The Right Triangle and the Circle

This problem was inspired by a problem from the Jiuzhang Suanshu: Gougu 16. I developed this problem to derive the Pythagorean theorem.

Problem

Consider a right triangle with three sides $a \le b < c$.

Part 1: What is the diameter of a circle inscribed in a right triangle? Find two formulas that include the three sides of the right triangle.

Part 2: Then use these two formulas to prove $a^2+b^2=c^2$.

Solution

Part 1

Formula 1

Let $a,b,c$ be the side lengths of the triangle, where $c$ is the hypotenuse. Let $r$ be the radius of the circle.

Figure out the fact that $(a-r)+(b-r) = c$.

$a+b = c+2r$

$a+b=c+d$

$a+b-c=d$

Formula 2

Figure out two formulas for the area of the triangle:

$A= \frac{1}{2} ab$

$A= \frac{1}{2} r(a+b+c)$.

Set the two formulas equal

$\frac{1}{2} ab = \frac{1}{2} r(a+b+c)$.

Therefore,

$d = \frac{2ab}{a+b+c}$.

Part 2

Equate the two formulas for diameter.

$a+b-c= \frac{2ab}{a+b+c}$

$(a+b-c)(a+b+c)=2ab$

$a^2+2ab+b^2-c^2 = 2ab$

$a^2+b^2=c^2$.