Ia Math
IA Task I
Introduction and purpose of task: The purpose of this task is to investigate the positions of points in intersecting circles and to discover the various relationships between said circles. Circle C1 has center O and radius r. Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has center A and radius r (therefore circles C1 and C3 are the same size). The point P’ (written P prime) is the intersection of C3 with OP. This is shown in the diagram below.
Analytically find OP’ using r=1 and OP=2, OP=3, and OP=4:
First, I created a line (see the dashed line in the above figure) between AP’ that creates the ΔAOP’. Because P’ is on the circumference of circle C3 and A is the center of circle C3, that means that AP’ is equal to the radius of C3, which is 1. We also know that because line AO connects the circumference of C1 with the center of C1 (O) and the circumference of C3 with the center of C3 (A), the radii of these circles is the same, which means that they are equivalent circles. Therefore, in the ΔAOP’, AO=AP. When a triangle has two equivalent sides, it is an isosceles triangle. By that logic, ∠O=∠P’.
Now, I looked at the triangle that is already drawn in the above figure, ΔAOP. We know that this triangle is also isosceles because OP=AP. By that logic, ∠A=∠O.
Using the law of cosines c^2=a^2+b^2-2abcos(C), which works for any triangle, I assigned θ to ∠O and determined that cos(θ)=1/(2*OP). Then, using the law of sines (insert law of sines here), sin(θ)/1=sin(180-2θ)/OP’
OP’=sin(180-2θ)/sin(θ)
OP’=sin(2θ)/sin(θ)
OP’=2cos(θ)
But because cos(θ)=1/2OP as earlier discovered;
OP’=1/OP
By using this equation, I derived the following answers analytically using r=1 and OP=2, OP=3, and OP=4.
OP 2 3 4
OP' 0.5 0.33 0.25
Behavior of intersecting circles and general statement describing interaction that occurs when value of OP is changed:
As OP changes, the resulting OP’ value
Introduction and purpose of task: The purpose of this task is to investigate the positions of points in intersecting circles and to discover the various relationships between said circles. Circle C1 has center O and radius r. Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has center A and radius r (therefore circles C1 and C3 are the same size). The point P’ (written P prime) is the intersection of C3 with OP. This is shown in the diagram below.
Analytically find OP’ using r=1 and OP=2, OP=3, and OP=4:
First, I created a line (see the dashed line in the above figure) between AP’ that creates the ΔAOP’. Because P’ is on the circumference of circle C3 and A is the center of circle C3, that means that AP’ is equal to the radius of C3, which is 1. We also know that because line AO connects the circumference of C1 with the center of C1 (O) and the circumference of C3 with the center of C3 (A), the radii of these circles is the same, which means that they are equivalent circles. Therefore, in the ΔAOP’, AO=AP. When a triangle has two equivalent sides, it is an isosceles triangle. By that logic, ∠O=∠P’.
Now, I looked at the triangle that is already drawn in the above figure, ΔAOP. We know that this triangle is also isosceles because OP=AP. By that logic, ∠A=∠O.
Using the law of cosines c^2=a^2+b^2-2abcos(C), which works for any triangle, I assigned θ to ∠O and determined that cos(θ)=1/(2*OP). Then, using the law of sines (insert law of sines here), sin(θ)/1=sin(180-2θ)/OP’
OP’=sin(180-2θ)/sin(θ)
OP’=sin(2θ)/sin(θ)
OP’=2cos(θ)
But because cos(θ)=1/2OP as earlier discovered;
OP’=1/OP
By using this equation, I derived the following answers analytically using r=1 and OP=2, OP=3, and OP=4.
OP 2 3 4
OP' 0.5 0.33 0.25
Behavior of intersecting circles and general statement describing interaction that occurs when value of OP is changed:
As OP changes, the resulting OP’ value