## Task B-1.4.

Angle $\varangle C B A$ of triangle $A B C$ is twice as large as angle $\varangle B A C$. If $|B C|:|A B|=4: 5$ and $|A C|=18$, determine the perimeter and area of triangle $A B C$.
## Solution.

Let $\varangle B A C=\alpha$. Then $\varangle C B A=2 \alpha$. Let $D$ be the point where the angle bisector of $\varangle C B A$ intersects the side $\overline{A B}$. With the labels as in the figure, we have $\varangle B D C=2 \alpha$ (external angle of triangle $A B D$), $a: c=4: 5, b=18, |B D|=x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-03.jpg?height=722&width=1002&top_left_y=1352&top_left_x=550)

According to the Angle Bisector Theorem in a triangle, we have

$$
\frac{b-x}{x}=\frac{a}{c}=\frac{4}{5}
$$

so we get

$$
\begin{aligned}
4 x & =5(18-x) \\
9 x & =90 \\
x & =10 \\
\Longrightarrow b-x & =8 .
\end{aligned}
$$

Triangles $A B C$ and $B D C$ are similar by the AA similarity criterion, so we have

$$
\frac{b-x}{a}=\frac{a}{b}=\frac{x}{c}
$$

from which we get $a^{2}=18 \cdot 8$, or $a=12$. From the given conditions (or similarity), we get $c=15$.

The perimeter of triangle $A B C$ is $12+15+18=45$. The area can be calculated using Heron's formula:

$$
\begin{aligned}
P & =\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{45}{2}\left(\frac{45}{2}-12\right)\left(\frac{45}{2}-15\right)\left(\frac{45}{2}-18\right)} \\
& =\frac{1}{4} \sqrt{45 \cdot 21 \cdot 15 \cdot 9}=\frac{135}{4} \sqrt{7} .
\end{aligned}
$$

Note: The length $x$ can also be calculated using only the similarity of the triangles mentioned above, instead of using the Angle Bisector Theorem.

The area of triangle $A B C$ can also be calculated as follows.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-04.jpg?height=725&width=1008&top_left_y=1074&top_left_x=544)

In the isosceles triangle $A D B$, the height to the base $\overline{A B}$ is $v_{D}=\sqrt{x^{2}-\left(\frac{c}{2}\right)^{2}}=\frac{5 \sqrt{7}}{2}$. Then, from the right triangle $A N D$, we get $\sin \alpha=\frac{\frac{5 \sqrt{7}}{2}}{10}=\frac{\sqrt{7}}{4}$, so the desired area is:

$$
P=\frac{b \cdot v_{B}}{2}=\frac{b \cdot c \sin \alpha}{2}=\frac{18 \cdot 15 \cdot \sqrt{7}}{8}=\frac{135 \sqrt{7}}{4}
$$