The three circles; O, P and A intersect to create an interesting investigation regarding circles. Since this is a Calculus course, the investigation does have to deal with Derivatives. The most important and the focus of this portfolio is the line segment, OP’. Using the given diagram above, this investigation consists of finding the general equation for discovering OP’.
The values that are given are that r, is the radius of C1 and C2. OP and AP are the radii of C3. This information allows for the information to be manipulated too create two isosceles triangles. The first triangle and the one that is given, ∆OPA is an isosceles triangle therefore it can be concluded, thanks to the Isosceles Triangle Theorem that angle O and A are congruent to each other in this triangle. ∆OPA is not the only triangle that can be created, ∆OP’A is the second triangle created with a radius from C2. Therefore ∆OP’A is also an isosceles triangle. Now in both the triangles stated above, they share a common angle, O. With the help of this information we can analyze the preliminary relationship between OP and OP’. We can find OP’, we have the tools. The two tools that we will be using mainly are the sine law and cosine law (respectively);
Sin Aa = Sin Bb = Sin Cca2= b2+ c22bc cosA
and the two triangles that are going to be used;
For the first calculations, r=1 and OP =2. By finding the ∠O in one triangle, I have found the ∠O in both triangles, allowing me a complete ration to perform the sine law. Side Note: All Final Answers are rounded to 3 Significant Figures. For the first calculations, r=1 and OP =2. By finding the ∠O in one triangle, I have found the ∠O in both triangles, allowing me a complete ration to perform the sine law. Side Note: All Final Answers are rounded to 3 Significant Figures. When a triangle has all three sides given and an angle needs to found, the Cosine Law can be used. By finding angle O in ∆OPA, a complete ratio...
...MATHS PORTFORLIO SL TYPE I
CIRCLES
In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page.
The following diagram shows a circle C1 with centre O and radius r, and any point P.
The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point Pʹ is the intersection of C3 with (OP). This is shown in the diagram below.
As shown on the assignment sheet, r=OA. We therefore need to find the values of OPʹ when r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4.
We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OPʹ. The circle C2 and triangle OPA are shown below with all side of OPA indicated.
OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated...
...Luna
Math IA (SL TYPE1)
CirclesCircles
Introduction
The objective of this task is to explore the relationship between the positions of points within circles that intersect.
The first figure illustrates circle C1 with radius r, centre O, and any point P. r is the distance between the centre O and any point (such as A) of circle C1.
Figure 1
The second diagram shows circle C2 with radius OP and centre P, as well as circle C3 with radius r and centre A. An intersection between C1 and C2 is marked by point A. The intersection of C3 with OP is marked by point P’.
Figure 2
Through this investigation I will examine how the r values correlate with the values of OP in determining the length of OP’ when r is held as a constant variable and the value of OP is the variable that is subject to change. I will then venture on to study the inverse, the relationship when the r values becomes the variable that is changed and the OP value is held constant.
r as a Constant
If we let the value of r be equal to 1, we can use that information to find the length of OP’ when OP=2, 3, and 4. The first thing one can deduce is that by using the points A, O, P’, and P two isosceles triangles can be formed; ∆AOP and ∆AOP’. To rationalize this...
...MathSLPortfolio – Tips and Reminders Checklist
Notation and Terminology
Check for the following:
• I did not use calculator notation. (I didn’t include things like ‘x^2’ for or Sn for Sn)
• I used appropriate mathematical vocabulary.
Communication
Check for the following:
• The reader will not need to refer to the list of questions in order to understand my work.
• My responses are not numbered.
• I have an introduction, conclusion, title page, and table of contents.
• All graphs are labeled – Each graph has a title, labeled axes, and appropriate scale.
• My graphs and tables are within the body of my work. They are not separate or in an appendix.
• I have explained why I made the choices I did when going through the task.
• I did not include key stroke sequences, e.g. “I pressed the 2nd key, then TRACE…”
• My tables do not straddle pages.
• My tables are labeled well, including my variable definitions in each column.
Use of Technology
Check for the following:
• I used technology to illustrate my points and ideas. I didn’t just “stick in” a graph.
• Each graph or table (or other piece of tech.) is accompanied by explanations and my ideas.
• I did not include too many graphs on the same axes – my graphs are easy to read.
Mathematical Process (Type 1)
Check for the following:
• I explicitly defined variables and parameters the first time I used them, even if they were...
...what extent Divorce Law in Brazil was fair to both sides of the divorce. Moreover, I also have been told the basics of the profession by uncles  who are bachelors in Brazil – and about how timedemanding and complex it can be. My uncle also told me, though, that as long as it is exerted for passion, and not for other reasons, the high charge of work and responsibility are not a burden, but a pleasure. So high was my interest in analyzing processes and things alike that I got to spend hours with these uncles, asking questions about their procedures and their opinion about processes.
I believe that my differential contribution to the university is based on the cultural interchange interchanged I have been through between the years 2011 and 2013, during which I have been living in Germany in a boarding school, understanding the functionality of a community in the sense that, unlike day schools, offers an intense interaction with other students and teacher, and the constant presence of academic pressure. Together with my coming to Europe in 2011 came the interest of studying in an International Baccalaureate school in order to, later on, fulfill my dream of studying Law in the UK....
...Taipei European SchoolMath Portfolio

VINCENT CHEN 
Gold Medal Heights
Aim: To consider the winning height for the men’s high jump in the Olympic games
Years  1932  1936  1948  1952  1956  1960  1964  1968  1972  1976  1980 
Height (cm)  197  203  198  204  212  216  218  224  223  225  236 
Height (cm)
Height (cm)
As shown from the table above, showing the height achieved by the gold medalists at various Olympic games, the Olympic games were not held in 1940 and 1944 due to World War II.
Year (1=1932, 2 = 1936 and so on)
Year (1=1932, 2 = 1936 and so on)
Using autograph, the graph above is a scatter graph showing the high jump results from the table.
The plot suggests that the high jump heights start off with a steep positive slope then coming to a decreasing negative slope, however without the 1940 and 1944 high jump competitions, it may not be certain. Then finally, it starts increasing again with a fairly steep positive slope.
However, it would not be realistic if the function has an infinitely increasing range, such as quadratic, exponential and linear because of the limitations that humans have due to natural forces like gravity. Therefore, narrowing down the options that may fit this graph to natural logarithm and logistics
Since the statistics given starts from year 1896, in order to make sure that calculations can be as simplified as possible, I have decided to rearrange the table with the...
...SL TYPE 1LACSAP’S FRACTIONS
* INTRODUCTION
This investigation is going to do research patterns relates to the Lacsap’s Fractions. For its external structure, Lacsap’s Fraction is analogous to Pascal’s Triangle. Lacsap’s Fraction presents the way of generating and organizing the binomial coefficients. Within this investigation, the work is planning to be divided into two parts. In the first part, the content will relate to the pattern of numerators. In the second part, I am going to do the research on the patterns of denominator and the general statement for. Admittedly, the technology of computing will be involved into this investigation (E.g. Geogebro and GSP5chs). The following figure 11 illustrates Lacsap’s Fraction.
Fig.11
* PART A  CALCULATIONS and ANALYSIS
Firstly, I am going to research the numerator patterns. By observing the numerators of these fractions, it is illustrated that the first row of numerator is 1, second row of numerator is 3, third row of numerator is 6, fourth row of numerator is 10 and fifth row of numerator is 15. Let’s present it into the mathematical way: (= numerator of the row)
Continued
I realize that , , , Thus it is easily to find the numerator of the sixth row which is getting from. In order to do further investigation, it is essential to make a data table.
 numerator 
1  1 
2  3 
3  6 
4  10 
5  15 
6  21 
…  … 
n  ? 
Table 21
According to the...
...Introduction:
The task of this assessment is to consider and investigate geometric shapes that lead to stellar numbers. One of the simplest forms of this would be of square numbers, terms 1, 2, 3, 4, and 5 in terms of geometric shapes lead to the special numbers of 1, 4, 9, 16, and 25. To find these outcomes, simply all that is needed to be done is to square the nth term of the sequence. For example, 12=1, 22=4, 32=9, 42=16, and 52=25, therefore the resulting formula for square numbers would be n2=Tn where T=total amount of n. For this investigation though, the geometric shapes of equilateral triangles are studied. Triangular numbers are the amount of dots that evenly fit inside the triangle. After solving the sequence of the triangular numbers and finding a formula to solve for all equilateral triangles, the knowledge attained from triangular numbers will then be used to find and solve the correlation between triangular numbers and stellar (star) numbers.
Here is a sample of the first 5 terms in the triangular numbers sequence:
The row of numbers above the triangles is the number of each term or n.
The rows of numbers below the triangles are triangular numbers of each term or n.
` n1 n2 n3 n4 n5
Here is a sample of the first 4 stages of pstellar shapes that are being investigated. P indicates the amount of vertices in the stellar shape. In this case the stellar shapes being investigated are 6stellar, however after deriving a formula, the stellar...