5.10: Pythagoras’ Theorem in three dimensions

Pythagoras’ Theorem belongs in 2-dimensions. But does it generalise to 3-dimensions? The usual answer is to interpret the result in terms of coordinates.

Problem 213

(a) Construct a right angled triangle that explains the standard formula for the distance from P = (a, b) to Q = (d, e).

(b) Use part (a) to derive the standard formula for the distance from P =(a, b, c) to Q = (d, e, f).

This extension of Pythagoras’ Theorem to 3-dimensions is extremely useful, but not very profound. In contrast, the next result is more intriguing, but seems to be a complete uke of limited relevance. In 2D, a right angled triangle is obtained by

• taking one corner A of a rectangle ABCD, together with its two neighbours B and D;
• then “cutting off the corner” to get the triangle ABD.

This suggests that a corresponding figure in 3D might be obtained by

• taking one corner A of a cuboid, together with its three neighbours B, C, D;
• then cutting off the corner to get a pyramid ABCD, with the right angled triangle ΔABC as base, and with apex D.

The obvious candidate for the “3D-hypotenuse” is then the sloping face BCD, and the three right angled triangles ΔABC, ΔACD, ΔADB presumably correspond to the ‘legs’ (the shorter sides) of the right triangle in 2D.

Problem 214 You are given a pyramid ABCD with all three faces meeting at A being right angled triangles with right angles at A. Suppose AB = b, AC = c, AD = d.

(a) Calculate the areas of ΔABC, ΔACD, ΔADB in terms of b, c, d.

(b) Calculate the area of ΔBCD in terms of b, c, d.

(c) Compare your answer in part (b) with the sum of the squares of the three areas you found in part (a).

More significant (e.g. for navigation on the surface of the Earth) and more interesting than Problem 214 is to ask what form Pythagoras’ Theorem takes for “lines on a sphere”.

For simplicity we work on a unit sphere. We discovered in the run-up to Problem 34 that lines, or shortest paths, on a sphere are arcs of great circles. So, if the triangle ΔABC on the unit sphere is right angled at A, we may rotate the sphere so that the arc AB lies along the equator and the arc AC runs up a circle of longitude. It is then clear that, once the lengths c, b of AB and AC are known, the locations of B and C are essentially determined, and hence the length of the arc BC on the sphere is determined. So we would like to have a simple formula that would allow us to calculate the length of the arc BC directly in terms of c and b.

Problem 215 Given a spherical triangle ΔABC on the unit sphere with centre O, such that ∠BAC is a right angle, and such that AB has length c, and AC has length b.

(a) We have (rightly) referred to b and c as ‘lengths’. But what are they really?

(b) We want to know how the inputs b and c determine the value of the length a of the arc BC; that is, we are looking for a function with inputs b and c, which will allow us to determine the value of the "output" a. Think about the answer to part (a). What kind of standard functions do we already know that could have inputs b and c?

(c) Suppose $c=0\ne b$ . What should the output a be equal to? (Similarly if $b=0\ne c$.) Which standard function of b and of c does this suggest is involved?

(d) (i) Suppose $\angle B=\angle C=\frac{\pi }{2}$ , what should the output a be equal to?

(ii) Suppose $\angle B=\frac{\pi }{2}$ , but ∠C (and hence c) is unconstrained. The output a is then determined - but the formula must give this fixed output for different values of c. What does this suggest as the “simplest possible” formula for a?

The answers to Problem 215 give a pretty good idea what form Pythagoras’ Theorem must take on the unit sphere. The next problem proves this result as a simple application of the familiar 2D Cosine Rule.

Problem 216 Given any triangle ΔABC on the unit sphere with a right angle at the point A, we may position the sphere so that A lies on the equator, with AB along the equator and AC up a circle of longitude. Let O be the centre of the sphere and let T be the tangent plane to the sphere at the point A. Extend the radii OB and OC to meet the plane T at Bʹ and Cʹ respectively.

(a) Calculate the lengths of the line segments ABʹ and ACʹ, and hence of BʹCʹ.

(b) Calculate the lengths of OBʹ and OCʹ, and then apply the Cosine Rule to ΔBʹOCʹ to find an equation linking b and c with ∠BʹOCʹ(=a).

When “solving triangles” on the sphere the same principles apply as in the plane: right angled triangles hold the key - but Pythagoras’ Theorem and trig in right angled triangles must be extended to obtain variations of the Sine Rule and the Cosine Rule for spherical triangles. The corresponding results on the sphere are both similar to, and intriguingly different from, those we are used to in the plane. For example, there are two forms of the Cosine Rule extending the result in Problem 216.

Problem 217 Given a (not necessarily right angled) triangle ΔABC on the unit sphere, apply the same proof as in Problem 216 to show (with the usual labelling) that:

$$$$cosa=cosb·cosc+sinb·sinc·cosA

The other form of the Cosine Rule is “dual” to that in Problem 217 (with arcs and angles interchanged, and with an unexpected change of sign) - namely:

$$$$cosA=−cosB·cosC+sinB·sinC·cosa

The next two problems derive a version of the Sine Rule for spherical triangles.

Problem 218 Let ΔABC be a triangle on the unit sphere with a right angle at A. Let Aʹ lie on the arc BA produced, and Cʹ lie on the arc BC produced so that ΔAʹBCʹ is right angled at Aʹ. With the usual labelling (so that x denotes the length of the side of a triangle opposite vertex X, with arc AC = b, arc BC = a, arc BCʹ = aʹ, and arc AʹCʹ = bʹ, prove that:

$$\frac{}{}$$sinb sina = sin b ′ sin a ′

Problem 219 Let ΔABC be a general triangle on the unit sphere with the usual labelling (so that x denotes the length of the side of a triangle opposite vertex X, and X is used both to label the vertex and to denote the size of the angle at X). Prove that:

$$\frac{}{}$$sina sinA = sinb sinB = sinc sinC .

It is natural to ask (cf Problem 32):

“If the three ratios in Problem 219 are all equal, what is it that they are all equal to?”

The answer may not at first seem quite as nice as in the Euclidean 2-dimensional case: one answer is that they are all equal to

$$\frac{\sin a·\sin b·\sin c}{\text{volume of the tetrahedron}}$$OABC .

Notice that this echoes the result in the Euclidean plane, where the three ratios in the Sine Rule are all equal to 2R, and

$$2R=\frac{}{}$$abc 2( area of ΔABC) .