- Level: International Baccalaureate
- Subject: Maths
- Word count: 1712
Mathematic SL IA -Circles (scored 17 out of 20)
Extracts from this document...
Introduction
Mathematics Standard Level Portfolio
Type 1- Circles
Candidate name: Sun Ha (Rucia). Park
Candidate number:
School: Beijing No.55 High School
INTRODUCTION
In this task, I will investigate positions of points in intersecting circles.
I will analyze and investigate the problem in two different situations with different conditions, to find a general statement that shows general situation of points in intersecting circles. I am going to use Microsoft word to analyze the task, an application called Geogebra to graph intersecting circles, and TI-84 graphing calculator to calculate the values.
There are three circles intersecting. Circle 1(C1) has center O, Circle 2 (C2) has center P and Circle 3(C3) has center A. C1 has radius OP, and let A be the intersecting point of C1 and C2. C3 has its radius, r. The point P’ is the intersection of C3 with OP.
Used knowledge:
I have used 3 mathematical theorem, or rules to proof my findings.
- Similar triangle theorem
This theorem is used where two or more triangles with the same size angles. If the triangles are in the same shape with the same three (in fact, two) angles, then they are called similar triangles in different ratio.
- The Pythagoras’s theorem
Sides which are opposite to the angles are labeled using small alphabets of the angles. When angle B is 90°, which means the triangle is a right triangle, then
a2 +c2=b2.
- Cosine rule
Middle
We can notice that the length of OP’ gets small as the length of OP gets longer. Be specific, we can know that the length of OP’ is the reciprocal of the length of OP which is that OP’ is
, when OP is 2. From here, we can set a general statement, OP’=
. However, we have to realize that the length of r is 1, and this means the numerator of OP’ is not a just form of reciprocal number. Therefore, the general statement can be interpreted to another form, OP’=
.
Let’s consider another condition.
In the second condition, let OP=2, fixed, and find OP’ depending on the different lengths of r.
- when r=2,
By the graph I’ve drawn, we can know that OP’ is on the same point of OP.
Therefore,
- When r=3,
As the cosine rule,
Link
∵
∴
is an isosceles triangle, and
is an isosceles triangle
∵
∴
∴
∴
- When r=4,
As the graph shown,
∵AO = 4 = r,
OP’ = 2AO = 2r
∴OP’ = 8
From the second assumption, we can find;
OP | 2 | 2 | 2 |
r | 2 | 3 | 4 |
OP’ | 2 | 8 |
As the length of OP is stable and the length of r is changeable, the length of OP’ gets longer, which means the length of OP’ changes only by the changes in the length of r. In the first concept, the general statement I gave was OP’=
, however, it doesn’t match in this condition. If we follow the first general statement, the length of OP’ should be 1 when r=2. The same for the others, it should be
, and 2 when r=3, and r=4 respectively. Let’s make a diagram shown clearly.
Length of OP’ following the first general statement in the second condition. | 1 | 2 | |
Actual length of OP’ | 2 | 8 |
From the diagram, we can figure out that the actual length of OP’ is calculated by multiplying one more time of r to the length of OP’ calculated which is followed the first general statement. Therefore, we can approach to the new general statement, which is OP’=
.
To test the validity of my general statement, I would like to change the values of OP and r. I will check up for the validity by using FOUR different conditions;
Before testing, I am going to estimate the results using the general statement. The lengths of OP’ are going to be 1,
,
and 9 respectively.
Length of OP' Condition | Hypothetic value (by using the general statement, OP’= ) | Real value |
1 | ? | |
? | ||
? | ||
9 | ? |
Conclusion
r. This tells us about the length of OP should be longer than
r. I assume one more situation which has the length of OP is 1 and the length of r is 2. In this situation, OP=
r. I tried to draw the circles, and it was possible to draw three circles.
Now, what I have is that the length OP should be longer than
r, but it’s possible to equal
r. Then, let K denote the coefficient of r, set OP=Kr and test what value of K would be, whether three circles can be drawn. I would like to find the value of K which is greater than
and smaller than
,
. I assume a situation which is OP of 5, r of 12.
The graph shows, if K=
, so OP=
r, then C1 and C2 couldn’t intersect at one point, therefore, it is impossible to draw three circles required.
Since OP and r are the radii of circles, so their values can’t be negative. Moreover, the length of OP should not be less than
r, the limitation of the general statement is
r.
As conclusion, I could find out the general statement by two different conditions which have different fixed value of OP and r, and testing the validity of the general statement I found. Thus, the general statement I found finally is OP’=
. However, the statement also has a scope, which is
r.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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