The other day I saw something interesting,
Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.
Hello,Originally Posted by ThePerfectHacker
I'm a little bit puzzled about your question: What are the segments of the hypotenuse and what exactly do you take as divisor? The radius, the perimeter or the area of the inscribed circle?
I've attached a drawing of a right triangle so you can check which names I use for the variables.
Heron of Alexandra had found that in a right triangle is $\displaystyle \rho=\frac{a \cdot b}{a+b+c}$
The three bisectors of a triangle divide it into three smaller triangles. The height in every of these smaller triangles is always ρ. That's true with every triangle.
So you get:
$\displaystyle A=\frac{1}{2} \cdot \rho \cdot a+\frac{1}{2} \cdot \rho \cdot b+\frac{1}{2} \cdot \rho \cdot c=\frac{1}{2} \cdot \rho \cdot (a+b+c)$
Heron set $\displaystyle \frac{1}{2} \cdot (a+b+c)=s$ and you get: $\displaystyle A=\rho \cdot s$
With a right triangle you've additional conditions:
$\displaystyle A=\frac{1}{2} \cdot c \cdot h=\frac{1}{2} \cdot a \cdot b=\frac{1}{2} \cdot \rho \cdot (a+b+c) $
and therefore $\displaystyle \rho=\frac{a\cdot b}{a+b+c}$
With $\displaystyle a^2=p \cdot c$ and $\displaystyle b^2=q \cdot c$ you can change $\displaystyle p \cdot q=\frac{a^2}{c} \cdot \frac{b^2}{c}=\frac{a^2 \cdot b^2}{c^2}$
Now you want to divide this term by a part of the inscribed circle. But I cann't help you any further because I don't know which part you want to take. But maybe you can use my remarks to get a little bit ahead.
Bye