# Maths Circles Ia Essay

Math Circles Aim: The aim of this task is to investigate positions of points in intersecting circles. Introduction: The above diagram shows that distance r is the distance between any point, such as A, and the center of the circle, O, of the circle C1. The circle C2 has centre P and radius OP. A is one of the points of intersection of C1 and C2.

Circle C3 has centre A, and radius r. The point P’ is the intersection of C3 with (OP). The r=1, OP=2, and P’=0. 5. This is shown in the diagram below. This experiment will explore the relationship between the OP values and the r values, when the r values are held constant.

It will also investigate the reverse, the relationship when the OP values are held constant and the r values are modified. In the first example, the r value is 1 unit. An analytic approach will be taken to find the length of OP’ when OP=2. Firstly, one can note that 2 isosceles triangle can be drawn by using the points A, O, P’, and P. It is shown in the diagram below. in ? AOP’, lines OA and AP’ have the same lengths because both points O and P’ are within the circumference of the circle C3, which means that OA and AP’ are its radius. Similiarly, ?AOP forms another iscosceles triangle where AP and OP are equal in length, because both OP and AP are within the circumference of the circle C2.

OA = r of C3 or C1= 1. Since the circles have been graphed, their points can be denoted as coordinates. For OA, the coordinates of O will be (0,0), because it lies in the origin of the graph.

The coordinates of the point are undefined, so they shall be denoted as (a,b). We need these coordinates because to find the value of P’. AP = OP = r of C2 = 2. From this, we can get the coordinated of AP.

The coordinated for the point P will be (2,0), for it lies on the x-axis and has a radius of . A is still undefined, so we will leave it as (a,b). Using the distance formula, (a-c)2+ (b-d)2, we can use an algebraic method of solving equations to solve for the unknown coordinates.

Using this, we can find the coordinates for A. OA = 1, A = (a,b), O = (0,0) By the distance formula: 1= (a-0)2+(b-0)2 1= a2+b2 (1=a2+b2)2 1= a2+ b2 and, AP= 2, A(a, b), P(2, 0) Then, by the distance formula: 2= (a-2)2+(b-0)2 (2= v(a2+ 4a+4+ b2 ))2 4= a2+ 4a+4+ b2 Using the elimination method, the value of a can be found. 1= a2+ b2 – 4= a2+ 4a+4+ b2 _________________________ 1= 4a a= 14 ubstituting this value of a in previous equation for finding the value of b: 1= a2+ b2 1= 142 + b2 b2 = 1- 116 = 1516 b = ±1516 = ±154 We will not take the value of negative b, however, because in the graph, b falls in quadrant one, where the y value is positive.

So, the coordinates for point A are, A = ( 14, 154) Now, we have to find the coordinates of point P’. for this, we will use the distance formula and take P’= (c,0). We know the length of AP’ because it is the radius of C3, which is 1. Since we know the coordinates of A, we sub in the numbers in the distance formula. AP’= 1 = c- 142+0-1542 = c- 142+1542 =c- 142+1542 ]2 1= (c-14)2 + (154)2 1- (154)2 = (c-14)2 c-14 = 116 c= 14 ± 14 c= 12 , 0 According to the graph, the point P’ does not lie on the origin, so the 0 can be neglected. This means that the coordinates of the point P’ are ( 12 , 0).

Now, the length of OP’ can be calculated by using the distance formula for the two points A and P’. So, (0-12)2+ (0-0)2 = (12)2= 12 ? OP = 2, r = 1, OP’= 12 Using the same method of calculation for the other values, the following is obtained: r| OP| OP’| 1| 2| 12| 1| 3| 13| 1| 4| 14| The relationship between OP and OP’, when r is held constant, is inversely proportional.So when the value of one of them increases, the value of the other decreases proportionally. Therefore, the general statement to represent this would be: OP’= 1n , (n=OP) For checking the validity of the general statement, other values were tested. The general statement (OP’= 1n , (n=OP)) still worked when r=1, OP= 10, OP’=110. In the following graph, r=1, OP= 10, OP’=110. The following graph is a close up of the graph above, showing the value of OP’=110 To make sure that the results were properly verified, a bigger number, 75, was tested, and it still showed the result derived from the general statement, OP’= 1n , (n=OP).In the following graph, r=1, OP= 75, OP’=175 The following graph is a close-up of the graph above, showing the value of OP’= 175 = 0.

013 Thus, is 1? OP, the general will be valid. To verify if the general statement is valid when the length of OP ? 1, smaller values were also tried. In this example, the length of OP= 34 , r= 1. Calculations resulted that OP’ still equaled to 43. The following graph displayed is when r=1, OP= 34 , and OP’=43 = 1. 3 In this example, the length of OP= 34 , r= 1.

Calculations showed that OP’ still equaled to 43. The following graph displayed is when r=1, OP= 34 , and OP’= 43 = 1. 3In the above graph, circle C3 can’t be drawn because there is no intersection point between C1 and C2. This is plausible, because if the radius of C2 is less than half the radius of C1, then the circle won’t be large enough to intersect with C1. The limitations in the general statement, OP’= 1n , (n=OP) is that OP ? 0. 5. The value 0. 5 will only work for the case when r=1 and OP values are changed.

In a broader sense, when radius is held constant, limitations to {OP| OP ? r2, OP, r? N, natural number}. The OP, the radius of C2, and radius must be a natural number, because it cannot be 0, or be a negative number, since it refers to length.Now, the second part of the investigation will be focused on finding a relationship of the length of OP’ when OP is held as constant and r is changed.

The circle C2 has a center P and radius OP. A is one of the points of intersection of C1 and C2. A is also the center of circle C3 with radius r. The point of P’ is the intersection of C3 with OP. When r=2, OP=2. Then, in this case, P and P’ have the same coordinates. As you can see, ? AOP is an equilateral triangle as all the circles have a radius of 2. This means, OP and OP’ would be equal in length and would, therefore, be 2.

But, we have to look at this through an analytical approach. We know that the radius of circle C3 is OA. The point O lies on the origin of the graph, so it center would be at (0,0). The value of point A is unknown, so it will be denoted with the variables (a,b).

AP is the radius of the circle C2. Since P lies on the origin, the point P can be denoted as (2,0). Point A is unknown here too, so it will be kept as (a,b). Then, by using the distance formula: (a-c)2+ (b-d)2 we can find out the value of point A.

So: 2= (a-0)2+(b-0)2 2= a2+b2 (2=a2+b2)2 4= a2+ b2 and, AP= 2, P(2, 0), A(a, b) 2= (a-2)2+(b-0)2 2= a2-4a+4+b2 2=a2-4a+4+b2)2 4= a2-4a+4+b2 then, by using the elimination method, 4= a2+ b2 -4= a2- 4a+ 4+ b2 __________________________________ 4= 4a a= 1 Now, by using the substitution method, we can find the value of b. We get, 4= a2+ b2 4= 12+ b2 b2= 4-1 b= ±3 The negative value of b will be neglected because the value of b is in quadrant 1, where the y-value must be positive. With these two values, we can find the value of point A which is (1, 3). Now we find the coordinate of the point P’ by using the distance formula. We will let P’= (c, 0). We know the length of AP’because it is the radius of C3, which is 2.Since we know the coordinates of A, let’s sub in the numbers in the distance formula. AP’= 2, A(1, 3), P’(c, 0) 2= c-12+3-02 (2= (c-1)2+(3-0)2)2 4= (c-1)2+ 3 0= -1+ c2- 2c+ 1 0= c2- 2c 0= c(c-2) c= 2 or 0 There are two values for c.

0 is rejected, because in the graph shown above, P’=P, and the point does not lie at the origin. So, P’(2, 0). Again, by using the distance formula, the length of OP’can be calculated. O(0, 0), P’(2,0) OP’= (2-0)2+(0-0)2 = 22 = 2 ? OP= 2, r= 2, OP’= 2 The following graph displayed is when OP= 2, r= 2. Through the same method of calculation as the above, OP’= 92.Using the same method of calculation, this table was made, r| OP. | OP’| 1| 2| 12| 2| 2| 42= 2| 3| 2| 92| 4| 2| 162= 8| The relationship between OP and OP’ can be seen to be exponential.

So, the general pattern that would arise out of this is: OP’ = n22 (n= r) To check the validity of the general statement, other values for r were also calculated. In the following graph, OP=2, r = 35 . Through the same method of calculation as the sample calculation, OP’ = 950 , which validates the general statement. The following graph is a close-up of the graph above, showing the value of OP’=950 = 0. 18.It is interesting to note that the general statement is valid up to r= 1, 2, 3, 4 but after that, when r ;gt; 4, there was no point of intersection between C1 and C2.

In addition, when r ? 12, there was a point of intersection between C1 and C2, so C3 could be drawn, but there was no point of intersection between OP and C3. Thus, the limitation for the general formula OP’ = n22 (n= r) would be 4 ? r ;lt; 0. 5. These values only work if OP = 2, so the general limitation would be { r |(OP)-1 ;lt; r ? (OP)2 OP, r? N, Natural Numbers}The radius must be a natural number, because the length of something cannot be 0 or have a negative value.Two general statements we can derive from this investigation are: OP’=1n , (n=OP) and OP’ = n22 (n= r). The first general statement is consistent with the second general statement. When you calculate the lengths of OP’, you can use the same method for both situations.

The derived general statement appears different because each one is dealing with a different constant. The first general statement is when the radius of C1 and C2 is held constant, with changing radius of C3, while the second general statement is when the radius of C1 and C2 are modified, while the radius of C3 remains constant. Conclusion: r| OP. OP’| 1| 310| Not defined| 1| 12| 2| 1| 35| 53| 1| 34| 43| 1| 45| 54| 1| 1| 1| 1| 2| 12| 1| 3| 13| 1| 4| 14| 1| 5| 15| 1| 100| 1100| 1| 1000| 11000| OP and OP’ (when r is held constant) are inversely related (OP’ ? 1OP). Therefore, the general statement to represent this would be: OP’= 1n , (n=OP) Limitations in this general statement, when radius is held constant, is {OP| OP ? r2, r? N, natural number}. If the radius of C2 is less than half the radius of C1, then the circle won’t be large enough to intersect with C1.

The radius must be a natural number, not zero, or be a negative number, because it refers to length. | OP. | OP’| 310| 2| Not defined| 12| 2| Not defined| 25| 2| 950| 1| 2| 12| 2| 2| 42= 2| 3| 2| 92| 4| 2| 162= 8| 5| 2| Not defined| OP and OP’ (when OP is held constant) is an exponential relationship (OP’? (OP)22). Therefore, the general statement to represent this would be: OP’ = n22 (n= r). Limitations in this general statement would be {r |(OP)-1 < r ? (OP)2 OP, r? N, Natural Numbers}. If the radius is greater than length of OP’s squared value, then there won’t be any point of intersection between C1 and C2. When the radius is less than the inverse of OP, C3 and OP do not intersect.