This is a classic problem, hence the empty source author and year.

The intersection of two cylinders with the same radii at right angles is called a bicylinder. The shape is a stack of infinitely many squares.

Imagine a sphere with the same radius inside the bicylinder. The sphere is a stack of infinitely many circles, and the ratio of the area of each circle to the square around it is $\frac{4}{\pi }$.

The volume of a sphere is $\frac{4}{3}\pi r^3$, the sum of the area of each circle in the stack. Suppose each area was labelled with a lowercase letter of the alphabet. In reality, there are infinitely many, but this helps us understand. The sphere volume is then:

$$V_{sphere}=a+b+c+...+z=\frac{4}{3}\pi r^3$$

The bicylinder is the same way with the sum of the area of each of its squares. Suppose these are labelled with uppercase letters.

$$V_{bicylinder}=A+B+C+...+Z=?$$

Because we know the ratio of the areas, we can make each circle area into the corresponding square area, the sum of which is the volume of the bicylinder, the intersection of the crossed cylinders.

$$V_{bicylinder}=\frac{4}{\pi }a+\frac{4}{\pi }b+\frac{4}{\pi }c...+\frac{4}{\pi }z$$

$$V_{bicylinder}=\frac{4}{\pi }(a+b+c...+z)$$

$$V_{bicylinder}=\frac{4}{\pi }{V}_{sphere}$$

$$V_{bicylinder}=\frac{4}{\pi }\left(\frac{4}{3}\pi r^3\right)=\frac{16}{3}{r}^3$$

Let $h$ be the number of hours of snowing before noon, meaning the snow started at hour $12-h$. Let $s$ be the constant rate of snowfall in inches per hour, and let r the constant rate of removal in cubic feet per hour.

At time $12+t$, or $t$ hours after noon, the snow depth is $s(t+h)$. To maintain the constant rate of removal, the speed of the snowplow in miles per hour is inversely proportional to the depth of the snow. The snowplow is travelling at a speed where the constant c converts inches times miles times width of the snowplow blade to cubic feet. Fortunately we don't need to know any of the constants c, r or s.

Between 12:00 and 1:00 the snowplow travels two miles.

$${\int }_0^1\frac{cr}{s(t+h)}dt=2$$

Between 1:00 and 2:00 with deeper snow, it travels only one mile.

$${\int }_1^2\frac{cr}{s(t+h)}dt=1$$

Integrating $\frac{1}{t+h}$ gives:

$\frac{cr(log(1+h)-logh)}{s}=\frac{2cr(log(2+h)-log(1+h))}{s}$

which then yields:

$$\frac{1+h}{h}={\left(\frac{2+h}{1+h}\right)}^2$$

$$(1+h{)}^3=(2+h{)}^{2}h$$

$$h^2+h-1=0$$

$$h\in \{\frac{\sqrt{5}-1}{2},\frac{-\sqrt{5}-1}{2}\}$$

$$h=\{0.618,-1.618\}$$

Because it started snowing before noon, $h$ is positive. $h=0.618$ hours translates to 37 minutes, so it started snowing at 11:23 AM.

Let $R$ be the radius of the grass circle, $L$ be the length of the leash, and $O$ be the circle center. Let $Q$ be the point where the leash is fastened.

Let $P$ and $P^{\prime }$ be the two points on the circumference of the grass circle at distance $L$ from $Q$. Let $B$ denote the measure of angle $PQO$ in radians, and $C=\pi -2B$ the measure of $POQ$.

Because $PQO$ is isosceles:

$$L=2R\cos B$$

The pie-shaped region emanating from $O$ and reaching from $P$ to $P^{\prime }$ has area:

$$\frac{1}{2}\cdot 2R(2C)=2RC$$

The pie-shaped region emanating from $Q$ and reaching from $P$ to $P^{\prime }$ has area $2LB$.

Together the regions cover the sheep's eating area, but they both cover the quadrangle $OPQ{P}^{\prime }$ so we must subtract its area $RL\sin B$:

$$R^{2}C+L^{2}B-RL\sin B=\frac{1}{2}\pi R^2$$

$R^2(\pi -2B)+4R^{2}B{\cos }^{2}B-2R^2\cos B\sin B=\frac{1}{2}\pi R^2$

$\pi -2B+4B{\cos }^{2}B-2\cos B\sin B=\frac{\pi }{2}$

We solve this numerically for $B$, and obtain $B=0.952848$ , $C=1.235897$, $L=1.158728R$.

The leash must be approximately 1.16 times the radius.

Let $t$ denote time in hours. Let $(x,y)$ be the goblin's current position, with our initial position being $(0,0)$. Let $u=60$ be our speed, $v$ the goblin's unknown speed, and $r=\frac{u}{v}$ their ratio. Let the goblin's initial position be $(0,g)$ where $g=30$.

Because the goblin is always pointing at our current position, the goblin's motion satisfies the differential equations:

$$(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2=v^2$$

$$x-y(\frac{dx}{dy})=ut$$

The initial conditions are $y=g$ and $x=t=0$. The goblin's position $(x,y)$ at time $t$ is given parametrically in terms of y:

$$t=\frac{g}{v(1-r^2)}-\frac{g}{2v}\cdot \frac{(\frac{y}{g}{)}^{1+r}}{1+r}+\frac{(\frac{y}{g}{)}^{1-r}}{1-r}$$

$$x=\frac{gr}{1-r^2}+\frac{g}{2}\cdot \frac{(\frac{y}{g}{)}^{1+r}}{1+r}-\frac{(\frac{y}{g}{)}^{1-r}}{1-r}$$

Calculate $\frac{dt}{dy}$ and $\frac{dx}{dy}$, and verify that the differential equations are satisfied.

When $y=g$ we get $t=0$ and $x=0$, which meets the initial conditions. When $y=0$ we get:

$$t=\frac{g}{v(1-r^2)}$$

$$x=\frac{gr}{1-r^2}$$

The last equation gives:

$$100=\frac{30r}{1-r^2}$$

$$r=\frac{\sqrt{409}-3}{20}$$

Finally we can use the ratio we set up at the beginning to solve for the goblin's speed:

$$v=\frac{60}{r}=3\sqrt{409}+3$$

$$v=69.67124525km/h$$

The answer is five moves.

Suppose we mark the positions of bottles as $U$, $D$, or $?$, for up, down or unknown, respectively. Suppose we also write the table's four bottle positions in clockwise order. At the start the positions are:

$$\begin{gathered} ?? \\ ?? \end{gathered}$$

First move: Take two opposite bottles and turn them up.

$$\begin{gathered} ?U \\ U? \end{gathered}$$

Second move: Take two adjacent bottles and turn them up.

$$\begin{gathered} ?U \\ UU \end{gathered}$$

If the unknown bottle was also pointed up the referee would stop the game. Therefore, if the game isn't stopped, the last bottle must be down.

Third move: Take two opposite bottles. We know that only one bottle is down. If one of the two opposite bottles is down, turn it up to win.

$$\begin{gathered} UU \\ UU \end{gathered}$$

If instead both opposite bottles are up, turn one down.

$$\begin{gathered} DD \\ UU \end{gathered}$$

Fourth move: Take two adjacent bottles and flip them. If they were both up, we win with all down.

$$\begin{gathered} DD \\ DD \end{gathered}$$

If the two adjacent were both down, we win with all up.

$$\begin{gathered} UU \\ UU \end{gathered}$$

If the two adjacent are one up and one down, the flip makes our table:

$$\begin{gathered} DU \\ UD \end{gathered}$$

Fifth move: Take two opposite bottles and turn them over, winning with all up:

$$\begin{gathered} UU \\ UU \end{gathered}$$

or all down:

$$\begin{gathered} DD \\ DD \end{gathered}$$