There is a “folk theorem” that a four-footed table can always be rotated into a stable position on an uneven floor. Prove the following mathematical formulation of this theorem.
Define four points in \(\mathbb R^2\), depending on an angle \(\theta\), by \[ P_1(\theta)=(\cos\theta,\sin\theta),\quad P_2(\theta)=(-\sin\theta,\cos\theta), \] \[ P_3(\theta)=(-\cos\theta,-\sin\theta),\quad P_4(\theta)=(\sin\theta,-\cos\theta). \] Show that given any continuous function \(h:\mathbb R^2\to\mathbb R\), there exists a value of \(\theta\) such that the four points \[ Q_i(\theta)=(P_i(\theta),h(P_i(\theta))) \] on the graph of \(h\) are co-planar in \(\mathbb R^3\).
Solution
Set \[ \widetilde h(\theta) \coloneqq h(\cos\theta,\sin\theta) \] and \[ g(\theta) \coloneqq \widetilde h(\theta)-\widetilde h(\theta+\pi/2) +\widetilde h(\theta+\pi)-\widetilde h(\theta+3\pi/2). \]
1. The function \(g\) is continuous and satisfies\[ g(\theta+\pi/2)=-g(\theta) \]for every \(\theta\).
Continuity follows from continuity of \(h\) and the trigonometric parametrization of the unit circle. Also \(\widetilde h\) is \(2\pi\)-periodic, so \[ \begin{aligned} g(\theta+\pi/2) &= \widetilde h(\theta+\pi/2)-\widetilde h(\theta+\pi) +\widetilde h(\theta+3\pi/2)-\widetilde h(\theta+2\pi)\\ &=-g(\theta). \end{aligned} \]
2. There exists \(\theta_0\in\RR\) such that \(g(\theta_0)=0\).
Fix any \(\theta\). If \(g(\theta)=0\), take \(\theta_0=\theta\). Otherwise, step 1 shows that \(g(\theta)\) and \(g(\theta+\pi/2)\) have opposite signs. The intermediate value theorem therefore gives a zero of \(g\) on the interval between these two angles.
3. For the angle \(\theta_0\) from step 2, the line segments\(Q_1(\theta_0)Q_3(\theta_0)\) and\(Q_2(\theta_0)Q_4(\theta_0)\) have a common midpoint.
Write \(P_i=P_i(\theta_0)\) and \(Q_i=Q_i(\theta_0)\). From the definitions, \[ P_3=-P_1, \qquad P_4=-P_2. \] Thus the midpoint of \(Q_1Q_3\) is \[ \left(0,0, \frac{h(P_1)+h(P_3)}{2} \right), \] while the midpoint of \(Q_2Q_4\) is \[ \left(0,0, \frac{h(P_2)+h(P_4)}{2} \right). \] The equality \(g(\theta_0)=0\) is exactly \[ h(P_1)+h(P_3)=h(P_2)+h(P_4), \] so these two midpoints coincide.
4. The four points \(Q_1(\theta_0),\ldots,Q_4(\theta_0)\) are coplanar.
By step 3, the line through \(Q_1,Q_3\) intersects the line through \(Q_2,Q_4\). Two intersecting lines lie in a common plane, and that plane contains all four endpoints. If the two lines happen to coincide, the four points are collinear and hence are also coplanar.
5. Q.E.D.
Step 4 gives the required angle and coplanarity conclusion.