Solution
This problem is a classic application of the “Brachistochrone” principle (shortest time) or the “Circle of Equal Descent”.
To find the path of shortest time for a particle sliding under gravity from a point A to a straight line (the inclined plane), we consider the “Circle of Equal Descent”.
The locus of all points reachable from A in a minimum time $t$ is a circle passing through A with its top at A.
For the time to be minimized to reach the plane, this circle must be tangent to the inclined plane at the target point B.
Let the angle of the inclined plane with the horizontal be $\theta$.
The normal to the inclined plane (the radius $R$ to the point of tangency B) makes an angle $\theta$ with the vertical.
The geometry of the tangent circle dictates that the chord $AB$ (the thread) must bisect the angle between the vertical and the normal to the plane at point B.
- Angle between Vertical and Normal (Radius $R$) = $\theta$
- Therefore, the angle the thread $AB$ makes with the vertical is $\alpha = \theta / 2$.
We solve the triangle formed by the vertical line from A, the inclined plane, and the thread $AB$.
From the geometric derivation using the circle properties and trigonometric identities for this specific configuration, the length of the chord is given by:
$$ L = h \frac{\cos \theta}{\cos(\theta/2)} $$Using the secant identity $\sec x = 1/\cos x$, we can rewrite this as:
$$ L = h \cos \theta \sec(0.5\theta) $$