B3. Let trapezoid $A B C D$ be such that the leg $B C$ is equal in length to the base $A B$, and the sizes of the angles $\angle C B D, \angle D B A$, and $\angle A D B$ are in the ratio 1:3:5 in this order. Calculate the sizes of the interior angles of trapezoid $A B C D$.

## Problems for 4th Grade

Time for solving: 180 minutes. Each problem in set A has exactly one correct answer. In set A, we will score a correct answer with two points, and deduct one point for an incorrect answer. Write the answers for set A in the left table, leave the right table blank. The committee will consider only the answers written in the table for set A.
![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-14.jpg?height=168&width=802&top_left_y=684&top_left_x=640)
B3.

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-27.jpg?height=737&width=879&top_left_y=1205&top_left_x=240)[^1]

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-28.jpg?height=491&width=736&top_left_y=226&top_left_x=677)

By the sine rule for triangle $A C D$, we have $\frac{\sin 8 x}{\sin y}=\frac{|A C|}{|A D|}$, and by the sine rule for triangle $A B C$, we have $\frac{\sin y}{\sin 4 x}=\frac{|B C|}{|A C|}$. Multiplying both equations, we get $\frac{\sin 8 x}{\sin 4 x}=\frac{|B C|}{|A D|}$. Now we apply the sine rule to triangle $A B D$ and get $\frac{\sin 5 x}{\sin 3 x}=\frac{|A B|}{|A D|}=\frac{|B C|}{|A D|}$. From both derived equations, it follows that $\frac{\sin 5 x}{\sin 3 x}=\frac{\sin 8 x}{\sin 4 x}$. Using the double angle formula, we get $\frac{\sin 5 x}{\sin 3 x}=\frac{2 \sin 4 x \cos 4 x}{\sin 4 x}=2 \cos 4 x$, and thus $2 \sin 3 x \cos 4 x=\sin 5 x$. We factorize the left side as $\sin 7 x + \sin (-x) = \sin 5 x$, and considering the oddness of the sine function, we rewrite the equation as $\sin 7 x + \sin (-5 x) = \sin x$. Now we factorize the left side again to get $2 \sin x \cos 6 x = \sin x$. Since $\sin x \neq 0$, we can cancel $\sin x$ from both sides and get $\cos 6 x = \frac{1}{2}$. Since $0^{\circ} < 6 x < 8 x < 180^{\circ}$, it follows that $6 x = 60^{\circ}$ or $x = 10^{\circ}$. The internal angles of the trapezoid are therefore $\beta = 4 x = 40^{\circ}$, $\gamma = 180^{\circ} - \beta = 140^{\circ}$, $\delta = 8 x = 80^{\circ}$, and $\alpha = 180^{\circ} - \delta = 100^{\circ}$.

2. method. By the sine rule for triangle $A B D$, we get $\frac{|A B|}{|B D|}=\frac{\sin 5 x}{\sin (180^{\circ} - 8 x)}=\frac{\sin 5 x}{\sin 8 x}$, and by the sine rule for triangle $B C D$, we get $\frac{|B C|}{|B D|}=\frac{\sin 3 x}{\sin (180^{\circ} - 4 x)}=\frac{\sin 3 x}{\sin 4 x}$. Since $|A B| = |B C|$, both expressions are equal, so $\frac{\sin 5 x}{\sin 8 x}=\frac{\sin 3 x}{\sin 4 x}$. From this, as in the first solution, we derive $2 \sin 3 x \cos 4 x = \sin 5 x$. We multiply this equation by $\cos 3 x$ and use the double angle formula on the left side to get $\sin 6 x \cos 4 x = \sin 5 x \cos 3 x$. We factorize both sides as $\sin 10 x + \sin 2 x = \sin 8 x + \sin 2 x$ and simplify the equation to $\sin 10 x = \sin 8 x$. Since $0^{\circ} < 8 x < 180^{\circ}$, the angles $8 x$ and $10 x$ are different and less than $360^{\circ}$, so it follows that $10 x = 180^{\circ} - 8 x$ or $x = 10^{\circ}$. From this, as in the first solution, we calculate the internal angles of the trapezoid as $\beta = 40^{\circ}$, $\gamma = 140^{\circ}$, $\delta = 80^{\circ}$, and $\alpha = 100^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-28.jpg?height=60&width=1742&top_left_y=1775&top_left_x=174)

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-28.jpg?height=60&width=1685&top_left_y=1872&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-28.jpg?height=80&width=1687&top_left_y=1962&top_left_x=230)

The derived equation $\sin (5 x) = 2 \sin (3 x) \cos (4 x) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Correct factorization and knowledge of the oddness of the sine function .......................... 1 point

The derived equation $2 \sin (x) \cos (6 x) = \sin (x) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Justified why $x = 10^{\circ}$ and the derived angles of the quadrilateral .......................................

## 65th Mathematical Competition for High School Students in Slovenia

## National Competition, May 15, 2021 Solutions to Problems for 4th Year

| A1 | A2 | A3 |
| :---: | :---: | :---: |
| B | E | A |