Directrix will Drive you Delirious

EDIT: John Haussmann has provided us with a slightly better solution. Read his simplification in the comments below.

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First, we can notice that if the directrix of $\mathbf{P}$ is tangent with $\mathbf{P'}$ , then the directrix of $\mathbf{P'}$ is tangent with $\mathbf{P}$ . Therefore, all we need to do is find the angle $\theta$ such that the directrix of $\mathbf{P}$ , when rotated $\theta$ degrees, is tangent with $\mathbf{P}$ .

We can create the equations of two lines: the equation of a line tangent to $\mathbf{P}$ , and the equation of a line that is achieved from rotating the directrix of $\mathbf{P}$ a total of $\theta$ degrees. Let's tackle the first one first, since it seems to be easier.

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Suppose that the x-coordinate of the point at which the line is tangent to $\mathbf{P}$ is $a$ . Because $\dfrac{\text{d}}{\text{d}x}x^2=2x$ , the slope of the line is $2a$ . We also know that the line passes through the point $(a,a^2)$ . Therefore, we can represent the equation in point-slope form: $\begin{aligned}y-a^2&=2a(x-a)\\ y&=2a(x-a)+a^2\\ y&=\boxed{2ax-a^2}\end{aligned}$ and we have found the equation of the first line.

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The second line is a bit harder to find the equation of. Call the focus of $\mathbf{P}$ $F$ , the intersection of the directrix and the line of symmetry of $\mathbf{P}$ as $H$ , and the image of $H$ after rotating it $\theta$ degrees about $F$ as $H'$ . We see that $\angle HFH'=\theta$ . Also, $FH=FH'=\dfrac{1}{2}$ . If we find the point $H'$ in terms of $\theta$ , and then find the slope of the line in terms of $\theta$ , then we have the equation of the line.

We see that the slope of the line is simply $\tan\theta$ , because the angle that the directrix of $\mathbf{P'}$ crosses the x-axis is $\theta$ . The point $H'$ is a little harder to define, but using our knowledge of the unit circle will solve it quickly. We can see that the x-coordinate of $H'$ is $\dfrac{1}{2}\sin\theta$ , and the y-coordinate of $H'$ is $\dfrac{1}{4}-\dfrac{1}{2}\cos\theta$ (note that the locus of $H'$ is a circle with center $\left(0,\dfrac{1}{4}\right)$ and radius $\dfrac{1}{2}$ ). Therefore, the equation of our line is

$\begin{aligned}y-\left(\dfrac{1}{4}-\dfrac{1}{2}\cos\theta\right)&=\tan\theta\left(x-\dfrac{1}{2}\sin\theta\right)\\ y&=\tan\theta\left(x-\dfrac{1}{2}\sin\theta\right)+\left(\dfrac{1}{4}-\dfrac{1}{2}\cos\theta\right)\\ y&= \boxed{(\tan\theta)x+\left(\dfrac{1}{4}-\dfrac{1}{2}\sin\theta\tan\theta-\dfrac{1}{2}\cos\theta\right)}\end{aligned}$

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We know that the two lines are in fact the same line, because the directrix of $\mathbf{P'}$ (equation 2) is tangent to $\mathbf{P}$ (equation 1). Therefore, we set these two equations equal. Setting coefficients equal, we have a system of equations: $\left\{\begin{array}{l}2a=\tan\theta \\ -a^2=\dfrac{1}{4}-\dfrac{1}{2}\sin\theta\tan\theta-\dfrac{1}{2}\cos\theta\end{array}\right.$

From the second equation, we see that $\begin{aligned}-a^2&=\dfrac{1}{4}-\dfrac{1}{2}(\sin\theta\tan\theta+\cos\theta)\\ &= \dfrac{1}{4}-\dfrac{1}{2}\left(\dfrac{\sin^2\theta}{\cos\theta}+\cos\theta\right)\\ &=\dfrac{1}{4}-\dfrac{1}{2}\left(\dfrac{1}{\cos\theta}\right) \\ a^2&=\dfrac{1}{2\cos\theta}-\dfrac{1}{4}\\ \cos\theta &=\dfrac{1}{2a^2+\dfrac{1}{2}}\\ \cos\theta &=\dfrac{2}{4a^2+1}\end{aligned}$

From the first equation, we have that $\dfrac{\sin\theta}{\cos\theta}=2a$ . Substituting our value of $\cos\theta$ in to this, we get that $\sin\theta=\dfrac{4a}{4a^2+1}$ .

By using the property $\sin^2\theta+\cos^2\theta=1$ , we see that: $\begin{aligned}\left(\dfrac{2}{4a^2+1}\right)^2+\left(\dfrac{4a}{4a^2+1}\right)^2&=1\\ \dfrac{16a^2+4}{(4a^2+1)^2}&=1\\ 16a^2+4&= 16a^4+8a^2+1\\ 16a^4-8a^2-3&=0\\ (4a^2-3)(4a^2+1)&=0\\ a^2&=\dfrac{3}{4}\\a&=\pm \dfrac{\sqrt{3}}{2}\end{aligned}$

We know we're on the right track now, because we know that the directrix of $\mathbf{P'}$ can be tangent to $\mathbf{P}$ at two places, and these two places are opposites of each other! We can see now that the equation of the line is $2\left(\dfrac{\sqrt{3}}{2}\right)x-\left(\dfrac{\sqrt{3}}{2}\right)^2=x\sqrt{3}-\dfrac{3}{4}$ .

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Now, the smallest possible x-coordinate of all the points on $\mathbf{P'}$ corresponds to the lowest possible line perpendicular to $y=x\sqrt{3}-\dfrac{3}{4}$ that still intersects $\mathbf{P}$ . Therefore, we want the line tangent to $\mathbf{P}$ and perpendicular to $y=x\sqrt{3}-\dfrac{3}{4}$ . Our desired answer would be the shortest distance from this new line to $FH'$ (remember our labels back there?) The equation of the new line has slope $-\dfrac{\sqrt{3}}{3}$ , the negative reciprocal of $\sqrt{3}$ .

We know that this line is tangent to $\mathbf{P}$ , and we know the slope of the line; therefore, we know the point of tangency. Since the slope of a point with x-coordinate $a$ is $2a$ , the x-coordinate of the point of tangency is $-\dfrac{\sqrt{3}}{6}$ and the y-coordinate is $\left(\dfrac{\sqrt{3}}{6}\right)^2=\dfrac{1}{12}$ . Our equation is therefore $\begin{aligned}y-\dfrac{1}{12}&=-\dfrac{\sqrt{3}}{3}\left(x+\dfrac{\sqrt{3}}{6}\right)\\ y=-\dfrac{\sqrt{3}}{3}x-\dfrac{1}{12}\end{aligned}$

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All we need to do is find the shortest distance between $y=-\dfrac{\sqrt{3}}{3}x-\dfrac{1}{12}$ and $FH'$ . We do not know the equation of line $FH'$ , but we do know a point on $FH'$ : namely, $F$ . It's coordinates are $\left(0,\dfrac{1}{4}\right)$ . Using the formula for the shortest distance between a line and a point: $\dfrac{\dfrac{1}{4}+\dfrac{\sqrt{3}}{3}\cdot 0+\dfrac{1}{12}}{\sqrt{\left(-\dfrac{\sqrt{3}}{3}\right)^2+1}}=\dfrac{\sqrt{3}}{6}$ . Our desired answer is therefore $3+6=\boxed{9}$ and we are done. $\Box$

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Problem Maker's Note: This solution is waaaay too long. There is almost definitely a shortcut somewhere in there, or an easier way, because $\theta$ turns out to equal $60^{\circ}$ , which can't be a coincidence. However, right now I can't really think anymore from the sheer LaTeX overload, so I will not revise this solution any further. Thanks for reading.

This problem drove me quite delirious, from the sheer effort of making, solving, rechecking, finding an error, resolving, rechecking, and writing a solution. It easily beat my previous record for longest time spent working on a problem, at around 2 to 2.5 hours. So it would be nice if you spread the word. :D

Call y=ax+b that is tangent of Parabola (P) at $(x_0, y_0)$ denoted by (TT): Focus point is F(0, 1/4), directrix line (D) of P is: y=-1/4; $y=2x_0x+b, y_0=2x_0^{2}+b=>b=y_0-2x_0^{2}=-x_0^{2}, y=2x_0x-x_0^{2}<=>2x_0x-y-x_0^{2}=0$ $distance(F,TT)=|-1/4-x_0^{2}|/ \sqrt{4x_0^{2}+1}=|1/4+x_0^{2}|/ \sqrt{4x_0^{2}+1}=\frac{\sqrt{4x_0^2+1}}{4}$ $distance(F,D)=1/2=>\sqrt{4x_0^{2}+1}/4=1/2=>4x_0^{2}+1=4=>x_0^{2}=3/4=>x_0=\sqrt{3}/2, y_0=3/4$ => P is rotated $60^{0}$ So coordinate x of P' are calculated by the formula: $x'=f(x)=1/2 x + \sqrt{3}/2 (x^{2}-1/4)$ f(x) has two roots: $x_1=1/2\sqrt{3}, x_2=-\sqrt{3}/2=> X_{min}(P')=\frac{x_1+x_2}{2}=-\frac{\sqrt{3}}{6}=>Result=\boxed{9}$

I propose an analytical solution for this problem.

We will first find information about the angle of rotation.

A parametric equation of P is : $x=t,y=t^{2}$

The focus of P is $\left(\begin{array}{c} 0\\ \frac{1}{4} \end{array}\right)$

The rotation clockwise about the focus of P a total of degrees $\theta$ can be writen as:

$\left(\begin{array}{c} X\\ Y \end{array}\right)=\left(\begin{array}{c} 0\\ \frac{1}{4} \end{array}\right)+\left[\begin{array}{cc} cos(\theta) & sin(\theta\\ -sin(\theta) & cos(\theta \end{array}\right]\left(\begin{array}{c} x\\ y-\frac{1}{4} \end{array}\right)$

Therefore :

$X=cos(\theta)+sin(\theta)(y-\frac{1}{4})=cos(\theta)+sin(\theta)y-\frac{1}{4}sin(\theta))$

$Y=-sin(\theta)x+cos(\theta)(y-\frac{1}{4})+\frac{1}{4}=-sin(\theta)x+cos(\theta)y+\frac{1}{4}[1-cos(\theta)]$

So that the equation of P' is :

$x=cos(\theta)t+sin(\theta)t^{2}-\frac{1}{4}sin(\theta)$

$y=-sin(\theta)t+cos(\theta)t^{2}-\frac{1}{4}[1-cos(\theta)]$

from the fact that P' is tangent o the directrix of P having the equation $y=-\frac{1}{4}$ , we conclude the following equation has one unique root:

$y=-sin(\theta)t+cos(\theta)t^{2}+\frac{1}{4}[1-cos(\theta)]=-\frac{1}{4}$

And then the discriminant is equal to zerro, $\Delta=sin(\theta)^{2}-4cos(\theta)(\frac{1}{4})[2-cos(\theta)]=0$

$\Delta=sin(\theta)^{2}+cos(\theta)(cos(\theta)-2)=0$

$\Delta=1-2cos(\theta)=0$

$cos(\theta)=\frac{1}{2}$ then $sin(\theta)=\frac{\sqrt{3}}{2}$

From here the the equation of P' is:

$x=\frac{1}{2}(t+\sqrt{3}(t^{2}-\frac{1}{4}))$

$y=\frac{1}{2}(-\sqrt{3}t+t^{2}-\frac{1}{4})$

$x_{0}$ is minimal when $\frac{dx}{dt}=0$ then $1+2\sqrt{3}t_{0}=0$ and $t_{0}=-\frac{1}{2\sqrt{3}}$

$x_{0}=\frac{1}{2}(-\frac{1}{2\sqrt{3}}+\sqrt{3}(\frac{1}{12}-\frac{1}{4}))$

After simplication we have $x_{0}=-\frac{\sqrt{3}}{6}$

We will first find information about the point of tangency and the angle of rotation. We will use this to find the point on $\mathbf{P}'$ with minimal $x$ -value.

Set $C=(0,\frac{1}{4})$ , which is the focus of $\mathbf{P}$ .

Set $f(x)=-\frac{1}{4}$ , which is the directrix of $\mathbf{P}$ .

Let $A=(a,a^2)$ be the point on $\mathbf{P}$ such that when $A$ is rotated clockwise by $\theta$ degrees around $C$ , it is the point $A'=(x_0,-\frac{1}{4})$ on $f$ .

The tangent at $A$ has slope $2a$ , and when this line is rotated clockwise by $\theta$ degrees around $C$ , it is the line $f$ (which has slope $0$ ). We use this information to determine the following about $\theta$ .

$\cos(\theta)=\frac{1}{\sqrt{4a^2+1}}\quad\sin(\theta)=\frac{2a}{\sqrt{4a^2+1}}$

Therefore,

$\begin{bmatrix} \tfrac{1}{\sqrt{4a^2+1}}&\tfrac{2a}{\sqrt{4a^2+1}}\\ -\tfrac{2a}{\sqrt{4a^2+1}}&\tfrac{1}{\sqrt{4a^2+1}} \end{bmatrix} \begin{bmatrix} a-0\\a^2-\tfrac{1}{4} \end{bmatrix}+ \begin{bmatrix} 0\\\tfrac{1}{4} \end{bmatrix}= \begin{bmatrix} x_0\\-\tfrac{1}{4} \end{bmatrix}$

$\begin{aligned} -2a^2+a^2-\tfrac{1}{4}&=-\tfrac{1}{2}\sqrt{4a^2+1}\\ a^4+\tfrac{1}{2}a^2+\tfrac{1}{16}&=a^2+\tfrac{1}{4}\\ (a^2-\tfrac{3}{4})(a^2+\tfrac{1}{4})&=0\\ a&=\frac{\sqrt{3}}{2} \end{aligned}$

$\Rightarrow\quad\cos(\theta)=\frac{1}{2}\quad\sin(\theta)=\frac{\sqrt{3}}{2}$

Let $B=(b,b^2)$ be the point on $\mathbf{P}$ such that when $B$ is rotated clockwise by $\theta$ degrees around $C$ , it is the point $B'$ on $\mathbf{P'}$ with minimum $x$ -value

The tangent at $B'$ is vertical, therefore the tangent at $B$ has slope $-\dfrac{1}{\sqrt{3}}$ and $B=\left(-\dfrac{1}{2\sqrt{3}},\dfrac{1}{12}\right)$ . Thus,

$\begin{bmatrix} \tfrac{1}{2}&\tfrac{\sqrt{3}}{2}\\ -\tfrac{\sqrt{3}}{2}&\tfrac{1}{2} \end{bmatrix} \begin{bmatrix} -\tfrac{1}{2\sqrt{3}}-0\\\tfrac{1}{12}-\tfrac{1}{4} \end{bmatrix}+ \begin{bmatrix} 0\\\tfrac{1}{4} \end{bmatrix}= \begin{bmatrix} \tfrac{-\sqrt{3}}{6}\\\tfrac{5}{12} \end{bmatrix}$

Therefore, the minimal value is $-\dfrac{\sqrt{3}}{6}$ .