You have 8 sticks. Three of them can form the sides of a right triangle with area 50.

But you can use all eight to form a cyclic (inscribable within a circle) octagon all of whose angles are equal. What would be the area of such an octagon?

first crack at a puzzle in liek what, 3 years?

seems that the puzzle constrains the sticks to be of exactly 2 sizes, each represented equally.  Otherwise, the octagon would wither not be cyclic, or not be equi-angular.  This forces the sticks to be 4x10 and 4x(10root2), which uses two 10's and one 10rrot2 to make the described triangle.

then the area of the octagon can be broken into 2 columns (10 wide, unknown height), 1 square (side 10) counted twice where the columns overlap, and 4 small triangles.  As it happens, due to symmetry, these are right triangles and convieiently enough, are the original size triangles from the puzzle description.  This makes calculation of the height of the column, the last unknown, relatively easy.

Columns are 10x30, area 300, area 100
Square is 10x10
Triangtles areas are each 50, as per problem statement

Breaks down to 300+300-100+50+50+50+50, or 700.

Posted by Cory Taylor on 2009-04-13 16:55:28