What do you think?

gAr · 2011-02-04 00:59:23

Yes!

bobbym · 2011-02-07 02:09:09

A real dilly?!

2,4,4,6,6,6,6,8,8,8,8,8,8,8,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10...

1) Find the 2012th term.

2) Sum the first 2012 terms.

3) What term has the last 2012?

gAr · 2011-02-07 06:44:22

Hi bobbym,

bobbym · 2011-02-07 13:50:23

Hi gAr;

gAr · 2011-02-07 17:50:08

cool!

Howardroark · 2011-02-11 15:15:08

Hi bobbym,

bobbym wrote:

A real dilly?!
2,4,4,6,6,6,6,8,8,8,8,8,8,8,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10...
1) Find the 2012th term.
2) Sum the first 2012 terms.
3) What term has the last 2012?

bobbym · 2011-02-11 15:24:12

Hi Howardroark;

That is correct, good work.

Howardroark · 2011-02-11 20:11:04

Hi bobbym,

Nice question

bobbym · 2011-02-12 23:08:16

Thanks for looking at the problem.

New problem!

A guy draws a rectangle. Near each vertex he draws a circle with radius 1 that is tangent to the adjacent sides. He fits a circle with radius 4 inside the rectangle and finds that the circle is tangent to all four of the other circles. See the diagram. What is the area of the rectangle?

A says) 81
B says) 81.57999
C says) 81.67677
D says) 81.67984

Who do you agree with?

Howardroark · 2011-02-13 00:32:24

Hi bobbym,

Howardroark · 2011-02-13 00:38:38

Hi bobbym,

gAr · 2011-02-13 01:07:21

Hi bobbym,

I don't think it has a unique solution. Under the same constraints, angle between the diagonals may be varied, isn't it?

bobbym · 2011-02-13 01:40:07

Hi gAr;

I do think it has a unique solution. The rectangle is bigger than the largest circle which has all four little circles touching it and the rectangle. Can you draw two different looking rectangles that work?

gAr · 2011-02-13 01:57:09

Hi bobbym,

Sorry for not reading the question properly. I thought we are drawing the circle first.
I'll try again.

bobbym · 2011-02-13 01:59:30

Hi;

That is okay, I welcome the input. I do not have a really good method on this problem so I am not sure either.

gAr · 2011-02-13 02:21:11

Hi,

But still, I'm wondering that the order of drawing either circle or rectangle doesn't matter. Since you haven't mentioned about the distance between the smaller circles, there is room to move those circles by a small distance. Hence the area is not fixed.
Hope I read the question properly this time!

I'm struggling to draw it, I'm no expert in geogebra!

bobbym · 2011-02-13 02:37:11

Hi gAr;

Howard has a solution very close to mine. But it is not an exact answer. I think because the unit circles are tangent to the rectangle and to the big circle that there is an exact answer.I will keep working on it.

gAr · 2011-02-13 02:53:13

Oh, ok, I thought of it again. They must have a unique solution indeed.
Sorry, I posted everything in haste!

phrontister · 2011-02-13 05:22:07

Hi Bobby,

From your drawing and the 1:4 radii it appeared to me at first glance that there might be a 5x4 grid of smaller circles within the rectangle...and so I produced a scaled drawing in Word.

After drawing the 5x4 grid with a centred large circle in it I drew a rectangle with sides that touch the outer edges of the outer circles.

The rectangle measures exactly 10x8 (area = 80). That's only graphically accurate, of course, and it doesn't work well with a couple of other circle dimensions I tried it on.

I couldn't find a formula from the drawing's info. I also tried some other ideas, including relating the areas of the small and large circles, but got nowhere.

Also, I don't know how to work out the length of the rectangle's diagonal, given that it doesn't pass through the centre of the corner circles.

gAr · 2011-02-13 06:14:40

Hi bobbym,

I thought the solution would be a square. Got the solution as: (2+5√2)² = 82.2843 sq. units.

But phrontister's solution appears to be good as well. Perhaps, there really are multiple solutions.

bobbym · 2011-02-13 06:19:13

Hi phrontister and gAr.

Yes, I asked for another example and phrontister has produced it. His is a proof without words so please post yours.

phrontister · 2011-02-13 14:06:04

Hi Bobby,

I don't see any solutions other than the two so far.

Also, my formula can't be the same as gAr's because the rectangle's shape relates to circle placement that differs in each case, resulting in different areas.

Off to work now...

bobbym · 2011-02-13 14:21:35

Hi Howardroark, gAr and phrontister;

Very nice solutions. More than I expected.

Howardroark · 2011-02-13 23:47:17

Hi bobbym,

Thank you..

phrontister · 2011-02-14 01:10:55

Thanks, Bobby.

Do you have a non-graphical solution for the area of the unequal-sided rectangle?

Math Is Fun Forum

#526 2011-02-04 00:59:23

Re: What do you think?

#527 2011-02-07 02:09:09

Re: What do you think?

#528 2011-02-07 06:44:22

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#529 2011-02-07 13:50:23

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#530 2011-02-07 17:50:08

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#532 2011-02-11 15:24:12

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#533 2011-02-11 20:11:04

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#534 2011-02-12 23:08:16

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#535 2011-02-13 00:32:24

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#536 2011-02-13 00:38:38

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#537 2011-02-13 01:07:21

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#543 2011-02-13 02:53:13

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#544 2011-02-13 05:22:07

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#545 2011-02-13 06:14:40

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#547 2011-02-13 14:06:04

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