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?

## Comments

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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. – Anish Puthuraya · 3 years, 3 months ago

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– Anirudha Nayak · 3 years, 3 months ago

YEAH DID THE SAMELog in to reply

– Anirudha Nayak · 3 years, 3 months ago

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. – Dhruv Bhasin · 3 years, 3 months agoLog in to reply

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... – Matt Enlow · 3 years, 3 months ago

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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? – Matt Enlow · 3 years, 3 months ago

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– Michael Mendrin · 3 years, 3 months ago

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

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. – Anish Puthuraya · 3 years, 3 months agoLog in to reply

semi perimeterisnotreal, it is complex. – Anish Puthuraya · 3 years, 3 months agoLog in to reply

A triangle with sides \((\alpha + \beta ; \; \alpha ; \; \beta)\) will always have no area. – Guilherme Dela Corte · 3 years, 3 months ago

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– Matt Enlow · 3 years, 3 months ago

Interesting that that is a property that continues to hold when we expand into the complex numbers: If one side is the sum of two others, then the area of the triangle is zero. It's clear why this is the case, though, if we examine Heron's Formula.Log in to reply

– Michael Mendrin · 3 years, 3 months ago

By using the Law of Cosines, we can also prove that for any sides a, b, c, real or complex, the sum of the angles always add up to 180°, even if the angles themselves are complex. It's a bit tedious, but it does work out.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! – Swaraj Mohapatra · 3 years, 1 month ago

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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) – Sundar R · 3 years, 3 months ago

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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 – Sundar R · 3 years, 3 months ago

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are you all saying that there exist a triangle whose sides dont exst but area exists? – Prakhar Nigotiya · 3 years, 3 months ago

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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. :-) – Matt Enlow · 3 years, 3 months ago

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– Michael Mendrin · 3 years, 3 months ago

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. – Michael Mendrin · 3 years, 3 months ago

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– Matt Enlow · 3 years, 3 months ago

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

– Michael Mendrin · 3 years, 3 months ago

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