Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip?
TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on...
Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are not always pointing in the same direction) would disturb it. The sun's tidal forces would disturb it. I could go on.
The equation of motion of a pencil tells us that as soon as you are off center by the smallest amount, the motion will build up. It is not a stable equilibrium.
And graphite cannot sustain the weight of a pencil on a mono-atomically sharp tip... According to this supplier of high quality graphite the compressive strength is about 25 ksi (~170 MPa - figure 5-2 from the reference). The smallest tip that can support the weight of 0.05 N would be a circle with a radius of 0.01 mm. That's a pretty sharp tip, for a pencil. It's not nearly "atomic".
Finally, even at absolute zero, the uncertainty principle requires that the position of the center of mass not be known perfectly. The (quantum mechanically required) fluctuations of the position of the center of mass should be sufficient to cause the pencil to fall over eventually.
UPDATE - the impact of a single photon
It is instructive to calculate how long it takes a pencil to drop for a given deviation from equilibrium (assuming for a moment a perfect pivot at the bottom - i.e. the only torque applied is due to gravity). I am showing it here for a pencil that was hit by a single green photon - and the result is a surprisingly short time.
Modeling the pencil as a uniform rod with mass $m$, length $\ell$, moment of inertia $I=\frac{1}{3}m\ell^2$, the torque $\Gamma$ when it is at an angle $\theta$ to the vertical is
$$\Gamma = \frac12 m g \ell \sin \theta$$
For small deflections, $\sin\theta = \theta$ and we will use that assumption below. Then the equation of motion becomes
$$I\ddot\theta = \frac12 m g \ell \theta\\ \frac13 m \ell^2 \ddot\theta = \frac12 m g \ell \theta \ddot\theta = \frac{3 g}{2\ell}\theta$$
This looks a lot like the equation for a simple harmonic oscillator, but with the wrong sign. We do indeed get a very similar solution, but with hyperbolic functions.
Putting $\frac{3 g}{2 \ell} = \alpha^2$, we can integrate this twice to get a
$$\theta = C_1 e^{\alpha t} + C_2 e^{-\alpha t}$$
If the pencil starts at equilibrium, we can set $\theta=0$ at $t=0$, which reduces the above to
$$\theta = 2C_1 \sinh{\alpha t}$$
Given an initial velocity $v_0$, we see that
$$v_0 = \frac{\ell}{2}\dot\theta = \ell C_1 \alpha \cosh{\alpha t}$$ so
$$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{3 g /2 \ell}} = \frac{v_0}{\sqrt{\frac32 g \ell}}$$
Now we have an expression for $\theta$, we can solve with a given initial velocity:
$$t = \frac{1}{\alpha}\sinh^{-1}\frac{\theta}{C_1} $$
$$= \sqrt{\frac{2\ell}{3 g}}\sinh^{-1}\left(\frac{\theta \sqrt{\frac32 g \ell}}{v_0}\right)$$
Now comes the fun part. Let's assume we hit the perfectly balanced pencil with a single photon of green light. The momentum of such a photon is approximately
$$p = \frac{E}{c} = \frac{h}{\lambda} \approx 10^{-27} Ns$$
Let's assume the pencil is black so the photon is not reflected. The mass of a pencil is about 0.005 kg, length 20 cm. The velocity of the pencil after the collision with the photon is roughly (yes, there are some factors to account for offset impact etc. I am ignoring all those - it doesn't change the basic answer):
$$v_0 = \frac{p}{m} = 2 \cdot 10^{-25} m/s$$
Let's assuming that "definitely falling" corresponds to an angle of 0.5 degrees, or about 0.01 rad. We can put the values in the above equation, and find t $\approx$ 6 s.
One photon. Six seconds. That is a shockingly short time... but it's about to get worse:
UPDATE 2 - the importance of the uncertainty principle
It is also interesting to see how long it would take for a pencil to fall given an initial deflection from the vertical - because we can then get rid of the one photon and use the uncertainty principle to estimate the maximum time the pencil will balance.
If the center of mass is off by $\Delta x$, then the angle is $\Delta \theta = \frac{2\Delta x}{\ell}$.
Using the same equations as before, we find $C_1 + C_2 = \Delta \theta$. Let's assume the initial velocity is zero - because "on average" it will be, given that the direction of initial velocity is equally likely to point back towards equilibrium and away from it - so we get $C_1 = C_2$, and the solution is a $\cosh$ function:
$$\theta = 2 C_1 \cosh(\alpha t)$$
Where $C_1 = \frac{\Delta\theta}{2} = \frac{\Delta x}{\ell}$.
We now have the time to fall (time to reach a certain $\theta$) as
$$t = \frac{1}{\alpha}\cosh^{-1}\left(\frac{\theta\ell}{2\Delta x}\right)$$
The equation we derived earlier for the time taken with a given initial velocity can be rewritten as
$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)$$
and we know that
$$\Delta x \Delta p = \hbar$$
Obviously the longest time to balance will be reached when the two times are the same - otherwise, one will be longer and the other shorter, and it's the shorter time that will dominate. To solve, we substitute for $\Delta x = \frac{\hbar}{m\Delta v}$ and get
$$\cosh^{-1}\left(\frac{\theta\ell m \Delta v}{\hbar}\right) = \sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)$$
If the term in the parentheses is sufficiently large, then we can (for the purpose of estimating) set
$$\frac{\theta\ell m \Delta v}{\hbar} = \frac{\theta\ell\alpha}{\Delta v}$$
From which it follows that
$$\Delta v = \sqrt{\frac{\theta \ell \alpha \hbar}{\theta \ell m }} = \sqrt\frac{{\alpha \hbar}}{m}$$
Substituting the values for the pencil, we find
$$\Delta v = 4\cdot 10^{-16}m/s$$
which is many orders of magnitude larger than the velocity due to the pencil being hit by a photon. The time to fall is then
$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 3.7 s$$
So a perfectly balanced, "theoretical" pencil that stands on its mono-atomic tip, will fall on average in a handful of seconds because of the uncertainty principle.
AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs himself as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got.
No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time.
Momentum and position form a conjugate pair. $$\Delta x \Delta p \geq \frac{\hbar}{2}$$.
Angular momentum and angular position form one too. $$\Delta L \Delta \Theta \geq \frac{\hbar}{2}$$
This doesn't guarantee that angular momentum and angular position will be non-zero. It is an uncertainty - The actual values can be anything, including $0$.
But it does prevent you from arranging them both so the pencil stays upright. Furthermore, if you ask what the probability of finding both values very close to $0$, you find that it is very small. In the limit, infinitely improbable.
If it turns out that $L = \Theta = \sqrt{\hbar}$, and you plug in reasonable values for the mass and length of the pencil, you will find it falls over in a few seconds.
Belated update
I was waiting until the weekend to add an update. By the time it got here, Floris had left very little to add. And he did a better job than I would have. Good answers.
A number of users felt that an ideal pencil sharpened to an atomic tip was not realistic. The pencil should have a flat on the bottom.
My own thought is that the pencil should be mounted on one of those massless, frictionless pulleys that seem to be so common in high school physics classrooms.
Never the less, a pencil with a flat can be treated semi classically. Because of the uncertainty principal, the pencil has an initial momentum, and therefore an initial energy. This will cause the pencil to tip. Which in turn will cause the pencil to rotate about an edge of the flat. The center of mass will rise until it is directly over the edge of the flat. If the initial "uncertainty" energy is larger than the energy needed to raise the center of mass, the pencil will tip over.
A quantum mechanical treatment would treat the region where the center of mass is over the interior of the flat as a potential well. There is a probability of tunneling out.
Both of these scenarios are treated in full detail (with diagrams in case my description is unclear) here. I found this link by following Floris' "interesting post that calculates the same thing." That post had some comments at the bottom. The very last comment contains the link.
No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's measured in GigaPascal)