Constructing a right-angled triangle given the half-perimeter $s$ and an altitude $h_c$

Question

I would like to receive some help about the next problem:

Problem:

Construct a right-angled triangle given the half-perimeter $s$ and an altitude $h_c$, where $\angle(BCA) = 90°$.

What i did:

1) Analysis:

Asume that the problem is solved.

Let the $ABC$ be the asked triangle. Now we have that $\angle(BCA) = 90°$.

Let $H_c$ be other end of the altitude from the $C$. $|CH_c| = h_c$.

Let $C_1$ be such point that it is $C_1 - A - B$ and $|C_1A| = |AC| = b$. Now $ACC_1$ is isosceles triangle with the apex in $A$. Now it is $\angle(CC_1A) = \angle(ACC_1) = \frac{\alpha}{2}$.

Let $C_2$ be such point that it is $A - B - C_2$ and $|BC_2| = |BC| = a$. Now $BC_2C$ is isosceles triangle with the apex in $B$. Now it is $\angle(BC_2C) = \angle(C_2CB) = \frac{\beta}{2}$.

$$ |C_1C_2| = a + b + c = 2s$$ $$\angle(C_1CC_2) = \angle(ACC_1) + \angle(BCA) + \angle(C_2CB) = 135°$$

So, the vertex $C$ belongs to the set of points from which the $C_1C_2$ is seen at the angle of $135°$.

Also, because $ACC_1$ and $BC_2C$ are isosceles triangles we have that it is:

$$ \{A\} = sim[CC_1] \cap C_1C_2,$$ $$ \{B\} = sim[C_2C] \cap C_1C_2,$$

($sim$ is for the axis of symmetry).

Additionaly, we have that it is $\angle(C_1CH_c) = 90°- \frac{\alpha}{2}$ and $\angle(C_2CH_c) = 90°- \frac{\beta}{2}$.

My question:

Please, could you give me some hints for next steps, beacuse i can't figure out how to merge everything.

I can construct $135°$ and i can construct $C_1C_2$ and the line, let say $p$, such that $d(p, C_1C_2) = h_c$, but i don't see a way to connect these two constructions into one.

Thank you, for your time and help!

Answer 1

Draw a skech $A'B'C$ in which segment $AB$ is extended across $A$ for $AC$ to get $A'$ and across $B$ for $BC$ to get $B'$. Now the angle $\angle A'CB' = 135$. You can draw triangle $A'B'C$ since you have $A'B' =2s$ hight $h_c$ and an angle at $C$.

To do this first draw $A'B'=2s$. Then draw isoscales triangle $A'SB'$ so that $S$ is on different side of line $A'B'$ then $C$ is and $\angle A'SB' =90$. Then $C$ lies on one arc (which one) of circle with center at $S$ and $r=SA'=SB'$. Finally draw a parallel to $A'B'$ at distance $h_c$ (the one on different side than $S$ is). $C$ is an intersection point between parallel and circle.

Answer 2

Hint about another way.-First ask yourself if the problem has a unique solution. You can see it because of the system $$\begin{cases}x^2+y^2=z^2\\x+y+z=2s\\xy=hz\end{cases}$$ from which easily you get $$z=\frac{2s^2}{h+2s}$$ easy to construct. Now draw the circle of diameter $z$ and the two parallel to the diameter of the obtained circle spaced by $h$. The result, by intersection of the two parallels with the circle, are four solutions all equivalents of course.