This exercise was given as preparation for my upcoming midterm and I've been struggling with it enormously. The problems states;
The altitude from the right angle of a right angled triangle divides it into two triangles of perimeters p and q. Compute in terms of p and q the perimeter of the triangle ABC.
I'm aware that within my triangle ABC there are three similar triangles and there are various proportions of side length that can be found, but yet after a weekend of work I cannot find the perimeter of ABC depending only on p and q. Any guidance or advice will be much appreciated!
The two triangles are similar to the original triangle.
Let $t$ be the original triangle.
The perimeter of $t$ is the sums of the areas of $p$ and $q$ minus twice the altitude which is common to both. If the altitude is $h$, then, with "per" meaning "perimeter", $per(p)+per(q)-2h =per(t) $.
If the sides of the original triangle are $a, b, c$, then $area(t) = ab/2 =ch/2$.
Therefore $h = ab/c$.
Therefore $per(t) =per(p)+per(q)-2ab/c =per(p)+per(q)-2ab/\sqrt{a^2+b^2} $.
From the similar triangles $\dfrac{per(p)}{a} =\dfrac{per(q)}{b} =\dfrac{per(t)}{c} $, so $a=\dfrac{c\,per(p)}{per(t)}$ and $b=\dfrac{c\,per(q)}{per(t)}$.
Multiplying, $ab =\dfrac{c^2\,per(p)per(q)}{per^2(t)} $ so that $h =ab/c =\dfrac{c\,per(p)per(q)}{per^2(t)} $.
I don't see how to get an expression for $per(t)$ involving only $per(p)$ and $per(q)$, so I'll leave it at this.
Maybe the areas come in, and the fact that they are proportional to the squares of the sides would be of use.