The 1854 Problems paper

Let $t = AB = AC$ be the length of the tangents from $A$ to our circle.

A fact about any incircle. Take a triangle $ADE$ and its incircle. From each corner the two tangents to the incircle are equal; call them $x$ from $A$, $y$ from $D$ and $z$ from $E$. Then $AD = x + y$, $AE = x + z$ and $DE = y + z$, so $$x = \frac{AD + AE - DE}{2}.$$

Two circles in the same corner. Both our circle and the incircle of $ADE$ touch the lines $AD$ and $AE$ on the inside of the angle at $A$. Such a circle has its center on the bisector of that angle, and it is fixed once we know the tangent length from $A$. So the two circles are the same circle exactly when $t = x$, that is, when $$2t = AD + AE - DE.$$

Since $D$ lies beyond $B$ and $E$ beyond $C$, we have $AD = t + BD$ and $AE = t + CE$. The condition becomes $$DE = BD + CE.$$

That is exactly what we are told. So our circle is the incircle of triangle $ADE$, and in particular it touches $DE$. (The argument also runs the other way: if $DE$ touches the circle, then $DE = BD + CE$.)

Let $OA = x$ and $AC = c$, so $OC = x + c$. Harmonic progression means $$\frac{2}{OB} = \frac{1}{x} + \frac{1}{x+c}, \qquad\text{so}\qquad OB = \frac{2x(x+c)}{2x+c}.$$ Then $$AB = OB - x = \frac{2x^2 + 2xc - 2x^2 - xc}{2x+c} = \frac{xc}{2x+c} = \frac{c}{2 + c/x}.$$ When $O$ moves towards $A$, the distance $x$ shrinks, so $c/x$ grows, so $AB$ shrinks. With $A$ fixed, that means $B$ moves towards $A$.

Three positive numbers $p, q, r$ are in geometric progression when $q^2 = pr$.

What the first condition says. Here it reads $b^{2/(ac)} = (ac)^{2/b}$. Take logarithms and multiply by $abc/2$: $$b \ln b = ac \ln(ac).$$ The function $t \ln t$ is strictly increasing for $t \ge 1$ (its derivative $\ln t + 1$ is positive there). Both $b$ and $ac$ are positive integers, so they are at least $1$, and the equation forces $$b = ac.$$ This is where we need integers: for fractions, two different numbers below $1/e$ can give the same $t \ln t$.

The second condition. We need $b^{2/(a^n c^n)} = (ac)^{2/b^n}$. With $b = ac$, the left side is $b^{2/b^n}$ and so is the right side. So the three numbers are in geometric progression for every $n$.

Since $3^4 = 81 = 80 + 1$, every power $3^{4r} = 81^r$ leaves remainder $1$ on division by $10$ (a product of numbers ending in $1$ ends in $1$). So $$3^{m+4r} - 3^m = 3^m\,(3^{4r} - 1)$$ is a multiple of $10$. The two quantities $1 + 3^m$ and $1 + 3^{m+4r}$ differ by this multiple of $10$, so if one is a multiple of $10$, so is the other.

Let $x = \tan\alpha$ and $y = \tan\beta$ (neither can be $0$, or the cotangents would not exist). The equations become $$\frac{x+y}{1-xy} = \frac{x^2+y^2}{xy}, \qquad \frac{x-y}{1+xy} = \frac{x^2-y^2}{xy}.$$

The second equation. Its right side is $\dfrac{(x-y)(x+y)}{xy}$. So either $x = y$, or, cancelling $x - y$, $$xy = (x+y)(1+xy).$$

Case $x = y$. The second equation holds ($0 = 0$). The first becomes $\dfrac{2x}{1-x^2} = 2$, that is $x^2 + x - 1 = 0$, so $$x = \frac{-1 \pm \sqrt5}{2}.$$ Neither value makes $1 - x^2$ zero, so both work.

Case $xy = (x+y)(1+xy)$. Write $s = x + y$ and $p = xy$. Then $s = \dfrac{p}{1+p}$. The first equation, cleared of fractions, is $ps = (s^2 - 2p)(1-p)$. Substituting $s$ and dividing by $p$ gives, after a little algebra, $$p^3 - p - 1 = 0.$$ Its only real root is $p \approx 1.3247$, which gives $s \approx 0.5698$. But then $s^2 - 4p < 0$, so $x$ and $y$, the roots of $t^2 - st + p = 0$, are not real. This case gives nothing.

So $\tan\alpha = \tan\beta = \tfrac12(-1 \pm \sqrt5)$, that is, about $0.618$ or $-1.618$. (The angles are equal, so $\tan(\alpha-\beta) = 0$, and $\tan 2\alpha = 2$.)

Let $x = \tan A$, $y = \tan B$, $z = \tan C$.

An identity. Since $A + B = 90^\circ - C$, we have $\tan(A+B) = \cot C$, so $$\frac{x+y}{1-xy} = \frac1z, \qquad\text{that is}\qquad xy + yz + zx = 1.$$

An inequality. For any real numbers, $$x^2 + y^2 + z^2 - (xy+yz+zx) = \tfrac12\big[(x-y)^2 + (y-z)^2 + (z-x)^2\big] \ge 0.$$ So $\tan^2A + \tan^2B + \tan^2C \ge 1$, with equality when $x = y = z$, that is, $A = B = C = 30^\circ$, where each $\tan^2$ is $\tfrac13$. The least value is $1$.

We use angles between lines, measured with a sign and up to multiples of $180^\circ$. Write $[\ell]$ for the angle a line $\ell$ makes with the major axis.

Step 1: the line $SI$. Since $\angle SYP = 90^\circ$, the point $Y$ is on the circle with diameter $SP$, and so is $I$. For four points on a circle, the chord $IP$ is seen at the same angle from $S$ and from $Y$: $$[SI] - [SP] = [YI] - [YP].$$ Here $YI$ is perpendicular to the major axis, so $[YI] = 90^\circ$, and $YP$ is the tangent $t$ at $P$. So $$[SI] = [SP] + 90^\circ - [t].$$ In the same way, using the circle on $HP$, $$[HJ] = [HP] + 90^\circ - [t].$$

Step 2: add. The tangent at an ellipse makes equal angles with the two focal radii, so the lines $PS$ and $PH$ are mirror images in the tangent: $[SP] + [HP] = 2[t]$. Adding the two lines above, $$[SI] + [HJ] = 2[t] + 180^\circ - 2[t] = 180^\circ \equiv 0.$$ So $[HJ] = -[SI]$.

Step 3: a reflection. Reflect the figure in the minor axis. This swaps $H$ and $S$ and turns a line at angle $\theta$ into a line at angle $-\theta$. So it carries the line $HJ$ to the line through $S$ at angle $-[HJ] = [SI]$, which is the line $SI$ itself.

Two lines that are mirror images in the minor axis meet on the minor axis (or are both parallel to it). So $IS$, $JH$ and $BC$ pass through one point.

Reduce to a circle. Stretch the plane in one direction so that the ellipse becomes a circle of radius $1$ (a stretch of this kind turns tangents into tangents). It multiplies every area by the same constant, and it keeps the ratio $CV : TV$, since $C$, $V$, $T$ stay on one line. So it is enough to prove that in a unit circle the area of $QCQ'$ is equal to $$\frac{1}{\sqrt{CV/TV} + \sqrt{TV/CV}}.$$

The circle. Let $CT = d > 1$. The radius $CQ$ is perpendicular to the tangent $TQ$, and $QV$ is perpendicular to $CT$. In the right triangle $CQT$, $CV = CQ^2/CT = 1/d$, so $$TV = d - \frac1d = \frac{d^2-1}{d}, \qquad QV = \sqrt{1 - \frac1{d^2}} = \frac{\sqrt{d^2-1}}{d}.$$ The triangle $QCQ'$ has base $QQ' = 2\,QV$ and height $CV$, so its area is $$CV \cdot QV = \frac{\sqrt{d^2-1}}{d^2}.$$

The expression. Put $k = d^2 - 1$. Then $CV/TV = 1/k$ and $$\sqrt{\frac{CV}{TV}} + \sqrt{\frac{TV}{CV}} = \frac{1}{\sqrt k} + \sqrt k = \frac{1+k}{\sqrt k} = \frac{d^2}{\sqrt{d^2-1}},$$ whose reciprocal is the area. For the ellipse with semi-axes $a$, $b$, the area is therefore $ab$ divided by the expression.

Put the square on axes with side $1$: $B = (0,0)$, $C = (1,0)$, $D = (1,1)$, $A = (0,1)$. The three sides have equal weight, with midpoints $(0,\tfrac12)$, $(\tfrac12,0)$, $(1,\tfrac12)$, so the center of gravity is $$G = \left(\tfrac12, \tfrac13\right).$$

Hung from $A$, the wire hangs with $AG$ vertical; hung from $B$, with $BG$ vertical. Going from one position to the other, the whole wire (and so $BC$) turns through the angle between the lines $GA$ and $GB$.

From $G$, the vectors to $A$ and $B$ are $(-\tfrac12, \tfrac23)$ and $(-\tfrac12, -\tfrac13)$. The tangent of the angle between them is cross product over dot product: $$\tan\theta = \frac{\tfrac16 + \tfrac13}{\tfrac14 - \tfrac29} = \frac{1/2}{1/36} = 18.$$ So $BC$ turns through $\tan^{-1} 18$, a little under $87^\circ$.

Let the circle have center $O$ and radius $R$, the fixed weights $m_1$, $m_2$ sit at $A$ and $B$, and the moving weight $m$ sit at $P$. With $M = m_1 + m_2 + m$, the center of gravity is $$G = \frac{m_1 A + m_2 B + m P}{M} = K + \frac{m}{M}(P - O), \qquad K = \frac{m_1 A + m_2 B + m\,O}{M}.$$ As $P$ runs round the circle, $P - O$ runs over all vectors of length $R$, so $G$ runs round the circle with center $K$ and radius $\rho = mR/M$.

The second part. Keep $m_1 + m_2$ fixed, so $M$ and $\rho$ stay the same. Let $d$ be the distance from $O$ to the chord $AB$. Points $A$ and $B$ are on the chord, so the distance of $K$ from the line $AB$ is $$\frac{m_1\cdot 0 + m_2 \cdot 0 + m\,d}{M} = \frac{m d}{M},$$ which does not depend on how the total is split between $m_1$ and $m_2$. A circle of radius $\rho$ whose center is at distance $h$ from a line cuts off a piece of length $2\sqrt{\rho^2 - h^2}$. Both $\rho$ and $h$ are fixed, so every one of these circles cuts off the same length on the line $AB$.

Take $x$ up the slope and $y$ out of the plane. Gravity gives accelerations $-g\sin\alpha$ along $x$ and $-g\cos\alpha$ along $y$. Let the speed of projection be $u$.

First flight. Across the plane the ball starts with $u\sin\beta$, so it lands after $$T = \frac{2u\sin\beta}{g\cos\alpha},$$ arriving with cross-speed $u\sin\beta$ and slope-speed $u\cos\beta - g\sin\alpha\,T$.

The bounce and second flight. The cross-speed becomes $e u\sin\beta$, so the second flight lasts $eT$. The slope-speed carries on.

Back to the start. Along the slope the acceleration is the same constant $-g\sin\alpha$ for the whole time $(1+e)T$, and the slope-velocity never jumps. So the position along the slope after both flights is $$u\cos\beta\,(1+e)T - \tfrac12 g\sin\alpha\,(1+e)^2T^2.$$ Setting this to zero: $$u\cos\beta = \tfrac12 g\sin\alpha\,(1+e)\,T = (1+e)\,\frac{u\sin\alpha\sin\beta}{\cos\alpha}.$$ Divide by $u\sin\alpha\sin\beta$: $\cot\alpha\cot\beta = 1 + e$.

Let the direction of projection be the unit vector $\mathbf d$ and the speeds $u_1$, $u_2$. After time $t$ the positions are $$u_1 t\,\mathbf d - \tfrac12 g t^2\,\mathbf k \quad\text{and}\quad u_2 t\,\mathbf d - \tfrac12 g t^2\,\mathbf k,$$ where $\mathbf k$ points straight up. Their difference is $(u_1 - u_2)t\,\mathbf d$. So the line joining them always points along the direction of projection, and its length grows steadily.

Let the masses be $1$ and the common speed $v$. Let $A$ move along $+y$, so $B$ and $C$ come from the upper right and upper left with velocities $(\mp\tfrac{\sqrt3}{2}v, -\tfrac12 v)$.

Case 1. By symmetry each ball is pushed straight back and, with no energy lost, leaves with speed $v$. So $V_1 = v$.

The common step. When $B$ and $C$ touch $A$ together and touch each other (or nearly), the three centers form an equilateral triangle, and the lines from $A$ to $B$ and to $C$ point along $\mathbf n_B = (\tfrac12, \tfrac{\sqrt3}{2})$ and $\mathbf n_C = (-\tfrac12, \tfrac{\sqrt3}{2})$. By symmetry $B$ receives an impulse $J\mathbf n_B$, $C$ receives $J\mathbf n_C$, and $A$ receives the opposite of their sum, $(0, -\sqrt3 J)$. Energy fixes $J$.

Case 2. First $B$ and $C$ meet each other. Their line of centers is horizontal, so they swap horizontal velocities: $B$ now has $(\tfrac{\sqrt3}{2}v, -\tfrac12v)$ and $C$ the mirror image. After the blow on $A$, $$\text{energy: } (v - \sqrt3 J)^2 + 2\left|\left(\tfrac{\sqrt3}{2}v + \tfrac12J,\ -\tfrac12v + \tfrac{\sqrt3}{2}J\right)\right|^2 = 3v^2,$$ which simplifies to $5J^2 - 2\sqrt3\,vJ = 0$. So $J = \tfrac{2\sqrt3}{5}v$ and $A$ leaves with $v - \sqrt3J = -\tfrac15 v$. So $V_2 = \tfrac15 V_1$.

Case 3. Now $B$ and $C$ still have their original velocities $(-\tfrac{\sqrt3}{2}v, -\tfrac12v)$ and its mirror image. The same energy equation becomes $5J^2 - 4\sqrt3\,vJ = 0$, so $J = \tfrac{4\sqrt3}{5}v$ and $A$ leaves with $v - \sqrt3J = -\tfrac75v$. So $V_3 = \tfrac75 V_1$.

In every case $A$ is thrown back the way it came. (For three balls touching at once, “perfectly elastic” is taken, as the problem intends, to mean that the symmetric impulses conserve energy.)

Put the ellipse as $x^2/a^2 + y^2/b^2 = 1$ with eccentricity $e$ and $S = (-ae, 0)$. The distance from $S$ to a point of the ellipse with abscissa $x$ is $a + ex$.

Angular speed. Equal areas in equal times means $r^2\omega = h$, a constant. So $\omega = h/r^2$.

The pair $P$, $P'$. The chord $PP'$ is parallel to the major axis, so $P'$ is the mirror image of $P$ in the minor axis. If $P$ has abscissa $x_P$, then $P'$ has $-x_P$, and $$\frac{1}{\sqrt{\alpha\alpha'}} = \frac{r_P\, r_{P'}}{h} = \frac{(a + e x_P)(a - e x_P)}{h} = \frac{a^2 - e^2x_P^2}{h}.$$ In the same way $\dfrac{1}{\sqrt{\beta\beta'}} = \dfrac{a^2 - e^2x_D^2}{h}$.

Conjugate semidiameters. If $P = (a\cos\theta, b\sin\theta)$ then $D = (-a\sin\theta, b\cos\theta)$, so $x_P^2 + x_D^2 = a^2$. Adding, $$\frac{1}{\sqrt{\alpha\alpha'}} + \frac{1}{\sqrt{\beta\beta'}} = \frac{2a^2 - e^2a^2}{h} = \frac{a^2(2 - e^2)}{h},$$ which is the same for every pair of conjugate semidiameters.

For an ellipse of semi-major axis $a$ described about a focus under an inverse-square attraction of strength $\mu$, the speed at distance $r$ from that focus is given by $v^2 = \mu\left(\frac2r - \frac1a\right)$, and the period is $2\pi a^{3/2}/\sqrt{\mu}$. So $\mu = 4\pi^2a^3/T^2$.

Let the point be at distances $r$ and $r'$ from the two foci, with $r + r' = 2a$. Note that $$\frac2r - \frac1a = \frac{2a - r}{ar} = \frac{r'}{ar}.$$ Equal speeds about the two foci, with periods $t$ and $t'$, means $$\frac{4\pi^2a^3}{t^2}\cdot\frac{r'}{ar} = \frac{4\pi^2a^3}{t'^2}\cdot\frac{r}{ar'}, \qquad\text{so}\qquad r'^2t'^2 = r^2t^2, \qquad rt = r't'.$$ With $r + r' = 2a$ this gives $r = \dfrac{2at'}{t+t'}$ and $r' = \dfrac{2at}{t+t'}$.

Treat each flame as a point of light and the third candle as a vertical stick that rises above both flames.

Take one flame, $L$. Every light ray that grazes the stick lies in the vertical plane through $L$ and the stick. So the shadow on the ceiling lies along the line where this vertical plane meets the ceiling. That line passes through the point $K$ of the ceiling straight above the stick, because $K$ is in the plane too.

The same holds for the other flame. So both shadow lines, produced, pass through $K$: they meet at the point of the ceiling directly above the third candle. (If the two flames and the candle stand in one vertical plane, the shadows lie along one line.)

One ray. A ray along a chord that cuts off a small arc of angle $\gamma$ (measured at the center) hits the circle and reflects into another chord of the same length, cutting off the next arc of angle $\gamma$. The chord from the point at angle $\theta_1$ to the point at angle $\theta_2$ points in the direction $\tfrac12(\theta_1 + \theta_2) + 90^\circ$. Moving on by one arc adds $\gamma$ to both ends, so each reflection turns the ray through exactly $\gamma$.

Our two rays. The first divides the circle in the ratio $3:1$, so its small arc is $\tfrac14 \cdot 360^\circ = 90^\circ$. The second divides it $8:1$, so its small arc is $\tfrac19\cdot 360^\circ = 40^\circ$. Both rays lie on the same side of the center and point the same way, so they turn in the same sense. After $n$ reflections their directions differ by $$n(90^\circ - 40^\circ) = 50n^\circ.$$

Parallel again. The lines are parallel when $50n$ is a multiple of $180$, that is, when $5n$ is a multiple of $18$. The least such $n$ is $18$. At that moment the rays point in opposite directions ($50 \times 18 = 900 = 5 \times 180$). If “parallel” must mean pointing the same way, we need $50n$ to be a multiple of $360$, and the answer is $36$.

The plane of the hyperbola is vertical (it holds a horizontal asymptote and the depth direction). Use the two asymptotes as axes: $X$ along the one in the surface, $Y$ along the immersed one. In these slanting axes the hyperbola is $XY = k$.

A horizontal line is parallel to the surface asymptote, so it is a line $Y = \text{const}$, and its depth is $z = Y\sin\phi$, where $\phi$ is the angle between the asymptotes. Along that line the region runs from the immersed asymptote ($X = 0$) to the curve ($X = k/Y$), a width proportional to $1/z$. So a thin strip at depth $z$ has area proportional to $dz/z$.

The pressure there is proportional to $z$, so the strip carries a force proportional to $z \cdot dz/z = dz$: the same for every strip. The force is spread evenly over the depths from $z_1$ to $z_2$, and its center is at the middle: $$\bar z = \frac{\int_{z_1}^{z_2} z\,dz}{\int_{z_1}^{z_2} dz} = \frac{z_1 + z_2}{2}.$$

The forces. The total force of the fluid on the cone’s closed surface is the buoyancy: a force $W = \rho g V$ straight up. The force on the flat base is $Q = \rho g h A$, where $h$ is the given depth of the base’s center and $A$ the area of the base; it acts along the axis, pushing in. Neither $W$ nor $Q$ changes as the cone tilts.

So the force on the curved surface is $\mathbf F = W\mathbf k - Q\mathbf a$, where $\mathbf k$ points up and $\mathbf a$ is the unit vector along the axis from the base towards the vertex. Then $$F^2 = W^2 + Q^2 - 2WQ\,(\mathbf k\cdot\mathbf a).$$ Here $\mathbf k\cdot\mathbf a$ is the sine of the inclination of the axis to the horizon, taken positive when the vertex is above the base and negative when below. So $F^2 = \lambda + \mu s$ for constants $\lambda$, $\mu$.

The identity. With $P^2 = \lambda + \mu s$ and so on, $$\sum P^2 (s' - s'') = \lambda\sum(s'-s'') + \mu\sum s(s'-s'') = 0 + 0,$$ because $(s'-s'') + (s''-s) + (s-s') = 0$ and $s s' - s s'' + s' s'' - s' s + s'' s - s'' s' = 0$.

Let the floor with the opening be the plane at height $0$, the ceiling at height $H$ above it, and the lower floor at depth $D$ below it. Put $k = D/H$. Let the opening be an $a$ by $b$ rectangle and the disk have radius $r$.

Where a ray lands. A ray from a point $L$ of the disk through a point $Q$ of the opening goes on for $D/H = k$ times the distance it has come, so it lands at $$X = Q + k(Q - L) = (1+k)\,Q - k\,L$$ (thinking of points as horizontal position vectors).

The patch. As $Q$ runs over the opening, $(1+k)Q$ runs over a rectangle with sides $(1+k)a$ and $(1+k)b$. As $L$ runs over the disk, $-kL$ runs over a disk of radius $kr$. The patch is every sum of the two: the rectangle with a disk of radius $kr$ swept around it. That is a rectangle with rounded corners: the big rectangle, four strips of width $kr$ along its sides, and four quarter-circles of radius $kr$ at the corners. Its area is $$(1+k)^2ab + 2kr(1+k)(a+b) + \pi k^2 r^2.$$

Moving the disk. Sliding the disk along the ceiling by a vector $\mathbf w$ changes every $X$ by $-k\mathbf w$. The whole patch slides without changing shape or area.

Let the gnomons have height $h$ and let $\delta$ be the Sun’s declination on that day. At local noon the Sun is in the meridian, at angle $\lambda - \delta$ from the vertical, so the noon shadow has (signed) length $c = h\tan(\lambda - \delta)$.

Write $\theta_i = \lambda_i - \delta$ and $t_i = \tan\theta_i$, so $c_i = ht_i$. Since $\lambda_2 - \lambda_3 = \theta_2 - \theta_3$, $$\tan(\lambda_2 - \lambda_3) = \frac{t_2 - t_3}{1 + t_2t_3}.$$ So the first term is $$ht_1\cdot\frac{h^2(t_2-t_3)^2(1+t_2t_3)}{t_2-t_3} = h^3\,t_1(t_2-t_3)(1+t_2t_3) = h^3\big[t_1(t_2 - t_3) + t_1t_2t_3(t_2 - t_3)\big].$$ The other two terms are the same with the indices moved round. Adding all three, $$h^3\Big[\big(t_1t_2 - t_1t_3 + t_2t_3 - t_2t_1 + t_3t_1 - t_3t_2\big) + t_1t_2t_3\big((t_2-t_3) + (t_3-t_1) + (t_1-t_2)\big)\Big] = 0.$$