PoentWeptgj

Answered

2022-07-22

3-point' curve

If you have a loop of string, a fixed point and a pencil, and stretch the string as much as possible, you draw a circle. With 2 fixed points you draw an ellipse. What do you draw with 3 fixed points?

If you have a loop of string, a fixed point and a pencil, and stretch the string as much as possible, you draw a circle. With 2 fixed points you draw an ellipse. What do you draw with 3 fixed points?

Answer & Explanation

autarhie6i

Expert

2022-07-23Added 18 answers

Step 1

Let length of sting be $>max(AB,BC,CA)$. If the given three points are A,B and C are not in a straight line, then

there are three possibilities to trace three ellipses and the locus is composed of the these three discontinuous ellipses.

Step 2

Center point B can be ignored when inter-focal distance is AC,

Center point C can be ignored when inter-focal distance is BA, and

Center point A can be ignored when inter-focal distance is BC.

So in effect each is only a 2-point curve.

Let length of sting be $>max(AB,BC,CA)$. If the given three points are A,B and C are not in a straight line, then

there are three possibilities to trace three ellipses and the locus is composed of the these three discontinuous ellipses.

Step 2

Center point B can be ignored when inter-focal distance is AC,

Center point C can be ignored when inter-focal distance is BA, and

Center point A can be ignored when inter-focal distance is BC.

So in effect each is only a 2-point curve.

Almintas2l

Expert

2022-07-24Added 6 answers

Step 1

Assume that three non-collinear points A, B, C in the plane are given, and that you have a loop of string of length $\ell >|AB|+|BC|+|CA|$. Slinging this string around the three points and a pencil you can draw a loop $\gamma $ around $\mathrm{\u25b3}(ABC)$ in the obvious way, keeping the string tight at all times. This loop will be a continuous curve. In order to describe $\gamma $ more precisely we draw the lines $A\vee B$, $B\vee C$, $C\vee A$ in full. In this way the plane is divided into 7 compartments, one of them the triangle $\mathrm{\u25b3}(ABC)$.

Step 2

The loop $\gamma $ traverses the 6 unbounded compartments, and at each intersection with one of the above lines it has a corner. Within a given compartment $\gamma $ is an arc of an ellipse, whereby two of the three vertices A, B, C act as foci. The points A, B are foci for two such arcs. One of these arcs belongs to an ellipse with major axis $\ell -|AB|$ and the other to an ellipse with major axis $\ell -|BC|-|CA|$.

Assume that three non-collinear points A, B, C in the plane are given, and that you have a loop of string of length $\ell >|AB|+|BC|+|CA|$. Slinging this string around the three points and a pencil you can draw a loop $\gamma $ around $\mathrm{\u25b3}(ABC)$ in the obvious way, keeping the string tight at all times. This loop will be a continuous curve. In order to describe $\gamma $ more precisely we draw the lines $A\vee B$, $B\vee C$, $C\vee A$ in full. In this way the plane is divided into 7 compartments, one of them the triangle $\mathrm{\u25b3}(ABC)$.

Step 2

The loop $\gamma $ traverses the 6 unbounded compartments, and at each intersection with one of the above lines it has a corner. Within a given compartment $\gamma $ is an arc of an ellipse, whereby two of the three vertices A, B, C act as foci. The points A, B are foci for two such arcs. One of these arcs belongs to an ellipse with major axis $\ell -|AB|$ and the other to an ellipse with major axis $\ell -|BC|-|CA|$.

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