The sides of a certain triangle form an arithmetic progression with a common difference of $d$. The area of the triangle is $T$. What are the sides and angles of the triangle? The same problem needs to be solved if $d=1$ and $T=6$.

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First solution.

The sides of the triangle are $a, a+d, a+2d$; the area of $\Delta$ is

$$
T=\sqrt{s(s-a)(s-b)(s-c)}
$$

where

$$
\begin{gathered}
s=\frac{a+b+c}{2}, \quad s=\frac{3}{2}(a+d) \\
{[s-(a+d)]=\frac{a+d}{2}, \quad[s-(a+2 d)]=\frac{a-d}{2}}
\end{gathered}
$$

Thus, applying the formula,

$$
\begin{gathered}
T=\sqrt{\frac{3}{2}(a+d)\left(\frac{a+d}{2}\right)\left(\frac{a-d}{2}\right)\left(\frac{a+3 d}{2}\right)} \\
T=\frac{a+d}{4} \sqrt{2(a-d)(a+3 d) .}
\end{gathered}
$$

Let us assume that $(a+d)=x$, then

$$
\begin{gathered}
T=\frac{x}{4} \sqrt{3(x-2 d)(x+2 d)} \\
4 T=x \sqrt{3\left(x^{2}-4 d^{2}\right)} \\
16 T^{2}=3 x^{4}-12 x^{2} d^{2}
\end{gathered}
$$

from which

$$
\begin{aligned}
& x= \pm \sqrt{\frac{+12 d^{2} \pm \sqrt{144 d^{4}+192 T^{2}}}{6}} \\
& x= \pm \sqrt{\frac{6 d^{2} \pm \sqrt{36 d^{4}+48 T^{2}}}{3}}=a+d \\
& a=-d \pm \sqrt{\frac{6 d^{2} \pm \sqrt{36 d^{4}+48 T^{2}}}{3}}
\end{aligned}
$$

Substituting the given values $d=1$ and $T=6$ into this equation, we get

$$
\begin{gathered}
a=-1 \pm \sqrt{\frac{6 \pm \sqrt{36+1728}}{3}} \\
a=-1 \pm \sqrt{\frac{6 \pm \sqrt{1764}}{3}} \\
a=-1 \pm \sqrt{\frac{6 \pm 42}{3}} \\
a=-1 \pm 4 \\
a=3 ; \quad a+d=4 ; \quad a+2 d=5
\end{gathered}
$$

Thus, the triangle is a right-angled triangle with a hypotenuse of 5.

For the angles,

$$
\cos \frac{\alpha}{2}=\sqrt{\frac{s(s-a)}{b c}}
$$

Since it is a right-angled triangle, we can use the formula

$$
\sin \alpha=\frac{a}{c}
$$

$$
\begin{gathered}
\sin \alpha=\frac{3}{5} \\
\log \sin \alpha=\log 3-\log 5 \\
\log 3=0,47712 \\
\log 5=\underline{0,69897}
\end{gathered}
$$

$$
\begin{gathered}
\log \sin \alpha=9,77815-10 \\
\alpha=36^{\circ} 52^{\prime} 10,71^{\prime \prime} ; \beta=90-\alpha=53^{\circ} 7^{\prime} 49,29^{\prime \prime}
\end{gathered}
$$

Mihály Seidner's solution, awarded the 1st b. Eötvös prize at the 1st math. and phys. society competition.

## Second solution.

Let one side of the triangle be $=b$, then $a=b-d$ and $c=b+d$, and

$$
s=\frac{a+b+c}{2}=\frac{3 b}{2} ; \quad s-a=\frac{b}{2}+d, \quad s-b=\frac{b}{2}, \quad s-c=\frac{b}{2}-d
$$

Since

$$
t=\sqrt{s(s-a)(s-b)(s-c)}
$$

then

$$
\frac{3 b^{2}}{4}\left(\frac{b}{2}+d\right)\left(\frac{b}{2}-d\right)=t^{2}
$$

and

$$
\frac{3 b^{2}}{4}\left(\frac{b^{2}}{4}-d^{2}\right)=t^{2}
$$

and

$$
b^{4}-4 b^{2} d^{2}=\frac{16 t^{2}}{3}
$$

and

$$
b=\sqrt{2 d^{2}+\sqrt{4 d^{4}+\frac{16 t^{2}}{3}}}=\sqrt{2\left(d^{2}+\sqrt{d^{2}+\frac{4 t^{2}}{3}}\right)}
$$

We use the + sign in front of the square roots because $b$ cannot be negative or imaginary. Based on the last formula,

$$
a=b-d=-d+\sqrt{2\left(d^{2}+\sqrt{d^{4}+\frac{4 t^{2}}{3}}\right)}
$$

and

$$
c=b+d=d+\sqrt{2\left(d^{2}+\sqrt{d^{4}+\frac{4 t^{2}}{3}}\right)}
$$

The unknown angles can be determined most conveniently using the formulas $t=\frac{b c}{2} \sin \alpha$ and $t=\frac{a c}{2} \sin \beta$ and $\gamma=180^{\circ}-\alpha-\beta$, from which

$$
\sin \beta=\frac{2 t}{a c}, \quad \sin \alpha=\frac{2 t}{b c}
$$

If $d=1$ and $t=6$, then

$$
\begin{gathered}
b=\sqrt{2\left(d^{2}+\sqrt{d^{4}+\frac{4 t^{2}}{3}}\right)}=\sqrt{2\left(1+\sqrt{1+\frac{4 \cdot 36}{3}}\right)}= \\
=\sqrt{2(1+\sqrt{49)}}=4
\end{gathered}
$$

$$
a=b-d=3, \quad c=b+d=5 \text {. Thus } a=3, \quad b=4, \quad c=5 \text {. }
$$

For the angles:

$$
\begin{gathered}
\sin \alpha=\frac{2 t}{b c}=\frac{2 \cdot 6}{20}=\frac{3}{5} \\
\log 3=0,477121 \\
\frac{-\log 5=0,698970}{\log \sin \alpha=9,778151-10}
\end{gathered}
$$

$$
\begin{gathered}
\frac{14}{32}: 2,81= \\
\alpha=36^{\circ} 52^{\prime} 12^{\prime \prime} \\
\sin \beta=\frac{2 t}{b c}=\frac{4}{5} \\
\log 4=0,602060
\end{gathered}
$$

$$
\begin{gathered}
\frac{\log 5=0,698970}{\log \sin \beta=9,903090-10} \\
\frac{19}{76}: 1,58=48 \\
\beta=53^{\circ} 7^{\prime} 48^{\prime \prime}
\end{gathered}
$$

Since $\alpha+\beta=90^{\circ}$, then $\gamma=180^{\circ}-(\alpha+\beta)=90^{\circ}$.

Pál Pap's solution, awarded the 2nd b. Eötvös prize at the 1st math. and phys. society competition.

The problem was also solved by: Alex Ernö, fg. VIII. Budapest; Hary Berczely, fg. VIII. Budapest; Domokos Berzenczey, fr. VII. Déva; Béla Bolemann, fg. VIII. Budapest; János Breznyik, lyc. VIII. Selmeczbánya; Arrmin Fried, fr. VII. Déva; Bernát Friedmann, fg. VI. S. A.-Ujhely; Ödön Gere, fg. VIII. Budapest; György Jankovich, fg. VIII. Losoncz; Gergely Jorga, Gilád; Jenő Kiss, fr. VIII. Budapest; Dezső Lauber, fr. VII. Pécs; Elemér Meitner, fr. VIII. Budapest; Gábor Schulhof, fr. VIII. Pécs; Ferencz Segesváry, fg. VIII. Budapest; Hugó Stern, fr. VII. Debrecen; Aladár Visnya, fr. VII. Pécs.