June 7th, 2014, 01:08 PM  #1 
Member Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4  find Area of ABC 
June 7th, 2014, 04:09 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 
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. 
June 7th, 2014, 05:46 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, 06:49 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970  
June 7th, 2014, 06:53 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, 07:44 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 
I'm ONLY saying that triangle ABC shown by the OP will be a very small triangle, and simply picked triangle with sides 467 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. 
June 8th, 2014, 08:16 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 08:21 AM. Reason: error... 
June 8th, 2014, 12:01 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 
You got area = 144? Not much help if you don't show HOW... I'm sure 144 is way too high... 
June 8th, 2014, 12:16 PM  #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 $\displaystyle A^2 = p(pa)(pb)(pc) = \frac{a^4 + b^4 + c^4}{16} + \frac{a^2b^2 + a^2c^2 + b^2c^2}{8}.$ 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, $\displaystyle A = 12$. 
June 8th, 2014, 12:49 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
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