Školjka

%V0 Construct a right triangle with given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

%V0 In tetrahedron $ABCD$, vertex $D$ is connected with $D_0$, the centrod if $\triangle ABC$. Line parallel to $DD_0$ are drawn through $A,B$ and $C$. These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_2, B_1,$ and $C_1$, respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$. Is the result if point $D_o$ is selected anywhere within $\triangle ABC$?

%V0 In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: $$(AB+BC+CA)^2\le6(AD^2+BD^2+CD^2).$$ When do we have equality?

%V0 All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$. a.) If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length; b.) If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.

%V0 Prove the following assertion: The four altitudes of a tetrahedron $ABCD$ intersect in a point if and only if $$AB^2 + CD^2 = BC^2 + AD^2 = CA^2 + BD^2.$$

%V0 Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.