- Level: International Baccalaureate
- Subject: Maths
- Word count: 2392
Maths IA. In this task I am asked to investigate the positions of points in intersecting circles.
Extracts from this document...
Introduction
CIRCLES
In this task I am asked to investigate the positions of points in intersecting circles.
Circle C1 has the midpoint of “O”, its radius crosses the midpoint of circle C3 which is “A”. Circle C2 crosses both points O and A which means they are both equal as they are the radiuses of circle C2, with the midpoint being “P”. All three circles are located on a diagonal line which holds the points of O, P’ and P, and is represented by the x-axis on the diagram above.
The distance from O to A depends on the radius of either C1 or C3, (it doesn’t matter which one because they are both of equal size). The distance from O to P is the same as the distance from A to P. The three points (A, O and P) form an isosceles triangle meaning that angles O and A will have the same angle size, as well as O to P and A to P having the same side length.
Because of the three points I am able to make a triangle (mentioned above), yet I have another point which is P’. Drawing a line from A to P’ I am able to divide the triangle into two forming another (smaller) isosceles triangle. Now the angles of sides O and P’ will be equal, as well as the side lengths of O to A and P’ to A.
Provided that r=1, I need to find the values of O to P’ when OP=2, OP=3 and OP=4.
Middle
A = 180 – 82.8 – 82.
A = 14.4
O = 82.8
O to P’ = 0 .25
General statement
Using the three different values for O to P (OP=2, OP=3, and OP=4) I was able to come up with a general statement as I saw a pattern developing in my results for the distance from O to P’. Each time that the length of O to P increased the fraction decreased:
When, O to P = 2
cosO =
When, O to P = 3
cosO =
Each time that the value of O to P is increased the fraction decreases because the denominator increases each time while the numerator stays the same.
The lower the fraction that I
, the larger the angle becomes:
O =
O = 75.5
O = 80.4
And so on…
This pattern continues as I keep increasing the distance from O to P.
As the side length O to P increases the side length of O to P’ decreases:
OP = 2 is 0.501 for OP’
OP = 3 is 0.334 for OP’
OP = 4 is 0.25 for OP’
From finding out more and more results I started to see a pattern, the numerator of the fraction stays the same. The numerator of the fractions above is 1, the radius is also 1.
,
,
,
,
…
Then I looked at the values of O to P that were increasing.
I took the radius and divided it by the length of OP:
Now to test this on a real example I will use OP = 2, OP = 3, and OP = 4.
=0.5
=0.333
=0.25
This method gives an accurate enough measurement of the length of the side O to P’.
OP=2 | OP=3 | OP=4 | |
OP’ | 0.501 | 0.334 | 0.25 |
Conclusion
as
will still be 1.
Testing the validity of the general statement
Find O to P’ when r=1, OP=6 and OP=7 Find O to P’ when OP=2 and r=6
OP’ =
= 0.167 therefore O to P’ is 0.167. OP’ =
= 18 therefore O to P’ is 18.
OP’ =
= 0.143 therefore O to P’ is 0.143.
The limitations of the general statement
When using cosine and sine to figure out the angles and side lengths the calculator gives of more numbers. I put those numbers to three significant figures cutting of the rest. Yet when I use the general statement formula the numbers I am given are more basic which could result in a small, (not very significant) calculation error. Apart from that I found my general statement to work well in solving these problems.
Explaining how I arrived at the general statement
From looking at my results for the first problem I created a general statement formula. I noticed that the radius stayed the same so I placed it on top of my equation and divided it by the distance from O to P giving my length of O to P’
Although when I reached the second problem the general statement did not work therefore it meant that I would have to modify it somehow to make it work. I squared the radius and divided it by the length O to P.
When I looked back on the first problem the modified general statement formula worked, as squaring 1 (
) will still give me 1 therefore not changing the values I received through the use of technology through sine and cosine.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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