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June 7th, 2014, 01:08 PM   #1
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find Area of ABC

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June 7th, 2014, 04:09 PM   #2
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Huh?!
Well, won't be a biggie, since 4^4 + 6^4 + 7^4 = 3953

I make a^2 + b^2 + c^2 = 100 , anybody agree?

Last edited by Denis; June 7th, 2014 at 04:31 PM.
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June 7th, 2014, 05:46 PM   #3
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man, please explain how that is no biggie, I need to understand all this, to find and answer of sum of 3 numbers to fourth powers I am missing something big here lol

and then what does you second part mean, sorry but I like to understand what is going on in these questions
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June 7th, 2014, 06:49 PM   #4
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Quote:
Originally Posted by nmenumber1 View Post
man, please explain how that is no biggie...
Problem states a^4 + b^4 + c^4 = 3848

I've stated that a triangle with sides a=4, b=6 and c=7 results in:
4^4 + 6^4 + 7^4 = 3953

So obviously a "small" triangle...
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June 7th, 2014, 06:53 PM   #5
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yeh, but how did you know that triple? cant be off the top of your head?

sorry don't want to be annoying, just want to understand
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June 8th, 2014, 07:44 AM   #6
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I'm ONLY saying that triangle ABC shown by the OP will be a very small
triangle, and simply picked triangle with sides 4-6-7 to illustrate.

Also, since 8^4 = 4096, then longest side has to be < 8.

Was simply making a "comment", nothing to do with solving.

My "a^2 + b^2 + c^2 = 100" is from using fact that the 3 medians are:
SQRT[(2(b^2 + c^2) - a^2] / 2
SQRT[(2(a^2 + c^2) - b^2] / 2
SQRT[(2(a^2 + b^2) - c^2] / 2
and the sum of the squares of those is given as 75.

Doing substitutions then leads to this quadratic:
a^4 + (b^2 - 100)a^2 + b^4 - 100b^2 + 3076 = 0

I'm stuck there !

Last edited by Denis; June 8th, 2014 at 08:02 AM.
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June 8th, 2014, 08:16 AM   #7
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I got $\displaystyle A = \frac{1057}{2}$.

//Edit: error... $\displaystyle A = 144$

Last edited by fysmat; June 8th, 2014 at 08:21 AM. Reason: error...
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June 8th, 2014, 12:01 PM   #8
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You got area = 144?

Not much help if you don't show HOW...
I'm sure 144 is way too high...
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June 8th, 2014, 12:16 PM   #9
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Well, errors are very much possible since I calculated this very quickly. So let's see more carefully:

It's given that

$\displaystyle \begin{cases} a^4 + b^4 + c^4 = 3848 \\ m_a^2 + m_b^2 + m_c^2 = 75 \end{cases}.$

First, one can use the Heron's formula and get

$\displaystyle A^2 = p(p-a)(p-b)(p-c) = -\frac{a^4 + b^4 + c^4}{16} + \frac{a^2b^2 + a^2c^2 + b^2c^2}{8}.$

The first fraction can be evaluated to be $\displaystyle -\frac{3848}{16}$. Once one writes the squares of medians and sum's them up, one get

$\displaystyle 4(m_a^2 + m_b^2 + m_c^2) = 3(a^2 + b^2 + c^2)$

and thus

$\displaystyle a^2 + b^2 + c^2 = 100$.

If one then takes the square of that, one gets

$\displaystyle 100^2 = (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + a^2c^2 + b^2c^2) = 3848 + 2(a^2b^2 + a^2c^2 + b^2c^2)$

from which the $\displaystyle a^2b^2 + a^2c^2 + b^2c^2$ part can be evaluated.

Now just pick up the terms:

$\displaystyle A^2 = -\frac{3848}{16} + \frac{100^2 - 3848}{16} = 144.$

And finally, by noticing one more error, $\displaystyle A = 12$.
Thanks from Denis
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June 8th, 2014, 12:49 PM   #10
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