I almost posed this as a proper "Problem," but just couldn't bring myself to actually do so. I like it better as a jumping-off point for a discussion.

**What is the area of a triangle whose sides have lengths \(3+i\), \(3+2i\) and \(4-3i\)?**

Yes, it's a little ridiculous, but that's intentional. After all, we mathematicians, like the White Queen in *Through the Looking-Glass*, sometimes like to believe "six impossible things before breakfast."

Is it possible to make any sense out of such a question? Is there an answer? What does it mean?

What other questions do you find yourself asking?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWell, it works out nicely if you use

Heron's Formula.Note that, the

Semi perimeter(\(\displaystyle s\)) is real, i.e,\(\displaystyle s = \frac{a+b+c}{2} = \frac{(3+i)+(3+2i)+(4-3i)}{2} = 5\)

Using Heron's Formula,

\(A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5(2-i)(2-2i)(1+3i)} = \sqrt{100} = \boxed{10}\) sq.units

I think that this is wonderful.

Maybe, this will lead to new

breedof complex numbers that have thisunusualproperty.I can't make more sense of it than that.

Log in to reply

YEAH DID THE SAME

Log in to reply

MADE A SLIGHT MISTAKE\( \sqrt{100}= \boxed{10}\)

Log in to reply

I don't think that the area of a triangle with complex numbers having non-zero imaginary part being its sides make any sense because in order to find the area, we basically use

Heron's Formula. And if you have proved it, you would know that all in all it demandsCosine Rulewhich in turn has to be proved geometrically. And in order to prove it geometrically, you need to now where the sides of triangle are, what's the angle between them, etc. But complex sides are against it. I think, this is enough to conclude that area of triangle formed by a random triple of complex numbers doesn't make any sense. Anyways, one can always find many triples which when put into the expression ofHeron's Formularesult in aReal number.Log in to reply

This doesn't convince me. It feels like the same logic could be applied to things like complex exponentiation: Since the concept of exponentiation first came about in the context of real numbers (or, really, positive integers), you could say that it doesn't make sense when applied to complex numbers. But we know that's not true; there is a way to "bestow" meaning upon things like \(2^i\), typically by way of motions in the complex plane.

So I would start thinking about this this way... Typically the side lengths entered in Heron's Formula represent distances. Is there a way to conceive of the notion of "distance" that allows for complex numbers? Perhaps we could start more conservatively and ask whether there is a way to conceive of "distance" so that it includes negative real numbers. And there

doesexist a notion ofdirected distance. Could that be extended further?And so on...

Log in to reply

Can anyone else find triples of complex side lengths that give a "triangle" with real area? Must the semi-perimeter be real for this to happen?

Log in to reply

Sure, (1+i) + (1+i) + (-1+i) = 1+3i, which is an imaginary perimeter, but has an area of (1/2)√5

Log in to reply

Alright, I found a triangle that has real area, but unfortunately it turned out to be

zero.Consider,

Sides : \(\displaystyle 3+2i,2+i\) and \(\displaystyle 1+i\)

Note that,

Semi perimeter\(\displaystyle = s = \frac{(3+2i)+(2+i)+(1+i)}{2} = 3+2i\)It is clear that,

since \(\displaystyle s = a\), we get,

Area\(\displaystyle = \sqrt{s(s-a)(s-b)(s-c)} = \boxed{0}\)Hence,

we can

alwaysfind 3 complex side lengths that give a triangle withrealarea.Log in to reply

Also note that, the

semi perimeterisnotreal, it is complex.Log in to reply

A triangle with sides \((\alpha + \beta ; \; \alpha ; \; \beta)\) will always have no area.

Log in to reply

Log in to reply

Log in to reply

The question is that, can lengths be imaginary. Not sure, but maybe this could be explained with a little bit of physics and higher maths (existence of a figure simultaneously in different dimensions and therefore the concept of complex). But first what is a complex point. How to "see" it? These fundamental questions could provide meaning to the answer. I don't know how rational I am while writing this comment but anyways, that's what I was thinking while trying to get some meaning out of the question!

Log in to reply

One application area that comes to mind is those of mechanisms , especially if we consider not the link lengths but the distance of the vertices from a fixed point. Other illustrative related areas include coupled oscillators, molecular chemistry (the length of the bonds etc)

Log in to reply

Could it represent a triangle with side lengths that oscillate in time or space ? Possibly, the alternative representation would have been random and without any pattern

Log in to reply

are you all saying that there exist a triangle whose sides dont exst but area exists?

Log in to reply

I guess it depends on what you mean by "exist"! After all, one could argue that, for example, "a 3-4-5 triangle" does not exist. We could draw or build physical representations of a 3-4-5 triangle, but they would all be inaccurate/imperfect. Yet we can still talk about the properties of such a triangle in a way that makes sense.

If by "exists" you mean "has a way to measure it that corresponds to a positive real number," then... yes. That is what we're saying. :-)

Log in to reply

In a way, this illustrates the fact we don't need diagrams to demonstrate geometrical theorems, provided it doesn't involve ordering. A blind man or an alien with no concept of sight or figures will still be able to understand geometrical theorems purely on algebraic grounds. I believe the mathematician Laplace preferred not using diagrams at all in his works, as he considered them to be misleading.

Log in to reply

The area would be 10. Everything that is true about geometry and algebra using real numbers is also true using complex numbers, except for ordering (such as inequalities). We simply use Heron's formula to finding the area of a triangle, given its sides.

Log in to reply

So are you saying that, in particular, there can be no "Triangle Inequality" for complex-sided triangles?

Log in to reply

That's what I'm saying. That's the one thing we'd have to give up, we wouldn't be able to say that x is greater than y, whatever x and y might be, lengths, areas, angles, etc.

Log in to reply