[Guest]

This problem has been inspired by SuperPrismatic’s “Trisecting the Triangle” thread, which was developed by several contributors.

Think of a piece of standard graph paper. One draws a line OX along the x-axis, from the origin O to wherever. One now draws a line OW from the origin to some other arbitrary point W on the piece of graph paper.

Now suppose that you have already constructed a rigid triangle ABC and are permitted to slide point A along OX while allowing point B to slide along OW.

What sort of curve will the point C describe?

[Guest]

my first instinct would be a line, but i'm too lazy to work it out. i'd be interested to see someone else's math

[Guest]

Half an oval

[Guest]

I would imagine a binomial curve

[Guest]

I would say a point. once A and B get on the lines, they won't be able to move.

(its a "rigid" triangle.)

Edited September 16, 2009 by phillip1882

[Guest]

an ellipse?

[Guest]

assuming there is only one answer, point C will make a straight Line Segment

my logic: OX is a line along the x axis, now lets say that point W is on the -x axis so that OX and OW form a 180 degree angle, the only possible way to move the triangle with A and B remaining on both assigned lines is horizontally, thus point C moves horizontally too since it is a rigid triangle, which would create a horizontal line segment

[Guest]

i can imagine k4d's line case and tamoo/ogden/carl's curve/elipse/oval (though only an arc section - not a whole one). so would it be some eliptical equation based on the angle between the axes you draw?

Edited September 16, 2009 by ljb

[Guest]

I believe it would be a quarter elliptical curve, if the two points were constrained to the lines extending in one direction from the origin. If they were both permitted to continue along the lines in the reverse direction beyond the origin, I believe it would be a full ellipse.

[Guest]

I agree with phillip1882

Edit: Nevermind

Edited September 16, 2009 by psycho

[Guest]

ogden got it

[Guest]

Since I tend to overcomplicate many of these problems, I ask forgiveness in advance if I'm way off.

Since the question generalizes, I thought to illustrate the curve by taking a simplified case.

Let the ray OW be the vertical axis and use as a triangle a isosoles right triangle with sides of 1.

Position the right triangle with point A at the origin and point C along the X axis at (1,0) and Point B on the y axis at (0,1) as a starting point.

The path for point C for the first 90 degrees of rotation can be expressed as:

x = sin t + cos t

y = sin t

At this juncture, point B is now on the origin and point A is still on the X axis. Point C is at (1,1)

For the next 90 degrees of rotation, the path of C can be expressed as:

x = sqrt(2) * sin (PI/4 - t)

y = sqrt(2) * sin (PI/4 + t)

The triangle ends up with Point b on the origin and point C on the y-axis.

A plot of the result is attached. I'm not sure if it has a name.

None of the other posts seemed to have a mathematical basis for the answers.

Good post.

[Guest]

I just drew this out quick and it is as ljb describes. It is part of an ellipse.

[Guest]

At this juncture, point B is now on the origin and point A is still on the X axis. Point C is at (1,1)

For the next 90 degrees of rotation, the path of C can be expressed as:

x = sqrt(2) * sin (PI/4 - t)

y = sqrt(2) * sin (PI/4 + t)

The triangle ends up with Point b on the origin and point C on the y-axis.

A plot of the result is attached. I'm not sure if it has a name.

None of the other posts seemed to have a mathematical basis for the answers.

Good post.

This part doesn't follow the constraints of the problem. Once A leaves the x-axis... But the first part of your graph is very nice and shows the quarter of an ellipse created by C.

[Guest]

I would say it would be a right angle

[Guest]

My first answer is 'technically' right

but, I also wrote a quick program that draws the curve I 'think' you were looking for

the curve is (obviously) the red line

and all that black stuff was me drawing the triangles i didn't feel like commenting out

[Guest]

Well done, anybody who said that the answer was an arc of an ellipse. The question has quite a pedigree, and is known as van Schooten's problem, though a simple variant had been discussed by Proclus, a Byzantine living in the 5th century AD. van Schooten was a 17th century Dutch mathematician whose work Isaac Newton valued highly when he was an undergraduate at Cambridge. I am so amazed that SuperPrismatic's question, mentioned in my original post, involved a different conic, namely the hyperbola.

[Guest]

This part doesn't follow the constraints of the problem. Once A leaves the x-axis... But the first part of your graph is very nice and shows the quarter of an ellipse created by C.

Thanks. I knew it was a "user problem".

[Guest]

I have just found a superb applet demonstrating the solution to this problem. Enjoy.

http://poncelet.math.nthu.edu.tw/disk3/cabrijava/vanschooten.html