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Romanian Mathematical Olympiad

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Romania

We consider the cube $ABCD EFGH$ with side $a$ cm, $a > 0$ and the points $K$ , $L$ on the segments $AC$ , respectively $EG$ such that $CK = EL = \frac{AC}{4}$ . Let $M$ be the midpoint of $AE$ and the points $N$ , $P$ on the segment $HF$ such that
$HN = FP = \frac{(\sqrt{2} - 1)a}{2} \text{ cm.}$
Knowing that lines $MC$ and $LK$ intersect in $Q$ , prove that $MNPQ$ is a regular tetrahedron.

Solution

Triangles $\Delta MEP$ and $\Delta MEN$ are congruent, therefore $MN = MP$ .
Denote by $O_1$ the midpoint of $HF$ . It follows that $MO_1 = \frac{AC}{2} = \frac{a\sqrt{3}}{2}$ and, since $NP = a$ , the triangle $MNP$ must be equilateral.
Let $O$ be the midpoint of $AC$ , $M_1$ the midpoint of $AO$ and $\{L_1\} = LK \cap MO_1$ . Since $MM_1 \parallel EO \parallel LK \parallel CO_1$ , we have $ML_1/MO_1 = M_1K/M_1C = 2/3$ so $L_1$ is the centroid of the triangle $MNP$ .

Notice that $AL = AG = \frac{3\sqrt{2a}}{4}$ thus $ALGK$ is a rhombus, which implies $AG \perp KL$ . We also have $CO_1LK$ parallelogram, therefore $LK \parallel CO_1$ and $KL \perp MO_1$ . $HF \perp (ACGE)$ so $KL \perp HF$ which implies $KL \perp (MNP)$ . It follows that the pyramid $MNPQ$ is regular.
From the similarity of the triangles $\Delta ML_1Q$ and $\Delta MO_1C$ we have $QL_1 = \frac{a\sqrt{6}}{3}$ forcing the tetrahedron $MNPQ$ to be regular.

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