Parahexagon and Geogebra

bobbym · 2013-05-19 08:28:00

Hi;

As usual this came up in another thread:

In hexagon ABCDEF,AB=ED=2 ,AF=CD=6 ,BC=EF=4 , and all the interior angles are equal. Find the area of hexagon ABCDE.

Is this a tough problem? Pretty tough for me but not for geogebra!

1) Create (0,0) and (0,2) and hide the xy axes.

2) Use the angle with a given size tool and click B then A and enter 120, clockwise. Point B' will be created.

3) Use the circle with center and radius tool with a radius of 6 and center at A.

4) Draw a line through A and B'.

5) Find the intersection of that line with the circle. There will be 2, so pick the one in the 4th quadrant.

6) Hide the circle, the line and rename C to z1 and then hide z1. Hide B' and then rename D to F.

7) Use the angle with a given size tool and click A then B and enter 120 counterclockwise. A' will be created and another 120°

8) Use the circle with center and radius tool with a radius of 4 and center at B. Draw a line through B and A'.

9) Find the intersection of that line with the circle. There will be 2, so pick the one in the 1st quadrant.

10) Rename C ( the other intersection ) to z2 and E to C. Then hide the circle, the line, A' and z2.

11) Use the angle with a given size tool and click B then C and enter 120, counterclockwise. Point B' 1 will be created.

12) Use the circle with center and radius tool with a radius of 6 and center at C. Draw a line through C and B'1.

13) Find the intersection of that line with the circle. There will be 2, so we want the one in the 1st quadrant.

14) Rename D ( the other intersection ) to z3 and E to D. Then hide the circle, the line, A' and z2. Then hide the circle, the line, B'1 and z3.

15) Enter in the input bar (x(D),-1) and point E will created.

16) For peace of mind, check out distances EF and FA and angles AFE, CDE and FED. Set working precision to 15 in options before you do. You will see we are very close except for a slight fungus.

17) Use the polygon tool and click points A,B,C,D,E,F,A. Poly1 will created in the algebra pane and read, poly1 = 38.10511776651535

Your drawing should look like the image on the right.

The methods of experimental math are now used to clean the decimals up and deliver exact forms.

Use RootApproximant at Wolfram or a PSLQ or the ISC to find the exact form of 22 √3 for the area.

How can we prove the above answer is not just a mere approximation? I can hear the purists screaming, "numerics are not a proof!!!!!"

Okay, here is how. We utilize this quote, "act as if ye had faith and faith will be given to thee." If that area is close then the points of those vertices will be close too. Let's use the same technique to get the exact forms for the vertices.

We get:

We have found one answer!

I urge you to plot the points and check the angles using vectors or the law of cosines and also to check the area using determinants.

The finished answer is the image on the left.

Geogebra came so close we were able to guess at an exact answer and find the area using ordinary methods. Anyway, we are done with the problem.

ElainaVW · 2013-05-19 09:05:17

Hi Bobbym;

What pslq will do that? The angles and lengths check but of course there are many other hexagons like that.:(

Last edited by ElainaVW (2013-05-19 09:06:49)

bobbym · 2013-05-19 10:37:03

Hi;

Yes, that is only one possible parahexagon.

For the pslq you could try the vector

c = 38.10511776651535

[0, c, c^2, c^3, c^4]

One of the answers will be

[0, -1452, 0, 1, 0]

which corresponds to the equation.

Repeat for the rest.

Is it not amazing that the methods employed here reduce this problem down to a simple cubic equation!?

ElainaVW · 2013-05-20 00:35:31

Hi Bobbym;

I used your pslq routine to find the rest. How did you come up with that?

bobbym · 2013-05-20 00:42:28

Hi;

The PSLQ? That is an interesting story. Some would call it a rant.

As you should know there are a couple of discoverers of the algorithm. There is an implementation of it on the net but as usual it does not work. My brother and I spent a long time trying to debug it but this was one time "the human debugger" failed.

So off I headed to M land to see what the experts knew. I posted a request for a PSLQ that worked and was written in M. I got one reply from you know who. Now this guy is a big brain who knows everything. He remembered me because I had posted bobbym's integral there and he had failed to solve it. This irked the mighty genius and he demanded that I grovel for the PSLQ.

I was willing to bow long and low until he asked me my favorite question... Then I snapped and replied in a smart alecky way. Needless to say I did not get the implementation that day.

My brother the matrix master was fooling around with a cross product routine that would sometimes solve the PSLQ and sometimes not. That is where I got the idea for mine. It works! I do not know why but I am not the type to argue with success. Pretty boring story...

Agnishom · 2013-05-20 01:53:35

How does the root approximant function in M work?

bobbym · 2013-05-20 02:35:52

Hi;

I would assume that it works like all the rest. It is some type of implementation of the PSLQ or LLL algorithm.

Go over to Alpha and put this in:

RootApproximant[38.10511776651535]

Agnishom · 2013-05-20 02:52:39

What is a PSLQ?

bobbym · 2013-05-20 02:58:33

It is an extension of the extended GCD. Where the extended GCD solves ax + by = c, the PSLQ solves

and it solves it in such a way that the

are integers. It was called the most important algorithm of the 20th century.

Agnishom · 2013-05-20 13:40:40

How does this work?

bobbym · 2013-05-20 14:08:37

It uses something called Lattice Reduction. Exactly how this works I can not explain. In more general terms it is a way to intelligently guess, with a high probability at what constants make up a decimal.

Agnishom · 2013-05-20 14:27:03

Okay but is it related to root approximant?

bobbym · 2013-05-20 14:28:28

Yes, post #3 is doing a rootapproximant using a PSLQ.

Agnishom · 2013-05-28 23:41:03

bobbym wrote:

Hi;
The PSLQ? That is an interesting story. Some would call it a rant.
As you should know there are a couple of discoverers of the algorithm. There is an implementation of it on the net but as usual it does not work. My brother and I spent a long time trying to debug it but this was one time "the human debugger" failed.
So off I headed to M land to see what the experts knew. I posted a request for a PSLQ that worked and was written in M. I got one reply from you know who. Now this guy is a big brain who knows everything. He remembered me because I had posted bobbym's integral there and he had failed to solve it. This irked the mighty genius and he demanded that I grovel for the PSLQ.
I was willing to bow long and low until he asked me my favorite question... Then I snapped and replied in a smart alecky way. Needless to say I did not get the implementation that day.
My brother the matrix master was fooling around with a cross product routine that would sometimes solve the PSLQ and sometimes not. That is where I got the idea for mine. It works! I do not know why but I am not the type to argue with success. Pretty boring story...

Voldemort?
What was your favorite question?

bobbym · 2013-05-29 02:03:19

Voldemort? Sounds like a French cheese.

Agnishom · 2013-05-29 02:06:56

I can't believe you don't know who he is

bobbym · 2013-05-29 02:15:38

That is because I have never read those books.

ShivamS · 2013-05-29 03:42:19

You do not know who -He who must not be named- is?

bobbym · 2013-05-29 03:47:15

Hi;

No, I do not know that fellow.

Agnishom · 2013-05-29 04:03:57

Tom Marvolo Riddle = I Am Lord Voldemort = Immortal Odd Lover

bobbym · 2013-05-29 04:05:46

Yep, I have never heard of him.

Agnishom · 2013-05-29 04:10:44

Lord Voldemort

Have you read any popular superhero stories?

bobbym · 2013-05-29 04:11:50

Many of them. Particulary the ones that have a mythological background.

Math Is Fun Forum

#1 2013-05-19 08:28:00

Parahexagon and Geogebra

#2 2013-05-19 09:05:17

Re: Parahexagon and Geogebra

#3 2013-05-19 10:37:03

Re: Parahexagon and Geogebra

#4 2013-05-20 00:35:31

Re: Parahexagon and Geogebra

#5 2013-05-20 00:42:28

Re: Parahexagon and Geogebra

#6 2013-05-20 01:53:35

Re: Parahexagon and Geogebra

#7 2013-05-20 02:35:52

Re: Parahexagon and Geogebra

#8 2013-05-20 02:52:39

Re: Parahexagon and Geogebra

#9 2013-05-20 02:58:33

Re: Parahexagon and Geogebra

#10 2013-05-20 13:40:40

Re: Parahexagon and Geogebra

#11 2013-05-20 14:08:37

Re: Parahexagon and Geogebra

#12 2013-05-20 14:27:03

Re: Parahexagon and Geogebra

#13 2013-05-20 14:28:28

Re: Parahexagon and Geogebra

#14 2013-05-28 23:41:03

Re: Parahexagon and Geogebra

#15 2013-05-29 02:03:19

Re: Parahexagon and Geogebra

#16 2013-05-29 02:06:56

Re: Parahexagon and Geogebra

#17 2013-05-29 02:15:38

Re: Parahexagon and Geogebra

#18 2013-05-29 03:42:19

Re: Parahexagon and Geogebra

#19 2013-05-29 03:47:15

Re: Parahexagon and Geogebra

#20 2013-05-29 04:03:57

Re: Parahexagon and Geogebra

#21 2013-05-29 04:05:46

Re: Parahexagon and Geogebra

#22 2013-05-29 04:10:44

Re: Parahexagon and Geogebra

#23 2013-05-29 04:11:50

Re: Parahexagon and Geogebra

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