Is there an elementary way to explain that a map of the earth cannot preserve distances?

Question

I am teaching a short "topics in geometry" course to future high school math teacher in France. I plan to cover some spherical geometry. I will be treating the following topics:

• volume of the ball and area of the sphere,
• length-minimizing property of great circles,
• intrinsic distance on the sphere
• Girard's formula for the area of spherical triangles.

From this I would like to address some elementary questions of cartography, in particular I would like to prove the following theorem:

Theorem : Given an open set $U$ of the sphere, there is no map from $U$ to the plane, which preserve distances (with $U$ endowed with the intrinsic distance on the sphere, and the plane endowed with the euclidean distance).

The proof I know uses Gauss Egregium Theorem: a (smooth) map which preserves intrinsic distances between two surfaces will preserve the first fundamental form, and will also preserve the Gauss curvature (by the Egregium Theorem). Since the plane has curvature $0$ and the sphere has curvature $1$, this is impossible.

I suspect that a more elementary proof should be available because the Girards formula actually expresses (via the local Gauss-Bonnet theorem) that the sphere has curvature $1$.

Question : Does there exists an elementary proof of the theorem above ?

(I guess by elementary here I mean understandable by an average undergraduate student without prior exposition to differential geometry and little knowledge in multivariate calculus.)

Answer 1

A simple proof by counter-example:

1) Draw a map that preserves distances between (0°E 0°N), (0°E 90°N = north pole), and (90°E 0°N). If you start with a sphere, these points are 90° apart from each other. A distance-preserving map will be an equilateral triangle, possibly with bulging sides.

2) Try to add the point (0°E 90°S = south pole). A distance-preserving map would need to simultaneously place it 90° from the two equator points, and 180° from the north pole. But with two equilateral triangles, it will only be 90° * sqrt(3) away from the north pole.

Thus, even for a very simple example, there is no way to make a distance-preserving flat map for just these four points.

A more general proof: As Matt F. suggests, we can do the same exercise for a small square around a point, instead of for an entire hemisphere.

1) We are trying to prove that we cannot make a flat, uninterrupted map of a sphere that preserves all distances. Formally, "Given an open set U of the sphere, there is no map from U to the plane, which preserve distances (with U endowed with the intrinsic distance on the sphere, and the plane endowed with the euclidean distance)."

This is an example of two important points for math teachers to teach their students:

• In proofs, start by clearly stating what you are assuming, and what you intend to prove. In story problems, start by clearly indicating what you know, and what you are trying to find.
• You might need to translate the detailed "formal" language of mathematics into simpler language that other people can understand.

2) Let 𝜃 be an angle between 0° and 180°. (In the previous proof, 𝜃 was 90°.) For ease of drawing, 15° or 30° is convenient.

This is an example of some important points for math teachers to teach their students:

• At the start of your problem, write out the names of your variables.
• Angle variables typically do not show a degree symbol (°).
• There are two common units for angles: 360° for a full circle = 2𝜋 radians for a full circle.
• Consistently use the units that are most convenient for a problem.
• If necessary, use unit conversions (such as multiplying by a fraction that equals 1) to convert between units. For example, (2𝜋 radians) / 360° and 360° / (2𝜋 radians) are both fractions that equal 1.

3) To keep the math simple, this proof implicitly assumes a unit sphere (with radius = 1). The proof can be extended to a sphere with arbitrary radius R by multiplying all distances on the surface of the sphere and the surface of the flat map by R. Alternatively, if the sphere represents the Earth, the student can assume that 1° = 60 nautical miles, or that 90° = 10,000 kilometers. Nautical miles are not the same as the more common "statute miles". Nautical miles and kilometers were originally defined so that these statements were true when measuring from the north pole to the equator via Paris. The error due to the Earth being an oblate spheroid instead of a perfect sphere is less than one percent.

This is an example of some more important points for math teachers to teach their students:

• When setting up a problem, sometimes it helps to make an approximation.
• It is good to know what those approximations are, and estimate how much they could affect the result.
• It is often easier to start with a special case, and then add extra variables as needed to find a more general answer.

4) We are going to draw a map around an arbitrary point on the sphere. To make the math easier, we are going to make this point be (0°E 0°N). But the proof applies to any point on the sphere, not just (0°E 0°N).

This is an example of three more important points for math teachers to teach their students:

• If possible, draw a picture of the problem. Label your variables and coordinate system.
• You can choose any coordinate system that is both convenient and consistent.
• On a sufficiently symmetrical object (like a sphere), it does not matter where you start.

5) On the sphere, label the point (0°E 0°N).

6) On a flat map, label the point (0°E 0°N) at the origin.

7) On the sphere, label the four points (𝜃 E 0°N), (0°E 𝜃 N), (𝜃 W 0°N), and (0°E 𝜃 S).

8) On the flat map, label the four points (𝜃 E 0°N), (0°E 𝜃 N), (𝜃 W 0°N), and (0°E 𝜃 S). Note that these four points are all a distance 𝜃 from the origin.

9) The four points (marked in Step 8) should form a square, because -- if it were possible to preserve distances -- any two adjacent corners should be the same distance from each other as any other two adjacent corners.

10) On the flat map, the length of a side of the square is (√2)𝜃.

11) Inside the sphere, the straight-line distance as the heat-resistant rock-tunnelling gopher travels between opposite corners of the square is 2sin𝜃.

12) Inside the sphere, the straight-line distance as the heat-resistant rock-tunnelling gopher travels between adjacent corners of the square is (√2)sin𝜃.

13) Along the surface of the sphere, the distance between adjacent corners of the square is 2arcsin( sin𝜃 / (√2) ).

14) Using Taylor series expansions, we can show that (for angles measured in radians):

2arcsin( sin𝜃 / (√2) ) ≈ (√2)𝜃 - (𝜃^3)/6(√2) - 7(𝜃^5)/240(√2) + O(𝜃^7) ≠ (√2)𝜃 (unless 𝜃 = 0°).

15) The flat map cannot have all of the distances match the distances on the sphere. QED.

This is another important point for math teachers to teach their students:

• Clearly label the solution to the problem. For many story problems, this is best done by writing out a sentence describing the answer, and circling the answer in a cloud. For many proofs, this is done by stating "QED", which means "This is what I set out to prove."

16) Fortunately, for a map of a small area, the approximation is close. For angles measured in radians, in the limit as 𝜃→0 radians, sin𝜃 = 𝜃. In the limit as 𝜃→0°, 2arcsin( sin𝜃 / (√2) ) ≈ 2arcsin( 𝜃 / (√2) ) ≈ 2( 𝜃 / (√2) ) ≈ (√2)𝜃

This is a sanity check: We have confirmed that a flat map approximately preserves distances for a small enough portion of a sphere. This is consistent with our real world experience that we can make useful flat maps of small parts of the world.

This is an example of two more important points for math teachers to teach their students:

• At the end of a problem, do a sanity check to confirm that the answer is reasonable.
• If possible, perform a check-by-substitution, to confirm that the answer matches the information that you knew at the beginning of the problem. The previous limit is both a sanity check, and a partial check-by-substitution.

17) As a further partial check-by-substitution, when 𝜃 = 180°, 2arcsin( sin𝜃 / (√2) ) = 2arcsin( 0 / (√2) ) = 2arcsin( 0 ) = 0°. We can verify this by "pushing" the square halfway around the sphere to where the points of the square join up again.

18) And as one last partial check-by-substitution, we will look at the special case in the previous proof. When 𝜃 = 90°, 2arcsin( sin𝜃 / (√2) ) = 2arcsin( 1 / (√2) ) = 2(45°) = 90° ≠ (√2)𝜃.

Answer 2

The accepted answer works well for the scope and goal of your course to give the students a solid understanding of why the theorem is true without getting too indepth. However, if they were to take this approach when they are teaching in the future i think that it would likely be way more advanced than the scope of a secondary geometry class. It may be useful to at least discuss (not necessarily actually perform) the following teaching strategy for them to implement when they have their own classrooms.

instead of trying to formally prove the theorem, it maybe more beneficial to their learning to have them explore the idea of "flattening a sphere" and come up with their own conjectures. Especially for students who do not have a strong math background, as per your question, digging into the complexities of a proof may be overwhelming. Instead, how about have them all bring in, or provide them with, various balls of different sizes (tennis ball, basketball, soccer ball, beach ball, etc), have them draw some shapes on the ball with markers and measure the shapes' dimensions (easily done with lengths of string), and then challenge them to both flatten the ball (via cutting it up, or any other method they may want to try) and preserve the dimensions of the shapes that they drew. As we already know, this task is impossible but they should have a lot of fun trying and experimenting with different methods of flattening and will eventually conclude that it is impossible. Especially for future high school math teachers, i think that this approach will provide a better learning experience than just going through a proof

EDIT: Edited my answer to reflect @BenCrowell astute comment that this method would be great to use to teach students but may be too elementary for a college level geometry course

Answer 3

On the unit sphere, construct a circle of radius $r$, i.e., the set of all points that lie at arc length $r$ from a given point. The circumference of this circle is $C=2\pi \sin r$, which is strictly less than the Euclidean value $2\pi r$. If there were a distance-preserving map, then it would have to preserve both $r$ and $C$, but we've just shown that if it preserves $r$, it can't preserve $C$.

Answer 4

Try a distance preserving flattening of an orange skin.