MATH PUZZLE... CAN YOU SOLVE IT?

[kingaaron08041991]

Suppose the circumfernce of a circle of radius r is divided into n equal pieces by n points, P1, P2, ... Pn, where P1 is adjacent to Pn and P2, P2 is adjacent to P1 and P3, etc. Let li be the length of the line segment that connect Pi and Pi+1 except ln is the length of the line segment that connects Pn to P1. Let L = l1+l2+...+ln. (a.) express L as a function of n. (hint: use trig functions/radians). (b.) find the lim L n->infinity.

 

[dleeper47]

Re: MATH PUZZLE/RIDDLE... CAN YOU SOLVE IT?

Suppose the circumfernce of a circle of radius r is divided into n equal pieces by n points, P1, P2, ... Pn, where P1 is adjacent to Pn and P2, P2 is adjacent to P1 and P3, etc. Let li be the length of the line segment that connect Pi and Pi+1 except ln is the length of the line segment that connects Pn to P1. Let L = l1+l2+...+ln. (a.) express L as a function of n. (hint: use trig functions/radians). (b.) find the lim L n->infinity.

From the description, the lines drawn will form a regular polygon of n sides inscribed inside the circle. From the end of each line segment in the polygon, draw a line to the center of the circle. Then we will have n isosceles triangles with equal sides = r and the angle between those equal sides will be 2pi/n radians.

Next draw of one the triangles and draw a line from the apex of the triangle to the center of the short line segment. Then we will have two right triangles, and using what you know about triangles, you can show that the length of the short line segment must be 2r sin(pi/n).

Since there are n such sides in the regular polygon, the circumference of the n-sided regular polygon must be 2nr sin(pi/n).

Now for very small x, sin(x) is approximately x. Therefore in the limit, as n->infinity, sin(pi/n)->pi/n. In that case, the polygon circumference approaches 2nr*pi/n = 2 pi r. This answer makes sense because as n->infinity, the polygon approaches a perfect circle, and we already know that the circumference of a circle is 2 pi r.

 

[kingaaron08041991]

Next draw of one the triangles and draw a line from the apex of the triangle to the center of the short line segment. Then we will have two right triangles, and using what you know about triangles, you can show that the length of the short line segment must be 2r sin(pi/n).

How would you show this?

Now for very small x, sin(x) is approximately x. Therefore in the limit, as n->infinity, sin(pi/n)->pi/n. In that case, the polygon circumference approaches 2nr*pi/n = 2 pi r. This answer makes sense because as n->infinity, the polygon approaches a perfect circle, and we already know that the circumference of a circle is 2 pi r.

I'm not quite sure how that would look... can you please elaborate on this idea?