Problem 1345

Let $P$ be an arbitrary point on the intersection line of planes $S$ and $T$, and let $E, F$ be points in space such that $PE$ is parallel to $e$, $PF$ is parallel to $f$, and the lengths of segments $PE$ and $PF$ are both unit length. We will show that for the angle $\omega = EPF$,

$$|\cos \omega| \leq 1 / 3$$

holds, which implies the statement of the problem.

!

Let the projections of points $E, F$ onto plane $S$ be denoted by $E_0$ and $F_0$, respectively. According to the problem, $E_0P$ is perpendicular to $F_0P$. First, applying the cosine rule and then the Pythagorean theorem, we get:

$$EF^2 = 2 - 2 \cos \omega = E_0F_0^2 + (E_0E - F_0F)^2 = (PE_0^2 + E_0E^2) + (PF_0^2 + F_0F^2) - 2E_0E \cdot F_0F = 2 - 2E_0E \cdot F_0F,$$

where the distances $E_0E$ and $F_0F$ are understood with their signs: they have the same sign if they are in the same direction. From the two equations, it follows that

$$\cos \omega = E_0E \cdot F_0F$$

Introduce a coordinate system in space such that its origin is at $P$, the $y$-axis lies on the intersection line of $S$ and $T$, the $x$-axis is in $S$, the $z$-axis is in $T$, and denote the coordinates of $E, F$ by $(e_1; e_2; e_3)$ and $(f_1; f_2; f_3)$, respectively. Then

$$\sum_{i=1}^{3} e_i^2 = \sum_{i=1}^{3} f_i^2 = 1$$

and applying (2) without change and then by swapping the roles of $S$ and $T$, we get

$$e_3 f_3 = e_1 f_1 = \cos \omega$$

In the $(x; y)$ plane, the coordinates of points $E_0, F_0$ are $(e_1; e_2)$ and $(f_1; f_2)$, respectively. Applying the Pythagorean theorem in the right triangle $E_0F_0P$, we get:

$$(e_1 - f_1)^2 + (e_2 - f_2)^2 = (e_1^2 + e_2^2) + (f_1^2 + f_2^2)$$

or

$$e_1 f_1 + e_2 f_2 = 0$$

From (4) and (5), it is clear that $\cos \omega$ is also equal to $-e_2 f_2$, so (1) follows from the inequality

$$|e_1 f_1 - e_2 f_2 + e_3 f_3| \leq 1$$

since in our case, the three quantities on the left side are equal.

Thus, only the proof of (6) remains. Let $g_1 = e_1$, $g_2 = -e_2$, $g_3 = e_3$, then (6) follows from the well-known

$$\left(\sum_{i=1}^{3} f_i g_i\right)^2 \leq \sum_{i=1}^{3} f_i^2 \sum_{i=1}^{3} g_i^2$$

inequality based on (3). The inequality (7) can be shown most simply by noting that the quadratic polynomial

$$\sum_{i=1}^{3} (f_i - t g_i)^2 = \sum_{i=1}^{3} f_i^2 - 2t \sum_{i=1}^{3} f_i g_i + t^2 \sum_{i=1}^{3} g_i^2$$

is non-negative for all real $t$, so its discriminant cannot be positive.

Remark. If we had not applied the Pythagorean theorem at the beginning of our solution, we would have obtained that

$$\cos \omega = \sum_{i=1}^{3} e_i f_i$$

It is customary to call the quantity on the right-hand side the dot product of the vectors $(e_1; e_2; e_3)$ and $(f_1; f_2; f_3)$. The name suggests that although we are multiplying vectors, the result is a number, a scalar quantity. It is clearly visible here the dual nature of the concept of multiplication: there are times when the product of two quantities is of the same type as the multiplicands, and there are times when the product is a single real number that expresses the relationship between the quantities, regardless of their type. In the latter case, if the product is 0, it is customary to say that the quantities are perpendicular, since in two and three dimensions, we call vectors perpendicular if they are geometrically perpendicular.