- Level: International Baccalaureate
- Subject: Maths
- Word count: 4050
Circles Portfolio. The purpose of this assignment is to investigate several positions of points in three intersecting circles in order to discover a general statement
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Introduction
Mathematics SL, Portfolio Type 1Circles
CIRCLES
The purpose of this assignment is to investigate several positions of points in three intersecting circles in order to discover a general statement. There are three circles: circle C1 has center O and radius r, circle C2 has center P and radius OP, and circle C3 has center A and radius r (the same radio as C3). Point P' is the intersection of C3 with (OP).
All diagrams in this portfolio were created with Geogebra. To give a clear display of the length of each segment the diagrams are presented in the Cartesian coordinate system using point O as origin and (OP) as x-axis.
At first I will try to find a general statement representing the relation of OP’, OP and r in three examples, in which r will remain constant and OP will have three different values.
Diagram 1: r=1, OP=2
The method I will use to find OP’ will include the cosine function used in two triangles I will sketch in the first three diagrams:
Middle
3
r
1
OP
2
3
4
OP’
0.5
0.33
0.25
Table 1: Results from first part of investigation
In the first part of the investigation with the help of the cosine function we find out that in all cases
and
are similar triangles, as in each case they share the same angle measurements (∢AOP=∢OAP’, ∢OPA =∢OAP’ and ∢OAP =∢AP’O) and the corresponding sides are in proportion (
).
When observing the results we find that if each number of OP is multiplied by their correspondent OP’ the result is 1, so this pattern can be generalized by:
=OP’.
In the following part of the investigation, I will change the roles of r and OP so that OP will remain constant and r will have three different values.
All values of OP’ were found with Geogebra.
P (0, 2)
OP’=2
is an equilateral triangle, as OP=P’A=AO=2
P’ (0, 4.5)
OP’=4.5
P’ (0, 8)
OP’=8
VALUES:
Diagram | 4 | 5 | 6 |
OP | 2 | ||
r | 2 | 3 | 4 |
OP’ | 2 | 4.5 | 8 |
Table 2: Results from second part of investigation
Verifying if the statement found in the previous part of the investigation is consistent:
Diagram 4: Diagram 5: Diagram 6:
=OP’
=OP’
=
Conclusion
-Example 9: Radius r is a negative number.
-Example 10:
Diagram 11: r=9, OP=4.49
Diagram 11 shows that point P’ is undefined. The combination of values of r=9 and OP=4.5 causes the two circles with radius r to form one single circle. This makes it impossible to define the intersection of C3 with (OP), as C3 does not exist.
The general statement OP’=
is valid only if r
If r
- (diagram 11), the value of OP’ is undefined.
- If r and OP are not bigger than 0 the value of OP’ is undefined.
If r = 2 · OP (diagram 7), the center of C3 is situated on the x-axis. 2·OP is the biggest value r can reach. Triangles
and
- cannot be sketched, but the general statement can still be used for finding OP’.
If r
- (diagrams 1, 2, 3, 4, 5, 6, 8, 9,10), OP’ can be found through the general formula.
The process of finding the general statement starts already in the first part of the investigation. With help of the cosine function we find out that in all cases
and
are similar triangles. The demonstration of the corresponding sides being in proportion can be represented by(
). By simplifying
we get to the general statement OP’=
.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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