×

# 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
1 year ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

- 1 year 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

- 1 year ago