Problem:
Regular octagon A1βA2βA3βA4βA5βA6βA7βA8β is inscribed in a circle of area 1. Point P lies inside the circle so that the region bounded by PA1ββ,PA2ββ, and the minor arc A1βA2ββ of the circle has area 71β, while the region bounded by PA3ββ,PA4ββ, and the minor arc A3βA4ββ of the circle has area 91β. There is a positive integer n such that the area of the region bounded by PA6ββ,PA7ββ, and the minor arc A6βA7ββ is equal to 81ββn2ββ. Find n.
Solution:
The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, 1, and assume the side length of the octagon is 2. Let r denote the radius of the circle, O be the center of the circle. Then r2=12+(2β+1)2=4+22β. Now, we need to find the "D" shape, the small area enclosed by one side of the octagon and 1/8 of the circumference of the circle:
which gives PU=(71ββ81β)Ο(4+22β)+2β+1 and PV=(91ββ81β)Ο(4+22β)+2β+1. Now, let A1βA2β intersects A3βA4β at X,A1βA2β intersects A6βA7β at Y,A6βA7β intersects A3βA4β at Z. Clearly, β³XYZ is an isosceles right triangle, with right angle at X and the height with regard to which shall be 3+22β. Now 2βPUβ+2βPVβ+PW=3+22β which gives PW=3+22ββ2βPUββ2βPVβ