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Math Ia circles

Extracts from this document...

Introduction

Aim: The aim of this task is to investigate positions of points in intersecting circles

image17.png

In the above visual, we can see that the triangle OAP is an isosceles triangle, where OP is a variable length, and OA is a constant and is the radius of C1. OA is also the base of the isosceles triangle, with a length of 1. OP is given to be 2, and seen as this is an isosceles triangle, AP must be 2 as well.  So we can state that OP = OA = 2 and OA = 1.

In the figure we have the isosceles triangle OAP. It can also be noted that if a line AP’ is drawn, we have another isosceles triangle: OAP’. This is can be seen because C1’s r = 1, and AP’ = OA, and in turn, it being an isosceles triangle AP’ = OA = 1.          

Seen as both OAP and OAP’ are

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Middle

, thus meaning that there is an exponential relationship between OP and OP’.

When we work at r = 2, instead, we have an equilateral triangle, seen as now the radius of C1 matches that of C2 and 3 and A, O and P are all at a same distance. So now, when r = 2, OAP is an equilateral triangle. In the case of an equilateral triangle segment AP = segment OA, and so, OA = AP = 2

image18.png

From the graph we can say that: when r =2, segment OP = segment OP’, and therefore segment OP’ = 2
image19.png

When r = 3, the graph looks like the above. OA = AP’. To find the measure angle POA, we use trigonometry. Angle POA =image10.png

 Since OP and AP are radiuses of C2, OP = AP =2. We also know that angle POA = angle PAO = OP’A = 41.41°. To find the length of OP’, we use the cosine function. In order to use the cosine function, we need to find the size of angle PAP’.

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Conclusion

class="c2" colspan="1" rowspan="1">

1

2

1/2

1

3

1/3

1

4

1/4

1

5

1/5

1

6

1/6

1

7

1/7

1

8

1/8

Instead when we apply a constant OP and a variable r to the general statement, this is the result:

r

OP

OP’

1

2

½

2

2

2

3

2

4 ½

4

2

8

5

2

12 ½

6

2

18

7

2

24 ½

The only limitation to this statement is that once the radius surpasses 8, there is no triangle anymore, and thus our postulation is invalid since it is based upon similar triangles. So our revised general statement will be: image16.png

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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