Forty standout Tripos problems

In two throws of one die, the chance of no ace is $\tfrac56\cdot\tfrac56 = \tfrac{25}{36}$, so the chance of at least one ace is $\tfrac{11}{36}$.

With two dice thrown once, each die is a one with chance $\tfrac16$, independently. The chance that neither is a one is again $\tfrac{25}{36}$, so the chance of an ace is again $\tfrac{11}{36}$.

The chances are equal. That makes sense: throwing one die twice and throwing two dice once are the same experiment, just done at different times.

Let the plane have length $L$ and inclination $\alpha$. From the top, the bottom is at horizontal distance $x = L\cos\alpha$ and depth $L\sin\alpha$.

Speed for a given angle. Launch at speed $v$ and angle $\phi$ above the horizontal. The path is $y = x\tan\phi - \dfrac{gx^2}{2v^2\cos^2\phi}$, and it must pass through $y = -L\sin\alpha$ at $x = L\cos\alpha$. Solving for $v^2$: $$v^2 = \frac{gL\cos^2\alpha}{2\cos\phi\,(\cos\alpha\sin\phi + \sin\alpha\cos\phi)} = \frac{gL\cos^2\alpha}{2\cos\phi\sin(\phi + \alpha)}.$$

Make the denominator largest. $2\cos\phi\sin(\phi+\alpha) = \sin(2\phi + \alpha) + \sin\alpha$, which is largest when $2\phi + \alpha = 90^\circ$, that is, $\phi = 45^\circ - \tfrac12\alpha$. Then $$v^2 = \frac{gL\cos^2\alpha}{1 + \sin\alpha} = gL(1 - \sin\alpha).$$ The best direction bisects the angle between the upward vertical and the line from the top down to the bottom. (On level ground, $\alpha = 0$, this is the familiar $45^\circ$.)

Let the height be $h$ and let each component of the launch velocity be $w$ (they are equal at $45^\circ$). In $5$ seconds:

• sideways: $5w = h$;
• up and down: the body ends $h$ below where it started, so $-h = 5w - \tfrac12 g\cdot 25$.

Substituting $5w = h$ gives $-h = h - \tfrac{25}{2}g$, so $h = \tfrac{25}{4}g$. With $g = 32.2$ ft/s², the tower is about $201$ feet high.

Let the tower have height $h$ and the speed be $v$. Launch at angle $\theta$; let $t$ be the time of flight and $R$ the range.

A vector picture. The landing velocity is $\mathbf V = \mathbf v + \mathbf g t$, and $|\mathbf V|^2 = v^2 + 2gh$ by energy. The area of the triangle with sides $\mathbf v$ and $\mathbf g t$ (drawn from one point) is $$\tfrac12|\mathbf v \times \mathbf g t| = \tfrac12 g t\,v\cos\theta = \tfrac12 gR.$$ The same triangle also has sides $\mathbf v$ and $\mathbf V$, so its area is $\tfrac12 vV\sin\gamma$, where $\gamma$ is the angle between launch and landing directions. So $$R = \frac{vV\sin\gamma}{g} \le \frac{vV}{g},$$ with equality when the launch and landing directions are perpendicular.

The best angle. Let the body land at angle $\theta'$ below the horizontal. Perpendicular launch and landing directions mean $\theta + \theta' = 90^\circ$. The horizontal part of the velocity never changes, so $v\cos\theta = V\cos\theta' = V\sin\theta$: $$\tan\theta = \frac vV = \frac{v}{\sqrt{v^2 + 2gh}}, \qquad R_{\max} = \frac{v}{g}\sqrt{v^2 + 2gh}.$$ With $h = 0$ this gives $45^\circ$; from a tower the best angle is lower.

Let $R$ be the Earth’s radius and $h$ the height. Gravity at distance $d$ from the center is proportional to $1/d^2$, so $\sqrt g$ is proportional to $1/d$, and the period of the pendulum, proportional to $1/\sqrt g$, is proportional to $d$.

So on the mountain each swing takes $\frac{R+h}{R}$ times as long, and in a true day the clock counts only $\frac{R}{R+h}$ of the $86400$ seconds. It loses $$n = 86400\cdot\frac{h}{R+h}, \qquad\text{so}\qquad h = \frac{nR}{86400 - n} \approx \frac{nR}{86400}.$$ With $R \approx 3960$ miles (about $20.9$ million feet), each second lost per day means about $240$ feet of height. (This ignores the pull of the mountain itself, which makes the real loss a little smaller.)

At speed $v$ and angle $\theta$, the range is $\dfrac{v^2\sin2\theta}{g}$ and the greatest height $\dfrac{v^2\sin^2\theta}{2g}$. The area under a parabolic arch is two-thirds of base times height, so $$A = \frac23\cdot\frac{v^2\sin2\theta}{g}\cdot\frac{v^2\sin^2\theta}{2g} = \frac{2v^4}{3g^2}\sin^3\theta\cos\theta.$$ Differentiate $\sin^3\theta\cos\theta$: $3\sin^2\theta\cos^2\theta - \sin^4\theta = \sin^2\theta(3\cos^2\theta - \sin^2\theta)$. This is zero when $\tan^2\theta = 3$, so the area is greatest at $\theta = 60^\circ$ — steeper than the $45^\circ$ that gives the longest range.

Each of the four lives or dies with chance $\tfrac12$, independently, so the $2^4 = 16$ patterns are equally likely. Exactly two deaths can happen in $\binom42 = 6$ ways, so the chance is $\tfrac{6}{16} = \tfrac38$.

If “two of them will die” is read as “at least two,” add the cases of three deaths ($4$ ways) and four deaths ($1$ way): $\tfrac{11}{16}$.

Let the height be $h$. The speed is $\sqrt{2gh}$, and the time to fall $h$ is $\sqrt{2h/g}$. Moving sideways at constant speed for that time, the body travels $$\sqrt{2gh}\cdot\sqrt{\frac{2h}{g}} = 2h.$$ It lands at twice the height of the tower from its base, whatever $g$ is.

Count in ounces: the moving masses total $16 + 16 + 1 = 33$, and the unbalanced pull is the weight of $1$ ounce. So the acceleration is $$a = \frac{g}{33} \approx \frac{32.2}{33} \approx 0.976 \text{ ft/s}^2.$$ To fall $12$ feet from rest takes $t = \sqrt{2\cdot 12/a} = \sqrt{792/g} \approx 4.96$ seconds, and the speed then is $at = \sqrt{2a\cdot12} \approx 4.84$ feet per second. This slow, measurable fall is the idea behind Atwood’s machine, built to study gravity at a manageable speed.

Let $T$ be the tension and $a$ the acceleration (masses $P/g$ and $W/g$). For each weight: $$\frac Pg a = P - T, \qquad \frac Wg a = T - W.$$ Adding gives $a = g\dfrac{P - W}{P + W}$, and then $T = \dfrac{2PW}{P+W}$.

Both strands pull down on the pulley with tension $T$, so the axle bears $$2T = \frac{4PW}{P + W}.$$ This is less than $P + W$ unless $P = W$, since $(P + W)^2 - 4PW = (P - W)^2$. The weights are accelerating, so they do not lean their full weight on the axle.

There are $6^3 = 216$ equally likely ordered throws.

Total 5. The sets of numbers are $\{1,1,3\}$ and $\{1,2,2\}$, each in $3$ orders: $6$ throws.

Total 10. The sets are $\{1,3,6\}$, $\{1,4,5\}$, $\{2,3,5\}$ ($6$ orders each), and $\{2,2,6\}$, $\{2,4,4\}$, $\{3,3,4\}$ ($3$ orders each): $18 + 9 = 27$ throws.

So the chances are $\tfrac{27}{216}$ and $\tfrac{6}{216}$, in the ratio $27 : 6 = 9 : 2$.

Let the radius be $r$ and the height $h$, with fixed volume $V = \pi r^2h$. The inside surface (base and side, no lid) is $$S = \pi r^2 + 2\pi rh = \pi r^2 + \frac{2V}{r}.$$ Then $\dfrac{dS}{dr} = 2\pi r - \dfrac{2V}{r^2}$, which is zero when $V = \pi r^3$, that is, $h = r$. Since $S$ grows without bound as $r \to 0$ and as $r \to \infty$, this is the least surface. The diameter is twice the height.

Unfolding. Each bounce reverses the component of velocity across the cushion, which is exactly what a mirror does. So instead of bouncing, imagine the ball going straight on into a reflected copy of the table. A path $A \to$ first side $\to$ second side $\to B$ becomes a straight line from $A$ to the image of $B$ reflected in the second side and then in the first.

The corner. The two sides are perpendicular and meet at a corner $O$. Reflecting in two perpendicular lines through $O$ is a half-turn about $O$: it sends $B$ to the point $B^*$ with $O$ the midpoint of $BB^*$.

So strike $A$ straight towards $B^*$. (The order of the sides doesn’t matter, and after both bounces the ball travels exactly opposite to its first direction.) This works when the straight line from $A$ to $B^*$ really crosses the two cushions near that corner.

Let the foci be $S$ and $H$, and let $P$ be any point of the ellipse. The tangent at $P$ makes equal angles with $PS$ and $PH$. (One proof: for points $X$ on the tangent line, $SX + XH$ is least at $X = P$, since every other point of the line is outside the ellipse where this sum is bigger; and the shortest path from $S$ to a line and on to $H$ makes equal angles with the line.)

A perfectly elastic ball bounces like light off a mirror, with equal angles to the tangent. So a ball struck from $S$ that hits the cushion at $P$ bounces off towards $H$. It passes through $H$, hits the cushion again at some point $Q$, and by the same property bounces back towards $S$. After two reflections it is back at its starting point.

In base $b$, the digits $30203$ mean $3b^4 + 0b^3 + 2b^2 + 0b + 3$. So $$3b^4 + 2b^2 + 3 = 803, \qquad 3b^4 + 2b^2 - 800 = 0.$$ This is a quadratic in $b^2$: $b^2 = \dfrac{-2 + \sqrt{4 + 9600}}{6} = \dfrac{-2 + 98}{6} = 16$, so $b = 4$. Check: $3\cdot256 + 2\cdot16 + 3 = 803$. (The digit $3$ is allowed in base $4$.)

When the water stands at height $x$, its surface is a circle of radius $y = (ax)^{1/4}$, with area $\pi y^2 = \pi\sqrt a\,\sqrt x$.

Water leaves through a hole of small area $\sigma$ at speed $\sqrt{2gx}$, so the volume falls at the rate $\sigma\sqrt{2gx}$. The same volume leaves the top layer, so $$\pi\sqrt a\,\sqrt x\,\frac{dx}{dt} = -\sigma\sqrt{2g}\,\sqrt x, \qquad \frac{dx}{dt} = -\frac{\sigma\sqrt{2g}}{\pi\sqrt a}.$$ The $\sqrt x$ cancels: the level falls at a constant speed. So the scale on the axis is graduated in equal parts, each part marking the same time. That is the point of the shape: in a plain cylinder the level would fall fast at first and slowly at the end.

Let $B$ be the given point and $B'$ its image in the mirror. For any point $M$ of the mirror, $MB = MB'$. A ray from $X$ that reflects at $M$ and reaches $B$ has the same length as the path $X \to M \to B'$. The law of reflection (equal angles) says exactly that $X$, $M$, $B'$ are in a straight line, so the path length is $XB'$.

So the path has length $\ell$ exactly when $XB' = \ell$. The locus is the part of the circle with center $B'$ and radius $\ell$ that lies in front of the mirror (with the line $XB'$ crossing the mirror itself).

Sliding. On a smooth plane of inclination $\alpha$ the acceleration is $g\sin\alpha$.

Rolling. Let friction $F$ act up the slope. Then $ma = mg\sin\alpha - F$, and the turning effect of friction spins the ball: $Fr = \tfrac25mr^2\cdot\dfrac ar$, so $F = \tfrac25ma$. Together, $\tfrac75ma = mg\sin\alpha$, so $a = \tfrac57g\sin\alpha$.

Times. From rest over a distance $L$, $t = \sqrt{2L/a}$, so time is proportional to $1/\sqrt a$: $$\frac{t_{\text{slide}}}{t_{\text{roll}}} = \sqrt{\frac57} = \frac{\sqrt5}{\sqrt7} \approx 0.845.$$

Fix the parabola $y^2 = 4ax$, with vertex $A$ at the origin, focus $S = (a, 0)$ and directrix $x = -a$.

The rolling copy is a mirror image. At the start the two parabolas touch at their vertices, mirror images of each other in the common tangent there. Rolling without slipping means the arcs measured from the two vertices to the point of contact are always equal. So the contact point is the same point of the curve on both, and the rolling parabola is always the reflection of the fixed one in the common tangent at the contact point $P$. Its focus and vertex are the reflections of $S$ and $A$ in that tangent.

The focus. Let $M$ be the foot of the perpendicular from $P$ to the directrix. Then $PS = PM$ (the defining property of the parabola), and the tangent at $P$ bisects the angle $SPM$ (the reflecting property). So the tangent is the perpendicular bisector of $SM$, and the reflection of $S$ is $M$, on the directrix. The rolling focus slides along the directrix of the fixed parabola.

The vertex. Take $P = (at^2, 2at)$, where the tangent is $x - ty + at^2 = 0$. Reflecting the origin in this line gives $$\left(-\frac{2at^2}{1+t^2},\ \frac{2at^3}{1+t^2}\right).$$ Writing $X$, $Y$ for these coordinates, one checks $Y^2(2a + X) = -X^3$. This is the cissoid of Diocles, with its cusp at $A$ and the asymptote $x = -2a$.

Let the ladder have weight $W$ and length $2l$, with the foot at the origin. Three forces besides the push $F$ act: the weight $W$ at the middle, the upward reaction $R$ of the ground, and the reaction $N$ of the slope, perpendicular to the slope, so at $60^\circ$ to the vertical and leaning back towards the foot.

Moments about the foot. The top is at $(2l\cos30^\circ, 2l\sin30^\circ)$ and $N = N(-\sin60^\circ, \cos60^\circ)$. Its moment is $$2lN(\cos30^\circ\cos60^\circ + \sin30^\circ\sin60^\circ) = 2lN\cos30^\circ,$$ and the weight’s moment the other way is $Wl\cos30^\circ$. So $N = \tfrac12W$.

Horizontal forces. The slope pushes the top back with horizontal force $N\sin60^\circ$, so the foot needs $$F = \tfrac12W\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{4}W \approx 0.433\,W$$ towards the slope.

Each die shows a six with chance $\tfrac16$. The number of sixes among $n$ dice is binomial.

$$P(A) = 1 - \left(\tfrac56\right)^6 \approx 0.6651,$$ $$P(B) = 1 - \left(\tfrac56\right)^{12} - 12\cdot\tfrac16\left(\tfrac56\right)^{11} \approx 0.6187,$$ $$P(C) = 1 - \left(\tfrac56\right)^{18} - 18\cdot\tfrac16\left(\tfrac56\right)^{17} - \binom{18}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^{16} \approx 0.5973.$$ Each player expects the same proportion of sixes, but $A$ does best. With more dice the count of sixes clusters more tightly around its average, and needing the full average (or more) gets harder.

Down the roof. A rolling solid sphere accelerates at $\tfrac57g\sin30^\circ = \tfrac{5}{14}g$ (see the 1825 problem on sliding and rolling). The roof is $2(H - h)$ long, so at the eaves $$v^2 = 2\cdot\tfrac5{14}g\cdot2(H-h) = \tfrac{10}{7}g(H - h).$$

The fall. It leaves at $30^\circ$ below the horizontal, with downward speed $\tfrac12v$ and sideways speed $\tfrac{\sqrt3}2v$, from height $h$. The time $t$ to the ground solves $h = \tfrac12vt + \tfrac12gt^2$: $$t = \frac{\sqrt{\tfrac14v^2 + 2gh} - \tfrac12v}{g}.$$ It lands $\tfrac{\sqrt3}{2}vt$ beyond the eaves (horizontally).

Example. With $H = 40$ ft, $h = 30$ ft and $g = 32$ ft/s²: $v \approx 21.4$ ft/s, $t \approx 1.08$ s, landing about $19.9$ ft out from the wall.

Let $k$ be the number of black balls in $A$; then $B$ holds $3 - k$ black and $k$ white. A swap raises $k$ by one when it takes a white ball from $A$ (chance $\tfrac{3-k}{3}$) and a black ball from $B$ (chance $\tfrac{3-k}{3}$).

To reach $k = 3$ in three swaps, every swap must raise $k$:

• from $k = 0$: chance $1\cdot 1 = 1$;
• from $k = 1$: chance $\tfrac23\cdot\tfrac23 = \tfrac49$;
• from $k = 2$: chance $\tfrac13\cdot\tfrac13 = \tfrac19$.

Multiplying, the probability is $\tfrac{4}{81} \approx 0.049$.

Look down the line where the mirrors meet, a point $O$. Measure angles from the first mirror: it lies along $0^\circ$, the second along $\theta$, and the object at angle $\phi$, with $0 < \phi < \theta$.

Positions. Reflecting in the first mirror sends angle $\psi$ to $-\psi$; reflecting in the second sends it to $2\theta - \psi$. Reflections keep distances from $O$, so every image lies on the circle through the object with center $O$. Starting with the first mirror, the images are at $$-\phi,\quad 2\theta + \phi,\quad -(2\theta + \phi),\quad 4\theta + \phi,\ \dots$$ and starting with the second mirror they are at $2\theta - \phi,\ -2\theta + \phi,\ 4\theta - \phi, \dots$. Each new image is formed only if the previous one lies in front of the mirror that reflects it.

How many. The images of each chain move round the circle in steps of $\theta$ (in angular distance from the nearer mirror), and a chain stops once an image passes into the region behind both mirrors. Counting the steps gives $$\left\lceil\frac{180^\circ - \phi}{\theta}\right\rceil \text{ images in the first chain}, \qquad \left\lceil\frac{180^\circ - (\theta - \phi)}{\theta}\right\rceil \text{ in the second},$$ and if the two chains end at the same point it is one image, not two.

Examples. At $90^\circ$ you see $3$ images, at $60^\circ$ you see $5$, and whenever $360^\circ/\theta = m$ is an even whole number you see $m - 1$, wherever the object is. That is how a kaleidoscope makes a perfect pattern. For $m$ odd you see $m - 1$ if the object is on the bisector of the angle and $m$ otherwise.

Deviation. A ray meets a spherical drop at incidence $i$ and bends to $r$, with $\sin i = \mu\sin r$. It turns through $i - r$ entering, $180^\circ - 2r$ at the reflection inside, and $i - r$ leaving: $D = 180^\circ + 2i - 4r$. The rainbow sits where $D$ is least, because the rays bunch up there.

Least deviation. Differentiating $\sin i = \mu\sin r$ gives $\cos i\,di = \mu\cos r\,dr$. At the least deviation $dD = 2di - 4dr = 0$, so $di = 2dr$ and $2\cos i = \mu\cos r$. Dividing $\sin i = \mu\sin r$ by this: $$\tan i = 2\tan r.$$

The radius. The bow is seen at angle $\rho = 180^\circ - D = 4r - 2i$ from the point opposite the Sun. So $i = 2r - \tfrac\rho2$, and $\tan(2r - \tfrac\rho2) = 2\tan r$. With $t = \tan r$ and $k = \tan\tfrac\rho2$, expanding the left side and clearing fractions gives $$2t^3 - 3kt^2 - k = 0.$$ This cubic has one positive root. From it, $r = \tan^{-1}t$, $i = 2r - \tfrac\rho2$ and $\mu = \sin i/\sin r$.

Red light. For $\rho = 42^\circ$: $t \approx 0.845$, $r \approx 40.2^\circ$, $i \approx 59.4^\circ$, and $\mu \approx 1.333$, the refractive index of water. A check: least deviation also requires $\cos^2 i = \tfrac13(\mu^2 - 1)$, and these numbers satisfy it.

Any three of the points that are not in a line fix a plane. Count the triples: there are $\binom n3$. A plane containing exactly three of the points comes from exactly one triple. The special plane contains $\binom p3$ triples but is only one plane. So the number of planes is $$\binom n3 - \binom p3 + 1.$$ (We take it, as intended, that no three of the points are in a straight line.)

Each coin drawn, taken alone, is equally likely to be any of the nine coins, so its expected value is the average value in the bag. With the four coins worth $x$ shillings each, that average is $\dfrac{5\cdot20 + 4x}{9}$. For two coins, $$2\cdot\frac{100 + 4x}{9} = 24 \quad\Longrightarrow\quad 100 + 4x = 108 \quad\Longrightarrow\quad x = 2.$$ The other coins are worth $2$ shillings each: florins.

Take the rod to have length $1$. Three lengths adding to $1$ make a triangle exactly when each is less than $\tfrac12$ (each side must be shorter than the other two together).

First part. Let the marks be at $u$ and $w$, chosen independently and uniformly, so the point $(u, w)$ is uniform in the unit square. Take $u < w$ (half the square; the other half is the same by symmetry). The pieces are $u$, $w - u$, $1 - w$, and we need $$u < \tfrac12, \qquad w > \tfrac12, \qquad w - u < \tfrac12.$$ In the triangle $0 < u < w < 1$ (area $\tfrac12$) these cut out a small triangle with corners $(0, \tfrac12)$, $(\tfrac12, \tfrac12)$, $(\tfrac12, 1)$, of area $\tfrac18$. The probability is $\tfrac18 \big/ \tfrac12 = \tfrac14$.

Second part. Let the first piece have length $x$, uniform on $(0, 1)$, and cut the rest at a uniform fraction $y$: the pieces are $x$, $(1-x)y$, $(1-x)(1-y)$. We need $x < \tfrac12$, and then $(1-x)y < \tfrac12$ and $(1-x)(1-y) < \tfrac12$, that is, $$1 - \frac{1}{2(1-x)} < y < \frac{1}{2(1-x)},$$ an interval of length $\dfrac{1}{1-x} - 1$. So the probability is $$\int_0^{1/2}\left(\frac{1}{1-x} - 1\right)dx = \ln 2 - \frac12 \approx 0.193.$$

Everything settles down. A move “half way to $A$” is $P \mapsto \tfrac12(P + A)$. If two starting points differ by a vector $\mathbf d$, after one move they differ by $\tfrac12\mathbf d$. So whatever the start, after many moves the path is as close as we like to the path from any other start.

The limiting cycle. Look for points $X_A$, $X_B$, $X_C$ (the places reached just after moving towards $A$, $B$, $C$) with $$X_A = \tfrac12(X_C + A), \quad X_B = \tfrac12(X_A + B), \quad X_C = \tfrac12(X_B + C).$$ Substituting round the cycle, $X_A = \tfrac12A + \tfrac14C + \tfrac18B + \tfrac18X_A$, so $$X_A = \frac{4A + B + 2C}{7}, \quad X_B = \frac{2A + 4B + C}{7}, \quad X_C = \frac{A + 2B + 4C}{7}.$$ Every path approaches the triangle $X_AX_BX_C$, going round its perimeter.

Its area. The area of a triangle whose corners are these weighted averages of $A$, $B$, $C$ is the determinant of the weights times the area of $ABC$: $$\frac{1}{7^3}\det\begin{pmatrix}4&1&2\\2&4&1\\1&2&4\end{pmatrix} = \frac{49}{343} = \frac17.$$

Let $H = l\sin\alpha$ be the height of the top, and write $s = \sin\alpha$, $c = \cos\alpha$.

Leaving the top. The plane is smooth, so at the top the speed is $u$ with $u^2 = 2g(h - H)$, and the body flies off in the direction of the plane, at angle $\alpha$ above the horizontal, from height $H$.

The flight. Its horizontal run $X$ before landing solves $\dfrac{gX^2}{2u^2c^2} - X\dfrac sc - H = 0$. Taking the positive root and using $u^2/g = 2(h-H)$, $$X = 2c\left[(h-H)s + \sqrt{(h-H)(hs^2 + Hc^2)}\right].$$ (Inside the square root, $(h-H)^2s^2 + H(h-H) = (h-H)(hs^2 + Hc^2)$.)

The distance from the foot. The top is $H\cot\alpha$ from the foot, so $z = Hc/s + X$, and $$\frac zc = H\,\frac{\cos2\alpha}{s} + 2hs + 2\sqrt{(h-H)(hs^2 + Hc^2)},$$ using $\tfrac1s - 2s = \tfrac{\cos2\alpha}{s}$. The square root of a positive product of two linear factors with opposite slopes is a concave function, so $z$ is greatest where its derivative in $H$ vanishes: $$\frac{\cos2\alpha}{s} + \frac{h\cos2\alpha - 2Hc^2}{\sqrt{(h-H)(hs^2 + Hc^2)}} = 0.$$

The best plane. Try $H = h\cos2\alpha/c^2$. Then $h - H = hs^2/c^2$ and $hs^2 + Hc^2 = hc^2$, so the square root is $hs$, and the derivative is $\frac{\cos2\alpha}{s} - \frac{h\cos2\alpha}{hs} = 0$. This $H$ is positive only when $\alpha < 45^\circ$, which is why the problem needs that. The length of the plane is $$l = \frac Hs = \frac{h\cos2\alpha}{sc^2} = 2h\,\frac{\cot2\alpha}{\cos\alpha},$$ since $\sin2\alpha = 2sc$.

The greatest distance. Putting this back, $\dfrac zc = \dfrac{h\cos^22\alpha}{c^2s} + 4hs$, so $$z = \frac{h\left(\cos^22\alpha + 4s^2c^2\right)}{sc} = \frac{h\left(\cos^22\alpha + \sin^22\alpha\right)}{sc} = \frac{h}{\sin\alpha\cos\alpha}.$$

Let $B$ be at distance $R$ from $A$ along a line at elevation $\theta$, so $x = R\cos\theta$, $y = R\sin\theta$. Launch at speed $v$ and elevation $\phi$, with $T = \tan\phi$ and $k = \dfrac{gx^2}{2v^2}$.

The two directions. Passing through $B$ means $y = xT - k(1 + T^2)$, or $kT^2 - xT + (k + y) = 0$. Its two roots satisfy $T_1 + T_2 = \dfrac xk$ and $T_1T_2 = 1 + \dfrac yk$, so $$\tan(\phi_1 + \phi_2) = \frac{T_1 + T_2}{1 - T_1T_2} = \frac{x/k}{-y/k} = -\cot\theta = \tan(90^\circ + \theta).$$ So $\phi_1 + \phi_2 = 90^\circ + \theta$. Then the angle between the first direction and the vertical is $90^\circ - \phi_1 = \phi_2 - \theta$, which is the angle between the second direction and the line $AB$.

The sight. A sight setting fixes the angle $\delta$ between the barrel and the line of sight $AB$. If $B$ is hit with the barrel at angle $\delta$ to $AB$, then by the first part the other possible aim makes angle $\delta$ with the vertical. So $B$ lies on the path of a bullet fired at angle $\delta$ from the vertical, which is the path you get by sighting straight up at the zenith with that sight. The direction around the vertical can be anything, so the points form the surface swept out when that path is spun about the vertical through $A$. And each step reverses, so every point of that surface needs that sight.

Let the ring’s center move sideways by $X$ and the bead sit at angle $\theta$ from the bottom, small.

Momentum. The smooth plane pushes only upward, so the total sideways momentum stays zero: $\mu\dot X + m(\dot X + a\dot\theta) = 0$ (to first order), giving $\dot X = -\dfrac{ma}{m + \mu}\dot\theta$. The bead’s sideways speed is then $\dot X + a\dot\theta = \dfrac{\mu a}{m+\mu}\dot\theta$.

Energy. For small $\theta$, $$\text{KE} = \tfrac12\mu\dot X^2 + \tfrac12m(\dot X + a\dot\theta)^2 = \tfrac12\,\frac{m\mu a^2}{m + \mu}\,\dot\theta^2, \qquad \text{PE} = \tfrac12mga\,\theta^2.$$ This is simple harmonic motion with $$\omega^2 = \frac{mga}{m\mu a^2/(m+\mu)} = \frac{g(m + \mu)}{\mu a}.$$ A pendulum of length $L$ has $\omega^2 = g/L$, so $L = \dfrac{\mu a}{m + \mu}$. A very heavy ring ($\mu \gg m$) gives $L = a$, the ordinary bead in a fixed ring; a light ring swings much faster.

Put the line along the $x$-axis, the common point of contact at the origin, and the small circles’ centers at $(0, r)$ and $(0, -r')$. Let the big circle have center $(h, k)$. Its half-chord on the $x$-axis is $\sqrt{R^2 - k^2}$.

The small circles lie inside the big one and touch it, so the distances between centers are $R - r$ and $R - r'$: $$h^2 + (k - r)^2 = (R - r)^2, \qquad h^2 + (k + r')^2 = (R - r')^2.$$ Expanding, $h^2 + k^2 = R^2 - 2r(R - k)$ and $h^2 + k^2 = R^2 - 2r'(R + k)$. So $r(R - k) = r'(R + k)$, which gives $$k = R\,\frac{r - r'}{r + r'}, \qquad R^2 - k^2 = R^2\,\frac{4rr'}{(r + r')^2}.$$ The chord is $2\sqrt{R^2 - k^2} = \dfrac{4R\sqrt{rr'}}{r + r'}$. Dividing top and bottom by $\sqrt{rr'}$ gives the stated form.

Take $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ with center $C$, let $c = \sqrt{a^2+b^2}$, so the focus is $S = (c, 0)$, and use the asymptote $y = \tfrac bax$. Its point $(a\lambda, b\lambda)$ is at distance $c\lambda$ from $C$, and the asymptote makes an angle with $CS$ whose cosine is $a/c$. Let $P = (a\sec\phi, b\tan\phi)$, and write $\sigma = \sin\phi$, $\kappa = \cos\phi$.

The point $T$. The tangent at $P$ is $\dfrac{x\sec\phi}{a} - \dfrac{y\tan\phi}{b} = 1$. On the asymptote this gives $\lambda(\sec\phi - \tan\phi) = 1$, so $$\lambda_T = \frac{\kappa}{1 - \sigma}, \qquad \lambda_T^2 - 1 = \frac{\kappa^2 - (1-\sigma)^2}{(1-\sigma)^2} = \frac{2\sigma}{1-\sigma}.$$

The point $Q$. $(a\lambda, b\lambda)$ is on the line $SP$ when $(a\lambda - c)\,b\tan\phi = b\lambda\,(a\sec\phi - c)$. Solving and multiplying through by $\kappa$, $$\lambda_Q = \frac{c\,\sigma}{c\kappa - a + a\sigma}.$$

Compare the lengths. By the cosine rule in triangle $SCQ$, $SQ^2 = c^2\left(\lambda_Q^2 - \tfrac{2a}{c}\lambda_Q + 1\right)$, and $QT^2 = c^2(\lambda_T - \lambda_Q)^2$. So $$\frac{SQ^2 - QT^2}{c^2} = 2\lambda_Q\left(\lambda_T - \frac ac\right) - (\lambda_T^2 - 1).$$ Now $\lambda_T - \dfrac ac = \dfrac{c\kappa - a + a\sigma}{c(1-\sigma)}$, so $2\lambda_Q\left(\lambda_T - \frac ac\right) = \dfrac{2\sigma}{1-\sigma} = \lambda_T^2 - 1$. The difference is zero: $SQ = QT$. For the other focus, replace $c$ by $-c$ throughout; the same lines go through.

Squaring the sum $1 + 2 + \cdots + n$ gives every square once and every product of two different numbers twice: $$\Big(\sum_{i=1}^n i\Big)^2 = \sum_{i=1}^n i^2 + 2\sum_{i<j} ij.$$ So $$\sum_{i<j} ij = \frac12\left[\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}\right] = \frac{n(n+1)\big[3n(n+1) - 2(2n+1)\big]}{24} = \frac{n(n+1)(3n^2 - n - 2)}{24},$$ and $3n^2 - n - 2 = (n - 1)(3n + 2)$.

Let $I$ be the incenter, $I_a$ the excenter opposite $A$, $r_a$ its radius, and $O$, $R$ the circumcenter and circumradius.

A key point. The bisector of angle $A$ passes through $I$ and $I_a$ and meets the circumcircle again at $M$, the middle of arc $BC$. We know $MB = MC = MI$ (see the 1878 paper, problem 1). The bisectors $BI$ and $BI_a$ of the inside and outside angles at $B$ are perpendicular, so the circle on $II_a$ as diameter passes through $B$, and its center, the midpoint of $II_a$, is equally far from $I$ and $B$. Only one point of the line $II_a$ is equally far from $I$ and $B$, and $M$ is such a point. So $M$ is the midpoint of $II_a$, and $MI_a = MI = MB$.

Power of $I_a$. Along the line, the order is $A$, $I$, $M$, $I_a$. The power of $I_a$ with respect to the circumcircle is $$OI_a^2 - R^2 = I_aM\cdot I_aA.$$ Now $I_aM = MB = 2R\sin\tfrac A2$ (a chord facing an angle $\tfrac A2$ at the circumference) and $I_aA = r_a/\sin\tfrac A2$ (from the right triangle formed by $A$, $I_a$ and the point where the excircle touches $AB$ extended). So $$OI_a^2 = R^2 + 2Rr_a > R^2.$$ Every excenter lies outside the circumcircle.

Equal distances. The distance from $I_a$ to the circle is $OI_a - R = \sqrt{R^2 + 2Rr_a} - R$, which increases with $r_a$. So the three excenters are equally far from the circle only if $r_a = r_b = r_c$. Since $r_a = \dfrac{\Delta}{s - a}$ (area over semiperimeter minus the side), this means $s - a = s - b = s - c$, so $a = b = c$.

Take the incircle to have radius $1$ and center $O$; then the corners are at distance $2$ from $O$, at angles $120^\circ$ apart. A tangent line has a unit normal $\mathbf n$ with the line at $\mathbf n\cdot X = 1$. The signed distance of a corner at angle $\psi$ from $\mathbf n$ is $$d = 1 - 2\cos\psi,$$ positive on the same side as $O$. The claim is that $\dfrac1{d_1} + \dfrac1{d_2} + \dfrac1{d_3} = 0$, with exactly one $d$ negative.

A cubic. With $\psi_k = \psi + 120^\circ k$, the numbers $c = \cos\psi_k$ are the three roots of $4c^3 - 3c = \cos3\psi$. Put $c = \tfrac12(1 - d)$: $$\tfrac12(1-d)^3 - \tfrac32(1-d) = \cos3\psi \quad\Longrightarrow\quad d^3 - 3d^2 + 2 + 2\cos3\psi = 0.$$ This cubic has no term in $d$, and for a cubic $d^3 + pd^2 + qd + s$ the reciprocals of the roots add to $-q/s$. So $\sum 1/d_k = 0$.

Signs. The product of the roots is $-(2 + 2\cos3\psi) < 0$ (when no corner is on the tangent), so an odd number of them are negative, and they cannot all be negative since the line touches the incircle between the corners. So one corner lies on the far side, and the reciprocals of the other two add up to the reciprocal of its distance.

Let the force be $k^2$ times the distance, towards $O$.

Motion on one chord. Let $N$ be the foot of the perpendicular from $O$ to the chord. The force splits into a part along $ON$, which the chord holds back, and a part along the chord, equal to $k^2$ times the distance from $N$. So the particle does simple harmonic motion about $N$ with angular frequency $k$, whatever the chord: starting from rest at distance $s_0$ from $N$, it is at distance $s_0\cos kt$ (on the same side) at time $t$.

Chords through $D$. Let $DE$ be the diameter through $O$ and $DA$ a chord. The angle $DAE$ is a right angle (angle in a semicircle), so $ON \perp DA$ means $ON \parallel EA$. By similar triangles, $$\frac{DN}{NA} = \frac{DO}{OE},$$ the same for every such chord.

The time. Starting from rest at the end farther from $N$ — say $D$, if $DO \ge OE$ — the particle reaches $A$ when $\cos kt = -\dfrac{NA}{ND} = -\dfrac{OE}{DO}$, so $$t = \frac1k\cos^{-1}\left(-\frac{OE}{DO}\right),$$ the same for all chords from $D$. The same holds for the chords from $E$.

Let $y$ be the distance from the plane, so $\mu = c/y$, and let $\theta$ be the angle between the ray and the plane (in the plane of the ray). The medium is layered parallel to the plane, so by Snell’s law, layer after layer, $\mu\cos\theta$ is constant: $$\frac{\cos\theta}{y} = \text{constant}, \qquad\text{so}\qquad y = \rho\cos\theta$$ for some constant $\rho$.

A circle does exactly this. On a circle of radius $\rho$ with center in the plane ($y = 0$), the point at height $y$ has its radius at angle $\beta$ to the plane with $y = \rho\sin\beta$, and the tangent there is perpendicular to the radius, so it makes the angle $\theta = 90^\circ - \beta$ with the plane. Then $y = \rho\cos\theta$.

The ray’s direction is fixed by its height through $y = \rho\cos\theta$, and the circle through the starting point with center in the plane and the right direction obeys the same rule. So the ray follows that circle. (Such a medium is a model of the hyperbolic plane, the Poincaré half-plane, where these arcs play the part of straight lines.)

Let the circle have center $C$ and radius $a$, the center of force $O$ be on the circle, and the bead (mass $1$) be at $P$ with $OP = r$. The force is $\mu/r^5$ towards $O$.

Speed. The potential energy is $-\dfrac{\mu}{4r^4}$, so energy conservation gives $v^2 = 2E + \dfrac{\mu}{2r^4}$ for a constant $E$.

Geometry. $CO = CP = a$, so triangle $OCP$ is isosceles; with $\psi$ the angle $OPC$, we have $r = 2a\cos\psi$. The part of the force pointing towards $C$ is $$\frac{\mu}{r^5}\cos\psi = \frac{\mu}{r^5}\cdot\frac{r}{2a} = \frac{\mu}{2ar^4}.$$

The push of the wire. Moving on a circle needs a total force $v^2/a$ towards the center. With $N$ the push of the wire towards $C$, $$N + \frac{\mu}{2ar^4} = \frac{v^2}{a} = \frac{2E}{a} + \frac{\mu}{2ar^4}, \qquad N = \frac{2E}{a},$$ a constant. The pull of the force and the need for more speed cancel exactly.