MathDB

524

Part of 1990 All Soviet Union Mathematical Olympiad

Problems(1)

ASU 524 All Soviet Union MO 1990 <XDY = \pi /2n, in a 2n-gon, angle bisector

Source:

8/14/2019
A,B,CA, B, C are adjacent vertices of a regular 2n2n-gon and DD is the vertex opposite to BB (so that BDBD passes through the center of the 2n2n-gon). XX is a point on the side ABAB and YY is a point on the side BCBC so that XDY=π2nXDY = \frac{\pi}{2n}. Show that DYDY bisects ∠XYC\angle XYC.
geometryregular polygonanglesangle bisector