In triangle $ABC$, with the usual notations, $\gamma=3\alpha, a=27, c=48$. Determine the length of side $b$.
I. solution. Let $D$ be the point on side $AB$ such that $\angle ACD = \alpha$. Then $\angle CDB$ is the exterior angle of $\triangle ADC$, so $\angle CDB = 2\alpha$. The triangles $\triangle ADC$ and $\triangle CDB$ are isosceles, with $CB = BD = 27$, and thus $AD = AB - DB = 21 = DC$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_b67c115f9b3c8cab4c9dg-1.jpg?height=344&width=542&top_left_y=305&top_left_x=778)

Applying the cosine rule in $\triangle CDB$, $\cos 2\alpha$ can be expressed as:

$$
\cos 2\alpha = \frac{DC^2 + DB^2 - CB^2}{2 \cdot DC \cdot DB} = \frac{21^2 + 27^2 - 27^2}{2 \cdot 21 \cdot 27} = \frac{7}{18}
$$

Writing the cosine rule in $\triangle ADC$ and using that $\cos(180^\circ - 2\alpha) = -\cos 2\alpha$:

$$
\begin{aligned}
b^2 = AC^2 & = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos(180^\circ - 2\alpha) = \\
& = 2 \cdot 21^2 + 2 \cdot 21 \cdot 21 \cdot \cos 2\alpha = 2 \cdot 21^2 \left(1 + \frac{7}{18}\right) = \frac{2 \cdot 21^2 \cdot 5^2}{2 \cdot 9} = 7^2 \cdot 5^2
\end{aligned}
$$

From which the length of side $b$ is 35 units.

Remark. In the figure similar to the 1st figure, a relationship known as Stewart's theorem holds among the segments. If $D$ is any point on side $AB$, then with the labeling of the 2nd figure,

$$
c \cdot (d^2 + c_1 \cdot c_2) = a^2 \cdot c_1 + b^2 \cdot c_2
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_b67c115f9b3c8cab4c9dg-1.jpg?height=336&width=545&top_left_y=1426&top_left_x=779)

II. solution. Draw the angle trisectors of the $3\alpha$ angle at vertex $C$, intersecting side $AB$ at points $D$ and $E$. According to the previous solution, $AD = 21$, $DB = 27$, and $\angle CEB = 3\alpha$ (3rd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_b67c115f9b3c8cab4c9dg-1.jpg?height=331&width=542&top_left_y=1930&top_left_x=781)

Two pairs of similar triangles can be found in the figure:

(1) $\triangle ABC \sim \triangle CBE$, since both have an angle of $\alpha$ and an angle of $3\alpha$, and

(2) $\triangle AEC \sim \triangle CED$, since both have an angle of $\alpha$ and an angle of $2\alpha$.

From the first similarity, $\frac{AC}{CE} = \frac{BC}{BE}$, so $AC = \frac{CE \cdot BC}{BE} = \frac{27 \cdot CE}{BE}$. From the second, $\frac{CE}{AE} = \frac{DE}{CE}$, so $CE^2 = AE \cdot DE$.

Using that $AE = 21 + DE$ and $BE = 27 - DE$, we get

$$
AC = \frac{27 \cdot \sqrt{(21 + DE) \cdot DE}}{27 - DE}
$$

The length of segment $DE$ can be obtained from $\triangle BCD$ using the angle bisector theorem:

$$
\frac{DE}{EB} = \frac{DE}{27 - DE} = \frac{DC}{CB} = \frac{21}{27}, \quad \text{so} \quad DE = \frac{189}{16}
$$

Substituting into (1), $AC = 35$.

III. solution. Write the sine rule for sides $a$ and $c$: $\frac{48}{27} = \frac{\sin 3\alpha}{\sin \alpha}$. Using the addition theorem:

$$
\begin{aligned}
& \sin 3\alpha = \sin(2\alpha + \alpha) = \sin 2\alpha \cdot \cos \alpha + \cos 2\alpha \cdot \sin \alpha = \\
& \quad = 2 \sin \alpha \cdot \cos^2 \alpha + \cos^2 \alpha \cdot \sin \alpha - \sin^3 \alpha = 3 \cos^2 \alpha \cdot \sin \alpha - \sin^3 \alpha \\
& \frac{3 \cos^2 \alpha \cdot \sin \alpha - \sin^3 \alpha}{\sin \alpha} = \frac{48}{27} = \frac{16}{9}
\end{aligned}
$$

We can simplify by $\sin \alpha \neq 0$, so using the Pythagorean identity, we get the quadratic equation: $36 \cos^2 \alpha = 25$, hence $\cos \alpha = \pm \frac{5}{6}$. Due to the condition $\gamma = 3\alpha$, $\cos \alpha$ cannot be negative.

Then write the cosine rule for side $a$:

$$
a^2 = b^2 + c^2 - 2bc \cdot \cos \alpha
$$

Substituting the known values:

$$
27^2 = b^2 + 48^2 - 2b \cdot 48 \cdot \frac{5}{6}
$$

Performing the operations, we get the quadratic equation $b^2 - 80b + 1575 = 0$. From this, $b = 35$, or $b = 45$. It is easy to show that $b = 45$ is not a solution to the problem. Since $\alpha$ is the angle opposite the smaller of sides $a$ and $b$, during the construction, we get two triangles, but in the case of $b = 45$, the condition $\gamma = 3\alpha$ is not satisfied.

Remark. The discussion problem that arose at the end of the last solution can be avoided if, after determining $\cos \alpha = \frac{5}{6}$, we also use the equations:

$$
\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt{11}}{6}
$$

and

$$
\sin \beta = \sin(180^\circ - 4\alpha) = \sin 4\alpha = 2 \sin 2\alpha \cdot \cos 2\alpha = 4 \sin \alpha \cdot \cos \alpha \cdot (2 \cos^2 \alpha - 1)
$$

and apply the sine rule again:

$$
\frac{b}{27} = \frac{\sin \beta}{\sin \alpha} = 4 \cdot \cos \alpha \cdot (2 \cos^2 \alpha - 1) = 4 \cdot \frac{5}{6} \cdot \frac{14}{36}
$$

from which

$$
b = 27 \cdot 4 \cdot \frac{5}{6} \cdot \frac{14}{36} = 35
$$