∢AOP is shared by both triangles, which means that ∢OP’A is 75.52ᵒ as well. From this we can assume that ∢OAP’=180ᵒ-2·75.52ᵒ=28.96ᵒ=A₂
Now I can easily find the length of OP’:
a₂²= (OP’) ²=
)
a₂=OP’=
a₂= OP’=
a₂= OP’=0.50
In diagram 2:
Triangle OPA Triangle OP’A a₁=OP=3 a₂=OP’=? o=PA=3 o₁= P’A=1 p=p’=AO=1 p’=p=AO=1
Finding ∢OAP (I will use the procedure as in Diagram 1):
a₁²=o²+p²-2·o·p·cos A
A=cosˉ¹ (
)
A= cosˉ¹ (
)
A=80.41ᵒ
∢OAP=∢AOP=80.41ᵒ ∢OAP’=180ᵒ-2·80.41ᵒ=19.18ᵒ=A₂
a₂²= (OP’) ²=
)
a₂=OP’=
a₂= OP’=
a₂= OP’=0.33
Diagram 3: r=1, OP=4
In diagram 3:
Triangle OPA Triangle OP’A a₁=OP=4 a₂=OP’=? o=PA=4 o₁= P’A=1 p=p’=AO=1 p’=p=AO=1
Finding ∢OAP (I will use the procedure as in Diagram 1 and 2) :
a₁²=o²+p²-2·o·p·cos (A)
A=cosˉ¹ (
)
A= cosˉ¹ (
)
A=82.82ᵒ
∢OAP=∢AOP=82.82ᵒ
∢OAP’=180ᵒ-2·82.82ᵒ=14.36ᵒ=A₂
a₂²= (OP’) ²=
)
a₂=OP’=
a₂= OP’=
a₂= OP’=0.25
VALUES:
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:
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’
=OP’
= 1
=0.67
=2
1≠ OP’ 0.67≠ OP’ 2≠ OP’
The results do not correlate with the values in table 2.
If the results are observed, you can spot a pattern which leads to the general statement for this part of the investigation:
OP’=
Checking the general statement:
Diagram 4 Diagram 5 Diagram 6
OP’=
OP’=
OP’=
= 2
= 4.5
= 8
2= OP’ 4.5= OP’ 8= OP’
The results correlate with the values from table 2.
Now I will investigate further different values of r and OP, not keeping any of them constant.
Again I will find all the values of OP’ with Geogebra.
P’ (0, 2.28)
OP’= 2.28
P’ (0, 7.2)
OP’=7.2
P’ (0, 0.17)
OP’=0.17
P’ (0, 8.33)
OP’=8.33
OP’=3.06v
OP’=3.06 P’ (0, 3.06)
OP’=3.06
VALUES:
Table 3: Results from third part of investigation
Verifying if the statement found in the previous part of the investigation is consistent:
Diagram 7 Diagram 8 Diagram 9 Diagram 10 Diagram 11
OP’=
OP’=
OP’=
OP’=
OP’=
= 2.28
= 7.2
= 0.17
= 8.33
= 3.06
2.28= OP’ 7.2=OP’ 0.17=OP’ 8.33=OP’ 3.06=OP’
The general statement discovered in the previous part of the investigation OP’=
can be applied to these results as well, showing that it is not only for some cases, such as the erroneous general statement OP’=
which was correct only if using those values from the first part of the investigation.
EXAMPLES
1) r =
, OP=
7) r =
, OP= е
2) r =
, OP=
8) r = cos75ᵒ, OP= sin49ᵒ
3) r = 0, OP= 0 9) r = tan150ᵒ, OP=cot98ᵒ
4) r =log4ᵉ, OP=
10) r =9, OP=4.49
5) r = -5, OP= -7
6) r = 4.5, OP= 7.3
*COMMENTS
Undefined values of OP’:
-Example 3: If the lengths of both radius are 0, then point P’ does not exist as there are no existing circles with r=0.
-Example 5: The concept of a negative radius does not exist for an ordinary circle.
-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’=
.
by
lareinadermundo (student)
