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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Feb 7, 2012

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Changing coordinates if necessary, we can assume that the middle one of the three inflection points is at the origin. Let the other two inflection points be at $x=-a$ and $x=b$, where $a,b>0$.

Then $f''(x) = kx(x+a)(x-b) = k(x^3 + (a-b)x^2 - abx)$. The value of the constant $k$ does not affect the calculations except to cause clutter, so I will assume that $k=1$. Then $f(x) = \frac1{20}x^5 + \frac1{12}(a-b)x^4 - \frac1{6}abx^3 + cx$ for some constant $c$. (The constant term in $f(x)$ is zero because the curve passes through the point of inflection at the origin.)

Next, $f(-a) = -\frac1{20}a^5 + \frac1{12}(a-b)a^4 + \frac1{6}a^4b - ca = \frac1{30}a^5 + \frac1{12}a^4b - ac$, and similarly $f(b) = -\frac1{30}b^5 - \frac1{12}a^4b + bc$. But the points $(-a,f(-a))$ and $(b,f(b))$ lie on a straight line through the origin. Therefore $\dfrac{f(-a)}{-a} = \dfrac{f(b)}b$, so that $\frac1{30}a^5 + \frac1{12}a^4b = \frac1{30}b^5 + \frac1{12}a^4b,$ which simplifies to $(b^2 - a^2)(12a^2 + 30ab + 12b^2) = 0.$ Since the second term in that product is positive, it follows that $b^2 - a^2=0$ and so $b=a$. Therefore $f(x) = \frac1{20}x^5 - \frac1{6}a^2x^3 + cx$.

\begin{tikzpicture}

[scale=3]\draw [help lines, ->] (-1.75,0) -- (1.75,0) ;

\draw [help lines, ->] (0,-1) -- (0,1) ;

\draw [help lines] (-1,0) -- (-1,0.5) ;

\draw [help lines] (1,-0.5) -- (1,0) ;

\draw[ domain=-1.75:1.75, samples=100] plot (\x,0.6*\x^5 - 2*\x^3 + \x);

\draw (-1.75, 0.7) -- (1.75,-0.7) ;

\draw (-1,-0.1) node {$-a$} ;

\draw (1,0.1) node {$a$} ;

\draw (-1.3,0.7) node {$\color{red}A$} ;

\draw (-0.5,0) node {$\color{red}B$} ;

\draw (0.5,0) node {$\color{red}C$} ;

\draw (1.3,-0.7) node {$\color{red}D$} ;

\end{tikzpicture}

The difference between the quintic polynomial and the straight line is $\frac1{60}x(3x^4 - 10a^2x^2 + 7a^4) = \frac1{60}x(x^2-a^2)(3x^2 - 7a^2)$. The points of intersection are at $x = \pm a$ and $x = \pm\sqrt{\frac73}a$. Integrating over the appropriate intervals, I found the ratios of the areas $A:B:C:{D}$ to be $32:81:81:32$. But my arithmetic is unreliable, so please check those numbers.