## Task 2 - 130822

Construct a right-angled triangle $ABC$ with the right angle at $C$ from $\rho=2.5 \text{ cm}$ and $\alpha=50^{\circ}$! Here, $\rho$ is the inradius and $\alpha$ is the measure of the angle $BAC$.

Describe and justify your construction! Determine whether the given pieces uniquely determine a triangle!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0661.jpg?height=385&width=646&top_left_y=2029&top_left_x=248)

(I) Suppose $\triangle ABC$ is a triangle that meets the conditions of the problem, $M$ is the intersection of its angle bisectors, i.e., the center of its incircle, and $E, F$ are the feet of the perpendiculars from $M$ to the sides $BC, CD$.

Then the quadrilateral $CFME$ has right angles at $E, C$, and $F$ and is therefore a square with side length $\rho$.

The bisector of the angle $BAC$ passes through $M$; it holds that $\angle FMA = 90^{\circ} - \frac{\alpha}{2}$. Point $A$ lies first on the ray from $C$ through $F$ and second on the free leg of an angle of size $90^{\circ} - \frac{\alpha}{2}$ drawn at $M$ on $MF$ on the side of the line through $M$ and $E$, on which $C$ does not lie.

Point $B$ lies first on the ray from $C$ through $E$ and second on the free leg of an angle of size $\alpha$ drawn at $A$ on $AC$ on the side of the line through $A$ and $C$, on which $E$ lies.

(II) It follows from this that a triangle $ABC$ meets the conditions of the problem only if it can be obtained by the following construction:

1. We construct the square $CFME$ with side length $\rho$.
2. We draw the ray from $C$ through $F$.
3. We draw an angle of size $90^{\circ} - \frac{\alpha}{2}$ at $M$ on $MF$ on the side of the line through $M$ and $F$, on which $C$ does not lie. If its free leg intersects the ray from $C$ through $F$, the intersection point is called $A$.
4. We draw an angle of size $\alpha$ at $A$ on $AC$ on the side of the line through $A$ and $C$, on which $E$ lies.
5. We draw the ray from $C$ through $E$. If it intersects the free leg of the angle constructed in (4), the intersection point is called $B$.

(III) Proof that every triangle $ABC$ constructed in this way meets the conditions of the problem: According to the construction, the angle at $C$ is a right angle. Similarly, the angle $BAC$ has the size $\alpha$ according to the construction. $M$ has the distance $\rho$ from $AC$ and $BC$ according to the construction. Since furthermore, according to the construction,

$$
\angle CAM = \angle FAM = \frac{\alpha}{2} \quad \text{and} \quad \angle MAB = \angle CAB - \angle CAM = \frac{\alpha}{2},
$$

$AB$ is also a tangent to the circle with radius $\rho$ around $M$, just like $AC$. Therefore, $M$ is the center of the incircle of the triangle $ABC$.

(IV) Construction step (1) is uniquely constructible up to congruence. Similarly, construction step (2) is uniquely constructible. Furthermore, because $0^{\circ} < 90^{\circ} - \frac{\alpha}{2} < 90^{\circ}$, (3) is also uniquely constructible. After that, (4) and because $0^{\circ} < \alpha < 90^{\circ}$, (5) is also uniquely constructible.

Therefore, the triangle $ABC$ is uniquely determined by the given pieces up to congruence.

# Task 3 - 130823

Determine all rational numbers $r$ with the following property:

If you subtract $r$ from the numerator of the fraction $\frac{3}{4}$ and add $r$ to its denominator, you get a fraction that is half as large as $\frac{3}{4}$.
Assume there is a rational number $r$ that satisfies the conditions of the problem. Then we have:

$$
\frac{3-r}{4+r}=\frac{3}{8}
$$

This implies $24-8r=12+3r$. Therefore, $r=\frac{12}{11}$ can at most satisfy the conditions of the problem. Indeed,

$$
\frac{3-\frac{12}{11}}{4+\frac{12}{11}}=\frac{\frac{21}{11}}{\frac{56}{11}}=\frac{3}{8}
$$

which means $r=\frac{12}{11}$ satisfies the conditions.

#### Abstract