Let's construct a right-angled triangle, given one of its legs and the difference between the sum of the legs and the hypotenuse.
I. solution: Imagine the problem is solved and set the difference $a+b-c=d$ on the side line $CA$ of $\triangle ABC$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_285385c7c4e2168d3924g-1.jpg?height=325&width=718&top_left_y=290&top_left_x=671)

Figure 1

By rotating the vertex $B$ around $C$ and $A$ to the extension of $AC$ beyond $C$, we get $B'C=a$, $B''A=c$, and $B'B''=B'C+CA-B''A=a+b-c=d$. Now, based on our data, we can determine the relative positions of the points $B, C, B', B''$, and from these, $A$: On the perpendiculars from $C$, mark $B$ and $B'$ such that $CB=CB'=a$, and mark $B''$ on $B'C$ such that $B'B''=d$ (from $B'$ towards $C$, because $AB''=c=AC$, the condition for constructibility is that $B''$ falls within the segment $B'C$, i.e., $B'B''=d < B'C=a$, in which case there is one solution.

István Markács (Pécs, Mining Industry Secondary School II. Class)

II. solution: The segment $a+b-c=2(s-c)$ in any triangle gives twice the distance from the point of tangency of the incircle with sides $CB$ and $CA$ to the vertex $C$. Based on this, for a given $\angle BCA = \gamma$, after placing the segment $CB=a$ and $\gamma$, the kite $CA'OB'$ and the incircle can be constructed, and finally, the intersection of the second tangents from $B$ and $C$ gives $A$. In our case, $\gamma=90^\circ$, the kite specializes to a square, and $d$ gives the diameter of the incircle.

The second tangent from $B$ intersects the line $CB'$ on the same side of $BC$ as $B'$ (at a single point) if and only if $d=2\rho < a$.

Ottó Szekeres (Pécs, Mining Industry Secondary School I. Class)

III. solution: According to the Pythagorean theorem:

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

Here, from $a+b-c=d$, we have $c-b=a-d$, so the only unknown $c+b=e$ can be constructed as a line segment using a geometric mean construction method, and subsequently, the missing sides can also be constructed:

![](https://cdn.mathpix.com/cropped/2024_05_02_285385c7c4e2168d3924g-1.jpg?height=464&width=845&top_left_y=1622&top_left_x=629)

Figure 2

The initial steps of Figure 2 are the same as those of Figure 1; the perpendicular to $BB''$ at $B$ intersects the point $D$ on $B'C$ such that $B''D=(c-b)+(c+b)=2c$, and finally, we get $A$ at the midpoint of $B''D$, because $AC=AB''-CB''=c-(c-b)=b$ (or $AC=CD-AD=c+b-c=b$).

András Török (Nagykőrös, J. Arany Secondary School I. Class)

IV. solution: Eliminate $c$ from the equation $a+b-d=c$ using the Pythagorean theorem: $a+b-d=\sqrt{a^2+b^2}$. From here, squaring and rearranging gives the expression for the side $b$:

$$
b=\frac{d(2a-d)}{2(a-d)}
$$

which can be constructed as a fourth proportional.