At the National Assessment for VIII-th grade graduates, the students received the following subjects: see the section "Subiecte si bareme" (Subjects and scales).
Regarding the Mathematics test, the subject is in the
"ENVIII_Matematica_2024_var_07_LRO.pdf"
file, and the correction scale in the
"ENVIII_Matematica_2024_bar_07_LRO.pdf"
file. There was also, like every year, a reserve option : see "Matematică rezervă - 27 iunie 2024" section; see files "ENVIII_Matematica_2024_var_02_LRO.pdf" and "ENVIII_Matematica_2024_bar_02_LRO.pdf".
Two problems were considered more difficult. Problem III.4 and problem III.5.
In translation:
"The attached figure shows the isosceles triangle $ABC$ with $AB=AC$.
The height from the top $A$ intersects the side $BC$ at point $D$ and $AD=BC$.
The height from vertex $B$ intersects side $AC$ at point $E$. Heights $AD$
and $BE$ intersect at point $H$.
a) Show that angles $DAC$ and $EBC$ have the same measure.
b) Prove that $AH=3 \cdot HD$."
In translation:
"The attached figure shows the circle with center $O$, in which $CD$ is
the diameter. Point $B$ belongs to the circle so that lines $BO$ and $CD$
are perpendicular. Point $M$ belongs to the small arc $BC$, lines $DM$
and $BO$ intersect at point $N$, $DN=2 \cdot MN$ and $MN=4\;cm$.
a) Show that the measure of the angle $CMD$ is $90^{\circ}$.
b) Calculate the area of the triangle $DON$."
The full Subject and Scale are in the Images below.
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Solution and comments CiP
The solution to exercise III.4.a) is standard, noting that the angle $\measuredangle DAC$ is, in the right triangle $ACD$, the complement of the angle $\measuredangle C$, and the angle $\measuredangle EBC$ is, in the right triangle $BCE$, the complement of the same angle $\measuredangle C$.
In exercise
III.4.b) the key to the solution is the consideration of similar triangles $ACD$ and $BDH$.
The order of the vertices in the relation of similarity is very important: respectively equal angles are opposed to respectively proportional sides. In more detail,
first, $BD=DC=\frac{BC}{2}=\frac{AD}{2}$;
secondly $\measuredangle BDH =90^{\circ}=\measuredangle ADC$ and
$$\measuredangle HBD \underset{same\;angle}{=}\measuredangle EBC \underset{a)}{=}\measuredangle CAD.$$
Then from the last row above we have based on the "
AA" criterion that
$$\Delta \;BDH \sim \Delta \;ADC$$
hence
$$\frac{BD}{AD}=\frac{DH}{DC}=\frac{BH}{AC}. \tag{$\sigma$}$$
But $\frac{BD}{AD}=\frac{\frac{BC}{2}}{BC}=\frac{1}{2}$, so from ($\sigma$) $\frac{HD}{DC}=\frac{1}{2}$. From here $HD=\frac{DC}{2}=\frac{BC}{4}=\frac{AD}{4}$.
We get $AH=AD-HD=AD-\frac{AD}{4}=3 \cdot \frac{AD}{4}=3 \cdot HD.$
$\square$
The key to exercise III.5.b) consists in determining the radius of the circle. For this, all the solutions I have come across are based on writing segments ratios. The most common is $\frac{OD}{ND}=\frac{MD}{CD}$. In the official scale, it results from writing the cosine of angle $\measuredangle C$ in two different triangles. Or one could use, less common with our students, the similarity of the triangles $DNO$ and $DCM$.
But a simple idea can solve the problem without resorting to the most elementary knowledge of geometry. Join points C and M.
$BO$ is the median of the segment $[CD]$, so the point $N$ on it verifies $NC=ND$. Hence
$$NC=8.$$
The right triangle $CMN$ has the leg $MN$
half of the hypotenuse $CN$ therefore it is a "
30-60-90" triangle. (I have marked this on the figure with
blue lines.) We still have $\measuredangle CND=180^{\circ}-60^{\circ}=120^{\circ}$. But in the isosceles triangle $CDN$, the median $NO$ from the vertex $N$ is also a bisector, so $\measuredangle CNO =\measuredangle DON =120^{\circ}\;:\;2=60^{\circ}$ (marked with
red). It follows easily that the triangles $CNO$ and $DNO$ are of the same type. So $NO=4$ and we now have enough data to finish the problem.
$\blacksquare$
