Q. circle geometry (1 Viewer)

[marsenal]

I've got two questions,

(1) An n-sided regular polygon is inscribed in a circle. Let the ratio of the perimeter of the polygon to the circumference of the circle be p:1, and let the ratio of the area of the polygon to the area of the circle be a:1.
Find expressions of p and a as functions of n.


(2) KL is a fixed chord of length a on a circle centre O, and the point P varies on the major arc KL. Let y be the sum of the lengths of PK and PL.
Also angle KPL= b, angle KLP= c
Prove that y is a maximum when c= 0.5(pi - b), then find and simplify the maximum value.

For the second one, is y maximum when it's an isosceles triangle?

 

[spice girl]

Originally posted by marsenal
I've got two questions,

(1) An n-sided regular polygon is inscribed in a circle. Let the ratio of the perimeter of the polygon to the circumference of the circle be p:1, and let the ratio of the area of the polygon to the area of the circle be a:1.
Find expressions of p and a as functions of n.


(2) KL is a fixed chord of length a on a circle centre O, and the point P varies on the major arc KL. Let y be the sum of the lengths of PK and PL.
Also angle KPL= b, angle KLP= c
Prove that y is a maximum when c= 0.5(pi - b), then find and simplify the maximum value.

For the second one, is y maximum when it's an isosceles triangle?

Q1:
if you draw the nth fraction of the circle out:
length of chord = 2r * sin(pi/n)
length of arc = r * 2pi/n
thus p = {nsin(pi/n)} / pi

area of triangle = (1/2)r^2 * sin(2pi/n)
area of sector = r^2 * (pi/n)

so a = {nsin(2pi/n)} / 2pi

Q2: Using sine rule
PK/ sinc = KL / sinb
PK = a*sinc / sinb

note that angle PKL = pi - b - c
PL / sin(pi - b - c) = PL / sin(b + c) = KL / sinb
PL = asin(b+c)/sinb

PK + PL = (a/sinb)(sinc + sin[b+c]) = y

Take a, b, as constants (because they are), and differentiate y w.r.t c
dy/dc = (a/sinb)(cosc + cos(b+c))

when cosc + cos(b+c) = 0
(c) + (b+c) = pi
2c = pi - b
c = 0.5(pi - b)

max value y = 2asinc/sinb

 

[marsenal]

Re: Re: Q. circle geometry

I don't quite follow aspects of your solution...

Originally posted by spice girl


length of chord = 2r * sin(pi/n)

What do you use to get the above?

Originally posted by spice girl

area of triangle = (1/2)r^2 * sin(2pi/n)
area of sector = r^2 * (pi/2n)

so a = {2nsin(2pi/n)} / pi

For some reason I keep getting a= {nsin(2pi/n}/pi


With the second I think I understand it, what is the purpose of taking the derivative?
Otherwise thank you very much for your help once again.

 

[ezzy85]

what book/set is this from? and what topic is this? it seems to be more than just circle geometry...

 

[spice girl]

Re: Re: Re: Q. circle geometry

Originally posted by marsenal
I don't quite follow aspects of your solution...



What do you use to get the above?



For some reason I keep getting a= {nsin(2pi/n}/pi


With the second I think I understand it, what is the purpose of taking the derivative?
Otherwise thank you very much for your help once again.

ahh my mistake, i've edited my post now...

with the second, it's a max-min problem, so you differentiate and find when dy/dc = 0 to find turning pt. it's clearly the max there.

 

[marsenal]

Oh yeah of course it's max/min. It must be the lateness that I ain't thinking straight.

Ezzy, the questions are from Cambridge 3U Year 12, circle geo chapter, both extensions. First one 9A and second one 9C.