$ for $B$ by the end of the time, and define the other five numbers the same way.

Each town balances. Town $A$ loses $n_{AB} + n_{AC}$ families and gains $n_{BA} + n_{CA}$, and its size is unchanged, so these are equal. Then $$n_{AB} + n_{AC} + n_{BA} + n_{CA} = 2(n_{BA} + n_{CA})$$ is even. In words: the moves that touch town $A$ are even in number.

Subtract from the total. All six numbers add up to an even total. Taking away the moves that touch $A$ leaves the moves between $B$ and $C$: $$n_{BC} + n_{CB} \text{ is even.}$$ The same argument with town $C$ in place of $A$ shows that $n_{AB} + n_{BA}$ is even.

So $n_{AB}$ and $n_{BA}$ are both even or both odd. If $n_{AB}$ is even, so is $n_{BA}$.

Let $u$ be the sum of the two cube roots, so $3a = u - 1$ and $b = \tfrac{2}{81}(3a)^4 = 2a^4$.

An equation for $a$. The two numbers under the cube roots add to $\tfrac{23}{2}$ and multiply to $\tfrac{529 - 513}{16} = 1$, so the cube roots multiply to $1$. Cubing a sum $u = p + q$ gives $u^3 = p^3 + q^3 + 3pq(p+q)$, so $$u^3 = \tfrac{23}{2} + 3u.$$ Put $u = 3a + 1$: $27a^3 + 27a^2 + 9a + 1 = \tfrac{23}{2} + 9a + 3$, which is $$a^3 + a^2 = \tfrac12. \tag{1}$$

First pair. With $b = 2a^4$, we have $b/a = 2a^3$ and $a - b = a - 2a^4 = a(1 - 2a^3)$. By (1), $1 - 2a^3 = 2a^2$, so $a - b = 2a^3 = b/a$.

Second pair. We need $a^2 + 4a^8 = 2a^3$. Use (1) to lower powers: $a^3 = \tfrac12 - a^2$, then $$a^4 = \tfrac12a - \tfrac12 + a^2, \qquad a^6 = (a^3)^2 = \tfrac14 - a^2 + a^4 = \tfrac12a - \tfrac14,$$ $$a^8 = a^2\cdot a^6 = \tfrac12a^3 - \tfrac14a^2 = \tfrac14 - \tfrac34a^2.$$ So $a^2 + 4a^8 = a^2 + 1 - 3a^2 = 1 - 2a^2 = 2a^3$, using (1) once more. All three quantities equal $2a^3 \approx 0.3611$.

Let $D = (x-y)(y-z)(z-x)$.

First brace. Over the common denominator $xyz$ it is $\dfrac{yz(y-z) + zx(z-x) + xy(x-y)}{xyz}$. The numerator vanishes when any two of $x, y, z$ are equal, so it is a multiple of $D$; comparing one term (try $x = 0$, $y = 1$, $z = 2$) shows it equals $-D$. So the first brace is $-D/(xyz)$.

Second brace. Over the common denominator $D$ it is $\dfrac{\sum x(z-x)(x-y)}{D}$. Now use $y + z = -x$: $$x(z-x)(x-y) = x\big(xz + xy - x^2 - yz\big) = x\big(-x^2 - x^2 - yz\big) = -2x^3 - xyz.$$ Adding the three such terms and using $x^3 + y^3 + z^3 = 3xyz$ gives $-6xyz - 3xyz = -9xyz$. So the second brace is $-9xyz/D$.

Product. $\left(-\dfrac{D}{xyz}\right)\left(-\dfrac{9xyz}{D}\right) = 9.$

Pair the equations as vectors. Let $\mathbf u = (z', y, x')$ and $\mathbf v = (y', x', z)$. The second and third equations on the left say $|\mathbf u| = b$ and $|\mathbf v| = c$, and the first on the right says $\mathbf u\cdot\mathbf v = y'z' + yx' + x'z = bc$. By the Cauchy–Schwarz inequality $\mathbf u\cdot\mathbf v \le |\mathbf u||\mathbf v|$, with equality only when one vector is a positive multiple of the other. So $\mathbf u = \tfrac bc\,\mathbf v$: $$z' = \tfrac bc\,y', \qquad y = \tfrac bc\,x', \qquad x' = \tfrac bc\,z.$$ The other two pairs work the same way: $(x', z, y') = \tfrac ca\,(z', y', x)$ and $(y', x, z') = \tfrac ab\,(x', z', y)$, so $$z = \tfrac ca\,y', \qquad y' = \tfrac ca\,x, \qquad x = \tfrac ab\,z', \qquad z' = \tfrac ab\,y.$$

Everything in terms of $x'$. From these, $y = \tfrac bc x'$, $y' = \tfrac ab x'$, $z' = \tfrac bc y' = \tfrac ac x'$, $z = \tfrac ca y' = \tfrac cb x'$ and $x = \tfrac ab z' = \tfrac{a^2}{bc}x'$. Put $x' = bc/k$; then $$x = \frac{a^2}{k},\quad y = \frac{b^2}{k},\quad z = \frac{c^2}{k},\quad x' = \frac{bc}{k},\quad y' = \frac{ca}{k},\quad z' = \frac{ab}{k}.$$

Find $k$. The first equation gives $\dfrac{a^4 + a^2b^2 + a^2c^2}{k^2} = a^2$, so $k^2 = a^2 + b^2 + c^2$. Every equation is unchanged if all six unknowns change sign, so $k = \pm\sqrt{a^2+b^2+c^2}$ gives the two real solutions, and the argument shows there are no others.

Since $0 < a < b$, write $a = b\cos\phi$ with $0 < \phi < \tfrac\pi2$.

The pattern. Using $1 + \cos\phi = 2\cos^2\tfrac\phi2$: $$a_1 = \tfrac12 b(1 + \cos\phi) = b\cos^2\tfrac\phi2, \qquad b_1 = \sqrt{a_1 b} = b\cos\tfrac\phi2.$$ Then $a_2 = \tfrac12 b\cos\tfrac\phi2\,(1 + \cos\tfrac\phi2) = b\cos\tfrac\phi2\cos^2\tfrac\phi4$ and $b_2 = b\cos\tfrac\phi2\cos\tfrac\phi4$. In general, by induction, $$b_n = b\,\cos\tfrac\phi2\cos\tfrac\phi4\cdots\cos\tfrac\phi{2^n}, \qquad a_n = b_n\cos\tfrac{\phi}{2^n}.$$

The limit. Repeatedly using $\sin 2\theta = 2\sin\theta\cos\theta$, $$\cos\tfrac\phi2\cos\tfrac\phi4\cdots\cos\tfrac\phi{2^n} = \frac{\sin\phi}{2^n\sin(\phi/2^n)}.$$ As $n\to\infty$, $2^n\sin(\phi/2^n) \to \phi$, so $b_n \to \dfrac{b\sin\phi}{\phi}$, and $a_n$ has the same limit because $\cos(\phi/2^n) \to 1$. Finally $b\sin\phi = \sqrt{b^2 - a^2}$ and $\phi = \cos^{-1}(a/b)$, which is the formula.

Computing $\pi$. Take $a = 0$, $b = 1$ (the steps still make sense, with $\phi = \tfrac\pi2$). The limit is $\dfrac{1}{\pi/2} = \dfrac2\pi$. So run the iteration from $0$ and $1$ and take $2$ divided by the limit. The error shrinks by about a factor of $4$ at each step, so twenty steps already give $\pi$ to about twelve decimal places.

Coordinates. Write each point as $xA + yB + zC$ with $x + y + z = 1$, and record it as $(x, y, z)$; any multiple of the triple names the same point. Then $$A' \sim (0,\ q_1,\ p_1), \qquad B' \sim (p_2,\ 0,\ q_2), \qquad C' \sim (q_3,\ p_3,\ 0).$$ (For instance $A' = \frac{q_1B + p_1C}{p_1 + q_1}$, which divides $BC$ in the ratio $p_1 : q_1$.)

The three lines. A point $(x, y, z)$ is on $AA'$ when $y : z = q_1 : p_1$, on $BB'$ when $x : z = p_2 : q_2$, and on $CC'$ when $x : y = q_3 : p_3$.

The three crossings. Solving these in pairs, $$A'' \sim (p_2q_3,\ p_2p_3,\ q_2q_3), \quad B'' \sim (q_1q_3,\ p_3q_1,\ p_3p_1), \quad C'' \sim (p_1p_2,\ q_1q_2,\ p_1q_2).$$ The sums of these triples are exactly the three factors in the denominator of the claimed ratio: $p_2p_3 + q_2q_3 + p_2q_3$, and so on.

The area. Divide each triple by its sum to get true coordinates. Since the map from $(x, y, z)$ to the point $xA + yB + zC$ is linear, the area of $A''B''C''$ is the absolute value of the determinant of the three normalized rows, times the area of $ABC$. The determinant of the unnormalized rows works out to $$\det\begin{pmatrix} p_2q_3 & p_2p_3 & q_2q_3 \\ q_1q_3 & p_3q_1 & p_1p_3 \\ p_1p_2 & q_1q_2 & p_1q_2 \end{pmatrix} = (p_1p_2p_3 - q_1q_2q_3)^2,$$ and dividing the rows by their sums gives the stated ratio. (When $p_1p_2p_3 = q_1q_2q_3$ the three lines meet in a point, which is Ceva’s theorem.)

Take the parabola $y^2 = 4ax$ with focus $(a, 0)$, and write points as $(at^2, 2at)$.

Focal chords. A line through the focus at angle $\theta$ to the axis meets the parabola at distances $\dfrac{2a}{1\mp\cos\theta}$ from the focus (the polar equation from the focus), so its length is $$\frac{2a}{1-\cos\theta} + \frac{2a}{1+\cos\theta} = \frac{4a}{\sin^2\theta}.$$ If a chord of length $L$ rises by $h$, then $\sin\theta = h/L$, so the focal chord in that direction has length $\dfrac{4aL^2}{h^2}$.

Apply it twice. Let $P = (ap^2, 2ap)$ and $Q = (aq^2, 2aq)$. The point $R$ is on the line through $Q$ parallel to the axis, so $R$ has the same height $2aq$ as $Q$. Both $PQ$ and $PR$ rise by $h = 2a(q - p)$.

$PQ$ is itself a focal chord, so $PQ = \dfrac{4a\,PQ^2}{h^2}$, which gives $PQ = \dfrac{h^2}{4a}$. The focal chord parallel to $PR$ has length $$\frac{4a\,PR^2}{h^2} = \frac{PR^2}{PQ}.$$

Use the ellipse $x = a\cos\phi$, $y = b\sin\phi$ with eccentricity $e$, foci $S = (ae, 0)$ and $H = (-ae, 0)$, and recall that the point at eccentric angle $\phi$ is at distance $a(1 \mp e\cos\phi)$ from $S$ and $H$.

Where $O$ is. Let $P$ and $Q$ have eccentric angles $m + d$ and $m - d$. The tangent at $\phi$ is $\frac xa\cos\phi + \frac yb\sin\phi = 1$, and the point $O = \left(\frac{a\cos m}{\cos d}, \frac{b\sin m}{\cos d}\right)$ is on both tangents, because $\cos(\phi - m) = \cos d$ for $\phi = m \pm d$.

Tangent lengths. The tangent at $P$ points along $(-a\sin\phi, b\cos\phi)$, and the semidiameter parallel to it ends at the point of angle $\phi + 90^\circ$, so $CP'^2 = a^2\sin^2\phi + b^2\cos^2\phi$. A short computation with $\phi = m + d$ gives $$O - P = \tan d\,\big(a\sin\phi,\ -b\cos\phi\big), \qquad\text{so}\qquad OP = |\tan d|\;CP'.$$ Likewise $OQ = |\tan d|\,CQ'$. Hence $$OP\cdot OQ + CP'\cdot CQ' = (1 + \tan^2 d)\,CP'\cdot CQ' = \frac{CP'\cdot CQ'}{\cos^2 d}.$$

Semidiameters and focal distances. Since $b^2 = a^2(1 - e^2)$, $$CP'^2 = a^2 - a^2e^2\cos^2\phi = a(1 - e\cos\phi)\cdot a(1 + e\cos\phi) = SP\cdot HP.$$ So $CP'\cdot CQ' = \sqrt{SP\cdot SQ\cdot HP\cdot HQ}$.

The focal distances of $O$. Expanding, $$OS^2\cos^2 d = (a\cos m - ae\cos d)^2 + b^2\sin^2 m = a^2\big[1 - 2e\cos m\cos d + e^2(\cos^2 d - \sin^2 m)\big],$$ $$SP\cdot SQ = a^2\big(1 - e\cos(m+d)\big)\big(1 - e\cos(m-d)\big) = a^2\big[1 - 2e\cos m\cos d + e^2(\cos^2 m - \sin^2 d)\big].$$ These agree, because $\cos^2 d - \sin^2 m = \cos^2 m - \sin^2 d$. So $OS^2\cos^2 d = SP\cdot SQ$, and in the same way $OH^2\cos^2 d = HP\cdot HQ$.

Put together. $OS\cdot OH\cos^2 d = \sqrt{SP\cdot SQ\cdot HP\cdot HQ} = CP'\cdot CQ'$, so $$OP\cdot OQ + CP'\cdot CQ' = \frac{CP'\cdot CQ'}{\cos^2 d} = OS\cdot OH.$$

The shape. A smooth peg pushes at right angles to its rod. The supported rods are parallel, so the two reactions are parallel, and together they balance the weight. So they are vertical, and the supported rods are horizontal. Call the upper supported rod $AB$, resting on the peg $P$ just beneath it, and the lower one $DC$, resting on the peg $Q$; the unsupported rods are $DA$ and $CB$. Take $D$ as origin, $x$ across and $y$ up: $$D=(0,0),\quad C=(a,0),\quad A=(d,k),\quad B=(a+d,k),\qquad d^2+k^2=a^2 ,$$ since the gap between the horizontal rods is the vertical distance $k$ between the pegs. Let $W$ be the weight of each rod and let $Q=(q,0)$.

Each peg carries $2W$. Let rod $DA$ push rod $AB$ at $A$ with force $(X_A,Y_A)$, and rod $CB$ push it at $B$ with $(X_B,Y_B)$. For rod $AB$, horizontally $X_A+X_B=0$, and vertically $Y_A+Y_B+R_P=W$. Rod $DA$ feels $(-X_A,-Y_A)$ at $A$; taking moments about $D$, $$kX_A-dY_A-\tfrac12Wd=0,$$ and in the same way, for rod $CB$ about $C$, $kX_B-dY_B-\tfrac12Wd=0$. Adding, and using $X_A+X_B=0$, gives $Y_A+Y_B=-W$ (the rhombus is not a square, so $d\ne0$). Hence $R_P=2W$, and so $R_Q=4W-2W=2W$.

The pegs are symmetric about the centre. Let $O$ be the centre of the rhombus. The weights of opposite rods act at points symmetric about $O$, so their moments about $O$ cancel. The two equal reactions must then have cancelling moments too: $P$ and $Q$ are equally far from $O$ horizontally, on opposite sides. So $x_P+x_Q=a+d$.

The product. Name the pegs so that $P$ is the one to the right, $x_P-x_Q=h$. Then $$q=x_Q=\tfrac12(a+d-h),$$ which is left of $O$, so the unsupported rod on $Q$’s side is $DA$, with ends at horizontal positions $0$ and $d$. The product of the horizontal distances, counted with sign (negative if $Q$ lies horizontally between the ends), is $$q\,(q-d)=\tfrac14\big((a-h)+d\big)\big((a-h)-d\big)=\tfrac14\big((a-h)^2-d^2\big)=\tfrac14\big(k^2-2ah+h^2\big),$$ using $d^2=a^2-k^2$. A half turn about $O$ maps the frame to itself and swaps $P$ with $Q$ and $DA$ with $CB$, so the same holds for $P$.

Bending moment. For rod $DC$, let $Y_D$ and $Y_C$ be the vertical forces on it at $D$ and $C$. Vertically $Y_D+Y_C+2W=W$, and moments about $D$ give $aY_C+2Wq-\tfrac12Wa=0$. So $$Y_C=\tfrac12W-\frac{2Wq}{a},\qquad Y_D=-\frac{W(3a-4q)}{2a}.$$ Take a point of the rod at distance $x>q$ from $D$, and add the moments about it of the forces on the piece between $D$ and the point. Horizontal forces act along the rod and give nothing. An upward force $F$ at position $x_F$ gives $(x_F-x)F$, so the bending moment is $$-xY_D+\frac{Wx^2}{2a}+2W(q-x)=\frac{W}{2a}\big(x^2-ax-4qx+4aq\big)=\frac{W}{2a}\,(x-a)(x-4q).$$ It vanishes at $x=4q$, the point at distance $3q$ beyond the peg. Here $q$ is the distance from the peg to $D$, where the unsupported rod $DA$ is hinged. When $q$ is the shorter of the peg’s two distances to the hinges $D$ and $C$ (measured along the rod), this is the stated point, and it lies on the rod when $q\le\tfrac14a$. If the peg is nearer $C$, the same work from the other end puts the point at distance $3(a-q)$ beyond the peg, toward $D$. The upper rod follows by the half turn about $O$.

The condition. The moving lamina touches the fixed one at a single point $T$. Friction can supply any force along the tangent, so the only condition is that the weight, acting at the centre $G$ of the moving ellipse, has no moment about $T$: the line $TG$ is vertical.

Axes. Use axes along the major and minor axes of the fixed ellipse, centre $O$: $$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1,\qquad B^2=A^2(1-e^2).$$ The major axis rises at the angle $\alpha$ above the horizon, so in these axes the upward vertical has direction $(\sin\alpha,\cos\alpha)$.

The point of contact. Let the common tangent have direction $\mathbf t=(\cos\psi,\sin\psi)$ and normal $\mathbf n=(-\sin\psi,\cos\psi)$. The point of the ellipse where the tangent has this direction is $$T=\lambda\,(-A^2\sin\psi,\ B^2\cos\psi),\qquad \lambda=\big(A^2\sin^2\psi+B^2\cos^2\psi\big)^{-1/2},$$ since the normal there is along $(x/A^2,\,y/B^2)=\lambda(-\sin\psi,\cos\psi)$. Then $$T\cdot\mathbf n=\frac1\lambda,\qquad T\cdot\mathbf t=-\lambda(A^2-B^2)\sin\psi\cos\psi=-\lambda A^2e^2\sin\psi\cos\psi .$$

The moving ellipse. It is the mirror image of the fixed one in the tangent line. Its centre is the image of $O$, namely $G=2(T\cdot\mathbf n)\,\mathbf n$. Its major axis is the image of a line at angle $0$ in a line at angle $\psi$, so it lies at angle $2\psi$ in our axes, which is $\alpha+2\psi$ above the horizon. So $\theta=\alpha+2\psi$, that is, $\psi=\tfrac12(\theta-\alpha)$.

$TG$ vertical. We have $G-T=-(T\cdot\mathbf t)\,\mathbf t+(T\cdot\mathbf n)\,\mathbf n$. The vertical has components $\sin(\alpha+\psi)$ along $\mathbf t$ and $\cos(\alpha+\psi)$ along $\mathbf n$. The two are parallel when $$(T\cdot\mathbf t)\cos(\alpha+\psi)+(T\cdot\mathbf n)\sin(\alpha+\psi)=0 .$$ Multiply by $\lambda$, and note $\lambda^2A^2=\dfrac{1}{\sin^2\psi+(1-e^2)\cos^2\psi}=\dfrac1{1-e^2\cos^2\psi}$: $$\sin(\alpha+\psi)\,(1-e^2\cos^2\psi)=e^2\sin\psi\cos\psi\cos(\alpha+\psi),$$ $$\sin(\alpha+\psi)=e^2\cos\psi\big(\cos\psi\sin(\alpha+\psi)+\sin\psi\cos(\alpha+\psi)\big)=e^2\cos\psi\,\sin(\alpha+2\psi).$$ With $\psi=\tfrac12(\theta-\alpha)$ we have $\alpha+\psi=\tfrac12(\theta+\alpha)$ and $\alpha+2\psi=\theta$, so $$\sin\tfrac12(\theta+\alpha)=e^2\sin\theta\cos\tfrac12(\theta-\alpha).$$

A flat picture. Everything is symmetric about the vertical plane midway between the strings, so work in the cross-section: a square of side $s$ with one side against the wall. Let $w$ be each hanging weight. The pulleys are smooth, so each string has tension $w$, and together they pull the far top corner $E$ with force $2w$. The cube weighs $4w\sin\beta$ and its weight acts at the centre. Let the slant strings run from $E$ toward the wall at the angle $\chi$ above the horizontal ($\chi<0$ if the pulleys are below $E$). The wall pushes with a normal force $R$ at some point of the contact face, a depth $\delta$ below the top edge with $0\le\delta\le s$, and with friction $F$, counted upward.

Equilibrium.

• Across: $R=2w\cos\chi$.
• Up: $F=4w\sin\beta-2w\sin\chi$.
• Moments about the point where $R$ acts (friction acts along the wall, through that point): $2w\,(s\sin\chi+\delta\cos\chi)=4w\sin\beta\cdot\tfrac12s$, so $$\delta=s\,\frac{\sin\beta-\sin\chi}{\cos\chi}.$$

Friction. With $\mu=\tan\beta$, the condition $F\le\mu R$ reads $2\sin\beta-\sin\chi\le\tan\beta\cos\chi$. Multiplying by $\cos\beta$, this is $$\sin(\chi+\beta)\ \ge\ \sin2\beta .$$ Because $\tfrac14\pi<\beta<\tfrac12\pi$, the angle $2\beta$ lies between $\tfrac12\pi$ and $\pi$, so this holds exactly when $\pi-2\beta\le\chi+\beta\le2\beta$, that is, $$\pi-3\beta\ \le\ \chi\ \le\ \beta .$$ (The other condition, $F\ge-\mu R$, always holds: it reads $\sin(\chi-\beta)\le\sin2\beta$, and $\sin(\chi-\beta)<\cos\beta<2\sin\beta\cos\beta$.)

The wall reaction stays on the face. We need $0\le\delta\le s$. The first part is $\chi\le\beta$. At $\chi=\beta$, $\delta=0$: the reaction is at the top edge. At $\chi=\pi-3\beta$, we have $\sin\chi=\sin3\beta$ and $\cos\chi=-\cos3\beta$, so $\delta=s$ exactly when $\sin3\beta-\sin\beta=\cos3\beta$, that is, $2\sin\beta\cos2\beta=\cos3\beta$. The given condition says the same thing: $$\tan3\beta+\tan\beta=2\iff\sin4\beta=2\cos3\beta\cos\beta\iff\sin2\beta\cos2\beta=\cos3\beta\cos\beta\iff2\sin\beta\cos2\beta=\cos3\beta .$$ So at the lower end the reaction is exactly at the bottom edge. (The condition gives $\beta\approx62.1^\circ$, so $\pi-3\beta\approx-6.3^\circ$.) In between, $\delta\le s$ means $\sin\beta\le\sin\chi+\cos\chi=\sqrt2\sin(\chi+\tfrac14\pi)$. The right side is concave on this range, so it is smallest at an end, and it holds at both ends.

The two limits. So the cube rests exactly when $\pi-3\beta\le\chi\le\beta$. Raising or lowering the cube changes $\chi$ steadily, so the two limiting positions have the slant strings at $\chi=\beta$ above the horizontal and at $\chi=\pi-3\beta$, which is $3\beta-\pi$ below it. In both the cube is on the point of slipping down. The angle between the two directions is $$\beta-(\pi-3\beta)=4\beta-\pi .$$

Measure from the starting point. The particle is at $$x = Vt\cos\beta, \qquad y = Vt\sin\beta - \tfrac12gt^2.$$ The telescope sits at $(X, 0)$ and points at elevation $\alpha$, so the particle stays in view exactly when $y = (x - X)\tan\alpha$, that is, $$X = x - y\cot\alpha = Vt(\cos\beta - \sin\beta\cot\alpha) + \tfrac12 g t^2\cot\alpha.$$ Differentiating, the platform’s velocity at the start is $$V(\cos\beta - \sin\beta\cot\alpha) = V\,\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\sin\alpha} = V\frac{\sin(\alpha - \beta)}{\sin\alpha},$$ and its acceleration is the constant $g\cot\alpha$.

Let the common angle of projection be $i$ and $T = \tan i$. Write $k_j = \dfrac{ga_j}{2v_j^2}$ and $t_j = \tan\alpha_j$.

Two quadratics. A projectile with speed $v$ reaches horizontal distance $a$ at height $a\tan i - \dfrac{ga^2}{2v^2\cos^2 i}$. For $P$ that height is $a_1t_1$. Dividing by $a_1$ and using $\sec^2 i = 1 + T^2$: $$k_1T^2 - T + (k_1 + t_1) = 0, \qquad k_2T^2 - T + (k_2 + t_2) = 0.$$

Eliminate $T$. Subtracting gives $(k_1 - k_2)T^2 = (t_2 - t_1) + (k_2 - k_1)$. Taking $k_2$ times the first minus $k_1$ times the second gives $(k_1 - k_2)T = k_1t_2 - k_2t_1$. Squaring the second and using the first: $$(k_1t_2 - k_2t_1)^2 + (k_1 - k_2)^2 = (k_1 - k_2)(t_2 - t_1).$$

Back to angles. The left side is $k_1^2(1 + t_2^2) + k_2^2(1 + t_1^2) - 2k_1k_2(1 + t_1t_2)$. Multiply the whole equation by $\cos^2\alpha_1\cos^2\alpha_2$, using $(1 + t_2^2)\cos^2\alpha_2 = 1$, $(1 + t_1t_2)\cos\alpha_1\cos\alpha_2 = \cos(\alpha_2 - \alpha_1)$ and $(t_2 - t_1)\cos\alpha_1\cos\alpha_2 = \sin(\alpha_2 - \alpha_1)$. Then divide by $\cos\alpha_1\cos\alpha_2$: $$k_1^2\frac{\cos\alpha_1}{\cos\alpha_2} + k_2^2\frac{\cos\alpha_2}{\cos\alpha_1} = 2k_1k_2\cos(\alpha_2 - \alpha_1) + (k_1 - k_2)\sin(\alpha_2 - \alpha_1).$$ Finally put back $k_j = ga_j/(2v_j^2)$ and multiply by $4v_1^4v_2^4/g$. The terms become exactly those of the stated identity.

Who carries what. Let the buckets have masses $M_P$ and $M_Q$, and let $n$ balls be in $Q$ at the start. A ball resting in a bucket moves with it; a ball hanging from its string is at rest and plays no part. The balls hang at intervals $a$, so each time the buckets move a distance $a$, one more ball is lifted out of $Q$ and, at the same moment, $P$ reaches its next ball and carries it up. At the first instant one ball leaves $Q$ and $P$ takes up the ball it was touching. So after $r$ balls have been lifted off $Q$, $Q$ holds $n-r$ balls and $P$ holds $r$.

The acceleration. The moving mass is $M=M_P+M_Q+nm$, the same in every interval, and the driving force is the difference of the two weights. So, counting the acceleration positive when $Q$ descends, $$f_r=g\,\frac{M_Q-M_P+(n-2r)m}{M_P+M_Q+nm}.$$

The jerks. At the end of each interval two things happen at once. A ball is snatched out of $Q$ by its string and stopped; it simply leaves the base, and does not act on the rest. And $P$ strikes a ball at rest and carries it on. The string over the pulley makes this a one-line impact, so momentum along the motion is shared: a speed $v$ just before becomes $$v'=\frac{(M-m)\,v}{M}=v\Big(1-\frac mM\Big)$$ just after. (To first order in $m/M$, which is all we use, the result does not depend on how the two events are timed.)

One interval. Let $D$ and $A$ be the total descending and ascending masses during an interval, so $f=g(D-A)/M$. Let the speed be $u$ at the start (just after a jerk) and $v$ at the end (just before the next jerk). Then $v^2-u^2=2fa$, and the mean velocity over the interval, for uniform acceleration, is $\bar w=\tfrac12(u+v)$. So $$v-u=\frac{2fa}{u+v}=\frac{fa}{\bar w}=\frac{ga\,(D-A)}{M\bar w}.$$ The condition in the problem is $\bar w^2:ga=(D-A):m$, that is, $ga(D-A)=m\bar w^2$. Then $$v-u=\frac mM\,\bar w .$$ The jerk then takes away $\dfrac mM\,v$. So the speed just after the jerk, where the next interval begins, is $$v'=u+\frac mM(\bar w-v)=u-\frac mM\cdot\frac{v-u}{2}=u-\frac12\Big(\frac mM\Big)^2\bar w .$$ The gain during the interval and the loss at the jerk are each about $\frac mM$ of the speed, and they cancel. What is left is only $\tfrac12(m/M)^2$ of it. So the velocity at the end of the interval is approximately the same as at the beginning, and the motion settles into a steady rhythm.

Setup. Take the centre $C$ of the ellipse as origin, with semi-axes $a$ and $b$, so a point of the ellipse is $(a\cos E,\ b\sin E)$. Let $O=(x_0,y_0)$, inside the ellipse. The polar of $O$ is the line $xx_0/a^2+yy_0/b^2=1$. Put $N=\sqrt{x_0^2/a^4+y_0^2/b^4}$. The distance of a point $(x,y)$ on the centre’s side of the polar is $$p=\frac{1-xx_0/a^2-yy_0/b^2}{N},\qquad\text{so}\qquad p_0=\frac1N .$$ On the ellipse, write $u=1-\dfrac{x_0}{a}\cos E-\dfrac{y_0}{b}\sin E=\dfrac p{p_0}$. It is positive: by Cauchy–Schwarz, $\frac{x_0}{a}\cos E+\frac{y_0}{b}\sin E\le\sqrt{x_0^2/a^2+y_0^2/b^2}<1$.

The speed of $E$. The force points to $O$, so twice the rate at which the radius from $O$ sweeps area is a constant $h$: $$(x-x_0)\dot y-(y-y_0)\dot x=h .$$ With $\dot x=-a\sin E\,\dot E$ and $\dot y=b\cos E\,\dot E$ the left side is $ab\,\dot E\,(1-\tfrac{x_0}{a}\cos E-\tfrac{y_0}{b}\sin E)=ab\,u\,\dot E$. So $$\dot E=\frac ku,\qquad k=\frac h{ab}.$$

The acceleration. Let $u'=du/dE=\frac{x_0}{a}\sin E-\frac{y_0}{b}\cos E$. Then $\ddot E=-\dfrac{k u'\dot E}{u^2}=-\dfrac{k^2u'}{u^3}$, and $$\ddot x=-a\cos E\,\dot E^2-a\sin E\,\ddot E=\frac{k^2a}{u^3}\big(-u\cos E+u'\sin E\big)=\frac{k^2a}{u^3}\Big(\frac{x_0}{a}-\cos E\Big)=-\frac{k^2}{u^3}(x-x_0).$$ In the same way $\ddot y=-\dfrac{k^2}{u^3}(y-y_0)$. So the acceleration points to $O$ and has size $$\frac{k^2r}{u^3}=\frac{h^2p_0^3}{a^2b^2}\cdot\frac r{p^3}.$$

The period. This must equal $\mu r/p^3$, so $h^2p_0^3=\mu a^2b^2$, and $h=ab\sqrt\mu\,/\,p_0^{3/2}$. The area $\pi ab$ is swept at the rate $\tfrac12h$, so the period is $$\frac{\pi ab}{\tfrac12h}=\frac{2\pi p_0^{3/2}}{\sqrt\mu}.$$

A check. If $O$ is a focus, its polar is the directrix and $p=r/e$, so the law is $\mu e^3/r^2$, the inverse square law. Here $p_0=a/e$, and the period is $2\pi a^{3/2}/\sqrt{\mu e^3}$, which is Kepler’s law for that strength.

Let the radius be $r$ and measure angles at the center from the surface. A point at distance $s$ from the center and angle $\theta$ is at depth $s\sin\theta$, where the pressure is $\rho g s\sin\theta$. So the force on the sector between $\theta_1$ and $\theta_2$ is $$\rho g\int_{\theta_1}^{\theta_2}\int_0^r s\sin\theta\; s\,ds\,d\theta = \frac{\rho g r^3}{3}\,(\cos\theta_1 - \cos\theta_2).$$ So the force depends only on the change in $r\cos\theta$, which is the position of the edge of the sector when you drop it straight onto the diameter.

Construction. Divide the diameter into $n$ equal parts. Through the points of division draw lines perpendicular to the diameter, meeting the semicircle. Join these points of the arc to the center. The $n$ sectors carry equal forces, since each gives the same change in $r\cos\theta$. (The sectors near the ends of the diameter, close to the surface, come out wider than those in the middle.)

One degree of freedom. Water does not compress, so the diaphragm $w_1$, the water and the diaphragm $w_2$ move round the tube together, the water filling an arc of angle $\gamma$. The air fills the rest of the tube, one connected space whose volume does not change, so its pressure is the same everywhere and stays fixed.

Moments about the centre. Let the tube have radius $a$ and centre $O$. Measure angles at $O$ from a horizontal radius, anticlockwise, and let $w_1$ sit at angle $\phi$, the water fill the arc from $\phi$ to $\phi+\gamma$, and $w_2$ sit at $\phi+\gamma$. Look at the piece made of the diaphragms and the water, and take moments about the axis of the tube through $O$:

• The air pushes the two diaphragms with equal forces along the tube, in opposite senses, at the same distance $a$ from the axis. Their moments cancel.
• The smooth walls of the tube push at right angles to the tube’s direction, so they have no moment about the axis.

So the weights alone have no moment about $O$. A weight acting at horizontal position $X$ has moment proportional to $X$. The water’s centre of gravity is at horizontal position $\dfrac a\gamma\int_\phi^{\phi+\gamma}\cos t\,dt=\dfrac a\gamma\big(\sin(\phi+\gamma)-\sin\phi\big)$. Hence $$w_1\cos\phi+w_2\cos(\phi+\gamma)+\frac{w_3}{\gamma}\big(\sin(\phi+\gamma)-\sin\phi\big)=0 .$$ Multiply by $\gamma$ and expand: $$\cos\phi\,\big(w_1\gamma+w_2\gamma\cos\gamma+w_3\sin\gamma\big)=\sin\phi\,\big(w_3(1-\cos\gamma)+w_2\gamma\sin\gamma\big),$$ $$\tan\phi=\frac{w_1\gamma+w_2\gamma\cos\gamma+w_3\sin\gamma}{w_3(1-\cos\gamma)+w_2\gamma\sin\gamma}.$$ The radius to $w_1$ makes the angle $\phi$ with the horizontal, so this is the stated inclination.

Which position is stable. Write the equation as $A\cos\phi=B\sin\phi$. The height of the centre of gravity of the piece is proportional to $$V(\phi)=w_1\sin\phi+w_2\sin(\phi+\gamma)+\frac{w_3}{\gamma}\big(\cos\phi-\cos(\phi+\gamma)\big)=\frac1\gamma\big(A\sin\phi+B\cos\phi\big),$$ and the equation above is $V'(\phi)=0$. Since $V$ is a sine curve in $\phi$, it has one lowest and one highest position, half a turn apart. Both give the same line through $O$, at the angle found. The stable one is the lowest, where $\sin\phi=-A/\sqrt{A^2+B^2}$ and $\cos\phi=-B/\sqrt{A^2+B^2}$.

Pressure of the vapour. The total pressure in the space above the mercury balances the atmosphere minus the column of mercury raised in the tube: $b - h$, in millimetres of mercury. Some of it comes from the mercury’s own vapour, $e$. So the vapour of our liquid has pressure $$p = b - h - e.$$

Air under the same conditions. Air at $0^\circ$C and $760$ mm has density $0.001293$ g/cm³. By Boyle’s law density is proportional to pressure, and by Charles’s law, at fixed pressure, inversely proportional to the absolute temperature, $273 + t$. So air at temperature $t$ and pressure $p$ has density $$0.001293\cdot\frac{p}{760}\cdot\frac{273}{273 + t}.$$

The ratio. The vapour has density $m/v$. Dividing, $$\text{vapour density} = \frac{m}{0.001293\,v}\cdot\frac{273+t}{273}\cdot\frac{760}{b - h - e}.$$ The factor $\frac{273+t}{273}$ is often written $1 + 0.00366\,t$. (We ignore the small expansions of the glass and of the mercury column with heat; a careful experimenter corrects for them.)

Let $S$ be the focus, and let all the incident rays make the angle $\psi$ with the axis.

A fixed angle at the focus. At a point $P$ of the parabola, the tangent makes equal angles with $PS$ and with the line through $P$ parallel to the axis (that is why rays along the axis reflect through $S$). So reflection in the tangent swaps those two lines. By the law of reflection it also swaps the incident line and the reflected line. A reflection keeps the size of angles but reverses their sense. So the reflected ray at $P$ makes the angle $\psi$ with $PS$, turned the opposite way from the incident ray’s turn from the axis. This holds at every point of incidence.

Two rays: the construction. Let the two rays strike at $P_1$ and $P_2$, and let the reflected rays $r_1$ and $r_2$ meet at $X$. Measure angles between lines with sign, up to multiples of $180^\circ$. Then $$\angle(r_1,r_2)=\angle(r_1,P_1S)+\angle(P_1S,P_2S)+\angle(P_2S,r_2)=\angle(P_1S,P_2S),$$ because the first and last terms are equal and opposite. So the lines $XP_1$, $XP_2$ make the same angle as $SP_1$, $SP_2$, and $X$ lies on the circle through $S$, $P_1$ and $P_2$.

Construction: draw the circle through $S$, $P_1$, $P_2$. At $P_1$ draw the line that makes the angle $\psi$ with $P_1S$, turned the opposite way from the incident ray’s turn from the axis; this is the reflected ray. It meets the circle again at the required point $X$.

The caustic. Let $P_2$ slide toward $P_1=P$. The circle through $S$, $P_1$, $P_2$ becomes the circle through $S$ that touches the parabola at $P$. The meeting point of two neighboring reflected rays becomes the point where the reflected ray at $P$ touches the caustic (the caustic is the envelope of the reflected rays). So:

The reflected ray at $P$ touches the caustic where it meets again the circle through $S$ that touches the parabola at $P$.

How far along the ray. In a circle of radius $\rho$, a chord from $P$ making the angle $\epsilon$ with the tangent at $P$ has length $2\rho\sin\epsilon$ (the angle between tangent and chord equals the angle in the opposite segment). The chord $PS$ makes with the tangent the angle $\chi$ between the tangent and the axis. The chord $PX$ makes with the tangent the angle $\epsilon$ between the tangent and the reflected ray, which by the law of reflection is also the angle between the tangent and the incident ray. So $$PX=SP\,\frac{\sin\epsilon}{\sin\chi}.$$ For the parabola $y^2=4ax$, at the point $(at^2,2at)$ the tangent has slope $1/t$, so $\sin\chi=1/\sqrt{1+t^2}$, and $SP=a(1+t^2)$. So $\sin\chi=\sqrt{a/SP}$ and $$PX=\frac{SP^{3/2}\sin\epsilon}{\sqrt a}.$$ For rays parallel to the axis, $\epsilon=\chi$ and $PX=PS$: every reflected ray passes through the focus, and the caustic shrinks to that point, as it should.

Let $AB = u$ and the radius be $r$. Measure distances along the line, positive in the direction the light travels (from $A$ towards the sphere), from the surface where the refraction happens. For refraction from index $n_1$ to $n_2$ at a surface of radius $R$, the image distance $v$ of an object at distance $s$ satisfies $$\frac{n_2}{v} - \frac{n_1}{s} = \frac{n_2 - n_1}{R}.$$

First surface ($s = -u$, $R = r$, from $1$ to $\mu$): $\dfrac{\mu}{v_1} = \dfrac{\mu - 1}{r} - \dfrac1u$.

Second surface (a distance $2r$ further on; $s = v_1 - 2r$, $R = -r$, from $\mu$ to $1$): $\dfrac1{v_2} = \dfrac{\mu - 1}{r} + \dfrac{\mu}{v_1 - 2r}$.

Working these out, the final image $C$ lies on $BA$ produced (beyond $A$) when $(2-\mu)(u + r) > \mu u$, at distance $$AC = \frac{2(\mu - 1)\,w^2}{(2 - \mu)\,w - \mu u}, \qquad w = u + r.$$

Minimize. As a function of $w$, the derivative of $\dfrac{w^2}{\alpha w - \beta}$ is $\dfrac{w(\alpha w - 2\beta)}{(\alpha w - \beta)^2}$, which is zero at $w = 2\beta/\alpha$, where the quotient changes from falling to rising. Here $\alpha = 2 - \mu$ and $\beta = \mu u$, so the least $AC$ comes at $$u + r = \frac{2\mu u}{2 - \mu}, \qquad r = \frac{2\mu - (2-\mu)}{2-\mu}\,u = \frac{3\mu - 2}{2 - \mu}\,AB.$$ For glass with $\mu = 1.5$ that is a sphere of radius $5\,AB$.

Let Venus be at distance $\Delta$ from the Earth, and let $\psi$ be the angle Sun–Venus–Earth. The lit fraction of the visible disk is $\tfrac12(1 + \cos\psi)$, so in this model the brightness is proportional to $(1 + \cos\psi)/\Delta^2$.

Brightness in terms of $\Delta$. By the cosine rule in the triangle Sun–Venus–Earth, $\cos\psi = \dfrac{v^2 + \Delta^2 - e^2}{2v\Delta}$, so $$1 + \cos\psi = \frac{(v + \Delta)^2 - e^2}{2v\Delta}, \qquad \text{brightness} \propto f(\Delta) = \frac{\Delta^2 + 2v\Delta - p}{\Delta^3},$$ where $p = e^2 - v^2$. In terms of $x = 1/\Delta$, $f = x + 2vx^2 - px^3$.

Two positions at one elongation. By the cosine rule at the Earth, the distances at elongation $\varepsilon$ solve $\Delta^2 - 2e\cos\varepsilon\,\Delta + p = 0$. Let $s = \Delta_1 + \Delta_2 = 2e\cos\varepsilon$; also $\Delta_1\Delta_2 = p$. With $x_j = 1/\Delta_j$ we have $x_1 + x_2 = s/p$ and $x_1x_2 = 1/p$.

Equal brightness. $f(x_1) - f(x_2) = (x_1 - x_2)\big[1 + 2v(x_1 + x_2) - p(x_1^2 + x_1x_2 + x_2^2)\big]$. For distinct positions the bracket must vanish. Using $x_1^2 + x_1x_2 + x_2^2 = (x_1+x_2)^2 - x_1x_2 = s^2/p^2 - 1/p$: $$1 + \frac{2vs}{p} - \frac{s^2}{p} + 1 = 0 \quad\Longleftrightarrow\quad s^2 - 2vs - 2p = 0.$$ The positive root is $s = v + \sqrt{v^2 + 2p} = v + \sqrt{2e^2 - v^2}$, so $\cos\varepsilon = s/(2e)$, as stated.

Maxima and minima. At this elongation the nearer and farther positions have equal brightness. They lie on the same side of the Sun, so as Venus moves from one to the other, the distance $\Delta$ changes steadily from $\Delta_1$ to $\Delta_2$ and no conjunction comes in between. The brightness $f(\Delta)$ is a smooth function with equal values at the two ends, so by Rolle’s theorem its derivative vanishes somewhere in between. That is a maximum or minimum of brightness at a point that is not a conjunction. For the real orbits ($v \approx 0.72e$) this elongation is about $14^\circ$, well inside Venus’s greatest elongation of about $46^\circ$.