I’m not precisely sure how to embed this, so, let me just point over to the Five Triangles blog (on blogspot) where there’s a neat little puzzle. It starts with Pythagorean triplets, the sets of numbers (a, b, c) so that a2 plus b2 equals c2. Pretty much anyone who knows the term “Pythagorean triplet” knows the set (3, 4, 5), and knows the set (5, 12, 13), and after that knows that there’s more if you really have to dig them up but who can be bothered?
Anyway, the problem at Five Triangle’s “Inscribed Circle” here draws that second Pythagorean triplet triangle, the one with sides of length 5, 12, and 13, and inscribes a circle within it. The problem: find the radius of the circle?
I’m embarrassed to say how much time I took to work it out, but that’s because I was looking for purely geometric approaches, when casting it over to algebra turns this into a pretty quick problem. I do feel like there should be an obvious geometric solution, though, and I’m sure I’ll wake in the middle of the night feeling like an idiot for not having that before I talked about this.
9 thoughts on “About An Inscribed Circle”
Very interesting… The Pythagorean theorem is almost as important as Einstein’s findings… Wel, that’s my humble opinion.
Best wishes. Aquileana :D
Well, I’m not sure how I’d rate them in importance. Pythagoras’s result is certainly the more universally useful, though; you can barely do any real-world stuff without relying upon it. Relativity and quantum mechanics may make up the world but you rarely have to know anything about either.
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As ever I am intrigued by apparently straightforward problems. I took a geometric view and eventually arrived at the following:
sqr(a) + sqr(b) = sqr(c) is a right angled triangle – ANY
then the radius of the inscribed circle is ab/(a + b + c)
You want to know how I got it ? I’ll send you the diagram and derivation.
I did think “This result is too good to be true !”.
Oh, yes, yes. It’s obvious once … well, once you realize it’s obvious, isn’t it?
I have just posted the derivation. Check it out, and there’s a little conjecture, just for you !
Quite nicely done, and differently from the geometric approach I’d tumbled across. Thank you.
I’m confident your conjecture is right, by the way, and might write that up too.