Math Olympics Problem: Solving Equations with Fractions - Tips & Tricks

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SUMMARY

The Math Olympics problem involves solving the equation involving fractions: \(\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{a+b+d}+\frac{d}{a+b+c}=1\) and calculating the value of \(\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}\). Participants suggested various approaches, including simplifying the fractions into a single expression and leveraging symmetry in the variables. The consensus is that recognizing patterns and applying algebraic manipulation are key strategies for solving such problems efficiently.

PREREQUISITES
  • Understanding of algebraic fractions and their properties
  • Familiarity with symmetric sums and equations
  • Knowledge of mathematical problem-solving techniques
  • Ability to manipulate and simplify complex expressions
NEXT STEPS
  • Study techniques for simplifying algebraic fractions
  • Learn about symmetric sums in algebra
  • Explore advanced problem-solving strategies in competitive mathematics
  • Practice solving similar equations from math olympiads
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Students participating in math competitions, educators teaching algebraic concepts, and anyone interested in enhancing their problem-solving skills in mathematics.

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Homework Statement


I had this problem on the national math olympics of my country yesterday, I couldn't solve it... Knowing that the first statement is true, calculate the value of the second equation.

Homework Equations


[tex]\frac{a}{b+c+d}[/tex]+[tex]\frac{b}{c+d+a}[/tex]+[tex]\frac{c}{a+b+d}[/tex]+[tex]\frac{d}{a+b+c}[/tex]=1

Calculate the value of

[tex]\frac{a^2}{b+c+d}[/tex]+[tex]\frac{b^2}{c+d+a}[/tex]+[tex]\frac{c^2}{a+b+d}[/tex]+[tex]\frac{d^2}{a+b+c}[/tex]

The Attempt at a Solution



I lost a lot of time in this problem. I first tried to make one single, although huge, fraction of it, yet it didn't seem to simplify itself just to one value. Any better ways to approach this?

By the way, please excuse me if I posted this on the wrong forum, I just found this website and am still getting used to it. Also, since it isn't my homework or something I have to eventually give as homework, it doesn't matter if you show me directly the answer... I just posted it on this forum because it somewhat looked like the right one.
 
Last edited:
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Caculate the following:

[tex]\frac{a+b+c+d}{a+b+c+d}\left(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\right)=1[/tex]
 

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