Radius of the inscribed circle of a right angled triangle

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Pythagoras, triples, 3,4,5, a calculator.

How to generate Pythagorean triples (example: 3,4,5), well one way at least.

Starting with (x + y)² = x² + y² + 2xy and (x – y)² = x² + y² – 2xy

we can write the difference of two squares

(x + y)²  –  (x – y)² = 4xy

and if we write  x = A² and y = B² the right hand side is a square as well.

Thus:

(A²  +  B²) ² – (A² – B²) ² = 4A² B² = (2AB) ²

which can be written as

(A²  +  B²) ² = (A² – B²) ² + (2AB) ²

the Pythagoras form.

Now just put in some integers for A and B

2 and 1 gives 3,4,5

Conjecture1: This process generates ALL the Pythagorean triples.

Conjecture2: Every odd number belongs to some  Pythagorean triple.

Have fun…….

My next post will be about finding the radius of the inscribed circle in a right angled triangle…..

And now Pythagoras again, with bonus

I was attempting to solve a geometrical problem the other day, a problem which, due to my complete misunderstanding, had no solution, when this popped out. It is probably bog-standard, but new to me, and this time I don’t have the heart to check if it is one of the 100 proofs of The Pythagoras theorem.

Now let theta be the angle ACB. Angle ABD is then 2*theta.

Set  r = 1, then  a is  sin(2*theta)  and b is  cos(2*theta),  and so

sin(theta) = (1 – b)/a = 1/a -b/a = cosec(2*theta) – cot(2*theta)

and  cosec(theta) = (1 + b)/a = 1/a + b/a = cosec(2*theta) + cot(2*theta)

I’ve done most of the work,

so now you can show that  sin(2*theta) = 2*sin(theta)*cos(theta)

Pythagoras converse, proof from scratch

There is a website with 100 proofs of the famous theorem of Pythagoras, but when I trawled the net looking for a proof of the converse, they all assume the basic theorem.

Here’s how to do it from scratch, which is considerably more satisfying, and also a simple application of similar triangles and basic algebra: