Some Abstract Calculus

First, let's look at the problem qualitatively in order to make a plan of attack, since the question doesn't give a clear-cut direction to go.

If we make an arbitrary line in the $x,y$ plane, we can begin looking for general polynomials to satisfy the conditions:

• A line will only intersect another line once, so degree one is out
• A parabola won't work either, since if monic it will be concave up over all $x \in \mathbb{R}$ . Thus, if it is ever tangent to $L(x)$ for some $x=\alpha$ , $f(x)>L(x)$ for $x \neq \alpha$ , and will never intersect again.
• A cubic polynomial will work, since it does have a change in concavity. We can see by drawing a sketch that if $f(x)$ is monic, then $k<0$ in order to satisfy all conditions and $f(x)$ will touch $L(x)$ from above.

Condition I: By the given conditions, we know that $p,q$ are the solutions to $L(x)=f(x)$ . More importantly, for tangent curves to occur at $x=p$ , then $p$ must be a double root. We can prove this by looking at why tangency occurs; $f(p)-L(p)=0$ , and for $x$ right around $p$ , $f(p)-L(p)>0$ .

We know that $f(x)-L(x)=(x-p)(x-q)(x-z)$ for an unknown $z$ . As $x$ increases and crosses $p$ , $f(x)-L(x)$ will change sign unless $z,q=p$ , thus $p$ must be a double root for tangency to occur. It was stated that $p>q$ in the problem, so $z=p$ .

This gives $f(x)-L(x)=(x-p)^{2}(x-q)$ , or adding $L(x)=kx$ to both sides, $f(x)=(x-p)^{2}(x-q)+kx$ .

Condition II: For $f(x)$ to intersect $L(x)$ perpendicularly at $x=q$ , we know that $f(q)=L(q)$ and $f'(q)=-\frac{1}{L'(q)}$ .

The first condition is already met, since $f(q)=L(q)=kq$ .

The second condition can be solved for. $f'(x)=k+(p-x)(p+2q-3x)=\frac{-1}{L'(x)}=\frac{-1}{k}$ . Plugging in $q$ for $x$ gets $0=q^{2}-2pq+p^{2}+k+\frac{1}{k}$ . This is a quadratic in $q$ :

$q=\frac{2p \pm \sqrt{4p^{2}-(4p^{2}+4(k+\frac{1}{k})}}{2}=p \pm \sqrt{-(k+\frac{1}{k})}$

$q<p$ , so keep the minus solution. Note that $k<0$ , otherwise we don't get a real solution for $q$ . This supports our initial assessment that $k<0$ for the conditions to be met.

Now $p-q=\sqrt{-(k+\frac{1}{k})}$ . The minimum value of this is $\sqrt{2}$ , thus $d^{2}=\boxed{2}$

First, we have to figure out the least degree of the polynomial $f$ that satisfies the conditions. One way to do so is to consider the number of roots that it has. However, since the conditions involve the linear function $L(x) = kx$ , we don't have information that is purely about $f(x)$ .

Instead, if we consider the function $g(x) = f(x) - kx$ , we observe that the conditions can be rewritten as:

1. $g(x)$ is tangent to the x-axis at some value $x =p$ .
2. $g(q) = 0$ (along with some funky information that doesn't translate well as yet)

The first condition tells us that that $g(x)$ has a double root at $p$ , and a single root at $q$ . Hence, $g(x)$ must have degree at least 3, and is of the form $g(x) = h(x) (x-p)^2 (x-q)$ for some polynomial $h(x)$ . Since $f(x)$ is of minimial degree, hence $f(x) = g(x) + kx$ has degree exactly 3, and so $h(x)$ is a constant. Since $f(x)$ is monic, this implies that $h(x) = 1$ . Thus, $f(x) = (x-p)^2 (x-q) + kx$ .

Second, now that we know the form of $f(x)$ , the only missing information is the "funky information that doesn't translate well as yet". Let's dig into it. (Taken mostly from Brandon's solution)

For $f(x)$ to intersect $L(x)$ perpendicularly at $x=q$ , we know that $f(q)=L(q)$ (which is met by our construction of $f(x)$ and

$f'(q)=-\frac{1}{L'(q)} \Rightarrow k+(p-q)(p+2q-3q)=\frac{-1}{k} \\ \Rightarrow 0=q^{2}-2pq+p^{2}+k+\frac{1}{k}$

Treating this as a quadratic in $q$ , we obtain

$q=\frac{2p \pm \sqrt{4p^{2}-(4p^{2}+4(k+\frac{1}{k})}}{2}=p \pm \sqrt{-(k+\frac{1}{k})}.$

Since $q<p$ , so we take the minus sign. Note that we must have $k<0$ , otherwise we don't get a real solution for $q$ .

Now $p-q=\sqrt{-(k+\frac{1}{k})}$ . The minimum of this function is acheived at $k = - 1$ , and we can set $p = 0$ and $q = - \sqrt{2}$ which satisfy the original conditions. (In fact, the above shows that the conditions are satisfied whenever $k = -1, p-q= \sqrt{2}$ .)