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# RMO -1999

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

If someone could write a proof to the above question

Note by Akarsh Jain
8 months, 3 weeks ago

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Using 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: $r=\frac{a b }{a+b+c}=\frac{2 k^2 m n (m^2-n^2)}{k(2m^2 + 2 m n)}=k n(m-n)$

- 8 months, 3 weeks ago

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

- 8 months, 3 weeks ago