Above we see three regular polygons, the equilateral triangle, square and regular pentagon. Each has a side length of 2 units. Associated with any regular polygon are two circles: its incircle, which touches its sides, and its circumcircle, which passes through its vertices.
Surprisingly, the area between these two circles, shaded pink in the figures above, always has an area of π/4 square units, however many sides the regular polygon has. How do we know this?
Here we see two consecutive vertices of the polygon, painy=ts A and B, with M being the midpoint of AB and O the common centre of the incentre and the circumcentre.
The side length is 1, so MB is ½, and because AB is tangent to the incentre and OM a radius of the circle angle OMB is a right angle. Let the radiiusof the circumcentre be R and that of the incircle r, so OB = R and OM = r.
Then using Pythagoras in the irght-angled triangle OMB we have
OM2 + (½)2 = OM2, so r2 + ¼ = R2, or R2 - r2 = ¼.
But now considering the ares of the two circles we have
Area of circumcircle - Area incircle = πR2 - πr2 = π/4.
Notice we haven't used the number of sides anywhere in this! So the area is always π/4 irrespective of the number of sides, as claimed.
Coincidence or what?
but no proof was given there, so here is mine..
Above we see three regular polygons, the equilateral triangle, square and regular pentagon. Each has a side length of 2 units. Associated with any regular polygon are two circles: its incircle, which touches its sides, and its circumcircle, which passes through its vertices.
Surprisingly, the area between these two circles, shaded pink in the figures above, always has an area of π/4 square units, however many sides the regular polygon has. How do we know this?
Here we see two consecutive vertices of the polygon, painy=ts A and B, with M being the midpoint of AB and O the common centre of the incentre and the circumcentre.
The side length is 1, so MB is ½, and because AB is tangent to the incentre and OM a radius of the circle angle OMB is a right angle. Let the radiiusof the circumcentre be R and that of the incircle r, so OB = R and OM = r.
Then using Pythagoras in the irght-angled triangle OMB we have
OM2 + (½)2 = OM2, so r2 + ¼ = R2, or R2 - r2 = ¼.
But now considering the ares of the two circles we have
Area of circumcircle - Area incircle = πR2 - πr2 = π/4.
Notice we haven't used the number of sides anywhere in this! So the area is always π/4 irrespective of the number of sides, as claimed.