XXXI Brazilian Math Olympiad

The solution is based on the following Theorem. Let $P=(x_0,y_0,0)$ be a point in the xy-plane and $(a,b,1)$ a vector perpendicular to the vector $v=(x_0,y_0,0)$, that is, such that $a{x}_0+b{y}_0=0$. Then the set of points obtained by rotating the line $P+t\cdot v$, $t\in \mathbb{R}$, around the $z$-axis is the hyperboloid $$x^2+y^2-(a^2+b^2)z^2=x_0^2+y_0^2.$$ Moreover, each point of the hyperboloid belong to exactly one of the rotated lines. Proof. A generic point in the line $\ell :P+t\cdot v$ is $Q=(x_0+at,y_0+bt,t)$, $t\in \mathbb{R}$. Since $$\begin{gathered} (x_0+at{)}^2+(y_0+bt{)}^2-(a^2+b^2)t^2 \\ =x_0^2+y_0^2+2(a{x}_0+b{y}_0)t+a^2{t}^2+b^2{t}^2-(a^2+b^2)t^2=x_0^2+y_0^2, \end{gathered}$$ $Q$ belongs to the aforementioned hyperboloid. Since rotating the line around the $z$-axis does not change the $z$-coordinate of $Q$ and the section of the hyperboloid in the plane $z=t$ is a circle, the image of $Q$ obtained by rotating $\ell$ belongs to the hyperboloid as well. Conversely, if $(x_1,y_1,z_1)$ belongs to the hyperboloid, then it belongs to the plane $z=z_1$, whose section in the hyperboloid has equation $x^2+y^2=(a^2+b^2)z_1^2+x_0^2+y_0^2=(a^2+b^2)z_1^2+x_0^2+y_0^2+2a{x}_0{z}_1+2b{y}_0{z}_1=(x_0+a{z}_1{)}^2+(y_0+b{z}_1{)}^2$. So there exists a rotation $\sigma$ around the $z$-axis that maps $(x_1,y_1,z_1)$ to the point $Q=(x_0+a{z}_1,y_0+b{z}_1,z_1)$ from $\ell$. So $(x_1,y_1,z_1)$ can be obtained by applying the inverse rotation ${\sigma }^{-1}$ to $\ell$. Note that $\sigma$ is unique, so only one of the rotated lines passes through $Q$. Now we can solve the problem. In this case, since the hyperboloid has equation $3x^2+3y^2-z^2-1=0⟺x^2+y^2-\frac{1}{3}{z}^2=\frac{1}{3}$ we can use $(x_0,y_0,0)=(\frac{\sqrt{3}}{3},0,0)$ and $(a,b,1)=(0,\frac{\sqrt{3}}{3},1)$.

a. Since $(-a,-b,1)=(0,-\frac{\sqrt{3}}{3},1)$, which is not parallel to $(a,b,1)$, is also perpendicular to $(x_0,y_0,0)$, the hyperboloid can be obtained rotating either the line ${\ell }_1:(\frac{\sqrt{3}}{3},0,0)+t(0,\frac{\sqrt{3}}{3},1)$ or ${\ell }_2:(\frac{\sqrt{3}}{3},0,0)+t(0,-\frac{\sqrt{3}}{3},1)$. So each point from $H$ belongs to at least two lines. We will prove that there are no other lines by showing that $H$ does not contain a line that is neither parallel to $(0,\frac{\sqrt{3}}{3},1)$ nor $(0,-\frac{\sqrt{3}}{3},1)$. Suppose it does. Since the sections of $H$ by horizontal planes are circles, this new line is not parallel to the $xy$-plane. So it contains a point in this plane, which we can suppose, without loss of generality, that is $(\frac{\sqrt{3}}{3},0,0)$. So suppose the line $r:(\frac{\sqrt{3}}{3},0,0)+t(c,d,1)$ is contained in $H$. So, for all $t\in \mathbb{R}$, $$\begin{array}{rl}& 3{\left(\frac{\sqrt{3}}{3}+tc\right)}^2+3\cdot (td{)}^2-t^2-1=0\\ \iff & (3c^2+3d^2-1)t^2+\frac{2\sqrt{3}}{3}tc=0\\ \iff & \{\begin{array}{l}\frac{2\sqrt{3}}{3}c=0\\ 3c^2+3d^2-1=0\end{array}\iff \{\begin{array}{l}c=0\\ d=\pm \frac{\sqrt{3}}{3}\end{array}\end{array}$$ So the support vector of $r$ should be $(0,\frac{\sqrt{3}}{3},1)$ or $(0,-\frac{\sqrt{3}}{3},1)$, and we are done.

b. We need to compute the angle that the vectors $(0,\pm \frac{\sqrt{3}}{3},1)$ define with the $xy$-plane: $\arctan \frac{1}{\sqrt{3}/3}=60^{\circ }$.