Investigation:
In the first part of this investigation the radius will stay the same and the variable will be the length of OP.
When OP = 2
First a graph is drawn and AP’ is linked by a line and point G is introduced to make line AG which is perpendicular to OP.
AP’=OA
ΔOAP is an isosceles triangle. AG is perpendicular to OP so as a result of this,
AOP=
AP’O. With all this being said OG=GP =
OP’.
Assume that OG=x, this will make GP=2-x
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=
When OP=3
When OP=3 we will use the same method used to find OP’ when OP=2.
AP’=OA
ΔOAP is an isosceles triangle. AG is perpendicular to OP so as a result of this,
AOP=
AP’O. With all this being said OG=GP =
OP’.
Assume that OG=x, this will make GP=3-x.
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=
When OP=4
The same method will be used to find OP’ that was used for the previous two values of OP.
AP’=OA
ΔOAP is an isosceles triangle. AG is perpendicular to OP so as a result of this,
AOP=
AP’O. With all this being said OG=GP =
OP’.
Assume that OG=x, this will make GP=4-x.
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.
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
When r=4
After this graph is made it is seen that OA=AP’. Therefore, OP’=2OA=8.
To attempt to find a pattern with these results a table is made
After making the table and plugging in the results into the general statement that was found earlier it is concluded that the general statement is not consistent with the results that were found when the radius was the variable.
After analyzing both tables I come up with a new general statement that works for both sets of results. The general statement is
. Using a graphing calculator I investigate other values for r and OP. the results can be seen in the table below
To test the validity of this result I will use different values of r and OP:
Test 1:
r= 2, OP=3
First connect AP’
In
AOP’ and
PAO
OA=AP’=2
AP=OP=3
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=2,OP=3. When plugged into the general statement (
) 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
,
.
by
hahajuly1013 (student)
