The legs of a right triangle are $a$ and $b$. Find the distance from the vertex of the right angle to the nearest point of the inscribed circle.

Question

From Sharygin's Book:

The legs of a right triangle are $a$ and $b$. Find the distance from the vertex of the right angle to the nearest point of the inscribed circle.

My solution appears in this answer.

Thank you.

Answer 1

We know:

$$[ABC]=\frac 1 2 ba$$ ($[XYZ]$ to denote area)

Again we see, $$[ABC]=sr$$ Where $s$ denotes semi-perimeter and $r$ denotes inradius.

Certainly, $AC=\sqrt{a^2+b^2}$

Combining both equations gives: $$ \frac 1 2 ab=\frac 1 2(a+b+\sqrt{a^2+b^2})r$$ $$or, r=\frac {ab} {a+b+\sqrt{a^2+b^2}}$$

Clearly, $r\sqrt2$ is the value of $IB$ (the diagonal) of the square which is formed by the points of tangency due to the incircle near $B$.

The distance is their difference.

Subtracting and rationalising gives required distance: $$(\sqrt2-1)(a+b-\sqrt{a^2+b^2})$$

Answer 2

Hint: Imagine the triangle aligned with the right angle at the origin, and the two legs along the axes. Center of the inscribed circle is $(c,c)$. Equation of the hypotenuse can be taken as $y=-(b/a)x + b$. Also the distance between $(c,c)$ and the hypotenuse is also $c$, because of the inscription property. Now just apply the distance formula between a point and the line. This gets you the value of c, in terms of a and b. Now the required distance is simply $ \sqrt{2}c - c$.