Math Ia circles

by highondope (student)

Aim: The aim of this task is to investigate positions of points in intersecting circles

In the above visual, we can see that the triangle OAP is an isosceles triangle, where OP is a variable length, and OA is a constant and is the radius of C1. OA is also the base of the isosceles triangle, with a length of 1. OP is given to be 2, and seen as this is an isosceles triangle, AP must be 2 as well. So we can state that OP = OA = 2 and OA = 1.

In the figure we have the isosceles triangle OAP. It can also be noted that if a line AP’ is drawn, we have another isosceles triangle: OAP’. This is can be seen because C1’s r = 1, and AP’ = OA, and in turn, it being an isosceles triangle AP’ = OA = 1.

Seen as both OAP and OAP’ are isosceles, and have two of the same sides for each of the triangles, we can go on and prove that they are similar triangles. To prove this relationship between the two, we have to first determine the corresponding sides and angles in the triangle doing as such:

For the sides we have: AO is corresponding to OP’, where AO and OP’ are the bases, OP is corresponding to AP’ being the right sides, and for the left we have AP corresponding to AO.

Meanwhile, the angles are as such: angle OPA corresponds to angle P’AO, angle P’AO corresponds to angle AOP’, and AOP corresponds with OP’A.

Once we have found the corresponding components of the triangles, we prove that they are similar triangles. Aside from having corresponding angles and sides, they have to have ratios of pairs of corresponding triangles:

= 2:1 ratio between the two. And ...

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For the sides we have: AO is corresponding to OP’, where AO and OP’ are the bases, OP is corresponding to AP’ being the right sides, and for the left we have AP corresponding to AO.

Meanwhile, the angles are as such: angle OPA corresponds to angle P’AO, angle P’AO corresponds to angle AOP’, and AOP corresponds with OP’A.

= 2:1 ratio between the two. And we have

= which is again a 2:1 ratio. The two triangles are isosceles and have two congruent sides each. Also all their angles are at the same ratio, so it means that also the bases of triangle OAP and triangle OAP’ are corresponding and have the same ratio. All three sides have the same ratio so the triangles are similar. To find the base of triangle OAP’ we just set up

to equal

. Then the desired values are plugged into OP, and we solve for OP’.

When: OP = 2, r = 1

When: OP = 3, r = 1

When: OP = 4, r = 1

After three trials we can see a pattern emerging. Given that the radius, OA, is at a constant value of 1, OP’ =

. If we graph these values, we will have an exponential graph seen as

, thus meaning that there is an exponential relationship between OP and OP’.

When we work at r = 2, instead, we have an equilateral triangle, seen as now the radius of C1 matches that of C2 and 3 and A, O and P are all at a same distance. So now, when r = 2, OAP is an equilateral triangle. In the case of an equilateral triangle segment AP = segment OA, and so, OA = AP = 2

From the graph we can say that: when r =2, segment OP = segment OP’, and therefore segment OP’ = 2

When r = 3, the graph looks like the above. OA = AP’. To find the measure angle POA, we use trigonometry. Angle POA =

Since OP and AP are radiuses of C2, OP = AP =2. We also know that angle POA = angle PAO = OP’A = 41.41°. To find the length of OP’, we use the cosine function. In order to use the cosine function, we need to find the size of angle PAP’. To do this we simply subtract all the known angles in triangle OP’A from 180°, including angle OAP , which add up to 124, and end up with angle PAP’ = 56°.

Cosine Rule to Solve for PP’

c^2 = a^2 + b^2 – 2ab cos C.

PP’ =

PP’

OP’ = OP + PP’, therefore OP’ = 2.51 + 2.

OP’ = 4.41

When r = 4, OP’ = 4 .

This is the visual for C1, C2, C3 when r = 4. C2 and 3 are overlapping, and OP and PP’ are the same distance and are thus an equal length. OP’ = OP + PP’, OP = 2 and so does PP’, and so OP’ = 4. Now we know that: given that r = 2, segment OP’ = 2. Given that r = 3, segment OP’ = 4.51. Given that r = 4, segment OP’ = 4.

OP’ = 2 when r = 2 and OP = 2. So :

For our general statement, we must solve for OP’. Through simple algebra we have:

Segment AO is also the r, therefore we can re-write this as

Here are some further values of OP’ when we apply our general statement using a constant r of 1.

Instead when we apply a constant OP and a variable r to the general statement, this is the result:

The only limitation to this statement is that once the radius surpasses 8, there is no triangle anymore, and thus our postulation is invalid since it is based upon similar triangles. So our revised general statement will be: