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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Then I used a couple of more relationships and the triangle inequality to test the integers from 7 to 84. 83 was the biggest.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Well, the 83 comes directly from the upper bound.

anonimnystefy wrote:

And, what exactly are we doing now. We have the value, we have constructed the triangle. What else is there to do?

*Last edited by anonimnystefy (2013-05-30 05:32:37)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

The upper bound I showed is 84. 83 works but 84 does not.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Well, of course it doesn't work. The triangle inequality states a+b>c, not >=c.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It took a little M code to try from 7 to 84.

**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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**anonimnystefy****Real Member**- From: The Foundation
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There was no need to try them, but whatever.

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: 86,636

With what I dug up, I had to try them all. What did you have to shorten that?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Well, I had the fact that the lengths 1/12, 1/14 and 1/hc must form a triangle, where 1/hc is as small as possible. Because of the triangle inequality we have that 1/hc>1/12-1/14=1/84. So, hc<84. The maximum possible length is 83.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Okay but what about this?

1/hc>1/12-1/14

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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What about it?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Why the minus and not a plus?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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The triangle inequality states that 1/hc+1/14>1/12.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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What triangle? That inequality is for the sides. You have altitudes there.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Yes, but, as I already said, 1/ha=1/12, 1/hb=1/14 and 1/hc must be length of some triangle in order to be a valid set of altitudes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I do not get it but it works so the problem is done.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Well, it is because a*ha=b*hb=c*hc. From this we can get a:b:c=(1/ha):(1/hb):(1/hc), which means that, if a, b and c can form a triangle, 1/ha, 1/hb and 1/hc must be able to form a triangle as well.

*Last edited by anonimnystefy (2013-05-30 06:58:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Okay, see you later.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Okay, see you!

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

I have a different solution.

*Last edited by ElainaVW (2013-06-01 00:28:27)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

What is it, please post what you have.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

[Code fixed by admin]

```
Solve[{12 ==1/a Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)), 14 == 1/b Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)),
83 == 1/c Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)), 12 == (b*c)/(2 R)}, {a, b, c, R}] // N
```

Only had to try 84 and 83.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

That is very good. Nice work.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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