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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi;

This one is hard ( I am reminded that it is not for some ) but geogebra with a lot of work and some luck makes it easy:

Rhombus, ABCD, has sides of length 12. A circle with center A passes through C. Another circle with center B passes through D. The two circles are tangent to each other.

Can you find the area of the rhombus?

Let's see what geogebra can really do! Now I am not going to pretend that I got this immediately. It took a whole day of experimenting with circles before I realized what the correct structure is.

1) Make a slider, it will be called a. Set a to 0 in it.

2) Create points (a,2) and (20,2).

3) Draw line segment AB.

4) Get the midpoint of AB, it will be called C.

5) Draw a perpendicular line through C to AB.

6) Use the circle and radius tool to draw a circle with radius 12 with center at A.

7) Get the two points of intersection of the circle and the perpendicular line through C. They will be called E and D.

8) Use the polygon tool on A, E, B and D.

9) Hide the circle, the perpendicular, the line segment and point C. You are left with just the rhombus and the slider.

10) Rename B to C and E to B.

11) Use the circle with center through a point tool and click A and then C. A circle with A as the center passing through C is created. Do the same with B and D to create a circle with center at B and passing through D.

Now we need to get the smaller circle tangent to the bigger one. Now we can see why the slider a was created. The key is the following technique:

12) Find the points of intersection of the two circles by using the intersection tool. Points E and F will be created. To be tangent those points will have to merge into a single point.

13) Move a on the slider until the little circle looks tangent to the big circle or when E and F are undefined in the algebra. Get it as close as you can by eye.

14) Set the rounding to 15 digits in the option menu. If you have good eyes and are careful you will see that E and D are undefined ( meaning they are not contacting the larger circle ) and a = - 1.90000000000000.

15) Set the increment in the slider to .01 and use the shift right arrow until the E and F appear then back it off one click. It should say -1.875 for a. Repeat by setting the increment to .001.

16) Use smaller and smaller increments until you go mad or you reach a = -1.87450786700001

Now here is the payoff, look in the algebra pane at the value associated with poly 1.

poly1 = 107.99999998818478

Doesn't take too big a leap of faith to conjecture 108 is the answer. You would be right too.

Check your work with the drawing below.

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

I got 180.3 using Sketchpad.

Cannot see a geometric solution at the moment.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi Bob;

There is a analytical solution to this which gives 108. It involves vectors! Yecch!

Are you sure of the drawing you made with sketchpad?

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi bobbym,

My diagram has BD as the longer diagonal. Other than that, it is the same as yours. So if I re-name AC as BD and BD as AC, we have the same.

My problem is in making AB = 12. I did this by setting up a measure for the length and then stretching the line until it reached the right length. The closest I could get was 11.99 or 12.01. I could probably do the construction another way and fix the 12 but I was close enough to be satisfied I was on the right lines.

Vectors. Hmm, thinks....... Thanks for the hint; I'll have another go.

Bob.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi;

What are lengths of your diagonals?

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi bobbym,

I tried a new approach but I've hit the accuracy problem again.

I set up a square grid and chose 'snap points' so I could set AB to 'exactly' 12.

Then I made the rhombus by equal sides and parallel lines and then the circles.

I put point E on the centre B circle at the point where AB cuts it and F on the circle centre A where AB cuts it.

In theory the tangent is achieved when E and F coincide.

See diagram below. Even though I 'set' AB = 12, it measures as 11.98563. Huh!

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi;

The area of that rhombus is about 107.76, so you are close.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

bobbym wrote:

Hi Bob;

There is a analytical solution to this which gives 108. It involves vectors! Yecch!

Are you sure of the drawing you made with sketchpad?

I got it without vectors...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

I know that and I have been waiting.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Waiting? For my solution?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

It might help Bob to know what the correct answer is. I think once he gets the sides to 12 he will get it.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Well, I just noticed that the point where the circles touch is on the same line as A and B (because they are the centers of the two circles), and got from that that R=AB+r, that is, AC=AB+BD. Now we can use the Cosine Rule to get

,where alpha is the acute angle of the rhombus. We can get from that equation the cosine of the angle and thus also its sine. We then use the formula for calculating the area of parallelogram using its sides and internal angle.

*Last edited by anonimnystefy (2013-04-18 07:29:07)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi Stefy,

Sounds brilliant but too many steps missing for my little brain.

Which triangle(s) are you using for the cosine rule?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

ABD with the angle alpha and ABC with angle 180-alpha.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

Thanks. Working on that now.

B

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Okay. Post when you get the equations.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi Stefy,

Your idea and then I followed a different path:

Triangle DBA

Triangle ABC

Using R = 12 + r this becomes

Eliminate alpha by adding

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Hm, how did you get the second to last row?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi Stefy,

There are three equations ( two from cosine rule and the straight line equation) and there are three unknowns (R,r and alpha). So it should be possible to eliminate two unknowns to find the third. Both the cosine rule equations have the same term in cos alpha but in one it is negative. So adding the equations together (LHS1 + LHS2 = RHS1 + RHS2) gets alpha out straight away. But I eliminated R first before the adding so I had a quadratic in r straight away.

Thanks for giving me the start. I had stared at that diagram for ages without thinking of that. Too distracted trying to use angle properties.

I think the expression for r and R could be left in surd form and then the area obtained without any recourse to decimals. Might try it later. That way you get to show the answer is exactly 180.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Ah, I get it. Great! Do you use SketchPad to get the area?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi Stefy,

I just did 1/2 product of the diagonals. Sketchpad does have an area function, but I've never used it.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Okay. If you leave the surds as is without simplification you will also get the exact answer of 108.

*Last edited by anonimnystefy (2013-04-19 02:09:42)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Don't you mean 108?

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,840

Yes, sorry. I edited it.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

Just to finish it off:

QED.

Bob

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