From Newsgroup: rec.puzzles

I have a square (ABCD) pushed into a corner marked by the x and y axes. I rotate the square 90 degrees clockwise by sliding point D up the y axis
and point C left, along the x axis.

https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

[Additional description: the square is labeled clockwise from A at the top-left to D at the bottom left. The x and y axes are oriented conventionally, y up and x to the right and the square initially sits
flush with both axes with point D on the origin. The midpoint of the top
edge of the square AB is labeled a, the centre of the square is labeled b
and the midpoint of the bottom edge of the square DC is labeled c.]
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10di4jp$358ih$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

A nice puzzle.

Can you find any of them without trigonometry?

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 12:08:41 -0000 (UTC), Richard Tobin wrote:

Can you find any of them without trigonometry?

I'd say the nature, but not the length, of the path of b, "by inspection";
if that isn't equivalent to cheating.

As a retired engineer, all be it electrical rather than mechanical, I'm surprised that I don't know the answer to the movement of point c. I
suspect I know what it is, but will need to work to verify that. It feels
like something I should have known since my time at secondary school.

a has me a bit flummoxed, I'll be interested to work it out, but will
probably start with cardboard again...
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10diker$3bcuo$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

As a retired engineer, all be it electrical rather than mechanical, I'm >surprised that I don't know the answer to the movement of point c. I
suspect I know what it is, but will need to work to verify that. It feels >like something I should have known since my time at secondary school.

The shape of c's path can be determined using geometry that I learnt
at secondary school.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 12:08:41 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

In article <10di4jp$358ih$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

A nice puzzle.

Can you find any of them without trigonometry?

Interesting question. Have to think about it.

As for David's problem,
I first came across it in the following modified form.
The person who posed the problem said that he would
write down what my immediate answer would be and
reveal it after I said it. I did take a quick guess and
he showed me what he had written and said that it
was wrong and I better work it out. I was surprised
by the answer when I did.

The problem he gave was the following: A ladder
rests vertically against a wall. It then it let to
slide "down", i.e. the foot of the ladder slides on
the floor, away from the wall. What is the path
traced out by the ladder's midpoint?

I David's problem, it is the reverse of the locus of
point 'c'.

Being physicists, the real problem was whether the
upper end of the ladder will, at some point, leave
the wall. So the real issue was calculating the
normal force exerted on the wall by the upper
end of the ladder as it slides.

I guess I have extended David's problem to the
physical world!
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 13:48:12 -0000 (UTC), David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

As a retired engineer, all be it electrical rather than mechanical, I'm >surprised that I don't know the answer to the movement of point c. I
suspect I know what it is, but will need to work to verify that. It feels >like something I should have known since my time at secondary school.

It is rather unintutive and I fell for it when I was a year out of
secondary school (i.e. a year after entering college). Even
though, I was armed with the techniques, my first guess was
wrong .... and afer a little thought, idiotically so!

a has me a bit flummoxed, I'll be interested to work it out, but will >probably start with cardboard again...

Well, think about the trigonometric route as Richard hinted.


--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 10:55:00 -0400, Charlie Roberts wrote:

I guess I have extended David's problem to the physical world!

It is interesting how these things crop up, in the real world. I had
ladder, rooftop and climbing training, for work - being involved in
wireless installations. I don't think we covered where the two attachment points, for a ladder, should and shouldn't be. Sort of related - we also decommissioned some 15m x 75mm guyed aluminium poles from rooftops, used
to support antennas. The method statement and risk assessment would
generally be done by the rigging team responsible for carrying out the
work and then reviewed by me. We needed a couple of passes to get to
something we were all happy with.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10di4jp$358ih$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote: >https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

Don't follow the link below if you don't want to see all the answers!
There's no discussion of the problem there, just the curves calculated
as parametric equations using trig functions.

https://www.cogsci.ed.ac.uk/~richard/sliding-square.pdf

It shows the tracks of a, b, c, and also A and B - you can tell
which one is which from their start and end points.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 15:52:37 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

In article <10di4jp$358ih$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote: >>https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

Don't follow the link below if you don't want to see all the answers!
There's no discussion of the problem there, just the curves calculated
as parametric equations using trig functions.

https://www.cogsci.ed.ac.uk/~richard/sliding-square.pdf

It shows the tracks of a, b, c, and also A and B - you can tell
which one is which from their start and end points.

-- Richard

A SPOILER below, but if you have already seen
Richard's plot, there is nothing new.




Nice! I did not think of plotting the results. The most
intriguing part in the centre of the square. Sure, the
trigonometry gives y = x for all orientations of the
square, but does that mean that the locus is a
straight line or that the point sits still? Needed to
do some geometric calculations to resolve that.


--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 15:24:12 -0000 (UTC), David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On Sat, 25 Oct 2025 10:55:00 -0400, Charlie Roberts wrote:

I guess I have extended David's problem to the physical world!

It is interesting how these things crop up, in the real world. I had
ladder, rooftop and climbing training, for work - being involved in
wireless installations. I don't think we covered where the two attachment >points, for a ladder, should and shouldn't be. Sort of related - we also >decommissioned some 15m x 75mm guyed aluminium poles from rooftops, used
to support antennas. The method statement and risk assessment would >generally be done by the rigging team responsible for carrying out the
work and then reviewed by me. We needed a couple of passes to get to >something we were all happy with.

The calculation for finding if and where the top of the ladder leaves
the wall is not easy. It was a bit of challege. The link below is a
rather nice video.



SPOILER BELOW
SPOILER BELOW
SPOILER BELOW
SPOILER BELOW




SPOILER BELOW
SPOILER BELOW
SPOILER BELOW



https://www.youtube.com/watch?v=C0kdOMqjHQM&t=1325s

Enjoy!
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <rrtpfk1blp510sngk4tm8unrbdnjc9l43u@4ax.com>,
Charlie Roberts <croberts@gmail.com> wrote:

Nice! I did not think of plotting the results. The most
intriguing part in the centre of the square. Sure, the
trigonometry gives y = x for all orientations of the
square, but does that mean that the locus is a
straight line or that the point sits still? Needed to
do some geometric calculations to resolve that.

If you consider the half-way point, when the square is at 45 degrees
to the axis, it should be clear which it is.

"Half-way" of course depends on how you paremetrize the movement. And
some parametrizations have dubious physical consequences - if C moves
toward the origin at constant speed, D has to initially move up
infinitely fast.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10diro5$12ct5$1@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

Once again, don't follow the link below if you don't want to see all
the answers!

I've made an animation of it:

https://www.cogsci.ed.ac.uk/~richard/sliding-square.html

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 14:49:11 -0000 (UTC), Richard Tobin wrote:

The shape of c's path can be determined using geometry that I learnt at secondary school.

Ah, yes. Nice and easy.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10di4jp$358ih$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I have a square (ABCD) pushed into a corner marked by the x and y axes. I >rotate the square 90 degrees clockwise by sliding point D up the y axis
and point C left, along the x axis.

https://pasteboard.co/Gp2XToJnLgW9.png

What shapes do the track of points a, b and c follow?

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

Let's first see what we can do without any trigonometry.

Take the side of the square as 1 and call the origin O.

COD is a right angle, so CD is the diameter of a circle with O on its circumference (converse of the "angle in a semicircle" theorem,
sometimes called Thales's theorem). c is the midpoint of CD, the
centre of the circle, so cO is always half CD. So c lies on a circle
with centre O and radius 1/2 - it follows a quarter circle moving from
(1/2,0) to (0,1/2).

Call the distance OC p and the distance OD q, so q varies from 0 to 1
as p varies from 1 to 0.

We can see then that the coordinates of the corners are

C = (p,0)
D = (0,q)
A = (q,p+q)
B = (p+q,p)

(It may help to imagine an upright square containing the tilted
square, so that the tilted square is surrounded by 4 congruent
right-angled triangles.)

a is the midpoint of AB, c is the midpoint of CD, and b is the midpoint
of AC (and BD), so

a = (p+2q,2p+q) / 2
b = (p+q,p+q) / 2
c = (p,q) / 2

So b lies on the straight line x=y - by considering the situation when p=q
we can see it moves outwards from (1/2,1/2) to (sqrt(2)/2,sqrt(2)/2)
and back again.

a's movement is more complicated - I think we have to introduce
some trigonometry.

Call the angle OCD t, so p = cos t and q = sin t.

Now c = (cos t, sin t) / 2, confirming our conclusion that c moves
along a quarter circle.

b = (cos t + sin t, cos t + sin t) / 2

Remembering the standard trig indentity

a cos t + b sin t = sqrt(a^2+b^2) sin(t + atan(a/b)),

b's x and y coordinates are both sin(t + pi/4) / sqrt(2), confirming
our conclusion that it moves out to (sqrt(2)/2,sqrt(2)/2) and back.

a = (cos(t)/2 + sin(t), cos(t) + sin(t)/2)

Applying the same identity, we get

a = sqrt(5/4) (sin(t + atan(1/2)), sin(t + atan(2)))

Curves where x and y are both sines with the same frequency are
ellipses, varying from straight lines when they are in phase to
circles when they are 90 degrees out of phase.

So the path of a is part of an ellipse.

The same is true for A and B. In fact, all of A, B, C, D, a, b, and c
are of this form.

Perhaps someone more familiar with conics could deduce the fact -
clear from that symmetry when you see it - that c's ellipse has its
major axis along x=y.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

Aha!

In article <10dl2r9$13ned$1@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

[...]

a = (p+2q,2p+q) / 2
b = (p+q,p+q) / 2
c = (p,q) / 2

Let's switch to axes x' = x+y and y' = y-x, which are rotated 45
degrees clockwise from the original.

Then for a, x' = (3p+3q) / 2 and y' = (p-q) / 2.

(3y')^2 + x'^2 = 1/4 (9p^2 - 6pq + 9q^2 + 9p^2 + 6pq + 9q^2)
= 1/4 (18p^2 + 18q^2)

but p^2+q^2 = 1 (Pythagoras), so this is just 9/2.

So a follows an ellipse aligned with the new axes, with its axis in
the x' direction 3 times as long as its axis in the y' direction.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10dlco7$13rl6$1@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

And finally (I hope)...

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

[...]

a = (p+2q,2p+q) / 2
b = (p+q,p+q) / 2
c = (p,q) / 2

Let's switch to axes x' = x+y and y' = y-x, which are rotated 45
degrees clockwise from the original.

Then for a, x' = (3p+3q) / 2 and y' = (p-q) / 2.

(3y')^2 + x'^2 = 1/4 (9p^2 - 6pq + 9q^2 + 9p^2 + 6pq + 9q^2)
= 1/4 (18p^2 + 18q^2)

but p^2+q^2 = 1 (Pythagoras), so this is just 9/2.

So a follows an ellipse aligned with the new axes, with its axis in
the x' direction 3 times as long as its axis in the y' direction.

The extreme point of the ellipse in the x' direction is x'^2 = 9/2
or x' = 3/sqrt(2), y' = 0.

x = (x' - y') / 2, y = (x' + y') / 2, so we have

x = y = 3 / (2 sqrt(2)) = sqrt(9/8) ~= 1.06. The distance from the
origin is sqrt(18/8) = 3/2.

In the y' direction, the extreme point is (3y')^2 = 9/2 or y' = 1/sqrt(2),
x' = 0, so

x = -1 / (2 sqrt(2)), y = 1 / (2 sqrt(2)) = sqrt(1/8) ~= 0.35. The distance from the origin is sqrt(2/8) = 1/2.

In short, the path of a is an ellipse with major axis 3 and minor axis
one, with its major axis 45 degrees clockwise from the x axis. We didn't
need any trig functions to find this.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sat, 25 Oct 2025 18:07:10 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

In article <rrtpfk1blp510sngk4tm8unrbdnjc9l43u@4ax.com>,
Charlie Roberts <croberts@gmail.com> wrote:

Nice! I did not think of plotting the results. The most
intriguing part in the centre of the square. Sure, the
trigonometry gives y = x for all orientations of the
square, but does that mean that the locus is a
straight line or that the point sits still? Needed to
do some geometric calculations to resolve that.

If you consider the half-way point, when the square is at 45 degrees
to the axis, it should be clear which it is.

"Half-way" of course depends on how you paremetrize the movement. And
some parametrizations have dubious physical consequences - if C moves
toward the origin at constant speed, D has to initially move up
infinitely fast.

-- Richard

Yes, that is what resolved it for me as

sin (45) = cos (45) = 1/sqrt(2)

is all you need.
--- Synchronet 3.21a-Linux NewsLink 1.2