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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

Hi;

This problem appears in another thread.

The brown circle is inscribed in the quadrilateral. There are 4 distinct green areas. Can you get the one on the farthest left?

We will solve this with Geogebra. First we will draw the figure. To do that we will need to find the coordinates of D and E.

q = 32, so D = (-16,0), E=(16,0) that is enough to start.

1) Create points (0,12),(0,-12),(-16,0),(16,0)

2) Rename the left point D and the right point E. The top point A and the bottom point B.

3) Now use the polygon tool and join all 4 points to create poly1 and color it green.

4) Find the radius of the brown inscribed circle.

5) Use the circle with radius tool and enter 9.6 choosing the origin as the center and color it red.

6) Use the intersection tool and click on the circle and the x axis to create point F and G.

7) 6) Use the intersection tool and click on the circle and the side of the parallelogram that is in the 2nd quadrant. Point H will be created.

8) Go into M and enter:

`Solve[x^2 + y^2 == 92 + 16/100, y]`

9) Pick this form of the answer:

10) Enter into Geogebra f(x):=sqrt(2304 - 25*x^2)/5, a half circle will be drawn.

11) Hide the circle and your drawing should look like this.

12) Draw a perpendicular line through F and the x axis.

13) Get the point of intersection with that line and side AD using the intersection tool. Point I will be created.

14) Draw a line through AD with the line tool and hide the x and y axes.

Your drawing should look like this.

15) Use the polygon tool and click DFI and back to D to make poly2, a small triangle and color it black with an opacity of 100%

16) Enter into Geogebra, IntegralBetween[(3x + 48) / 4, f, -9.6, -5.76] and color it yellow with opacity 100%.

Looking at the drawing you should see that the area you want is 2 times the black plus the yellow. So

17) Enter 2 (poly2 + j) and read off the answer in k. I get k = 37.420472708971396, this is quite close to the analytical answer. We are done.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,843

Hi Bobby,

We can also find the area of the blue shape by deducting the area of sector CGH (brown) from the area of polygon CGDH (brown and blue shape...a kite).

The areas of the sector and the polygon display automatically in the Algebra window upon creating those shapes, and then by opening the Spreadsheet window the two areas can be dragged into it to do the arithmetic there: 122.88 - 85.45952729102859 = 37.4204727089714 (correct to 13 decimal places).

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

Hi;

That is true and it is easier. The reason I did it like this is because someone is trying to learn integration between 2 curves and I offer this as an example.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,843

Oh, I see. I had wondered why you went that way.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

When I originally solved the problem I used integration. I missed your idea completely. That is the drawback of The Teakettle Principle.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,843

I looked up TTP yesterday, and can understand the blinkered view one would get by going down that path.

If I understand its meaning correctly, would an example of it be Elaina's answer in post #23 when compared to my earlier answer in post #11? Not that she would have done that, of course!! It's just a convenient example.

*Last edited by phrontister (2014-06-30 00:14:46)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

I would hope that she did. Her solution looks to be totally done with M but I am not sure.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,843

My solution's the same as hers (though she used arccos and I used ArcSin and simplified further than she).

I used trig for all my initial calcs, ending up with a fairly lengthy jumble that I simplified by hand and with M.

It didn't occur to me at all to look at G's quick 'n easy way until this evening. All I'd been using G for was to produce a good drawing and to check that my trig calcs were working.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

Gebra is powerful idea. If they ever get off their butt and add a few things to it, M might go the way of the Dodo.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,843

M's still got a while to run, then.

A least, that's the conclusion I draw from the following quote that seems to imply that dodos were around for close to 100 years following their discovery:

The extinction of the dodo within less than a century of its discovery...

*Last edited by phrontister (2014-06-30 00:46:55)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,400

I heard about that.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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