olymmath

ILL 1969 Problem 26

    (GBR3)(GBR 3) A smooth solid consists of a right circular cylinder of height hh and base-radius rr, surmounted by a hemisphere of radius rr and center O.O. The solid stands on a horizontal table. One end of a string is attached to a point on the base. The string is stretched (initially being kept in the vertical plane) over the highest point of the solid and held down at the point PP on the hemisphere such that OPOP makes an angle α\alpha with the horizontal. Show that if α\alpha is small enough, the string will slacken if slightly displaced and no longer remain in a vertical plane. If then pulled tight through PP, show that it will cross the common circular section of the hemisphere and cylinder at a point QQ such that ∠SOQ=ϕ\angle SOQ = \phi, SS being where it initially crossed this section, and sin⁡ϕ=rtan⁡αh\sin \phi = \frac{r \tan \alpha}{h}.

    Dataset text AoPS wiki problems. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator ill/1969/p26 · Tag 4GQT4Z1RF0 · Edit this page on GitHub · Keys: j next, k previous