Solve the triangle if one angle $C$, the opposite side $c$, and the difference of the squares of the other two sides $p^{2}=b^{2}-a^{2}$ are given. For example, $C=82^{\circ} 49^{\prime} 9.2^{\prime \prime}, \quad c=75 \mathrm{~cm}, \quad p^{2}=1406.25 \mathrm{~cm}^{2}$.
1. Solution. Since $b=c \frac{\sin B}{\sin C}$ and $a=c \frac{\sin A}{\sin C}$, we can write:

$$
p^{2}=c^{2}\left(\frac{\sin ^{2} B}{\sin ^{2} C}-\frac{\sin ^{2} A}{\sin ^{2} C}\right)
$$

from which

$$
\begin{gathered}
\frac{p^{2} \sin ^{2} C}{c^{2}}=\sin ^{2} B-\sin ^{2} A \\
=(\sin B+\sin A)(\sin B-\sin A) \\
=2 \sin \frac{B+A}{2} \cos \frac{B-A}{2} \cdot 2 \sin \frac{B-A}{2} \cos \frac{B+A}{2} \\
=\sin (B+A) \sin (B-A)
\end{gathered}
$$

but

$$
=\sin (B+A)=\sin C
$$

thus

$$
\sin (B-A)=\frac{p^{2} \sin C}{c^{2}}
$$

Substituting the given values, we get $B+A=97^{\circ} 10^{\prime} 50.8^{\prime \prime}$ and $B-A=14^{\circ} 21^{\prime} 42.7^{\prime \prime}$, so $B=55^{\circ} 46^{\prime} 16.8^{\prime \prime}$, $A=41^{\circ} 24^{\prime} 34^{\prime \prime}, a=50 \mathrm{~cm}, b=62.5 \mathrm{~cm}$.

(Kántor Nándor.)

2. Solution. The problem can be solved by construction as follows:

Let the height of the triangle be $C D=m$, then $b^{2}=m^{2}+\overline{A D}^{2}$ and $a^{2}=m^{2}+\overline{B D}^{2}$; subtracting the second equation from the first: $p^{2}=b^{2}-a^{2}=\overline{A D}^{2}-\overline{B D}^{2}$. The perpendicular to $A B$ at $D$ is thus one geometric locus of the sought $C$ vertex. To determine the point $D$, we consider that $\overline{A D}^{2}-\overline{B D}^{2}=p^{2}$ and $A D+B D=c$. From these two equations, we get $A D=\frac{c^{2}+p^{2}}{2 c}$ and $B D=\frac{c^{2}-p^{2}}{2 c} ;$ at a distance of $A B=c$, we mark $A D=\frac{c^{2}+p^{2}}{2 c}$ and draw a perpendicular at the thus determined point $D$. The second geometric locus of the $C$ vertex is obtained by drawing a circular arc over the side $A B$ such that the inscribed angles (corresponding to the chord $A B$) are equal to the given angle $C$. From point $E$ we draw a perpendicular to $A B$ at its midpoint $O$, and at any point on this perpendicular, we draw the angle $C$; we draw a line parallel to one side of the thus obtained angle $C$ through point $A$, which intersects the perpendicular at $O$ - the other side of the angle $C$ - at point $P$. $P$ is the center of the circular arc that is the second geometric locus of the $C$ vertex.

(Grünhut Béla, Pécs.)

The problem was also solved by: Feuer Mór, Freund Antal, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Hofbauer Ervin, Riesz Frigyes, Schiffer Hugó, Szabó István, Szabó Károly, Thieringer Aurél.