Mathematics Diary posted the following identity this morning. If *a* + *b* + *c* = π then

tan(

a) + tan(b) + tan(c) = tan(a) tan(b) tan(c).

I’d never seen that before. It’s striking that the sum of three numbers would also equal their product. In fact, the only way for the product of three numbers to be equal to their sum is for the three numbers to be tangents of angles that add up to π radians. I’ll explain below why that’s true.

First, we can generalize the identity above slightly by saying it holds if *a* + *b* + *c* is a multiple of π. In fact, it can be shown that

tan(

a) + tan(b) + tan(c) = tan(a) tan(b) tan(c)

if and only if *a* + *b* + *c* is a multiple of π. (Here’s a sketch of a proof.)

Now suppose *x* + *y* + *z* = *x* *y* *z*. We can find numbers *a*, *b*, and *c* such that *x* = tan(*a*), *y* = tan(*b*), and *z* = tan(*c*) and it follows that *a* + *b* + *c* must be a multiple of π. But can we chose *a*, *b*, and *c* so that their sum is not just a multiple of π but exactly π? Yes. If *a* + *b* + *c* equaled *k*π for some integer *k*, pick a new value of *c* equal to the original *c* minus (*k*-1)π. This leaves the value of tan(*c*) unchanged since the tangent function has period π.

( What fascinates me about math is that it always has a next amazement in store. ) You can find the identity in the beautiful book ‘Trig or Treat’ an Encycopledia of Trigonometric Identity Proofs by Y.E.O. Adrian.

I remember this one. I first saw it when my graduate advisor and I were trying to work out a manifestly symmetric proof of Heron’s formula. It turns out that this identity is actually equivalent to Heron’s formula. I’ll leave it as an exercise to supply the two proofs.

John,

I just came across this excellent post from the new Carnival: Math Teachers at Play #3. Interestingly, I had recently published the simpler version for two numbers some time ago and even that ‘easy’ question generated great interest. For your problem, my first reaction was to consider three equal numbers which of course leads to the trivial solution a = b = c = 0 as well as a = b = c = +-sqrt(3). The sqrt(3) might lead one to consider tan(pi/3) which of course is a special case of your general result. I would hope motivated high school students would make that connection and be interested in seeing hte general case!

We miss the trivial 0,0,0 solution.

Jonathan,

I think John is saying that, even in the trivial case, one could choose (that is, there always exists) angles whose sum is pi. Thus, we can use a = pi, b = pi, c = -pi. Then a+b+c = pi. In other words, we don’t have to use angles 0,0 and 0.

Another example :

1+2+3 = 1*2*3

tan inverse 1 = Pi/4 = 45 degrees

tan inverse 2 = 63.43 degrees

tan inverse 3 = 71.57 degrees

———————————–

180 degrees

i have started loving math after reading these articles..After reading this one, i am thinking whether we can give a geometrical proof for this one? for eg. half of a circle with unit radius is PI radians..just curious thats all..