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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.

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    $\begingroup$ The dynamics of a billiard in certain subsets of the plane are easier to analyze by looking at 2D surfaces which embed in 3D space, if that counts. $\endgroup$
    – Alex Becker
    Commented Sep 9, 2012 at 3:27
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    $\begingroup$ There are some geometry problems such as Desargues implies Pappus and common cotangents of three pairs of circles have intersections which are collinear. But do you want geometric examples? Gerhard "Ask Me About System Design" Paseman, 2012.09.08 $\endgroup$
    – Gerhard Paseman
    Commented Sep 9, 2012 at 5:10
  • $\begingroup$ 1. Ehrenpreis conjecture (solved by Kahn and Markovic) is easier in 3d than in 2d. 2. Quasi-isometric rigidity of uniform lattices in O(n,1) is easier for $n=3$ than for $n=2$. 3. Classification of buildings is easier in rank 3 than in rank 2 (where it is still unknown). 4. Closely related to your question example: Poincar\'e conjecture (in all its forms) is easier for $n>4$ than for $n=4$ and $n=3$. However, your question should be CW, so voting to close until then. $\endgroup$
    – Misha
    Commented Sep 9, 2012 at 23:04

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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.

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Desargues' Theorem is a statement about triangles in the plane that is easier to prove using solid geometry.

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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 squared square with no equal sub-squares

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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

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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 :)

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    $\begingroup$ In certain cases of a small disk between two larger ones you would not be able to have a plane tangent to all three spheres , but it is still an amazing proof. $\endgroup$
    – Aaron Meyerowitz
    Commented Apr 27, 2013 at 21:00
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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.

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    $\begingroup$ I'd say level set methods are not really adding a dimension; they trade a parameter for an extra "spatial" dimension. I admit, though, that I'm not very familiar with them and my goal is not so clear from my question. $\endgroup$
    – bobuhito
    Commented Sep 9, 2012 at 15:58
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Can you cover a planar disk of diameter 100 with 99 rectangles (possibly intersecting) of size $100\times 1$?

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  • $\begingroup$ Can I cut one of the rectangles into a w by 1 rectangle and a (100 - w) by 1 rectangle, with 33 < w < 67? Gerhard "Not Ready For Three Dimensions" Paseman, 2012.09.11 $\endgroup$
    – Gerhard Paseman
    Commented Sep 11, 2012 at 16:46
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    $\begingroup$ If you could do that, the solution would be easy. $\endgroup$
    – Dror Bar-Natan
    Commented Sep 11, 2012 at 18:32
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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

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  • $\begingroup$ Erm, not exactly. The manifold you get this way is not finite volume, so Mostow does not apply... $\endgroup$
    – Igor Rivin
    Commented Sep 10, 2012 at 0:07
  • $\begingroup$ @Igor Rivin My reference is library.msri.org/books/gt3m chapter 13, Corollary 13.6.2. Have I made a mistake? $\endgroup$
    – John Wiltshire-Gordon
    Commented Sep 10, 2012 at 14:27
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    $\begingroup$ @J W-G: you have committed the sin of omission. The Thurston trick is to consider the graph TOGETHER with the dual graph. Then, the polyhedron you are constructing is right-angled, and compact, so you can use Mostow (but this is a bit of overkill - a completely elementary argument for a much more general [in the polyhedral world] statement is in my paper Euclidean Structures on simplicial surfaces and hyperbolic volume [Annals, 1994]) $\endgroup$
    – Igor Rivin
    Commented Sep 10, 2012 at 15:52
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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.

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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?

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    $\begingroup$ When one generalizes to more than four cars, an interesting sort of collapse occurs. A. Bogomolny has some analysis (and spoilers) of this four traveller problem on his cut-the-knot.org website. Gerhard "Ask Me About System Design" Paseman, 2012.09.11 $\endgroup$
    – Gerhard Paseman
    Commented Sep 11, 2012 at 16:34
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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.

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