IB SL Math Type 1

Circles Math Portfolio

Sharat Ramamani

Ms. Alsop

Due: June 4, 2012

MPM-3UW

TABLE OF CONTENTS

Introduction ___________________________________________________________ #

Method _____________________________________________________________ #

Case 1 ______________________________________________________________ #

Case 2 ______________________________________________________________ #

Validity of General Statement _____________________________________________ #

Conclusion __________________________________________________________ #

Introduction

Method

To find chord , a variety of circular and trigonometric properties shall be applied, to arrive

at the general statement representing chord .

Figure 2: C2, with ∆AGO as a circumscribed right angle triangle within the circle, ∆AP’O as

an isosceles triangle within ∆AGO, and ∆AO’O and ∆AP’O’ as right angle triangles within ∆AP’O,

generated by Wolfram Mathematica.

Several claims must be proven in order to derive the general statement representing chord :

Claim 1: Side OG represents the diameter of C3, or twice the length of radius OP.

The radius of C2 is stated in the problem as OP. Side OG represents 2OP, or the diameter of C2

(see Figure 2).

Equation 1:

Claim 2: Side OA represents the radius of C3.

Figure 1 depicts the intersection of the three circles. The center of C3, point A, is connected to

point O by side OA. Point O is on the circumference of C3, therefore Side OA represents the

radius of C3. The radius of C1 has the same value as r.

Equation 2:

In this investigation, lowercase r will represent the radius of C3.

Claim 3: Chord segment OO’ is half the length of chord .

Figure 3: C1, C2 and C3 with side AP’ drawn on C3

If a line segment is drawn, attaching point A to point P’, then chord becomes a part of

∆AP’O, which is within C3 (see Figure 3). Since both side AO and side AP’ extend from the center

of C3 out to any point on C3’s circumference, they have the same side length, as they both

represent the radius of C3. Due to the fact that AO and AP’ have the same side length, ∆AP’O is

an isosceles triangle. Furthermore, if a line segment AO’ is drawn from the center of C3, and

bisects ∆AP’O to a point O’ on chord , then two right angle triangles are created; ∆AO’O

and ∆AP’O’. ∠O’ on both right angle triangles contain the 90° right angle. Since ∆AP’O is

bisected, that means point O’ bisects chord , which also means that chord segment OO’ is

half the length of chord .

Equation 3: 2

Claim 4: ∆AGO is a right angle triangle

To prove this claim, Thales’ Theorem must be applied. Thales’ Theorem states that

the diameter of a circle always subtends a right angle to any point on the circle. Figure 2

contains C2, which circumscribes ∆AGO. Side OG represents the diameter of the circle, as

proven by Claim 1, and point A represents a point on the circumference on C2. Therefore,

according to Thales’ Theorem, ∆AGO is a right angle triangle, and ∠A is a right angle.

Using these four proven claims, it is possible to derive the general statement that represents

chord .

Equation 4: ( )

O’

The initial equation of cos(o) =

is used to define cos(o), as

and will be

substituted into other equations later on. Now that cos(o) has been defined, it is possible to

work with the triangles inside ∆AGO. It should be noted that r cannot be equal to or greater

than 2OP, or else ∠O will not exist. If r is equal to 2OP, then ∠O is 0°, which cannot exist with

the geometric figure given, so it is a limiting case. If r is greater than 2OP, then ∠O does not

exist, as there is no intersection point between C1 and C2, where the normal intersection

point is Point A, which is non-existent. Figure 4 and 5* compare ∠O when 2OP>r and 2OP<r.

Furthermore, r cannot be equal to 0. Mathematically speaking, r and 2OP can be negative

numbers, if both are concurrently negative, but negative side lengths do not exist, thus they are

extraneous. Therefore {xƐR||0< r< 2OP}.

*Created using the geometrical program Geogebra

Figure 4: Case when OG (or 2OP) > r Figure 5: Case when OG (or 2OP) < r

Equation 5: ( )

( ( ))( )

(

) ( )

In these set of equations, ∆AO’O is being analyzed, in Figure 3. Chord segment OO’ is

being solved for, by using the trigonometric ratio of cosine. Cos(o) is then substituted with

,

as it is defined like such in Equation 4, and thus resulting in chord segment OO’ =

Equation 6: (

)

Equation 6 combines the proven claims and simplifies the other equations to produce

the general statement for chord , that is,

.

Case 1

In the first case, r=1 , and OP’ must be solved for when OP = 2, 3 and 4. The general

statement that models OP’ can be used, but can also be modified to better fit the case. Instead

of

,substitute r for 1, as it represents a constant in this case.

Equation 7:

when r=1 and OP=1

when r=1 and OP=2

when r=1 and OP=3

when r=1 and OP=4

In case 1, is the inverse of OP, due to the fact that r stays at a constant 1. There is no

maximum value of OP, but the minimum value of OP is

. This is because when OP becomes

larger , chord becomes smaller; but when OP <

,then C3 which contains the point P’

ceases to exist, meaning that chord will not exist. Figure 5 models this limiting case, where

C3 does not exist.

Figure 6: Graph of OP’ vs. OP when r=1

Figure 6 is a representation of values of OP ranging from 1-10, and the consequent values of

OP’ when r=1. It can be used to justify the modified general statement of when r=1, that is,

. This graph represents an inverse relationship between and OP. There is a clear

horizontal asymptote at y=0, depicting how as the value of OP gets larger, the value of

decreases but never reaches 0. The graph also shows the minimum value of OP and the

maximum value of OP’, which is 0.5 and 2 respectively. The domain and range for Figure 6 is

{xƐR||x>0.5} and {yƐR||2>y>0}, where x and y represent the value of OP and the value of ,

respectively

Case 2

In the second case, OP is set as a constant, where OP = 2, and OP’ must be solved for

when r=2, 3 and 4. Another modification to the general statement of

must be made

to model this case better. OP is substituted for 2, as it is a constant. Thus the modified general

statement to fit Case 2 is

.

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10

Val

ue

of

OP

'

Value of OP

Length of OP' vs. Length of OP when r=1

y=0

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4

Val

ue

of

OP

'

Value of r

Length of OP' vs. Length of r when OP=2

Equation 8:

when r=1 and OP=2

when r=2 and OP=2

when r=3 and OP=2

when r=4 and OP =2

In Case 2, is larger than r for every value of r except for when r=1. The same limitation of r

< 2OP is evident in Case 2, as C3 becomes non-existent when r>2OP. If C3 is non-existent, then

is also non-existent, because point P’ lies on C3.

Figure 7: Graph of OP’ vs. r when OP=2

Figure 7 depicts the relationship between r and when OP = 2. The maximum value

of r is 2OP because point A ceases to exist when the limit of 2OP is surpassed; therefore C3 also

ceases to exist. Since OP is held at a constant 2, 2OP=4, thus the maximum value of r in case 2 is

4. A minimum value of r does exist. The value of r must be greater than 0 due to the fact that a

radius of 0 does not produce a value of ,and also C1 and C3 are not formed. The minimum

and maximum values of are related to that of the minimum and maximum values of r. The

minimum value of is 0, and the maximum value of is half the square of the maximum

value of r, that is,

= 8.

Validity of General Statement

To test the validity of the general statement, different values of r and OP will be put into

the general statement, to produce a value of ; and the same values of r and OP will be

applied to C1, C2, and C3 on Geogebry graphing software, to virtually compare the lengths of OP’.

Example 1: OP=4, r=3

Figure 8 depicts C1, C2, and C3, with

the radius of C1 and C3 being 3, and

the radius (OP) of C3 being 4. The

general statement correctly

determined the value of ,as

validated by the measurement tool

on Geogebra. Both the general

statement and the software

determined the length of to be

2.25 when r=3 and OP=4.

Figure 8: C1, C2, and C3 of when OP=4 and r=3

Figure 9: C1, C2, and C3

and OP’ of when OP=2

and r=1.5

Example 2: OP=2, r=1.5

Conclusion

Chord is represented by the general statement

,where r is the radius of C1 and

C3, OP is the radius of C2, and is a chord on C3 and on the radius of C2. The general

statement was derived by utilizing basic trigonometric and chord properties, and also using

Thames’ Theorem. The general statement was validated by inputting different values of r and

OP into the general formula to generate a value for ; and graphing the three circles along

with the same values of r and OP, and comparing the values of outputted by both

mechanisms. The resulting outcome was that both the software and the general statement

came up with the same values for , meaning that the general statement was proven to be

valid. Chord however, only exists when 0<r<2OP, because otherwise all three circles would

not be present, and chord would not exist.

𝑂𝑃 𝑟

𝑂𝑃

Yet again the general

statement produced a

value of OP’ identical

to the value measured

by Geogebra, using the

same r and OP values

in both mechanisms. It

is reasonable to say

that the general

statement

is valid.