All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Red Area (Posted on 2014-02-16) Difficulty: 3 of 5
Let ABCD be a square of side length 3. P is a point on the plane such that each of angle APB, BPC, CPD and DPA is at least 60. If each possible position of P is painted red, find the area of the red region.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution After playing with Geometers' Sketchpad Comment 1 of 1

The red area is the overlap of four circles, each of which is circumscribed about an equilateral triangle internal to the square with its base as one side of the square.

This area looks like a "puffed up square" as the intersections of the four circles form a square with side length equal to 3/sqrt(3)=sqrt(3).  The radius of curvature for the arcs defining the area is also sqrt(3) and each of the four arcs subtends 60.

The central square therefore has area 3, to which we have to add the areas of the four segments of circles with the edges of the square as base (chord).

The area of a full circle would be pi*3, but 60 contains only 1/6 of that or pi/2 square units.  The triangle that needs to be subtracted to get the segment area has base sqrt(3) and height 3/2 for an area of 3*sqrt(3)/4. Therefore the area of each circle segment is pi/2 - 3*sqrt(3)/4. Four of these then add to 2*pi - 3*sqrt(3).

To that add the area 3 of the central square and get 3 + 2*pi - 3*sqrt(3) ~= 4.08703288447294.

  Posted by Charlie on 2014-02-16 12:30:22
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information