My Math Forum find Area of ABC

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 June 7th, 2014, 12:08 PM #1 Member     Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4 find Area of ABC
 June 7th, 2014, 03:09 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,280 Thanks: 931 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 03:31 PM.
 June 7th, 2014, 04:46 PM #3 Member   Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 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
June 7th, 2014, 05:49 PM   #4
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Quote:
 Originally Posted by nmenumber1 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...

 June 7th, 2014, 05:53 PM #5 Member   Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 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
 June 8th, 2014, 06:44 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,280 Thanks: 931 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 07:02 AM.
 June 8th, 2014, 07:16 AM #7 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics I got $\displaystyle A = \frac{1057}{2}$. //Edit: error... $\displaystyle A = 144$ Last edited by fysmat; June 8th, 2014 at 07:21 AM. Reason: error...
 June 8th, 2014, 11:01 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,280 Thanks: 931 You got area = 144? Not much help if you don't show HOW... I'm sure 144 is way too high...
 June 8th, 2014, 11:16 AM #9 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics 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
 June 8th, 2014, 11:49 AM #10 Global Moderator   Joined: Dec 2006 Posts: 19,698 Thanks: 1804 Please don't post links to images hosted at mathmontada.net.

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