- Level: International Baccalaureate
- Subject: Maths
- Word count: 2974
Math IA Type 1 Circles. The aim of this task is to investigate the positions of points in intersecting circles.
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Introduction
Math IA (SL Type I)CirclesMariya Lupandina
Aim: The aim of this task is to investigate the positions of points in intersecting circles.
Introduction
This investigation will examine the length of OP’ as it changes with different values of r and OP. A general statement will be deciphered to represent this relationship, using trigonometry, and the validity of this statement will be tested. In the first part of the investigation, r will be held constant, OP will vary and a general statement will be analytically developed. In the second part of the investigation, a general statement will be established for when OP is constant and r varies. Finally, in the third part of the investigation, technology will be used to test the two general statements developed in the first two parts, thus determining the general statement for OP’. For this investigation the following technology will be used: TI-Nspire Student Software, Geometer’s Sketchpad, and Microsoft Excel.
Figure 1 shows circle C1with the center O and radius r, and any point P.
Figure 1
The circle C2 has center P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3has a center A and a radius r. The point P’ is the intersection of C3 with (OP). This is shown in Figure 2, below.
Figure 2
It is observed that (OP) is the segment of line between the center of circle C1 and the center of circle C2.
Middle
, when r = 1.
Part II.
For Part Two, let OP =2, and an analytic approach will be used to find OP’, when r = 2, r = 3, and r = 4. In this case, OP is the constant, r is the independent variable and OP’ is the dependent variable. Again, the Cosine and Sine Laws will be used to find OP’.
Using TI-Npsire Student Software, the following diagram for when OP=2 and r=2 can be created.
Figure 5
In Figure 5, it is observed that an equilateral triangle has formed, OP= 2 and both OA and AP are also equal to two, since they are both equivalent to r. It can also be observed that point P’ is in the same location as point P, therefore OP = OP’. Thus, OP’ = 2.
When r = 3 and OP = 2, the Cosine Law will be used to find θ.
When r = 3, a =OA=3, b=OP=2, c=AP=2 and C=θ.
cosθ=
cosθ=
cosθ=
cosθ=
θ= cos-1 (
)
Now ∠OAP in ∆AOP is known, and so Φ in ∆AOP’ can be found; since
∠OAP ≅∠AP’O ≅∠AOP ≅ θ
Since the sum of all angles in a triangle is equal to 180o and ∆AOP’ has two congruent angles, the following can be done:
Φ = 180-2(cos-1(
))
Now, we are able to use the Sine Law to find the length of OP’.
=
, where a=OP’=x, b=AO= 2, A=∠OAP=180-2(cos-1(
)) and
B=∠OP’A= cos-1 (
)
=
x sin[180-2(cos-1(
))]= 2sin 180-2(cos-1(
))
x =
x =
OP’ =
Upon completing all of the calculations, it is established that when
Conclusion
), OP is the denominator and the denominator can never equal zero because you cannot divide by zero. This has been satisfied by the previous limitation. As we saw in Part Four, OP must be greater or equal to half the radius {OP
R I OP ≥
}, because otherwise circles C1and C2 will not intersect.
Therefore,
Limitations of r: {r
RI 0 < r < OP2}
Limitations of OP: {OP
R I OP ≥
}.
Conclusion
The purpose of this investigation was to examine the length of OP’ as it changes with different values of r and OP. Through calculations completed in Part I and Part II, two general statements were formulated. From Part I, where r =1 and OP varied, the statement OP’=
was derived. From Part II, where OP=2 and r varied, the statement OP’ =
was analytically determined. From using technology and observing trends in Part III these two general statements were combined into one: OP’=
. The use of programs such as TI-Nspire Student Software, Excel, and Geometer’s Sketchpad allowed for the efficient collection of great amounts of data. TI-Nspire Student Software and Geometer’s Sketchpad, were also used to create visual representation of the different situations that were presented.The use of these two programs was especially important when checking the validity of the general statement. Upon examining limitations of the general statement it has been concluded that OP’=
is limited to {OP
RIOP ≥
,} and {r
R I r ≤ OP2}, where OP, r, and OP’ must be Real numbers greater than zero.
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