Grapher in Mac OS X

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# Grapher in Mac OS X

Submitted by SKungen on Tue, 11/02/2010 - 15:34

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Grapher in Mac OS X

If you have not seen it, I recommend it!

I am not able to compare it to other mathematic tools.

Best Regards,

/Peter

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## Re: Grapher in Mac OS X

Dear Pedro

The solution for this month's ponder is published.

There are120 (five factorial) ways for the order of the edges and two ways to place the right angles (either adjacent or not).

Once we define the order and place the right angles, we can compute the area using the Pythagorean theorem and Heron's formula.

Reviewing all these possibilities for the lake, we can verify that there is only one possible area for a convex pentagon.

The smallest solution we found is 2=4=12-6-5 (where "=" is a right angle) with an area of â†33.99.

The smallest lake we know of that has an integer for its area is 7=5=22-3-13 (or 3=5=22-7-13), which has an area of 80.

The 22=19=81-37-25 lake has a nice round area: 1000.

Strange thing is I still wonder, how they have ruled out all other possibilities for a convex pentagon.

Thanks & Regards

Dave

## Re: october ponder this

Hey Steve,

you don't say which is the only area of the lake.

Heron formula is clearly unnecessary to calculate the area of the pentagon.

perucho

## Re: october ponder this

Dear Pedro

Yes, Herron's formula is not needed and moreover I still feel that there are more than 2 solutions for convex pentagon.

Thanks & Regards

Dave

## Re: october ponder this

No Dave, there is a lot of solutions and I'm sorry for IBM. Please take any of the pentagons that I wrote here in PM (my short table) and you will realize that these pentagons accomplish with all the conditions of this problem, i.e.

1.- They are convex

2.- They have 2 right angles

3.- a < b < c < d < e, integers

I agree about that there exist 120 different ways to put the edges. I just only have considered those cases where the right angles are not adjacent. Since you and I already knew that the solution is not unique, I was concentrated just only on the more simple configuration, that is, three right triangles.

Sheers,

perucho

PS. Still you don't have given me that unique solution that IBM is claiming.

## Re: october ponder this

Dear Pedro

Here is the answer they have published

For non-integer area: 2=4=12-6-5 (where "=" is a right angle) with an area of 33.99.

Smallest integer area: 7=5=22-3-13 (or 3=5=22-7-13), which has an area of 80.

I think the better way they could have opted for posting this question is "minimal possible non-integer area", I think then it would have been unambigious. As we know there are amny possibile areas, we could have chosen the minimal possible out of 120 options.

Thanks & Regards

Parashar Dave

## Re: october ponder this

Totally correct Dave. You may take any of my pentagons, and you will realizzze that they are solutios of this problem.

Greetings,

perucho

## Re: Grapher in Mac OS X

I have 10.6, Snow Leopard, but I knew earlier versions have Grapher too.

I also have Mathematica in an old version, but It is a bit cumbersome to use on my laptop, although I'm glad it works at all!

Best Wishes,

/Peter

## Re: Grapher in Mac OS X

Peter my friend,

are you talking about Leopard?

Pedro