B3. The angle bisector of angle at $A$ and the altitude and median from vertex $A$ of triangle $A B C$ divide the angle at $A$ into 4 equal parts. Determine the sizes of the angles of triangle $A B C$.

## Problems for 4th grade

Time for solving: 120 minutes. In section A, we will score a correct answer with two points, and deduct one point for an incorrect answer. Write the answers of section A in the left table, leave the right table empty.
![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-14.jpg?height=166&width=802&top_left_y=651&top_left_x=641)
B3. $A$

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-24.jpg?height=383&width=806&top_left_y=611&top_left_x=288)

Let $D$ be the midpoint of side $BC$, $E$ the foot of the altitude from $A$, and $S$ the intersection of the angle bisector at $A$ with side $BC$. Since the angle bisector divides the angle into two equal parts, the angle bisector must lie between the altitude and the median. Therefore, point $S$ lies between points $E$ and $D$. We have two possibilities: either $D$ lies between $B$ and $S$ and $E$ lies between $S$ and $C$, or $E$ lies between $B$ and $S$ and $D$ lies between $S$ and $C$. Due to symmetry, we can consider only the first case.

Let the angle $\angle BAC$ be $4\alpha$, where $0 < \alpha < \frac{\pi}{4}$. Then we have

$$
|BE| = |AE| \tan 3\alpha, \quad |DE| = |AE| \tan 2\alpha \quad \text{and} \quad |CE| = |AE| \tan \alpha
$$

From this, we derive

$$
|BD| = |BE| - |DE| = |AE| (\tan 3\alpha - \tan 2\alpha)
$$

and

$$
|CD| = |CE| + |DE| = |AE| (\tan \alpha + \tan 2\alpha)
$$

Since $D$ is the midpoint of side $BC$, it follows that $\tan 3\alpha - \tan 2\alpha = \tan \alpha + \tan 2\alpha$ or

$$
\tan 3\alpha - \tan \alpha - 2 \tan 2\alpha = 0
$$

The left side can be rearranged as

$$
\begin{aligned}
\tan 3\alpha - \tan \alpha & - 2 \tan 2\alpha = \frac{\sin 3\alpha}{\cos 3\alpha} - \frac{\sin \alpha}{\cos \alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \\
& = \frac{\sin 3\alpha \cos \alpha - \sin \alpha \cos 3\alpha}{\cos \alpha \cos 3\alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{\sin 2\alpha}{\cos \alpha \cos 3\alpha} - 2 \frac{\sin 2\alpha}{\cos 2\alpha} = \\
& = \frac{\sin 2\alpha (\cos 2\alpha - 2 \cos \alpha \cos 3\alpha)}{\cos \alpha \cos 2\alpha \cos 3\alpha} = -\frac{\sin 2\alpha \cos 4\alpha}{\cos \alpha \cos 2\alpha \cos 3\alpha}
\end{aligned}
$$

where in the fourth equation we used the addition formula for sine

$$
\sin 2\alpha = \sin (3\alpha - \alpha) = \sin 3\alpha \cos \alpha - \sin \alpha \cos 3\alpha
$$

and in the last equation the formula for the product of cosines

$$
2 \cos \alpha \cos 3\alpha = \cos (3\alpha + \alpha) + \cos (3\alpha - \alpha) = \cos 4\alpha + \cos 2\alpha
$$

From the derived equation, it follows that $\sin 2\alpha = 0$ or $\cos 4\alpha = 0$, so $2\alpha = k\pi$ or $4\alpha = \frac{\pi}{2} + k\pi$, where $k \in \mathbb{Z}$. Since $0 < \alpha < \frac{\pi}{4}$, the only possibility is $\alpha = \frac{\pi}{8}$. Therefore, $\angle BAC = 4\alpha = \frac{\pi}{2}$, $\angle ACB = \frac{\pi}{2} - \alpha = \frac{3\pi}{8}$, and $\angle CBA = \frac{\pi}{8}$.

Considering both symmetric cases, we have determined that the angles of triangle $ABC$ are either $\angle BAC = \frac{\pi}{2}$, $\angle ACB = \frac{3\pi}{8}$, and $\angle CBA = \frac{\pi}{8}$, or $\angle BAC = \frac{\pi}{2}$, $\angle ACB = \frac{\pi}{8}$, and $\angle CBA = \frac{3\pi}{8}$.

A clearly drawn and labeled diagram .1 point

Lengths of segments $|BE|, |DE|$, and $|CE|$ expressed in terms of $\alpha$ and $|AE|$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-25.jpg?height=54&width=1570&top_left_y=710&top_left_x=277)

Rearrangement of the equation $\mathbf{v} - \frac{\sin 2\alpha \cos 4\alpha}{\cos \alpha \cos 2\alpha \cos 3\alpha}$ 2 points

Written solution with justification that only one possibility is valid 1 point

## Solutions for 4th Grade

In section A, a correct answer will be awarded 2 points, while an incorrect answer will result in the deduction of 1 point. To avoid potential negative final scores, each contestant is given 3 initial points.

| 1 | 2 | 3 |
| :--- | :--- | :--- |
| $\mathrm{D}$ | $\mathrm{C}$ | $\mathrm{C}$ |

Justifications: