Editing Hilbert's fourth problem (section)

===Crofton formula===
{{main|Crofton formula}}
Consider a set of all oriented lines on a plane. Each line is defined by the parameters <math>\rho</math> and <math>\varphi,</math> where <math>\rho</math> is a distance from the origin  to the  line, and <math>\varphi</math> is an angle between the  line and the ''x''-axis. Then the set of all oriented lines is homeomorphic to a circular cylinder of radius 1 with the area element <math>dS = d\rho \, d\varphi </math>. Let <math>\gamma</math> be a rectifiable curve on a plane. Then the length of <math>\gamma</math> is
<math display="block">L = \frac{1}{4} \iint_\Omega n( \rho, \varphi) \, dp \, d\varphi</math>
where <math>\Omega</math> is a set of lines that intersect the curve <math>\gamma</math>, and <math>n(p, \varphi)</math> is the number of intersections of the line with <math>\gamma</math>.
Crofton proved this statement in 1870.<ref>{{cite book
| last1=Santaló | first1=Luís A. | authorlink1=Luis Santaló
| chapter=Integral geometry
| title=Studies in Global Geometry and Analysis
| editor-last1=Chern | editor-first1=S. S.
| publisher=Mathematical Association of America, Washington, D. C.
| pages=147–195
| date=1967}}</ref>

A similar statement holds for a projective space.