Dear All...

Isn't x√2 is the diagonal of a square with x for its lengths? So... would it helps if the square added to the original diagram? Just a silly thinking...

Hi, Mip...

I believe the 'outer circle' is added at last time using 3 points circle.. And in your example, if you make the circle radius bigger, then the bird house with those lengths will be possible!

I believe here's the key:
1. Create a circle with B as center, and the radius should get through A2 and C... let's call this circle HELPER
2. Rotate this circle with A2 as pivot point, in particular ways, so that IF we draw 3 point circle from A1, A2, and B, then the circle WILL intersect HELPER along with the V Axis and C.

BUT, this will take several steps before we get the wanted result after zooming execisevely to the intersection point mentioned in number 2.

Just a crazey idea.

Hi,
Attached a nod file that allows manual adjustment to get what is needed.
Maybe possible to automate rather than eyeball ?
Cheers
Barry

Hi Elang,
I may have misunderstood the problem with an assumption error on my part

I thought the question was about orientation and how to rotate the items like magenta lines. so i made the video (nephew in school meant very young to me).

I didnt see massive math.

I have assumed that x√2 is a result of A2 to C?

The angle of x defined by a circle of radius x. Thats how i start with circle and orient.

If you have random angle of x, i don't think the problem solves.

If i have made an error in reading the initial post, i must drop off as it is much deeper than i am capable

“””””” Isn't x√2 is the diagonal of a square with x for its lengths? ””””

I used 2 triangles. C and B to Center of Circle and B and A2 to center of Circle. The diagonal of A2 to C

I dont think you are looking for 45-45-90

I think you want 60-60-60

That nephew must have access to a CAD program with constraint feature :)

x is equivalent to a scale factor. So x = 1 or 10, or even 100 is good.

The base chord, (x*sqrt(2)), can be generated as the hypotenuse of a 45 45 90 triangle with two sides of x*1 unit(s), by pythagoran.

The fact that 4 chords are of equal length x*1 units, implies that their central angles are equal, so the rotation angle from one to the other is equal,
with a value of approximately 114.81 degrees +/-. There are 3 of these equal angles, I am calling phi.

The angle at A2, of approximately 97.78, (I'm calling it theta), must be adjusted so that all vertices fit on the 3point circle.
Every time angle theta is adjusted, the angle phi also changes, by the simple formula previously given. So all of the x*unit chords keep changing as the candidate theta value changes. So they need to be redrawn over and over.
Constraining the 3 angles phi to a function of theta, would solve the puzzle, if the Cad program had constraints, and correct snap(s).

I guess a MoI script or a node could be written with an epsilon of the error of intersecting the centerline at C with chordBC?
Keep running the adjustment of theta, until epsilon reaches zero?

- Brian

My 2018 Alibre program is in a zip file.

I was able to run Barry's node program. (After replacing Display node. Either my copy or Barry's must be old?)
It draw's close approximations to the solution, subject to visual adjustment to approach a solution. (As far as I can tell.)

I do not have enough knowledge of MoI to go much further.

Trigonometry can replace phi with a function of theta, but the resulting equation would still likely require numerical methods to solve. (?)

It seems like people are solving the geometry kernels job.

But integrityware already does all that.

If you specify X or x√2 or the circle radius, then the rest is a few seconds in MoI.

Same if you want to calc from any given value of those 3 objects.

No need for constraints of any kind.

Unless you are trying to program automation of the process. Then i’ll step out and watch.

Hi, Barry.

This could be exactly what we need!... Thank you very much.
Now I just need to browse the forum to study how NOD works in Moi.

Cheers!

PS: This could be a prove that constraints is possible within Moi! How about integrates constrains to Moi, Michael? :)

Hi BurrMan,

- The horizontal line of A1-A2, with x√2 length comes first in the quiz.
- Followed by the 2 magenta lines with x length connected each other
- Arranged (move and rotate) the magenta lines in particular way
- ... so that if we draw 3 points circle (RED colored) through point A1,A2, and B, then it would also 'touch' C
- C should be on both Red Circle and V axis.

Cheers.

Hi, Brian

Indeed it needs a constrain feature like i mentioned in the original post. But nephew only have Moi and love it so much that I won't 'spoil' him with other app which made we want to solve the quiz ONLY in Moi. Also, seems like you're much more smarter than me in math, so I can't understand much of your 'math talks' in your posts. But I appreciate much for the effort and please continue to do so if you please. :)

Cheers!

Hi BurrMan,

Yes it is... Only in the quiz, the red circle comes in the last order which make it tricky. ^_^

Hi All

I just found the video which very close to the quiz above, and how to solve it... only, it's done via Inventor using constraints like my friend sugested:

https://youtu.be/1xsWKAoXYk4?t=21

(it is in 0:21 through 1:08)

cheers!

Hi Elang,
""""""""
- The horizontal line of A1-A2, with x√2 length comes first in the quiz.""""""""""""

Here is a video I made creating a random x√2 value.



It works with any of the other 3 values given at any value.

The V axis is also irrelevant. All the rotation in the video is just to express it. The 3 point arc is valid. If you had geometry in some random world coords that you didnt want to move and rotate around, you could create a CPlane in that area on the Axis of your choice.

They use constraints in inventor because thats how inventor works!

Also, notice the 60x60x60 triangles the guy in inventor makes @1.08

He’s really just making random connection lines then telling them to be 60x60x60 triangles (his constraints)

But we want your nephew to understand trig

Also, he wants to do it in MoI! Its a simple solution with MoI’s toolset.

Here is another way to build a regular pentagon without using calculations or constraints.
While not the fastest way, it shows an easy method to get a 36 degree angle.
Hope you will find it interesting. Trig without inTrigue...

Radius of circumscribed circle of an irregular polygon

https://math.stackexchange.com/questions/2799954/finding-diameter-of-circumscribed-circle-of-an-irregular-polygon

Uses the Law of Cosines.

This can be used for the PentaChord puzzle.

m = number of sides with length A
n = number of sides with length B

The formula is:
m * arccos( 1 - ( A*A / 2*r*r ) + n * arccos( 1 - ( B*B / 2*r*r ) = 360 degrees.

For our puzzle, using length scale of x = 10:
m = 1
A = 10 * sqrt(2)
A*A = 100 * 2 = 200
n = 4
B = 10
B*B = 100

m * arccos( 1 - ( A*A / 2*r*r ) + n * arccos( 1 - ( B*B / 2*r*r ) = 360 degrees

arccos( 1 - ( 200 / 2*r*r ) + 4 * arccos( 1 - ( 100 / 2*r*r ) = 360 degrees

So the function to solve for is:
arccos( 1 - ( 100 / r*r ) + 4 * arccos( 1 - ( 50 / r*r ) - 360 degrees = 0

Radius "r" must be solved for. (with 40 year old calculator ? :-)

or solved with an Excel spreadsheet. (Somehow)

I think that the 360 is subtracted on the left side, setting the equation to 0?
Also see if the arccos is dealing with degrees or radians???

Increment "r" until some tolerance is achieved.
For scale x = 10, the approximate value of "r" is 9.2814, which is probably less that the actual value of "r".
(Some other values of approximate "r" are 9.2822, 9.2826.)
Might try an online calculator to test approximate "r" of 9.2814, to see if it is close to 0?

- Brian
With some learning I could probably do the Excel "macro".

https://en.wikipedia.org/wiki/Bisection_method
I suppose this could be done in Javascript?

Yes, this is breaking the "No Math" rule :-)

I used the computers calculator, manually with the formula, with test "r" = 9.2814, which resulted in 360.02368331... minus 360, which is close to zero.
Which is encouraging.

I have not yet tried to do a function solve...

- Brian

Actually, a quick Excel spreadsheet can replicate the Radius function equation, for a multitude of candidate radii. Maybe tomorrow.

- Brian