MHB Can a Straight Line Intersect a Curve in Four Distinct Points?

[Ackbach]

Here is this week's POTW:

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For which real numbers $c$ is there a straight line that intersects the curve $x^4+9x^3+cx^2+9x+4$ in four distinct points?

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[Ackbach]

Congratulations to MarkFL, kiwi, Opalg, and kaliprasad for their correct solutions. kiwi's solution follows:

Spoiler

Hypothesis: The required line can be formed iff the curvature of p(x) changes sign twice.

Proof:
Suppose we can draw the required line through p(x). We can rotate and translate the axes until this line coincides with the x axis. By inspection in this frame p(x) has either two maxima separated by a minima or it has two minima separated by a maxima. At the maxima the curvature has opposite sign to that at the minima so curvature of p(x) changes sign twice.

Conversely, suppose the curvature of p(x) changes sign twice. We divide p(x) into three parts each with curvature of just one sign. We now draw a line that intersects the centre part twice. The remaining (left and right) parts each intersect the line once and we are done.

The curvature of p(x) is the inverse of the radius of curvature and is given by:
\[\frac 1R = \frac{p''(x)}{(1+[p'(x)]^2)^{\frac 32}}\]

The curvature changes sign twice so $p''(x)$ must have two distinct roots.

\[p(x)=x^4+9x^3+cx^2+9x+4\]

\[\therefore p''(x)=12x^2+54x+2c\]

The discriminant is \(54^2-96c\)

which is zero when \(c=30\frac{3}{8}\).

Therefore, there are two changes of curvature and hence the required line can be drawn iff \(c \lt 30\frac 38\)