## Earth Tunnel 8/3/11

August 4, 2011 at 8:13 am | Posted in problem of the day solution | Leave a comment

### Problem

You dig a straight tunnel between two points on the surface of the Earth. You drop a ball down. How long does it take to go through? Assume the Earth isn’t spinning and that it’s a perfect sphere of uniform density, and that the tunnel is frictionless and very small compared to the Earth.

### Solution

Let the tunnel have an impact parameter $b$. This is the minimum distance from the tunnel to the center of the Earth, and also the distance from the equilibrium point to the center of the Earth. Let the distance of the ball from this equilibrium point be $x$. The kinetic energy of the ball is $\frac{1}{2}m\dot{x}^2$, with $m$ the ball’s mass. The potential energy of the ball is $\frac{4\pi}{3}G\rho(b^2 + x^2)$. Thus, the only effect of $b$ is to add a constant to the potential, which does not affect the dynamics. Therefore we can let $b=0$ and obtain a harmonic oscillator. The half-frequency is $\tau/2 = \sqrt{\frac{3\pi}{G\rho 8}}$

## Penguin on a Rubber Band 7/3/11

August 4, 2011 at 7:56 am | Posted in problem of the day solution | Leave a comment

### Problem

Suppose there is a penguin on the right hand side of a rubber band of length $L$. A polar bear grabs the right hand side of the rubber band and begins pulling it with speed $v$. The penguin simultaneously begins waddling towards the left at speed $w$. The left hand side of the rubber band is fixed. Does the penguin ever reach the left hand side of the rubber band? How long does it take?

### Solution

Denote the fraction of the rubber band the penguin has traversed as $f$. In a short time $dt$, the penguin increases $f$ by $df = \frac{wdt}{L + vt}$, which is the distance waddled divided by the length of the rubber band. Integrating gives $f = \frac{w}{v}\ln(L+vt)$, so the penguin will always reach $f=1$. Solving for $f= 1$ gives $t = \frac{e^{(v/w)}-L}{v}$

## Penguin on a Precipice 8/2/11

August 4, 2011 at 7:49 am | Posted in problem of the day solution | Leave a comment

### Problem

A penguin stands on the edge of a cliff. It waddles in steps that are either forward, towards the cliff (or over it, as the case may be), or backwards, away from the cliff. It always has a 2/3 chance to step backwards and a 1/3 chance to step forward. What is the probability of the penguin’s survival in the limit as the number of steps taken goes to infinity?

### Solution

In the interest of morbid curiosity, we’ll find the penguin’s probability of death, rather than its probability of living. Denote its probability of death under the initial conditions of the problem (starting on the edge of the cliff) by $D_1$.

Then imagine starting the penguin one step away from the edge of the cliff (potentially two steps from death). In this case, there is some new probability for death, $D_2$. Similarly, let $D_n$ denote the probability for the penguin to die after an infinite number of steps, given that the penguin starts $n$ steps from the death.

Next, we’ll argue that $D_n = D_1^n$. Suppose the penguin starts $n$ steps from death. The probability it will ever get within $n-1$ steps of death is $D_1$. Once it’s there, the probability it will die is $D_{n-1}$. Thus, $D_n = D_1D_{n+1}$, which justifies $D_n = D_1^n$.

Imagine the penguin starting on the precipice. If it takes one step, it can either die (with probability 1/3), or else its new probability to die will be $D_2$ (with probability 2/3). Thus $D_1 = \frac{1}{3} + \frac{2}{3} D_2 = \frac{1}{3} + \frac{2}{3} D_1^2$. This is a quadratic with two roots, and we can tell that either $D_1 = 1$ or $D_1 = 1/2$.

Finally, we argue that it is impossible for $D_1 = 1$. If we repeat the above analysis with $D_1 = (1-p) + p D_2$, corresponding to an arbitrary probability $p$ of stepping away from the cliff instead of the 2/3 in our problem, we find the solutions $D_1 = 1$ and $D_1 = \frac{1-p){p}$.

The two roots coincide when $p = 1/2$, so in that case the penguin is certain to die. When $p = 1$, we are certain to walk away from the cliff and cannot die, so $D_1 = 0$ in that case. In between, if we assume $D_1$ is continuous as we vary $p$, we must choose $D_1 = \frac{1-p}{p}$, and the penguin’s chance to survive is $1/2$.

If the assumption of continuity bother you, another argument is that if we assume that $D_1 = 1$ and penguins are certain to die, then by killing many penguins in this manner, we can sample independent events from a binomial distribution with p = 1/3 and still generate an expectation value of about 1/2 as the limit of the number of independent events goes to infinity. This is impossible.

### Note on other solutions

Sam found a nice solution in which he wrote down $D_n = 1/3 D_{n-1} + 2/3 D_{n+1}$ and then explicitly performed the sum to infinity. You may want to try it out. Moor also solved the problem by summing over the Catalan numbers using a generating function. You may want to avoid that.

## Perambulatory Bear 8/1/11

August 2, 2011 at 10:26 am | Posted in problem of the day solution | Leave a comment

### Problem

A bear walks one mile south, one mile east, and one mile north. It winds up where it started. What are all the possible starting locations?

### Solution

It could start at the north pole, which is easy to check.

The other solutions are harder to see, but there is an infinite series of them. The first one is close to the south pole, just a bit more than a mile away. The bear walks one mile south so that it’s now just a fraction of a mile from the pole. Then it walks one mile east, going all the way around back to its starting point because it’s on the line of latitude whose length is only one mile. Then it walks a mile north to return to its starting point.

Similarly, there is a line of latitude whose length is 1/2 a mile. The bear could start a mile north of this, walk around this line of latitude twice, and return home. Etc.

## Tug of War 8/1/11

August 2, 2011 at 10:19 am | Posted in problem of the day solution | Leave a comment

### Problem

Suppose a penguin and a polar bear and playing tug-of-war. To even things out, they wrap the rope part way around a pole with coefficient of friction $\mu$. If the polar bear pulls with force $F_b$ and the penguin pulls with force $F_p$, how far does the rope need to be wrapped around the pole so that there is no winner?

### Solution

Let’s fix the penguin’s strength and find the maximum strength of the bear as a function of the wrapping angle. We’re looking for $F_b(\theta)$. We ask, “Suppose we wrap the rope around the pole an amount $\textrm{d}\theta$ more. How much more strongly can the bear pull? What is $\textrm{d}F_b$?”

The tension in this bit of rope is $F_b$. This tension pulls on both the back and the front of the little piece of rope, and the two tension forces mostly cancel. They don’t completely cancel because the rope is bent – the two tension forces both make and angle $\theta/2$ with the tangent to the point half way down the little extra segment. Therefore, the force from the pole on the segment is $2 F_b \textrm{d}\theta/2 = F_b\textrm{d}\theta$. The maximum friction force is then $\mu F_b \textrm{d}\theta$. Thus

$\frac{\textrm{d}F_b}{\textrm{d}\theta} = \mu F_b$

$F_b = F_p e^{\mu \theta}$

Where $F_p$ has entered due to the initial condition $F_b(0) = F_p$, which says that with no wrapping, the penguin and polar bear must be equally strong.

Solving gives

$\theta = \ln(\frac{F_b}{F_p})\frac{1}{\mu}$

## Chessboard of Death 7/31/11

August 1, 2011 at 10:37 pm | Posted in problem of the day solution | Leave a comment

### There is an $n*n$ chessboard. Each square has an arrow on it pointing to one of the eight surrounding squares. Every time you move between adjacent squares (those sharing an edge), the arrow rotates plus or minus 45 degrees, or stays the same.A penguin in dropped on the board and begins following the arrows. If the penguin gets to the edge of the board, it falls to its death. Is it ever possible for the penguin to survive indefinitely.

I am going to describe an answer, rather than write out full detail.

Since there are finitely-many squares, if the penguin survives indefinitely, it must be going in a loop. We need to prove no loops exist.

Suppose a loop exists. It cannot cross itself, because the crossing place would not follow the 45-degree rotation rule. (You can draw this out for yourself.) Thus any loop must have a well-defined inside section.

Take the hypothetical board with a loop and rotate every arrow 45 degrees clockwise (assuming the loop itself circulates clockwise. Otherwise, rotate the other way). Now let the penguin start inside the place where the old loop was. Every time it gets to the old loop, the rotated arrows push it back to the inside region. Thus, the penguin will now be forced to find a smaller loop than before.

So if a loop exists, a smaller loop exists. But by examining small cases, we can see there are no small loops. Thus we get a contradiction and no loops exist.

## Lumpy Hill 7/31/11

August 1, 2011 at 10:26 pm | Posted in problem of the day solution | Leave a comment

### Problem

You slide a spherical ball down a frictionless lumpy hill and trace its bath.

Now you turn on friction and do it again so that the ball rolls down the hill without slipping. (Note that rolling without slipping implies no energy dissipation.)

Does the ball follow the same path as before?

Assume the ball always contacts the hill in exactly one spot.

## Nim 7/30/11

August 1, 2011 at 10:19 pm | Posted in problem of the day solution | Leave a comment

### Problem

Two players play a game, alternating turns. The original setup is $n$ boxes with one penny in each box. A turn consists of either removing a penny or moving a penny to a lower-numbered box. The player who removes the last penny wins. If both players play perfectly, who wins, and what’s their strategy.

### Solution

This problem is the famous game Nim. You can read about it on Wikipedia here. However, note that instead of presenting the game in its usual form, I found an isomorphism of the game to throw off anyone who had heard of Nim before.

August 1, 2011 at 10:13 pm | Posted in problem of the day solution | Leave a comment

### Problem

There is a lily pad floating on the surface of a lake. There is a penguin on the edge of the lily pad. The penguin walks around the edge of the lily pad until it returns to the same point on the lily pad’s surface. How much does the lily pad rotate relative to the surface of the water? Note that the lily pad can both rotate and translate.

## Tricky Calculus Problem 7/29/11

August 1, 2011 at 10:03 pm | Posted in problem of the day solution | Leave a comment

### Problem

The polynomial

$f(x) = x^6 - 9x^2 - 6x$

has exactly three critical points. Find a parabola through these critical points.

### Solution

Critical points occur when $f'(x) = 6x^5 - 18x - 6 = 0$.

Note that we can write

$f(x) = x\frac{f'(x)}{6} - 6x^2 -5x$

If $f'(x) = 0$, then $f(x) = -6x^2 - 5x$.

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