olymmath

HMMT 2008 F8

  • General 2
  • Answer

A piece of paper is folded in half. A second fold is made such that the angle marked below has measure ϕ(0∘<ϕ<90∘)\phi\left(0^{\circ}<\phi<90^{\circ}\right), and a cut is made as shown below.

When the piece of paper is unfolded, the resulting hole is a polygon. Let OO be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at OO, and also that ∠XOY=144∘\angle X O Y=144^{\circ}, where XX and YY are the the vertices of the hole adjacent to OO. Find the value(s) of ϕ\phi (in degrees).

Dataset text NuminaMath olympiads. Quoted with source.

Show the answer

81^{\circ}

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 hmmt/2008/f8 · Tag 49YACZ5CGT · Edit this page on GitHub · Keys: j next, k previous