Just saw this little video…
3-4-5 Triangles and Pi (youtube.com)
and attempted the same math to another Pythagorean Triangle…5-12-13…and, came up with 2 pi as the area of the inscribed circle.
Is it safe to say all Pythagorean triangles have an inscribed circle of pi times some integer?
That is, a nice, whole number, and not a nasty fraction or decimal?
It’s always going to be true for any Pythagorean triple.
Going with the same logic, but with sizes of a, b, and c, we have:
(a - r) + (b - r) = c
And so:
r = \frac{a + b - c}{2}
If (a + b - c) is an even integer, then it’s divisible by 2, and r is an integer. Note that we’re most of the way there since the whole point of a Pythagoran triple is that a, b, and c are all integers.
If a and b are both even, then a^2 and b^2 are both even, and therefore so is c^2 and thus c, and so (a + b - c) is even.
If a and b are both odd, then a^2 and b^2 are also odd, and c^2 and c are then both even, so (a + b - c) is even.
If one of a or b is odd, then one of a^2 and b^2 is odd, and so is c^2 and thus c again. So again, (a + b - c) is even.
So in all cases (a + b - c) can be divided by 2, which means r is an integer, and thus \pi r^2 is a nice multiple of \pi.
Quora, for once, answers the question quickly.
As I’ve already remarked on this Board, every odd number is the short leg of a Pythagorean triple in which the long leg and the hypoteneuse differ by one. You can write the length of the short leg as
2n + 1., where “n” can be any integer – 1, 2, 3, 4…
In that case the long leg is 2nsquared + 2n, and the hypoteneuse is obviously given by
2nsquared+2n+1
In all of these cases, the radius of the inscribed circle is given by n, which is always an integer.
Of course, the areas of the inscribed circles will always be multiples of pi, because they’ll all be pin squared.
Moreover, the points at which the inscribed circles touch the sides of the triangle divide those sides into two segments each with integral lenmgth.
This formulation, which I’m sure isn’t original with me, but which I’ve never seen noted before, isn’t as complete as the formulation of 2mn, m squared plus n squared, m squared minus n squared, which generates triples this method doesn’t (pretty clearly not all Pythagorean triples don’t have the long side and hypoteneuse differing by one) , but that broader formulation will generate all of this more restricted class of Pythagorean triples.
Just FYI, you can get nicely formatted math output using the LaTeX plugin. Just write:
$ 2n^2 + 2n + 1 $
to get
2n^2 + 2n + 1
Some of the advanced constructs can get tricky, but basic expressions like these are easy. Use \pi for \pi.
While that does seem to be true, the converse is not. There are primitive* triples where the smallest leg is not an odd number, and the two longer legs are not one apart. The smallest such example is (8, 15, 17).
*Not multiples of other triples, e.g. (6, 8, 10) is a multiple of (3,4,5)
Just to summarize, let a,b,c be a primitive PT (no common divisor), then one of a,b is even and the other odd. For if they are both even, then so is c and they it is not primitive. If they are both odd, then both a^2 and b^2 leave a remainder of 1 when divided by 4 (as do all odd squares since (2k+1)^2=4k^2+4k+1), so a^2+b^2 has a remainder of 2 when divided by 4, while c^2 is divisible by 4, so that is impossible. But then if, say, a is even and b,c odd, a+b-c is even.
If it is not primitive, just divide by the greatest common divisor; rinse and repeat.
I occurred to me: if we take this concept to the third dimension and employ a sphere inside a rectangular solid, could we use that analog to use as a test to find an Euler Brick?
3rd Power Diophantine Generator - Factual Questions - Straight Dope Message Board
The LaTex MathML thing is acting up. I’m seeing a red [Math Processing Error] for each of @Dr.Strangelove’s & @Hari_Seldon’s equations.
Oops.
I would say that this goes without saying, except I already said it in my post.
Yes, there are plenty of cases where the short leg is not an odd number and the other two sides don’t differ by one. In fact, that’s the majority of cases.
But the cases I brought up are interesting, and pretty clearly show that there is an infinite number of “primitive” cases where the long side and hypoteneuse differ by one (since any time the sort leg is a prime number that’s a nother priimitive case.
I will point out that for any positive integer k, the triple (2k+1,\frac{(2k+1)^2-1}2,\frac{(2k+1)^2+1}2) is a PT with consecutive integers. When k=1, you get (3,4,5), k=2 gives (5,12,13), k=3 gives (7,24,25), etc. Note that the sum of the hypotenuse and long leg is the square of the short leg. Which gives a short way of getting them. Take an odd square n and the floor and the two nearest integers to n^2/2. E.g. (17,144,145).
And I’ll just point out that your formula is identical to mine (quoted upstream), only with k as the variable instead of n.
Actually, I’m intrigued by the formulas. They make your statement about the squares obvious. I admit that, although I noticed that the short side is the nth odd number and the long side is four times the nth triangular number (and the hypoteneuse therefore one more than four times the nth triangular number), I hadn’t noticed that if you took half the square of the short side then the long side and hypoteneiuuse were the two nearest whole numbers.
(The short side m being odd, its square m^2 is also odd, and so taking half gets you to a point halfway between two whole numbers. I certainly hadn’t noticed that these were (m^2)/2 + 1/2 and (m^2)/2 - 1/2.
m = 2n +1 or 2kl +1, depending upon whose formula you use.
And, as I pointed out, the radius of the inscribed circle is n ( or k)
Sorry about that. I missed it.
Here’s another amusment. If n is even then (n,\frac{n^2}4-1,\frac{n^2}4+1) is a PT in which the long leg and hypotenuse differ by exactly 2. It is primitive if and only if n is divisible by 4. That’s because that is what forces the last two to be both odd.
Aside: Partly inspired by this thread, I asked my Discrete Math class to prove that, in any primitive Pythagorean triple, exactly one of a and b must be odd.
The proof that they’re not both even, of course, is that then c would also be, and it wouldn’t be primitive. And the proof that they’re not both odd is quite similar to the proof (which we saw in class) that \sqrt{2} is irrational.
Nitpick: It doesn’t work for n = 4, even though 4 is even. Plugging 4 in gives 4, 3, 5, and the long leg differs from the hypoteneuse by 1.
In all other cases the long leg and the hypoteneuse are given by the second and third expressions, and the difference between them is necessarily 2. But in the case of n = 4 the long leg is given by n, not the second expression. The difference between the last two expressions is still (5 - 3 = 2), but 3 isn’t the long leg.
Did anyone do it? It of course depends on congruence mod 4. Once you’ve done that it is straightforward to go on and show that all primitive PTs are giving by choosing two positive integers m>n that have no common divisor and one is even, the other odd. Then (2mn,m^2-n^2,m^2+n^2) is the corresponding PT. Someone mentioned this above, but didn’t know whether they all arise that way. The argument is based on the following. Given a PT (a,b,c), let a=2d be the even one. Then d^2=\frac{c+b}2\frac{c-b}2. The two factors on the right can have no common divisor (else b and c would) and the only way a product of two relatively prime terms can be a perfect is for each factor to be one. The rest is trivial. That last statement, incidentally, is equivalent to unique prime factorization.
Two (out of six) of the students did. The rest overlooked the “exactly one of them is odd” in the instructions, and proved that they couldn’t both be even, but neglected to prove that they couldn’t both be odd.