- Level: International Baccalaureate
- Subject: Maths
- Word count: 1375
MATH IA: investigate the position of points in intersecting circles
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Introduction
Julio Francati
Investigate the position of points in intersecting circles
IB analytical Geometry
Aim:
The aim of this task is to investigate positions of points in intersecting circles.
Introduction:
The following diagram shows a circle C1 with center O and radius r, and any point P.
Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1and C2. Circle C3has center A, and radius r. The point P’ is the intersection of C3 with (OP). This is shown in the diagram below
In order to address the aim of this Internal Assessment, the different situations of C1, C2, C3 need to be discussed and investigated. The knowledge used in this investigation is the Pythagorean Theorem. This theorem states that in a right triangle the hypotenuse is equal to the sum of the legs of the triangle squares. The equation used for this theorem is a2+b2=c2. The next form of knowledge used is a modified version of the cos θ rule. The formula for the
Middle
OAG and ΔAGP are both right triangles that share the side AG. To solve for x we will use the Pythagorean triangle and set the two legs of each triangle equal to each other. This can be seen as:
Get rid of the radicals.
Add x2 to both sides
Solve for x
OP’=2GP=
After OP’ is found I’m going to attempt to find a pattern so a general statement can be found.
OP | 2 | 3 | 4 |
OP’ | |||
r | 1 | 1 | 1 |
After looking at the results it can be seen that the general statement is:
.
In the second part of this investigation r is now the variable and OP will have a controlled value at 2.
When r=2
After making an accurate graph point P and P’ lie on the same point. As a result
OP=OP’=2.
When r=3
After making the graph a line is drawn to link AP’ and AP. Since OA and AP’ are in the same circle they are equal to each other making
AOP=
AP’O.
OA=3
OP=2=AP
Because PO and PA are in the same circle
AOP=
. As a result OP=OA,
POA=
OAP
41.41
PAP’=180
41.41
Thus, PP’
OP=2+PP’=2+2.51=4.51
Conclusion
) the value matches the value of OP’ making the general statement valid for this set of values.
Test 2:
When r=3,OP=4
First connect AP’
In
AOP’ and
PAO
OA=AP’=3
AP=OP=4
AOP’=
AP’O
PAO=
AOP’
Therefore
AOP’=
AP’O=
PAO
In
AOP’ and
PAO, two angles are equal so the last angle must be equal as well.
Therefore,
AOP’=
PAO
AP’O=
AOP’
OAP’=
OPA
Consequently,
AOP’ and
PAO are similar triangles.
The Similarity ratio of the two triangles is
OP’=
and the given values are r=3,OP=4. When plugged into the general statement (
) the value matches the value of OP’ making the general statement valid for this set of values.
In conclusion this general statement has both limitations and scope. One limitation that can be seen is that for every trial OP
. Another factor that must be taken into consideration is that OP
, this is the only way that allows C2 and C3 to have at least one intersection. Therefore the overall scope of this investigation is OP
. The formation of the general statement is seen below.
AOP’ and
PAO are similar triangles. Therefore:
Cross multiply: OP’
OP’=
OA=r, AP=OP
After making the proper replacements and implementing the scope the general statement is
,
.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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