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

#Geometry

Note by Akarsh Jain
4 years, 3 months ago

No vote yet
1 vote

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Comments

Using Euclid's formula we have a=k(m2−n2),b=2kmn,c=k(m2+n2)a=k(m^2-n^2), b=2 k m n , c = k(m^2+n^2) and formula for incircle radius r=△sr=\frac{\triangle}{s} where ss is semiperimeter s=(a+b+c)/2s=(a+b+c)/2. We get the following: r=aba+b+c=2k2mn(m2−n2)k(2m2+2mn)=kn(m−n)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)

Maria Kozlowska - 4 years, 3 months ago

You can draw a right triangle with sides aa and bb and hypotenuse cc with a circle inscribed in it.
Then;
By using the property that tangents from same external points are equal; the hypotenuse equals a+b−2ra+b-2r where rr is the inradius.
a+b−2r=ca+b-2r=c
  ⟹  r=a+b−c2\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−ca+b-c will remain even and hence; inradius will be integer

Yatin Khanna - 4 years, 3 months ago
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