What relationship exists between the sides of a triangle if $\beta-\alpha=3 \gamma$?
I. Solution: According to the condition,

$$
\beta=\alpha+3 \gamma=180^{\circ}-\beta-\gamma+3 \gamma=180^{\circ}-\beta+2 \gamma
$$

from which

$$
\beta=90^{\circ}+\gamma
$$

and thus

$$
\alpha=180^{\circ}-\beta-\gamma=90^{\circ}-2 \gamma
$$

According to (1), the angle at $B$ in the triangle is obtuse. Let's complete our triangle by drawing a perpendicular from $A$ to side $b$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_223caba0941674cf7fdeg-1.jpg?height=374&width=655&top_left_y=729&top_left_x=716)

Let this line intersect side $a$ at $D$. The size of $\angle DAB$ is $2 \gamma$ according to (2). If we draw the altitude $AE$ in $\triangle ADB$, the angle $\angle DAE$, being a right angle, is also $\gamma$. Therefore, $\triangle ADB$ is isosceles because its altitude bisects the $2 \gamma$ angle at $A$. Thus, $AD = AB = c$.

From the Pythagorean theorem,

$$
DC = \sqrt{b^2 + c^2}
$$

From the diagram,

$$
a = EC - EB = EC - ED
$$

The distances $EC$ and $ED$ can be calculated using the relationship between the legs and their projections in a right triangle:

$$
EC = \frac{b^2}{\sqrt{2 + c^2}}, \quad ED = \frac{c^2}{\sqrt{b^2 + c^2}}
$$

Substituting these into (3):

$$
a = \frac{b^2 - c^2}{\sqrt{b^2 + c^2}}
$$

This establishes a relationship among the sides of the triangle.

Elbert Árpád (Kaposvár, Közgazd. techn. III. o. t.)

II. Solution: In the previous solution, we proved that $\beta = 90^{\circ} + \gamma$.

![](https://cdn.mathpix.com/cropped/2024_05_02_223caba0941674cf7fdeg-1.jpg?height=342&width=585&top_left_y=2095&top_left_x=759)

If we draw a perpendicular from $B$ to side $a$, the angle at $B$ in the resulting $\triangle ADB$ is $\gamma$ (see Figure 2 in I.), and because two angles are equal:

$$
\triangle ABC \sim \triangle ADB
$$

Therefore (using the notation from Figure 1),

$$
c : AD = b : c \quad \text{and} \quad c : BD = b : a
$$

from which

$$
AD = \frac{c^2}{b} \quad \text{and} \quad BD = \frac{ac}{b}
$$

Applying the Pythagorean theorem to the right triangle $\triangle CBD$,

$$
(b - AD)^2 = a^2 + BD^2,
$$

substituting the values of $AD$ and $BD$ and multiplying through by $b^2$:

$$
\left(b^2 - c^2\right)^2 = a^2 b^2 + a^2 c^2 = a^2 \left(b^2 + c^2\right)
$$

This establishes a relationship among the sides of the triangle. (It can be seen that we obtained the same relationship as in the first solution.)

Endrödy Tamás (Bp. III., Árpád g. II. o. t.)