## Subject IV. (30 points)

Around a point $O$, $n$ angles are constructed, the first having a measure of $x^{0}$, the second $(2 x)^{0}$, the third $(3 x)^{0}$, and so on, the $n$-th having a measure of $120^{0} .\left(n, x \in \mathbf{N}^{*}\right)$

a) Determine the number of angles constructed;

b) Calculate the measure of the penultimate angle constructed.

Prof. Ioan Balica, Lic. in Informatics Tiberiu Popoviciu Cluj-Napoca

All questions are mandatory. 10 points are awarded by default.

Time allowed for work - 2 hours.

## Grading Scale for 6th Grade (OLM 2013 - local stage)

Default: $10 \mathrm{p}$
## Subject IV.

a) The last angle constructed has a measure of $(n x)^{0}$, so $n x=120$.(1)

(5 points) The sum of the measures of the constructed angles is $360^{\circ}$, thus

$$
x+2 x+3 x+\ldots+n x=360 \Rightarrow \frac{n(n+1)}{2} \cdot x=360 \Rightarrow n(n+1) \cdot x=720
$$

(10 points)

Dividing relation (2) by relation (1) yields $n+1=6 \Rightarrow n=5$.

(5 points)
b) $n x=120 \Rightarrow 5 x=120 \Rightarrow x=24$

In total, there are 5 angles, the penultimate one will have a measure of $(4 x)^{0}=96^{\circ}$.

(10 points)