Let's construct a right-angled triangle, given the leg $AC = b$ and the perimeter $AC + BC + AB = 2s$. What are the conditions for the possibility of this construction?

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Let's construct a right-angled triangle, given the leg $AC = b$ and the perimeter $AC + BC + AB = 2s$. What are the conditions for the possibility of this construction?
Let's now consider the circle inscribed in the right-angled triangle $ABC$ - between the legs of the right angle - and its points of tangency with the sides (or their extensions) of the triangle, denoted as $A', B', C'$.

Then,

$$
C A' = C B', \quad A C' = A B', \quad B C' - B A'
$$

and

$$
\begin{gathered}
2s = C A' + A C' + B C' + C B = \\
= C B' + C A' = 2 C B' \quad \text{so} \quad C B' = C A' = s
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_65cdb7596a335ed593e3g-1.jpg?height=350&width=499&top_left_y=519&top_left_x=802)

Based on this, the construction is as follows: on one leg of the right angle at $C$, measure the distances $C A = b$ and $C B' = s$, and on the other leg, measure the distance $C A' = s$. Then, construct the circle that is tangent to the legs of the right angle at $A'$ and $B'$. From point $A$, draw another tangent to this circle, which intersects $C A'$ at point $B$.

The condition for the possibility of this construction: $b < b$,

$$
2s = a + b + c > 2b, \quad \text{so} \quad s > b
$$

From this, we see that $s > b$ is a necessary condition; from the construction, we see that it is also sufficient.

Second Solution. Construct a right angle with vertex $C$. On one leg, measure the distance $C A = b$, and on the other leg, measure the distance $C D = 2s - b ( = a + c )$. Draw the perpendicular bisector of the hypotenuse $A D$; this intersects $C D$ at point $B$. Since $A B = B D$, the triangle $A B C$ indeed meets the requirements.

![](https://cdn.mathpix.com/cropped/2024_05_02_65cdb7596a335ed593e3g-1.jpg?height=453&width=309&top_left_y=1541&top_left_x=903)

The perpendicular bisector of $A D$ must intersect $C D$ between $C$ and $D$. For this to happen, the necessary and sufficient condition is that

$$
A C < C D, \quad \text{so} \quad b < 2s - b
$$

thus

$$
b < s
$$

(If $2s - b = b$, or $b = s$, then the perpendicular bisector of $A D$ passes through point $C$!)