Thursday, April 6, 2006

9th Degree? Topic: Algebra. Level: Olympiad.

Problem: (Math Olympiad Treasures - 1.4.8) Solve the equation

$x+a^3 = \sqrt[3]{a-x}$,

where $a$ is a real parameter.

Solution: We could cube and try to solve a cubic in $x$ or a $9$th degree polynomial in $a$, but neither of those sound particularly easy or fun. Consider the function $f(a) = x+a^3$ in $a$. We have $f^{-1}(a) = \sqrt[3]{a-x}$.

Our given equation becomes $f(a) = f^{-1}(a) \Rightarrow f(f(a)) = a$. We claim that $f(a) = a$.

Suppose $f(a) > a$. Since $f$ is a strictly increasing function, we have $f(f(a)) > f(a) > a$, contradiction. Similarly, if $f(a) < a$ then $f(f(a)) < f(a) < a$, another contradiction. Hence $f(a) = a$, as claimed.

So it remains to solve $x+a^3 = a$, which yields the only solution $x = a-a^3$. QED.

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Comment: This changing of variables required a little more ingenuity than the last one. Upon noticing that the two sides of the equation are in fact inverses of each other, proceeding from there is pretty simple. Discovering this fact is the key to solving the problem.

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Practice Problem: (Math Olympiad Treasures - 1.4.5) Let $a \in \left(0,\frac{1}{4}\right)$. Solve the equation

$\displaystyle x^2+2ax+\frac{1}{16} = -a+\sqrt{a^2+x-\frac{1}{16}}$.