2D problems which are easier to solve in 3D It sometimes happens that 1D problems are easier to solve by somehow adding a dimension.  For example, we convert linear differential equations for a real unknown to a complex unknown (to use complex exponentials), or we compute a power series' radius of convergence by thinking in the complex plane (or use complex analytic properties in path integrals), or we evaluate $\int^\infty_{-\infty} e^{-x^2}\ dx$ by squaring it...
So, are any 2D problems easier to solve in even higher dimensions?  I can't think of any.
 A: This is essentially the idea behind level set methods (cf https://en.wikipedia.org/wiki/Level_set_method ).
There are several situations where one needs to study the behavior of a dynamic surface with complicated topology. In fire simulation, one needs to track the motion of an air/fuel interface which is often not connected. In image segmentation, one needs to move the boundary between inner/outer regions until a steady state is reached. In fluid mechanics, breaking waves detach from the main body of water.
Solving differential equations on surfaces with complicated topology is difficult numerically. It turns out that when one represents these surfaces as level sets of a higher dimensional function these connectivity problems disappear and higher quality simulations are possible.
A: Given two disjoint disks of different radii, find the intersection of their common external tangents. For lack of a better name, call this the h-center of the pair (h- for homothety?).
Problem: Given three mutually disjoint disks, the h-centers of the three pairs are colinear.

The nicest solution involves adding one dimension and inflating the disks to balls (with centers in the original plane $\Pi$). The pairs of tangents become full-fledged cones with vertices in $\Pi$, and the proof is obtained by studying a plane tangent to all three balls. It is tangent to all three cones, so it contains their three vertices, but it also intersects $\Pi$ on a straight line :)

A: A striking example:
Consider arrangements of disks in the plane so that no two disks overlap (except on their boundaries) and their complement is a disjoint union of triangles (if we include a point at infinity).  You can imagine trying to build a particular finite triangulated planar graph by placing different sized coins on the table.
Here's the theorem: Any such graph may be obtained.  Further, the representation is unique up to Möbius (and anti-Möbius) transformations of the plane.
The proof of uniqueness is the striking bit.  You think of the plane as the boundary of hyperbolic upper half space!  Fill in each triangle of the original disks with a new disk tangent to them, and extend all the disks to half-balls.  We view the surface of each ball as a plane in hyperbolic space, and consider the group of reflections across them.  We then apply the Mostow rigidity theorem to the quotient manifold, and obtain the result.
This observation is due to Thurston. See
https://en.wikipedia.org/wiki/Circle_packing_theorem
A: Can you cover a planar disk of diameter 100 with 99 rectangles (possibly intersecting) of size $100\times 1$?
A: Of course, there is one such problem! This is the Cauchy problem for the wave equation
$$\frac{\partial^2u}{\partial t^2}=\Delta u,\qquad u(x,0)=f(x),\quad \frac{\partial u}{\partial t}(x,0)=g(x),$$
where $x\in{\mathbb R}^d$. To solve it, it is enough to know the case where $f\equiv0$.
If $d=3$, this problem is solved by using spherical means. We obtain
$$u(x,t)=tM_{t,x}[g],$$
where $M_{t,x}$ denotes the mean over the sphere of radius $t$ and center $x$.
The two-dimensional case is way more complicated. The formula can only be found by considering that a $2$D-solution is a special case of a $3$D-solution. Then the solution involves a complicated integral over the disk $D(x;t)$ instead of the circle. This is why the Huyghens principle holds true in $3$ space dimensions but not in $2$ space dimensions.
A: Voronoi diagrams in the plane can be described as the lower envelopes of wave-front surfaces in 3D. I'm not sure if this makes them 'easier', but it's a useful way of thinking about them.
A: Desargues' Theorem is a statement about triangles in the plane that is easier to prove using solid geometry.
A: There's a famous problem posed by Erdos that has an easy 3-D solution, but a very difficult 2-D solution. The problem is to prove the following: Given a decomposition of an n-cube into finitely many n-cubes $Q_1, ... Q_k$ ($k>1$), prove that there exist two distinct cubes $Q_i, Q_{i'}$, of equal size.
The above statement is certainly true for $n=3$ (this is a simple exercise), but it is in fact untrue for $n=2$. I think this is known as the "Squared square" problem, and you can read more about it here. Below is the first counter-example, due to Sprague, to the problem.
               
A: A not-so-serious answer; hopefully what it lacks in depth it makes up for by being elementary.
Suppose we forget Pythagoras's theorem and define a binary operation on positive reals by sending $(a, b)$ to the length of the hypotenuse of the right-angled triangle with side lengths $a, b$ forming the right angle.
The associativity of this operation is trivial in three dimensions but not so in two.
I came across this here:
D. Bell, "Associative Binary Operations and the Pythagorean Theorem", The Mathematical Intelligencer, Vol. 33, No. 1 (2011), 92-95, DOI: 10.1007/s00283-010-9171-6
Apparently it is also mentioned here:
L. Berrone, "The Associativity of the Pythagorean Law", The American Mathematical Monthly, Vol. 116, No. 10, Dec., 2009 https://www.jstor.org/stable/40391255
A: Four cars drive in the Sahara desert (an infinite plane) at constant and generic velocities and directions. It is known that car A at some point in time meets car B (though let's pretend they drive through each other without crashing). The exact same thing is also known for the pairs AC, AD, BC, and BD. Is it also true for the pair CD?
A: See: http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.aop/1176992821
A: Prove that one can always find 4 points on a smooth non-self-intersecting closed planar curve that are vertices of a rectangle.

Here's a very nice video with a solution.
