Inscribed Polygons
Imagine a circle with a radius of one unit. Inside the circle, inscribe the largest possible equilateral triangle. In the triangle, inscribe a circle. In the circle goes a square ... in the square a circle...in the circle a regular pentagon ... in the pentagon a circle ... etc. Each consecutive polygon has one more side ... alternating with circles. The circles get smaller and smaller.
Can you make a rough guess as to how small the circle will eventually become?
Can you make a rough guess as to how small the circle will eventually become?
Labels: geometrick
posted by Zaux at
8:09 PM
6 Comments:
Inscribed Polygons
Inscribe polygons on the circle have their corners touch the outer circle
the inner circle touches the sides of each polygon
Call the outer circle Radius Ro
So in the case of the triangle the distance from the center to each corner is R
The distance from the centre to the centre of each edge is R_3=Ro*cos(60)
R3 is the radius of the circle inside the triangle
So in the case of the square the distance from the center to each corner is R3
The distance from the centre to the centre of each edge is R[4]=R[3]*cos(45)
R4 is the radius of the circle inside the square
So in the case of the pentagon the distance from the center to each corner is R4
The distance from the centre to the centre of each edge is R[5]=R[4]*cos(36)
R5 is the radius of the circle inside the pentagon
So in the case of the n-sided shape the distance from the center to each corner is Rn-1
The distance from the centre to the centre of each edge is R[n]=R[n-1]*cos(180/n)
e.g. R[175]=R[174]*cos(180/175)
Rn is the radius of the circle inside the pentagon
R[n]=Ro*cos(180/3)*cos(180/4)*cos(180/5)*......*cos(180/(n-1))*cos(180/n)
as n-> inf (cos 180/n)=1
Not entirely sure about how to evaluate the series as n-> inf
P=cos(180/3)*cos(180/4)*cos(180/5)*......*cos(180/(n-1))*cos(180/n)
So I ended up using some brute force techniques (program that kept evaluating series until the change in P was sufficiently small). It appears to approach 0.114974
So R as n->inf = Ro*P=1*0.114974
Answer:
R=0.114974
Hopefully no mistakes above. Not exactly happy with evaluating the series using brute force. I may try to evaluate it for an exact solution at a later date.
Cam
Wow ... right again Cam ... what have I go to do to find a puzzle which stumps you? Nice work man!
Thanks Zaux,
Just as a side note:
I reevaluated R with some more iterations and it appears to be a bit closer to 0.114942
Cam
Hi cam ... got a problem for you.
Prove the Riemann Hypothesis! Maybe that will take a little longer! ... heh heh.
Best I've found is:
0.11494204485329620070104015746959874283079533720086
As far as I can determine, there is no exact expression for the infinite product. Mathematica didn't offer one, and the internet only came up with probable noes.
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