Prove that the inradius of a right triangle with sides of integer length is also an integer.

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TopNewestUsing Euclid's formula we have \(a=k(m^2-n^2), b=2 k m n , c = k(m^2+n^2)\) and formula for incircle radius \(r=\frac{\triangle}{s}\) where \(s\) is semiperimeter \(s=(a+b+c)/2\). We get the following: \[r=\frac{a b }{a+b+c}=\frac{2 k^2 m n (m^2-n^2)}{k(2m^2 + 2 m n)}=k n(m-n)\]

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You can draw a right triangle with sides \(a\) and \(b\) and hypotenuse \(c\) with a circle inscribed in it.

Then;

By using the property that tangents from same external points are equal; the hypotenuse equals \(a+b-2r\) where \(r\) is the inradius.

\(a+b-2r=c\)

\(\implies r = \frac{a+b-c}{2}\)

Now, the only thing that remains to be proved is that a+b-c will always remain even in a pythagorean triangle. This can be done by using parity of sides; either one or three of them will be even; in any case \(a+b-c\) will remain even and hence; inradius will be integer

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