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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

An equilateral triangle ** P Q R ** is inscribed in a square ** A B C D **. P is on AD, Q on AB and R on CD.

What is the side length of the triangle if the side length of the square = 10 cm and the distance AP=3 cm.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi hammana;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

Hi Bobbym

Hi I am sorry. I choose the wrong value. I take this opportunity to reedit this exercise and make it more interersting.

An equilateral triangle P Q R is inscribed in a square A B C D . P is on AD, Q on AB and R on CD.

What is the side length of the triangle if the side length of the square = 10 cm and the distance AP=3 cm.

You can start working on numerical values and use numerical method to find the answer. It is more rewarding

to start with AD=a, Ap=d, calculate the litteral value of the side lenth of the triangle. I hope I am right this time

By the way, my professor of mathematics defined a mathematician as one who thiks a, writes b, says c , while it sould be d.

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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

I am sorry again. Answers given above are the square of the sidelength

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

Hi Bobbym

I see that you calculated first the position of Q and R. I failed to do it that way. Iwas interested in this problem after seeing the same topic recently posted in the "Help me" section where I lefted post #4. If you have time i would appreciate to have your comments on my post there

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi hammana;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

I wanted to slove the 3 equations you wrote in your reply in the litteral form, replacing "10" by "a" the side of the square and "3" by "d", the distance AP, so that the values of x1 and x2 are obtained as functions of "a" and "d". I found by other approaches of the problem that x1 and x2 have very simple values:

x1=(2*a-d)/sqr(3) and x2=(a+d)/sqr(3)

which I could not deduce from the previously mentioned equations.

I was also surprised to see that the sum x1+x2 is independant of d, which means that when then vertex P moves along the side AD, the middle of the opposite side QR remains fixed

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi hammana;

There are many ways to a correct answer.

I do not suppose that you read "Commonsense, Brute Force and the Computer..."

Probably nobody did because the guy who wrote it never got it published. Although it deals with programming it is often applicable.

If someone writes a tremendously deep program that runs in under a second he would be happy. Someone else writes a program to do the same task but it takes 10 seconds. Is his program bad?

Not necessarily, when we find that the first program took 5 days to develope and the second person's 2 days.

The real running time of the two programs is 5 days and 1 second versus 2 days and 10 seconds. From this vantage point the second guy has the superior program. What I am getting at is solutions like programs have different ratings of simplicity and elegance based on how you view them.

I generally work towards a solution that can be solved using computational methods. I do not veer off much from that strategy. This way is natural and quick for me. If some other way is more natural to you then by all means use it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**hammana****Member**- Registered: 2012-03-02
- Posts: 48

Hi bobbym

Since I have plenty of free time, as a retired Telecom engineer, I can afford the luxury of looking for elegant solutions. My first and natural approach to this exercise was to introduce 3 variables: s=sidelength of the triangle, x1=DR, x2=AQ. I ended with 3 equations on which I spent much time without success. Then I thought to introduce 2 varaibles an the problem could be solved easisly

Yesterday I found that the solution is evident and perhaps the problem is not worth to be proposed as an exercise

I know that the running time of a program (or the solution of a problem) is inversely proportional to the time you spend for it, and I thank you for the time you spent for me.

I wanted to learn something about Latex to improve my presentation, but where the tutorial which appeared usually on top of the page has disappeared?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi hammana;

Thanks for providing your solution.

I use this place for all my latexing. No downloads, pull down menus, clean latex every time.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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