Can a right angle be divided into two acute angles, $\alpha$ and $\beta$, using a compass and a straightedge such that

$$
\sin \alpha - \sin \beta = \frac{\sqrt{2}}{2}
$$

holds?
I. solution: From the equation, we calculate $\sin \alpha$. Since $\alpha$ and $\beta$ are complementary angles,

$$
\sin \alpha - \cos \alpha = \frac{\sqrt{2}}{2}
$$

which means

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

Squaring and rearranging,

$$
2 \sin^2 \alpha - \sqrt{2} \sin \alpha - \frac{1}{2} = 0
$$

from which

$$
\sin \alpha = \frac{\sqrt{2} + \sqrt{6}}{4}
$$

(The other, negative root does not satisfy the problem, as $\sin \alpha > 0$.)

$\sqrt{2}$ and $\sqrt{6}$ can be constructed with a compass and straightedge. If we construct a right triangle where the hypotenuse is 4 and one leg is $\sqrt{2} + \sqrt{6}$, then the angle opposite the leg $\sqrt{2} + \sqrt{6}$ is $\alpha$, and the angle adjacent to it is $\beta$.

Heinemann Zoltán (Pécs, Bányaip. t. IV. o. t.)

II. solution: Since $\beta = 90^\circ - \alpha$, we have $\sin \beta = \cos \alpha$. Therefore, we need to find $\alpha$ such that

$$
\sin \alpha - \cos \alpha = \frac{\sqrt{2}}{2}
$$

Squaring,

$$
\sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha = \frac{1}{2}
$$

from which

$$
2 \sin \alpha \cos \alpha = \sin 2\alpha = \frac{1}{2}
$$

This equation is satisfied by two angles (less than $180^\circ$):

$$
2 \alpha_1 = 150^\circ, \quad \text{and} \quad 2 \alpha_2 = 30^\circ
$$

which means

$$
\alpha_1 = 75^\circ, \quad \text{and} \quad [\alpha_2 = 15^\circ]
$$

By substitution, we can verify that the latter value does not satisfy our equation (a false root introduced by squaring).

Thus, the solution to the problem is

$$
\alpha = 75^\circ
$$

which can be expressed as $45^\circ + 30^\circ$, and this can be constructed with a compass and straightedge.

Szatmáry Zoltán (Bp. VIII., Piarista g. IV. o. t.)

III. solution: Take two perpendicular lines with a unit arc, and divide the right angle into two parts: $\alpha$ and $\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_beb7f5349eabcb7fb6a2g-1.jpg?height=634&width=593&top_left_y=2074&top_left_x=755)

From the diagram, the values of $\sin \alpha$ and $\sin \beta$ can be read directly: $OB = \sin \alpha$, $AB = \sin \beta$. If we subtract $\sin \beta$ (i.e., $BA$) from $BO$, the resulting difference is

$$
OC = \sin \alpha - \sin \beta
$$

If we want $\sin \alpha - \sin \beta = \frac{\sqrt{2}}{2}$, we construct the value $\frac{\sqrt{2}}{2}$, measure it along one leg of the right angle to a length $OC (<1)$. The ray starting from $C$ and forming a $45^\circ$ angle with $CB$ intersects the unit circle at point $A$. The segment $OA$ provides the desired angles $\alpha$ and $\beta$.

Remark: It is clear that this construction method is not only applicable in this case but also always when $\alpha$ and $\beta$ need to be constructed such that their sum is $90^\circ$, and the difference of their sines is any number between 0 and 1 that can be constructed with a compass and straightedge. All that is needed is to construct the segment corresponding to the number and measure it as $OC$.

Frivaldszky Sándor (Bp. II., Rákóczi g. IV. o. t.)