Prove that the radius of the incircle of a right angles triangle with integer sides is an integer.
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This seems related to the following older question: http://math.stackexchange.com/q/1625657/137524 – Semiclassical May 26 '16 at 14:30
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Take any known Pythagorean triple and inscribe a circle. Draw three radii to meet perpendicular to each of the three sides. Now your triangle is split up into a square and two kites and the hypotenuse $c$ is equal to $a+b-2r$. Solve for $r$ using $a,b$, and $c$. – Leonidas Lanier May 26 '16 at 14:42
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Let a= m^2-n^2,b=2mn,c=m^2+n^2 then we can find the radius of the incentre. Then were should prove that it is an integer. – sufaidsaleel Jun 03 '16 at 01:42
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Suppose the right triangle has legs $\{a,b\}$ and hypotenuse $c$, and let $r$ denote the inradius. As usual we let $\{A,B,C\}$ denote the vertices opposite sides $\{a,b,c\}$ respectively. Dropping perpendiculars from the incenter to the legs and using the fact that two tangents from a fixed point to a fixed circle have the same length, we see that the two tangents from $A$ have length $b-r$ and the two tangents from $B$ have length $a-r$. It follows that $$c=a-r+b-r=a+b-2r\implies r=\frac 12\times (a+b-c)$$ All that remains is to note that $a+b-c$ is always even, but this follows at once from the standard parameterization of the Pythagorean triples.
lulu
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All Pythagorean triples can be generated by two integer numbers $m>n$ as:
$$ a=m^2-n^2 \qquad b=2mn \qquad c=m^2+n^2 $$
Dividing the triangle in three triangles with a common vertex in center of the incircle you can see that the radius of the incircle is:
$$r= \frac{2A}{a+b+c}$$
Where $A$ is the area of the triangle. So: $$ r=\frac{2ab}{a+b+c}=\frac{4mn(m^2-n^2)}{m^2-n^2+2mn+m^2+n^2}=n(m-n) $$ that is an integer
Emilio Novati
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